freshers career: 1/27/08

hi friends for every one securing a job is a challenging issue to face this challenge every one should be get prepared before facing it.

every company has its own pattern in conducting their interview for freshers written exam is the first step for securing job

written exam is conducted inorder to evaluate individual problem solving skills their verbal ability analytical reasoning and so on

The different questions that are invovled in written exam include the following

1.quantitavie aptitude

2.mental ability

3.logic reasoning

4.non verbal reasoning

5.technical questions

6.puzzling

7.verbal reasoning

QUANTITATIVE APTITUDE

NUMBERS

HCF AND LCM

DECIMAL FRACTIONS

SIMPLIFICATION

SQUARE AND CUBE ROOTS

AVERAGES

PROBLEMS ON NUMBERS

PROBLEMS ON AGES

SURDS AND INDICES

PERCENTAGES

PROFIT AND LOSS

RATIOS AND PROPORTION

PARTNERSHIP

CHAIN RULE

TIME AND WORK

PIPES AND CISTERNS

TIME AND DISTANCE

TRAINS

Numbers
Introduction:
Natural Numbers:
All positive integers are natural numbers.
Ex 1,2,3,4,8,......
There are infinite natural numbers and number 1 is the least natural number.
Based on divisibility there would be two types of natural numbers. They are
Prime and composite.
Prime Numbers:
A natural number larger than unity is a prime number if it
does not have other divisors except for itself and unity.
Note:-Unity i e,1 is not a prime number.
Properties Of Prime Numbers:
->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.
->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However,
if a number on being divided by 6 gives a remainder 1 or 5 need not be
prime.
->The remainder of division of the square of a prime number p>=5 divide by
24 is 1.
->For prime numbers p>3, p²-1 is divided by 24.
->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²
is composite.
->The remainder of the division of the square of a prime number p>=5
divided by 12 is 1.
Process to Check A Number s Prime or not:
Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number
will be prime.
Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.
Composite Numbers:
The numbers which are not prime are known as composite numbers.
Co-Primes:
Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11).....
Place value or Local value of a digit in a Number:
place value:
Example 689745132
Place value of 2 is (2*1)=2
Place value of 3 is (3*10)=30 and so on.
Face value:-It is the value of the digit itself at whatever
place it may be.
Example 689745132
Face value of 2 is 2.
Face value of 3 is 3 and so on.
Top
Tests of Divisibility:
Divisibility by 2:-A number is divisible by 2,if its unit's digit is
any of 0,2,4,6,8.
Example 84932 is divisible by 2,while 65935 is not.
Divisibility by 3:-A number is divisible by 3,if the sum of its digits is
divisible by 3.
Example 1.592482 is divisible by 3,since sum of its digits
5+9+2+4+8+2=30 which is divisible by 3.
Example 2.864329 is not divisible by 3,since sum of its digits
8+6+4+3+2+9=32 which is not divisible by 3.
Divisibility by 4:-A number is divisible by 4,if the number formed by last
two digits is divisible by 4.
Example 1.892648 is divisible by 4,since the number formed by the last
two digits is 48 divisible by 4.
Example 2.But 749282 is not divisible by 4,since the number formed by
the last two digits is 82 is not divisible by 4.
Divisibility by 5:-A number divisible by 5,if its unit's digit is either
0 or 5.
Example 20820,50345
Divisibility by 6:-If the number is divisible by both 2 and 3.
example 35256 is clearly divisible by 2
sum of digits =3+5+2+5+21,which is divisible by 3
Thus the given number is divisible by 6.
Divisibility by 8:-A number is divisible by 8 if the last 3 digits
of the number are divisible by 8.
Divisibility by 11:-If the difference of the sum of the digits in the
odd places and the sum of the digitsin the even places is zero or divisible
by 11.
Example 4832718
(8+7+3+4) - (1+2+8)=11 which is divisible by 11.
Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12.
Divisibility by 7,11,13:-The difference of the number of its thousands
and the remainder of its division by 1000 is divisible by 7,11,13.
BASIC FORMULAE:
->(a+b)²=a²+b²+2ab
->(a-b)²=a²+b²-2ab
->(a+b)²-(a-b)²=4ab
->(a+b)²+(a-b)²=2(a²+b²)
->a²-b²=(a+b)(a-b)
->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)
->a³+b³=(a+b)(a²+b²-ab)
->a³-b³=(a-b)(a²+b²+ab)
->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)
->If a+b+c=0 then a³+b³+c³=3a b c
DIVISION ALGORITHM
If we divide a number by another number ,then
Dividend = (Divisor * quotient) + Remainder
Top
MULTIPLICATION BY SHORT CUT METHODS
1.Multiplication by distributive law:
a)a*(b+c)=a*b+a*c
b)a*(b-c)=a*b-a*c
Example
a)567958*99999=567958*(100000-1)
567958*100000-567958*1
56795800000-567958
56795232042
b)978*184+978*816=978*(184+816)
978*1000=978000
2.Multiplication of a number by 5n:-Put n zeros to the right of the
multiplicand and divide the number so formed by 2n
Example 975436*625=975436*54=9754360000/16=609647500.
PROGRESSION:
A succession of numbers formed and arranged in a definite order according
to certain definite rule is called a progression.
1.Arithmetic Progression:-If each term of a progression differs from its
preceding term by a constant.
This constant difference is called the common difference of the A.P.
The n th term of this A.P is Tn=a(n-1)+d.
The sum of n terms of A.P Sn=n/2[2a+(n-1)d].
xImportant Results:
a.1+2+3+4+5......................=n(n+1)/2.
b.12+22+32+42+52......................=n(n+1)(2n+1)/6.
c.13+23+33+43+53......................=n2(n+1)2/4
2.Geometric Progression:-A progression of numbers in which every
term bears a constant ratio with ts preceding term.
i.e a,a r,a r2,a r3...............
In G.P Tn=a r n-1
Sum of n terms Sn=a(1-r n)/1-r
Problems
1.Simplify
a.8888+888+88+8
b.11992-7823-456
Solution: a.8888
888
88
8
9872
b.11992-7823-456=11992-(7823+456)
=11992-8279=3713
2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114
Solution: 5 P Q
3 R 7
2 Q 8
11 1 4
2+P+Q+R=11
Maximum value of Q =11-2=9 (P=0,R=0)
3.Simplify: a.5793405*9999 b.839478*625
Solution:
a. 5793405*9999=5793405*(10000-1)
57934050000-5793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.
4.Evaluate 313*313+287*287
Solution:
a²+b²=1/2((a+b)²+(a-b)²)
1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )
½(360000+676)=180338
Top
5.Which of the following is a prime number?
a.241 b.337 c.391
Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.
6.Find the unit's digit n the product 2467 153 * 34172?
Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.
7.Find the total number of prime factors in 411 *7 5 *112 ?

Solution: 411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29
8.Which of the following numbers s divisible by 3?
a.541326
b.5967013

Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.
9.What least value must be assigned to * so that th number 197*5462 is
divisible by 9?

Solution: Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
10.What least number must be added to 3000 to obtain a number exactly
divisible by 19?
Solution:On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.
11.Find the smallest number of 6 digits which is exactly divisible by 111?
Solution:Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
12.On dividing 15968 by a certain number the quotient is 89 and the remainder
is 37.Find the divisor?
Solution:Divisor = (Dividend-Remainder)/Quotient
=(15968-37) / 89
=179.
13.A number when divided by 342 gives a remainder 47.When the same number
is divided by 19 what would be the remainder?
Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient and 9 as
remainder.
Top
14.A number being successively divided by 3,5,8 leaves remainders 1,4,7
respectively. Find the respective remainders if the order of
divisors are reversed?
Solution:Let the number be x.
3 x 5 y - 1 8 z - 4 1 - 7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.
15.Find the remainder when 231 is divided by 5?
Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as
4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
16.How many numbers between 11 and 90 are divisible by 7?
Solution:The required numbers are 14,21,28,...........,84
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
17.Find the sum of all odd numbers up to 100?
Solution:The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.
18.How many terms are there in 2,4,6,8..........,1024?
Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.
19.2+22+23+24+25..........+28=?
Solution:Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.
20.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.The positive integer is?
a.1 b.3 c.5 d.7
Solution:1000+N>1000N
clearly N=1.
21.The sum of all possible two digit numbers formed from three
different one digit natural numbers when divided by the sum of the
original three numbers is equal to?
a.18 b.22 c.36 d. none
Solution:Let the one digit numbers x,y,z
Sum of all possible two digit numbers=
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
= 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of
one digit numbers gives 22.
22.The sum of three prime numbers is 100.If one of them exceeds another by
36 then one of the numbers is?
a.7 b.29 c.41 d67.
Solution:x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
23.A number when divided by the sum of 555 and 445 gives two times
their difference as quotient and 30 as remainder .The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
24.The difference between two numbers s 1365.When the larger number is
divided by the smaller one the quotient is 6 and the remainder is 15.
The smaller number is?
a.240 b.270 c.295 d.360
Solution:Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
25.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?
a.0 b.12 c.13 d.20
Solution:Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.

H.C.F and L.C.M
Facts And Formulae:
Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :
The H.C.F of two or more than two numbers is the greatest
number that divides each of them exactly.
There are two methods :
i.Factorization method: Express each one of the given numbers as
the product of prime factors. The product of least powers of common
prime factors gives HCF.
Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 ,
2*52 *72
Solution: The prime numbers given common numbers are 2,5,7
Therefore HCF is 22 * 5 *72 .
ii.Division Method : Divide the larger number by smaller one. Now
divide the divisor by remainder. Repeat the process of dividing
preceding number last obtained till zero is obtained as number. The
last divisor is HCF.
Example: Find HCF of 513, 1134, 1215

Solution:
1134) 1215(1
1134
----------
81)1134(14
81
-----------
324
324
-----------
0
-----------
HCF of this two numbers is 81.
81)513(6
486
--------
27)81(3
81
-----
0
---
HCF of 81 and 513 is 27.
Least common multiple[LCM] : The least number which is
divisible by each one of given numbers is LCM.
There are two methods for this:
i.Factorization method : Resolve each one into product of prime
factors. Then LCM is product of highest powers of all factors.
ii.Common division method.
Top
Problems:
1.The HCF of 2 numbers is 11 and LCM is 693.If one of numbers
is 77.find other.
Sol: Other number = 11 * 693/77=99.
2.Find largest number of 4 digits divisible by 12,15,18,27
Sol: The largest number is 9999.
LCM of 12,15,18,27 is 540.
on dividing 9999 by 540 we get 279 as remainder.
Therefore
number =9999 – 279 =9720.
3.Find least number which when divided by 20,25,35,40 leaves
remainders 14,19,29,34.
Sol: 20–14=6
25-19=6
35-29=6
40-34=6
Therefore number =LCM of (20,25,35,40) - 6=1394
4.252 can be expressed as prime as :
2 252
2 126
3 63
3 21
7
prime factor is 2 *2 * 3 * 3 *7
5.1095/1168 when expressed in simple form is
1095)1168(1
1095
------
73)1095(15
73
---------
365
365
---------
0
----------
So, HCF is 73
Therefore
1095/1168 = 1095/73/1168/73= 15/16
6.GCD of 1.08,0.36,0.9 is

Sol:
HCF of 108,36,90 36)90(2
72
----
18)36(2
36
----
0
----
HCF is 18.
HCF of 18 and 108 is 18
18)108(6
108
-------
0
--------
Therefore HCF =0.18
7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.
Sol:
Let the numbers be x.
Three numbers are x,2x,3x
Therefore
HCF is 2x)3x(1
2x
-----
x)2x(2
2x
--------
0
-------------
HCF is x so, x is 12
Therefore numbers are 12,24,36.
Top
8.The sum of two numbers is 216 and HCF is 27.
Sol: Let numbers are 27a + 27 b =216
a + b =216/27=8
Co-primes of 8 are (1,7) and (3,5)
numbers=(27 * 1 ), (27 * 7)
=27,89
9.LCM of two numbers is 48..The numbers are in ratio 2:3. The sum
of numbers is
Sol:
Let the number be x.
Numbers are 2x,3x
LCM of 2x,3x is 6x
Therefore
6x=48
x=8.
Numbers are 16 and 24
Sum=16 +24=40.
10.HCF and LCM of two numbers are 84 and 21.If ratio of two numbers
is 1:4.Then largest of two numbers is
Sol:
Let the numbers be x,4x
Then x * 4x = 84 * 21
x2 =84 * 21 /4
x = 21
Largest number is 4 * 21.
11.HCF of two numbers is 23,and other factors of LCM are 13,14.
Largest number is
Sol:
23 * 14 is Largest number.
12.The maximum number of students among them 1001 pens and 910
pencils can be distributed in such a way that each student gets
same number of pens and pencils is ?
Sol:
HCF of 1001 and 910
910)1001(1
910
------------
91)910(10
910
--------
0
---------
Therefore HCF=91
13.The least number which should be added to 2497 so that sum is
divisible by 5,6,4,3 ?
Sol: LCM of 5,6,4,3 is 60.
On dividing 2497 by 60 we get 37 as remainder.
Therefore number to added is 60 – 37 =23.
Answer is 23.
14.The least number which is a perfect square and is divisible by
each of numbers 16,20,24 is ?
Sol: LCM of 16,20,24 is 240.
2 * 2*2*2*3*5=240
To make it a perfect square multiply by 3 * 5
Therefore 240 * 3 * 5=3600
Answer is 3600.

Decimal Fractions
1.Decimal fractions: Fractionin which denominations are powers
of 10 are decimal fractions.
Example:1 /10 = 0.1, 1 / 100 = 0.01
2.Convertion of Decimal into fraction:-
Example: 0.25 = 25/100 = 1/4
3.i) If numerator and denominator contain same number of decimal
places, then we remove decimal sign. Thus, 1.84/2.99 =184/299
Problems
1.0.75 =75/100 =3/4
2.Find porducts= 6.3204*100
= 632.04
3.2.61*1.3=261*13=3393 some of decimal places 2 +1 =3
sol: 3.393
4.If 1/3.718 =0.2689,then find value of 1/0.0003718 ?
Sol: 10000/3.718 =10000*1/3.718
=10000*0.2689
= 2689
5.Find fractions :
i) 0.37 = 37/99
ii)3.142857 =3+0.142857
=3 +142857/999999
= 3 142857/ 999999
iii) 0.17=17-1/90 =16/90=8/45
iv)0.1254 =1254 -12/9900 =1242/9900=69/550
6.Fraction 101 27/100000
Sol: 101+27/100000
=101+0.00027
=101.00027
7.If 47.2506 =4A + 7/B +2C + 5/D + 6E then 40+7+0.2+0.05+0.0006
Sol: compairing terms
4A= 40 => A=10
7/B = 7 => B=1
2C= 0.2=> C=0.1
5/D= 0.05=>D=5/0.05 =>5*100/5 =100
6E= 0.0006=> E= 0.0001
5A + 3B+6C+ D+ 3E = 5*10+ 3*1+ 6*0.1 + 100+ 3*0.0001
=50+3+0.6+100+0.0003
=153.6003
Top
8.4.036 divided by 0.04
Sol: 4.036/0.04 =4036/4 =100.9
9.[ 0.05/0.25 + 0.25/ 0.05]3
Sol: =>[5/25 + 25/5]
= [1/5+ 5]3
=26/53
=5.23
= 140.603
10.The least among the following :-
a. 0.2 b.1/0.2 c. 0.2 d. 0.22
sol:10/2 =5 0.2222 0.04 0.04 < f=" 0.84181" 99000 =" 83340/99000" distence =" (550" 53 ="-5+1+0.53" 53 =" 4.53" 25="0.16" 93 ="2025" 3 ="">
(0.11) [1+2+--------+9]
=0.001331*2025
=2.695275
15.(0.96)3 – (0.1)3/ (0.96)2 +0.096 +(0.1)2
Sol: formula => a3 -b3/a2 +ab +b2 =a -b
(0.96-0.1)=0.86
16.3.6*0.48*2.50 / 0.12*0.09*0.5
Sol: 36*48*250/12*9*5=800
17.find x/y = 0.04/1.5
= 4/150 =2/75
find y-x/y+x
(1- x/y) / (1+ x/y)
1 - 2/75 /1 +2/75 =73/77
18.0.3467+0.1333
Sol: 3467 -34/9900 + 1333-13/9900
= 3433 +1320/9900
= 4753/9900
= 4801 -48/9900 =0.4301

Simplifications
Introduction:
'BODMAS' rule: This rule depicts the correct sequence in which
the operations are to be executed, so as to find out the value of
a given expression.
Here B stands for Bracket, O for Of, D for Division, M for
Multiplication, A for Addition and S for Subtraction.
First of all the brackets must be removed, strictly in the
order () , {} , [].
After removing the brackets, we want use the following operations:
1.Of 2. Division 3. Multiplication 4. Addition 5. Subtraction
Modulus of a real number:
Modulus of a real number is a defined as
a = a, if a>0 or -a, if a < 6=" ?" expression =" 5004/" 6 =" 36" 6 =" 30;" 2 =" 11)" 6 =" 11" 2 =" 11" 2 =" 12" 6 =" 12" 8888088 =" ?" 8888088 =" 808008" 10 =" [½" 15 =" 2/5"top
6.The value of 999 of 995/999* 999 is:
Sol: [1000- 4/1000]*999 = 999000-4
= 998996
7.Along a yard 225m long, 26 trees are planted at equal distance, one
tree being at each end of the yard. what is the distance between two
consecutive trees ?
Sol: 26 trees have 25 gaps between them.
Hence , required distance = 225/ 25 m= 9m
8.In a garden , there are 10 rows and 12 columns of mango trees. the
distance between the two trees is 2 m and a distance of one meter is
left from all sides of the boundary of the length of the garden is :
Sol: Each row contains 12 plants.
leaving 2 corner plants, 10 plants in between have 10 * 2 meters and
1 meter on each side is left.
length = (20 + 2) m = 22m
9.Eight people are planning to share equally the cost of a rental car,
if one person with draws from the arrangement and the others share
equally the entire cost of the car, then the share of each of the
remaining persons increased by?
Sol: Original share of one person = 1/8
new share of one person = 1/7
increase = 1/7 – 1/8 = 1/56
required fractions = (1/56)/(1/8) = 1/7
10.A piece of cloth cost Rs 35. if the length of the piece would
have been 4m longer and each meter cost Re 1 less , the cost
would have remained unchanged. how long is the piece?
Sol: Left the length of the piece be x m.
then, cost of 1m of piece = Rs [35 / x]
35/ x – 35 /x+4 = 1
x + 4 – x = x(x+ 4)/35
x2 + 4x – 140 = 0
x= 10
11.A man divides Rs 8600 among 5sons, 4 daughters and 2 nephews.
If each daughter receives four times as much as each nephew, and
each son receives five as much as each nephew. how much does each
daughter receive ?
Sol:
Let the share of each nephew be Rs x.
then, share of each daughter Rs 4x.
share of each son = 5x Rs
so, 5 *5x+ 4 * 4x + 2x =8600
2x + 16x + 25x= 8600
43x = 8600
x = 200
share of each daughter = 4 * 200 = Rs 800
12.A man spends 2/5 of his salary on house rent, 3/10 of his salary
on food, and 1/8 of his salary on conveyance. if he has Rs 1400 left
with him, find his expenditure on food and conveyance?
Sol: Part of the salary left = 1-[2/5 +3/10+1/9]
= 1- 33/40
=7/40
Let the monthly salary be rs x
then, 7/40 of x = 1400
x= [1400*40]/7
x= 8000
Expenditure on food = 3/10*8000 =Rs 2400
Expenditure on conveyance= 1/8*8000 =Rs 1000

Square and Cube Roots
Formula:
The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.
9 6
- 4
--------------
9 2
--------------
4*4=16
9 2 1 6
therefore square of the two numbers 96*96=9216.
Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604
9 8
- 2
-------------
9 6
-------------
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.
Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
------------
76
-----------
12*12=144
76
+ 144
--------------------
7744
--------------------

Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
----------------
4 2 as 50 = 100/2
42/2=21
now place 4*4=16 beside 21
2 1 1 6
Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.
3 7
1 3
-----------------
2 4 as 50 = 100/2
24/2=12
square 13* 13=169
1 2
+ 1 6 9
-----------------------
1 3 6 9
-------------------------
Top
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
---------------------
1 1 2
--------------------
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6
Square root: If x2=y ,we say that the square root of y
is x and we write ,âˆšy=x.
Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .
Examples:
1.Evaluate 60841/2 by factorization method.
Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.
Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.
2.what will come in place of question mark in each of the following
questions?
i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3
Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1
Answer is 8.1
ii)86.491/2 + âˆš(5+x2)=12.3
solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ âˆš(5+x1/2 )=12.3
=> âˆš(5+x1/2 ) =12.3-9.3
=> âˆš(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.
3.âˆš 0.00004761 equals:
Solution: âˆš (4761/108)
âˆš4761/âˆš 108
. 69/10000
0.0069.
Answer is 0.0069
4.If âˆš18225=135,then the value of
âˆš182.25 + âˆš1.8225 + âˆš 0.018225 + âˆš0.00018225.
Solution: âˆš(18225/100) +âˆš(18225/10000) +
âˆš(18225/1000000) +âˆš(18225/100000000)
=âˆš(18225)/10 + (18225)1/2/100 +
âˆš(18225)/1000 + âˆš(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.
5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?
Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.
6.If 0.13 / p1/2=13 then p equals
Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=âˆš(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1
Top
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:
Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.
8.If âˆš(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?
Solution: âˆš(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.
9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?
Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35
10.If 3a=4b=6c and a+b+c=27*âˆš29 then Find c value is:
Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2
11.If 2*3=131/2 and 3*4=5 then value of 5*12 is
Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.
12.The smallest number added to 680621 to make
the sum a perfect square is
Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.
13.The greatest four digit perfect square number is
Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.
14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.
Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.
15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:
Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.
16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.
17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?
Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.
18.3(4*12/125)1/2=?
Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.

Averages
Formula:
1.Average=Sum of quantities/Number of quantities.
2.Suppose a man covers a certain distance at x kmph
and an equal distance at y kmph ,then the average speed
during the whole journey is (2xy/x+y) kmph.
Examples:
1.Find the average of all these numbers.142,147,153,165,157.
Solution:
142 147 153 165 157
Here consider the least number i.e, 142
comparing with others,
142 147 153 165 157
+5 +11 +23 +15
Now add 5+11+23+15 = 52/5 = 10.8
Now add 10.8 to 142 we get 152.8
(Average of all these numbers).
Answer is 152.8
2.Find the average of all these numbers.4,10,16,22,28
Solution:
4,10,16,22,28
As the difference of number is 6
Then the average of these numbers is central one i.e, 16.
Answer is 16.
3.Find the average of all these numbers.4,10,16,22,28,34.
Solution:
Here also difference is 6.
Then middle numbers 16,22 take average of these
two numbers 16+22/2=19
Therefore the average of these numbers is 19.
Answer is 19.
4.The average marks of a marks of a student in 4 Examination
is 40.If he got 80 marks in 5th Exam then what is
his new average.
Solution:
4*40+80=240
Then average means 240/5=48.
Answer is 48.
5.In a group the average income of 6 men is 500 and that
of 5 women is 280, then what is average income of the group.
Solution:
6*500+5*280=4400
then average is 4400/11=400.
Another Method: here consider for 6 men
6 men â€“ each 500.
so 5th women is 280.
then 500-280=220.
then 220*6/11=120.
therefore 120+280=400.
Answer is 400.
6.The average weight of a class of 30 students is 40 kgs if the
teacher weight is included then average increases by 2 kgs then
find the weight of the teacher?
Solution:
30 students average weight is 40 kgs.
So,when teacher weight is added it increases by 2 kgs
so total 31 persons ,therefore 31*2=62.
Now add the average weight of all student to it
we get teachers weight i.e, 62+40=102 kgs.
Answer is 102 kgs.
7.The average age of Mr and Mrs Sharma 4 years ago is 28 years .
If the present average age of Mr and Mrs Sharma and their son
is 22 years. What is the age of their son.
Solution:
4 years ago their average age is 28 years.
So their present average age is 32 years.
32 years for Mr and Mrs Sharma then 32*2=64 years.
Then present age including their son is 22 years.
So 22*3 =66 years.
Therefore son age will be 66-64 = 2 years.
Answer is 2 years.
8.The average price of 10 books is increased by 17 Rupees when
one of them whose value is Rs.400 is replaced by a new book.
What is the price of new book?
Solution:
10 books Average increases by 17 Rupees
so 10*17= 170.
so the new book cost is more and by adding its cost average
increase,therefore the cost of new book is 400+170=570Rs.
Answer is 570 Rs.
9.The average marks of girls in a class is 62.5. The average marks
of 4 girls among them is 60.The average marks of remaining girls
is 63,then what is the number of girls in the class?
Solution:
Total number of girls be x+4.
Average marks of 4 girls is 60.
therefore 62.5-60=2.5
then 4*2.5 =10.
the average of remaining girls is 63
here 0.5 difference therefore 0.5*x=10(since we got from 4 girls)
(this is taken becoz both should be equal)
x=10/0.5
x=20.
This clear says that remaining are 20 girls
therefore total is x+4=20+4=24 girls
Answer is 24 girls.
Top
10.Find the average of first 50 natural numbers.
Solution:
Sum of the Natural Numbers is n(n+1)/2
therefore for 50 Natural numbers 50*51/2=775.
the average is 775/50=15.5
Answer is 15.5 .
11.The average of the first nine prime number is?
Solution:
Prime numbers are 2,3,5,7,11,13,17,19,23
therefore 2+3+5+7+11+13+17+19+23=100
then the average 100/9= 11 1/9.
Answer is 11 1/9.
12.The average of 2,7,6 and x is 5 and the average of and the
average of 18,1,6,x and y is 10 .what is the value of y?
Solution:
2+7+6+x/4=5
=>15+x=20
=>x=5.
18+1+6+x+y/5=10
=>25+5+y=50
=>y=20.
13.The average of a non-zero number and its square is 5 times the
number.The number is
Solution:
The number be x
then x+x2/2=5x
=>x2-9x=0
=>x(x-9)=0
therefore x=0 or x=9.
The number is 9.
14.Nine persons went to a hotel for taking their meals . Eight of
them spent Rs.12 each on their meals and the ninth spent Rs.8 then
the average expenditure of all the nine. What was the total money
spent by them?
Solution:
The average expenditure be x.
then 8*12+(x+8)=9x
=>96+x+8=9x.
=>8x=104
=>x=13
Total money spent =9x=>9*13=117
Answer is Rs.117
15.The average weight of A.B.C is 45 Kgs.If the average weight of
A and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?
Solution:
The weight of A,B,Care 45*3=135 Kgs.
The weight of A,B are 40*2=80 Kgs.
The weight of B,C are 43*2=86 Kgs.
To get the Weight of B.
(A+B)+(B+C)-(A+B+C)=80+86-135
B=31 kgs.
Answer is 31 Kgs.
16.The sum of three consecutive odd number is 48 more than the average
of these number .What is the first of these numbers?
Solution:
let the three consecutive odd numbers are x, x+2, x+4.
By adding them we get x+x+2+x+4=3x+6.
Then 3x+6-(3x+6)/3=38(given)
=>2(3x+6)=38*3.
=>6x+12=114
=>6x=102
=>x=17.
Answer is 17.
17.A family consists of grandparents,parents and three grandchildren.
The average age of the grandparents is 67 years,that of parents is 35
years and that of the grand children is 6 years . What is the average
age of the family?
Solution:
grandparents age is 67*2=134
parents age is 35*2=70
grandchildren age is 6*3=18
therefore age of family is 134+70+18=222
average is 222/7=31 5/7 years.
Answer is 31 5/7 years.
Top
18.A library has an average of 510 visitors on Sundays and 240 on
other days .The average number of visitors per day in a month 30
days beginning with a Sunday is?
Solution:
Here specified that month starts with Sunday
so, in a month there are 5 Sundays.
Therefore remaining days will be 25 days.
510*5+240*25=2550+6000
=8550 visitors.
The average visitors are 8550/30=285.
Answer is 285.
19.The average age of a class of 39 students is 15 years .
If the age of the teacher be included ,then average
increases by 3 months. Find the age of the teacher.
Solution: Total age for 39 persons is 39*15=585 years.
Now 40 persons is 40* 61/4=610 years
(since 15 years 3 months=15 3/12=61/4)
Age of the teacher =610-585 years
=>25 years.
Answer is 25 years.
20.The average weight of a 10 oarsmen in a boat is increases
by 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is
replaced by new man. Find the weight of the new man.
Solution: Weight of 10 oars men is increases by 1.8 Kgs
so, 10*1.8=18 Kgs
therefore 53+18=71 Kgs will be the weight of the man.
Answer is 71 Kgs.
21.A bats man makes a score of 87 runs in the 17th inning
and thus increases his average by 3. Find the average
after 17th inning.
Solution: Average after 17 th inning =x
then for 16th inning is x-3.
Therefore 16(x-3)+87 =17x
=>x=87-48
=>x=39.
Answer is 39.
22.The average age of a class is 15.8 years .The average age
of boys in the class is 16.4 years while that of the girls
is 15.4 years .What is the ratio of boys to girls in the class.
Solution: Ratio be k:1 then
k*16.4 + 1*15.4 = (k+1)*15.8
=>(16.4-15.8)k=15.8-15.4
=>k=0.4/0.6
=>k=2/3
therefore 2/3:1=>2:3
Answer is 2:3
23.In a cricket eleven ,the average of eleven players is
28 years .Out of these ,the average ages of three groups
of players each are 25 years,28 years, and 30 years
respectively. If in these groups ,the captain and the
youngest player are not included and the captain is
eleven years older than the youngest players ,
what is the age of the captain?
Solution: let the age of youngest player be x
then ,age of the captain =(x+11)
therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28
=>75+84+90+2x+11=308
=>2x=48
=>x=24.
Therefore age of the captain =(x+11)= 24+11= 35 years.
Answer is 35 years.
Top
24.The average age of the boys in the class is twice
the number of girls in the class .If the ratio of
boys and girls in the class of 36 be 5:1, what is
the total of the age (in years) of the boys in the class?
Solution: Number of boys=36*5/6=30
Number of girls =6
Average age of boys =2*6=12 years
Total age of the boys=30*12=360 years
Answer is 360 years.
25.Five years ago, the average age of P and Q was
15 years ,average age of P,Q, and R today is
20 years,how old will R be after 10 years?
Solution: Age of P and Q are 15*2=30 years
Present age of P and Q is 30+5*2=40 years.
Age of P Q and R is 20*3= 60 years.
R ,present age is 60-40=20 years
After 10 years =20+10=30 years.
Answer is 30 years.
26.The average weight of 3 men A,B and C is 84 Kgs.
Another man D joins the group and the average now
becomes 80 Kgs.If another man E whose weight is
3 Kgs more than that of D ,replaces A then the
average weight B,C,D and E becomes 79 Kgs.
The weight of A is.
Solution:Total weight of A, B and C is 84 * 3 =252 Kgs.
Total weight of A,B,C and Dis 80*4=320 Kgs
Therefore D=320-252=68 Kgs.
E weight (68+3)=71 kgs
Total weight of B,C,D and E = 79*4=316 Kgs
(A+B+C+D)-(B+C+D+E)=320-316 =4Kgs
A-E=4Kgs
A-71=4 kgs
A=75 Kgs
Answer is 75 kgs
27.A team of 8 persons joins in a shooting competition.
The best marksman scored 85 points.If he had scored
92 points ,the average score for the team would
have been 84.The team scored was.
Solution: Here consider the total score be x.
therefore x+92-85/8=84
=>x+7=672
=>x=665.
Answer is 665
28.A man whose bowling average is 12.4,takes 5 wickets
for 26 runs and there by decrease his average by 0.4.
The number of wickets,taken by him before his last match is:
Solution: Number of wickets taken before last match be x.
therefore 12.4x26/x+5=12(since average decrease by 0.4
therefore 12.4-0.4=12)
=>12.4x+2612x+60
=>0.4x=34
=>x=340/4
=>x=85.
Answer is 85.
29.The mean temperature of Monday to Wednesday was 37 degrees
and of Tuesday to Thursday was 34 degrees .If the
temperature on Thursday was 4/5th that of Monday.
The temperature on Thursday was:
Solution:
The total temperature recorded on Monday,Wednesday was 37*3=111.
The total temperature recorded on Tuesday,
Wednesday,Thursday was 34*3=102.
and also given that Th=4/5M
=>M=5/4Th
(M+T+W)-(T+W+Th)=111-102=9
M-Th=9
5/4Th-Th=9
Th(1/4)=9
=>Th=36 degrees.
30. 16 children are to be divided into two groups A and B
of 10 and 6 children. The average percent marks obtained
by the children of group A is 75 and the average percent
marks of all the 16 children is 76. What is the average
percent marks of children of groups B?
Solution: Here given average of group A and whole groups .
So,(76*16)-(75*10)/6
=>1216-750/6
=>466/6=233/3=77 2/3
Answer is 77 2/3.
31.Of the three numbers the first is twice the second and
the second is twice the third .The average of the reciprocal
of the numbers is 7/72,the number are.
Solution:Let the third number be x
Let the second number be 2x.
Let the first number be 4x.
Therefore average of the reciprocal means
1/x+1/2x+1/4x=(7/72*3)
7/4x=7/24
=>4x=24
x=6.
Therefore
First number is 4*6=24.
Second number is 2*6=12
Third number is 1*6=6
Answer is 24,12,6.
32.The average of 5 numbers is 7.When 3 new numbers
are added the average of the eight numbers is 8.5.
The average of the three new number is:
Solution: Sum of three new numbers=(8*8.5-5*7)=33
Their average =33/3=11.
Answer is 11.
33.The average temperature of the town in the first
four days of a month was 58 degrees. The average
for the second ,third,fourth and fifth days was
60 degree .If the temperature of the first and
fifth days were in the ratio 7:8 then what is
the temperature on the fifth day?
Solution :
Sum of temperature on 1st 2nd 3rd
and 4th days =58*4=232 degrees.
Sum of temperature on 2nd 3rd 4th
and 5th days =60*4=240 degrees
Therefore 5th day temperature is 240-232=8 degrees.
The ratio given for 1st and 5th days be 7x and 8x degrees
then 8x-7x=8
=>x=8.
therefore temperature on the 5th day =8x=8*8=64 degrees.

Problems on Numbers
Simple problems:
1.What least number must be added to 3000 to obtain a number
exactly divisible by 19?
Solution:
On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.
2.Find the unit's digit n the product 2467 153 * 34172?
Solution:
Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.
3.Find the total number of prime factors in 411 *7 5 *112 ?
Solution:
411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29
4.The least umber of five digits which is exactly
divisible by 12,15 and 18 is?
a.10010 b.10015 c.10020 d.10080
Solution:
Least number of five digits is 10000
L.C.Mof 12,15,18 s 180.
On dividing 10000 by 180,the remainder is 100.
Therefore required number=10000+(180-100)
=10080.
Ans (d).
5.The least number which is perfect square and is divisible
by each of the numbers 16,20 and 24 is?
a.1600 b.3600 c.6400 d.14400
Solution:
The least number divisible by 16,20,24 = L.C.M of 16,20,24=240
=2*2*2*2*3*5
To make it a perfect square it must be multiplied by 3*5.
Therefore required number =240*3*5=3600.
Ans (b).
6.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.
The positive integer is?
a.1 b.3 c.5 d.7
Solution:
1000+N>1000N
clearly N=1.
7.How many numbers between 11 and 90 are divisible by 7?
Solution:
The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
8.Find the sum of all odd numbers up to 100?
Solution:
The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.
9.How many terms are there in 2,4,6,8..........,1024?
Solution:
Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.
10.2+22+23+24+25..........+28=?
Solution:
Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.
11.Find the number of zeros in 27!?
Solution:
Short cut method :
number of zeros in 27!=27/5 + 27/25
=5+1=6zeros.
Top
Medium Problems:
12.The difference between two numbers 1365.When the larger
number is divided by the smaller one the quotient is 6 and
the remainder is 15.The smaller number is?
a.240 b.270 c.295 d.360
Solution:
Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
13.Find the remainder when 231 is divided by 5?
Solution:
210 =1024.
unit digit of 210 * 210 * 210 is 4
as 4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
14.The largest four digit number which when divided by 4,7
or 13 leaves a remainder of 3 in each case is?
a.8739 b.9831 c.9834 d.9893. Solution:
solution:
Greatest number of four digits is 9999
L.C.M of 4,7, and 13=364.
On dividing 9999 by 364 remainder obtained is 171.
Therefore greatest number of four digits divisible by 4,7,13
=9999-171=9828.
Hence required number=9828+3=9831.
Ans (b).
15.What least value must be assigned to * so that th number
197*5462 is divisible by 9?
Solution:
Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
16.Find the smallest number of 6 digits which is exactly
divisible by 111?
Solution:
Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
17.A number when divided by 342 gives a remainder 47.When
the same number is divided by 19 what would be the remainder?
Solution:
Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient
and 9 as remainder.
18.In doing a division of a question with zero remainder,a
candidate took 12 as divisor instead of 21.The quotient
obtained by him was 35. The correct quotient is?
a.0 b.12 c.13 d.20
Solution:
Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.
19.If a number is multiplied by 22 and the same number is
added to it then we get a number that is half the square
of that number. Find the number.
a.45 b.46 c.47 d. none
Solution:
Let the required number be x.
Given that x*22+x = 1/2 x2
23x = 1/2 x2
x = 2*23=46
Ans (b)
20.Find the number of zeros in the factorial of the number 18?
Solution:
18! contains 15 and 5,which combined with one even number
gives zeros. Also 10 is also contained in 18! which will
give additional zero .Hence 18! contains 3 zeros and the
last digit will always be zero.
21.The sum of three prime numbers is 100.If one of them
exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d67.
Solution:
x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
22.A number when divided by the sum of 555 and 445 gives
two times their difference as quotient and 30 as remainder .
The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:
Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
23.The difference of 1025-7 and 1024+x is divisible by 3 for x=?
a.3 b.2 c.4 d.6
Solution:
The difference of 1025-7 and 1024+x is
=(1025-7)-(1024-x)
=1025-7-1024-x
=10.1024-7 -1024-x
=1024(10-1)-(7-x)
=1024*9-(7+x)
The above expression is divisible by 3 so we have to
replace x with 2.
Ans (b).
Top
Complex Problems:
24.Six bells commence tolling together and toll at intervals
of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many
times do they toll together?
Solution:
To find the time that the bells will toll together we have
to take L.C.M of 2,4,6,8,10,12 is 120.
So,the bells will toll together after every 120 seconds
i e, 2 minutes
In 30 minutes they will toll together [30/2 +1]=16 times
25.The sum of two numbers is 15 and their geometric mean is
20% lower than their arithmetic mean. Find the numbers?
a.11,4 b.12,3 c.13,2 d.10,5
Solution:
Sum of the two numbers is a+b=15.
their A.M = a+b / 2 and G.M = (ab)1/2
Given G.M = 20% lower than A.M
=80/100 A.M
(ab)1/2=4/5 a+b/2 = 2*15/5= 6
(ab)1/2=6
ab=36 =>b=36/a
a+b=15
a+36/a=15
a2+36=15a
a2-15a+36=0
a2-3a-12a+36=0
a(a-3)-12(a-3)=0
a=12 or 3.
If a=3 and a+b=15 then b=12.
If a=12 and a+b=15 then b=3.
Ans (b).
26.When we multiply a certain two digit number by the
sum of its digits 405 is achieved. If we multiply the
number written in reverse order of the same digits
by the sum of the digits,we get 486.Find the number?
a.81 b.45 c.36 d. none
Solution:
Let the number be x y.
When we multiply the number by the sum of its digit
405 is achieved.
(10x+y)(x+y)=405....................1
If we multiply the number written in reverse order by its
sum of digits we get 486.
(10y+x)(x+y)=486......................2
dividing 1 and 2
(10x+y)(x+y)/(10y+x)(x+y) = 405/486.
10x+y / 10y+x = 5/6.
60x+6y = 50y+5x
55x=44y
5x = 4y.
From the above condition we conclude that the above
condition is satisfied by the second option i e b. 45.
Ans (b).
27.Find the HCF and LCM of the polynomials x2-5x+6 and x2-7x+10?
a.(x-2),(x-2)(x-3)(x-5)
b.(x-2),(x-2)(x-3)
c.(x-3),(x-2)(x-3)(x-5)
d. none
Solution:
The given polynomials are
x2-5x+6=0................1
x2-7x+10=0...............2
we have to find the factors of the polynomials
x2-5x+6 and x2-7x+10
x2-2x-3x+6 x2-5x-2x+10
x(x-2)-3(x-2) x(x-5)-2(x-5)
(x-3)(x-2) (x-2)(x-5)
From the above factors of the polynomials we can easily
find the HCF as (x-3)and LCM as (x-2)(x-3)(x-5).
Ans (c)
28.The sum of all possible two digit numbers formed from
three different one digit natural numbers when divided by
the sum of the original three numbers is equal to?
a.18 b.22 c.36 d. none
Solution:
Let the one digit numbers x,y,z
Sum of all possible two digit numbers
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z)
Therefore sum of all possible two digit numbers when
divided by sum of one digit numbers gives 22.
29.A number being successively divided by 3,5,8 leaves
remainders 1,4,7 respectively. Find the respective
remainders if the order of divisors are reversed?
Solution:
Let the number be x.
3 - x
5 y - 1
8 z - 4
1 - 7
z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now 8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.
30.The arithmetic mean of two numbers is smaller by 24
than the larger of the two numbers and the GM of
the same numbers exceeds by 12 the smaller of the numbers.
Find the numbers?
a.6,54 b.8,56 c.12,60 d.7,55
Solution:
Let the numbers be a,b where a is smaller and b
is larger number.
The AM of two numbers is smaller by 24 than the
larger of the two numbers.
AM=b-24
AM of two numbers is a+b/2.
a+b/2 = b-24
a+b = 2b-48
a = b-48...................1
The GM of the two numbers exceeds by 12 the smaller
of the numbers
GM = a+12
GM of two numbers is (ab)1/2
(ab) 1/2= a+12
ab = a2+144+24a
from 1 b=a+48
a(a+48)= a2+144+24a
a2+48a = a2+144+24a
24a=144=>a=6
Therefore b=a+48=54.
Ans (a).
31.The sum of squares of the digits constituting a positive
two digit number is 13,If we subtract 9 from that number
we shall get a number written by the same digits in the
reverse order. Find the number?
a.12 b.32 c.42 d.52.
Solution:
Let the number be x y.
the sum of the squares of the digits of the number is 13
x2+y2=13
If we subtract 9 from the number we get the number
in reverse order
x y-9=y x.
10x+y-9=10y+x.
9x-9y=9
x-y=1
(x-y)2 =x2+y2-2x y
1 =13-2x y
2x y = 12
x y = 6 =>y=6/x
x-y=1
x-6/x=1
x2-6=x
x2-x-6=0
x+2x-3x-6=0
x(x+2)-3(x+2)=0
x=3,-2.
If x=3 and x-y=1 then y=2.
If x=-2 and x-y=1 then y=-3.
Therefore the number is 32.
Ans (b).
Top
32.If we add the square of the digit in the tens place
of the positive two digit number to the product of the
digits of that number we get 52,and if we add the square
of the digit in the unit's place to the same product
of the digits we get 117.Find the two digit number?
a.18 b.39 c.49 d.28
Solution:
Let the digit number be x y
Given that if we add square of the digit in the tens place
of a number to the product of the digits we get 52.
x2+x y=52.
x(x+y)=52....................1
Given that if we add the square of the digit in the unit's plac
e to the product is 117.
y2+x y= 117
y(x+y)=117.........................2
dividing 1 and 2 x(x+y)/y(x+y) = 52/117=4/9
x/y=4/9
from the options we conclude that the two digit number is 49
because the condition is satisfied by the third option.
Ans (c)
33.The denominators of an irreducible fraction is greater
than the numerator by 2.If we reduce the numerator of the
reciprocal fraction by 3 and subtract the given fraction
from the resulting one,we get 1/15.Find the given fraction?
Solution:
Let the given fraction be x / (x+2) because given that
denominator of the fraction is greater than the numerator by 2
1 â€“ [(x â€“ 1/(x+2))/3] = 1/15.
1 â€“ (x2+2x-1) /3(x+2) = 1/15
(3x+6-x2-2x+1)/3(x+2) = 1/15
(7-x2+2x)/(x+2) = 1/5
-5x2+5x+35 = x+2
5x2-4x-33 = 0
5x2-15x+11x-33 = 0
5x(x-3)+11(x-3) = 0
(5x+11)(x-3) = 0
Therefore x=-11/5 or 3
Therefore the fraction is x/(x+2) = 3/5.
34.Three numbers are such that the second is as much
lesser than the third as the first is lesser than
the second. If the product of the two smaller numbers
is 85 and the product of two larger numbers is 115.
Find the middle number?
Solution:
Let the three numbers be x,y,z
Given that z â€“ y = y â€“ x
2y = x+z.....................1
Given that the product of two smaller numbers is 85
x y = 85................2
Given that the product of two larger numbers is 115
y z = 115...............3
Dividing 2 and 3 x y /y z = 85/115
x / z = 17 / 23
From 1
2y = x+z
2y = 85/y + 115/y
2y2 = 200
y2 = 100
y = 10
35.If we divide a two digit number by the sum of its digits
we get 4 as a quotient and 3 as a remainder. Now if we
divide that two digit number by the product of its digits
we get 3 as a quotient and 5 as a remainder .
Find the two digit number?
Solution:
Let the two digit number is x y.
Given that x y / (x+y)
quotient=4 and remainder = 3
we can write the number as
x y = 4(x+y) +3...........1
Given that x y /(x*y) quotient = 3 and remainder = 5
we can write the number as
x y = 3 x*y +5...............2
By trail and error method
For example take x=1,y=2
1............12=4(2+3)+3
=4*3+3
! =15
let us take x=2 y=3
1..............23=4(2+3)+3
=20+3
=23
2.............23=3*2*3+5
=18+5
=23
the above two equations are satisfied by x=2 and y=3
Therefore the required number is 23.
36.First we increased the denominator of a positive
fraction by 3 and then it by 5.The sum of the
resulting fractions proves to be equal to 2/3.
Find the denominator of the fraction if its numerator is 2.
Solution:
Let us assume the fraction is x/y
First we increasing the denominator by 3 we get x/(y-3)
Then decrease it by 5 we get the fraction as x/(y-5)
Given that the sum of the resulting fraction is 2/3
x/(y+3) + x/(y-5) = 2/3
Given numerator equal to 2
2*[ 1/y+3 + 1/y-5] =2/3
(y-5+y+3) / (y-3)(y+5) =1/3
6y â€“ 6 = y2-5y+3y-15
y2-8y-9 = 0
y2-9y+y-9 = 0
y(y-9)+1(y-9) = 0
Therefore y =-1 or 9.
37.If we divide a two digit number by a number consisting
of the same digits written in the reverse order,we get 4
as quotient and 15 as a remainder. If we subtract 1 from
the given number we get the sum of the squares of the
digits constituting that number. Find the number?
a.71 b.83 c.99 d. none
Solution:
Let the number be x y.
If we divide 10x+y by a number in reverse order
i e,10y+x we get 4 as quotient and 15 as remainder.
We can write as
10x+y = 4(10y+x)+15......................1
If we subtract 1 from the given number we get square of the digits
10x+y = x2+y2.....................................2
By using above two equations and trail and error method
we get the required number. From the options also we can
solve the problem. In this no option is satisfied so answer is d.
Ans (d)

Problems on Ages
Simple problems:
1.The present age of a father is 3 years more than three times
the age of his son.Three years hence,fatherâ€™s age will be 10
years more than twice the age of the son.Find the present age
of the father.
Solution: Let the present age be 'x' years.
Then father's present age is 3x+3 years.
Three years hence
(3x+3)+3=2(x+3)+10
x=10
Hence father's present age = 3x+3 = 33 years.
2. One year ago the ratio of Ramu & Somu age was 6:7respectively.
Four years hence their ratio would become 7:8. How old is Somu.
Solution: Let us assume Ramu &Somu ages are x &y respectively.
One year ago their ratio was 6:7
i.e x-1 / y-1 = 7x-6y=1
Four years hence their ratios,would become 7:8
i.e x-4 / y-4 = 7 / 8
8x-7y=-4
From the above two equations we get y= 36 years.
i.e Somu present age is 36 years.
3. The total age of A &B is 12 years more than the total age of
B&C. C is how many year younger than A.
Solution: From the given data
A+B = 12+(B+C)
A+B-(B+C) = 12
A-C=12 years.
C is 12 years younger than A
4. The ratio of the present age of P & Q is 6:7. If Q is 4 years
old than P. what will be the ratio of the ages of P & Q after
4 years.
Solution: The present age of P & Q is 6:7 i.e
P / Q = 6 / 7
Q is 4 years old than P i.e Q = P+4.
P/ P+4 = 6/7
7P-6P = 24,
P = 24 , Q = P+4 =24+4 = 28
After 4 years the ratio of P &Q is
P+4:Q+4
24+4 : 28+4 = 28:32 = 7:8
5. The ratio of the age of a man & his wife is 4:3.After 4 years this
ratio will be 9:7. If the time of marriage the ratio was 5:3,
then how many years ago were they married.
Solution: The age of a man is 4x .
The age of his wife is 3x.
After 4 years their ratio's will be 9:7 i.e
4x+4 / 3x+4 = 9 / 7
28x-27x=36-28
x = 8.
Age of a man is 4x = 4*8 = 32 years.
Age of his wife is 3x = 3*8 = 24 years.
Let us assume 'y' years ago they were married ,
the ratio was 5:3 ,i.e
32-y / 24-y = 5/ 3
y=12 years
i.e 12 years ago they were married
Top
6. Sneh's age is 1/6th of her father's age.Sneh's father's age will
be twice the age of Vimal's age after 10 years. If Vimalâ€™s eight
birthday was celebrated two years before,then what is Sneh's
present age.
a) 6 2/3 years b) 24 years c) 30 years d) None of the above

Solution: Assume Snehâ€™s age is 'x' years.
Assume her fathers age is 'y' years.
Snehâ€™s age is 1/6 of her fathers age i.e x = y /6.
Fatherâ€™s age will be twice of Vimal's age after 10
years.
i.e y+10 = 2( V+10)( where 'V' is the Vimal's age)
Vimal's eight birthday was celebrated two years before,
Then the Vimal's present age is 10 years.
Y+10 = 2(10+10)
Y=30 years.
Sneh's present age x = y/6
x = 30/6 = 5 years.
Sneh's present age is 5 years.
7.The sum of the ages of the 5 children's born at the intervals of
3 years each is 50 years what is the age of the youngest child.
a) 4 years b) 8 years c) 10 years d)None of the above

Solution: Let the age of the children's be
x ,x+3, x+6, x+9, x+12.
x+(x+3)+(x+6)+(x+9)+(x+12) = 50
5x+30 = 50
5x = 20
x=4.
Age of the youngest child is x = 4 years.
8. If 6 years are subtracted from the present age of Gagan and
the remainder is divided by 18,then the present age of his
grandson Anup is obtained. If Anup is 2 years younger to Madan
whose age is 5 years,then what is Gagan's present age.
a) 48 years b)60 years c)84 years d)65 years

Solution: Let us assume Gagan present age is 'x' years.
Anup age = 5-2 = 3 years.
(x-6) / 18 = 3
x-6 = 54
x=60 years
9.My brother is 3 years elder to me. My father was 28 years of
age when my sister was born while my father was 26 years of age
when i was born. If my sister was 4 years of age when my brother
was born,then what was the age my father and mother respectively
when my brother was born.
a) 32 yrs, 23yrs b)32 yrs, 29yrs c)35 yrs,29yrs d)35yrs,33 yrs
Solution: My brother was born 3 years before I was born & 4
years after my sister was born.
Father's age when brother was born = 28+4 = 32 years.
Mother's age when brother was born = 26-3 = 23 years.

SURDS AND INDICES
Simple problems:
1. Laws of Indices:

(i) am * an = a(m+n)
(ii) am / an = a(m-n)
(iii) (am)n = a(m*n)
(iv) (ab)n = an * bn
(v) (a/b)n = an / bn
(vi) a0 = 1
2.Surds :Let 'a' be a rational number & 'n' be a positive
integer such that a1/n = nth root a is irrational.Then nth
root a is called 'a' surd of 'n'.
Problems:-
(1)
(i) (27)2/3 = (33)2/3 = 32 = 9.
(ii) (1024)-4/5 = (45)-4/5 = (4)-4= 1/(4)4 = 1/256.
(iii)(8/125)-4/3 =((2/5)3)-4/3 = (2/5)-4 = (5/2)4 = 625/16
(2) If 2(x-1)+ 2(x+1) = 1280 then find the value of x .
Solution: 2x/2+2x.2 = 1280
2x(1+22) = 2*1280
2x = 2560/5
2x = 512 => 2x = 29
x = 9
(3) Find the value of [5[81/3+271/3]3]1/4

Solution: [5[(23)1/3+(33)1/3]3]1/4
[5[2+3]3]1/4
[54]1/4 => 5.
(4) If (1/5)3y= 0.008 then find the value of (0.25)y

Solution: (1/5)3y = 0.008
(1/5)3y =[0.2]3
(1/5)3y =(1/5)3
3y= 3 => y=1.
(0.25)y = (0.25)1 => 0.25 = 25/100 = 1/4

(5) Find the value of (243)n/5 * 32n+1 / 9n * 3 n-1

Solution: (35)n/5 * 32n +1 / (32)n * 3n-1
33n+1 / 33n-1 3
33n+1 * 3-3n+1 => 32 =>9.
(6) Find the value of (21/4-1)( 23/4 +21/2+21/4+1)

Solution: Let us say 21/4 = x
(x-1)(x3+x2+x+1)
(x-1)(x2(x+1)+(x+1))
(x-1) (x2+1) (x+1) [(x-1)(x+1) = (x2-1)]
(x2+1) (x2-1) => (x4-1)
((21/4))4 - 1) = > (2-1) = > 1.
(7) If x= ya , y = zb , z = xc then find the value
of abc.
Solution: z= xc
z= (ya)c [ x= ya ]
z= (y)ac
z= (zb)ac [y= zb]
z= zabc
abc = 1
(8)Simplify (xa/xb)a2+ab+b2*(xb/xc)b2+bc+c2*(xc/xa)c2+ca+a2
Solution:[xa-b]a2+ab+b2 * [xb-c]b2+bc+c2 * [xc-a]c2+ca+a2

[ (a-b)(a2+ab+b2) = a3-b3]

from the above formula
=> xa3-b3 xb3-c3 xc3-a3
=> xa3-b3+b3-c3+c3-a3
=> x0 = 1
(9) (1000)7 /1018 = ?
(a) 10 (b) 100 (c ) 1000 (d) 10000

Solution: (1000)7 / 1018
(103)7 / (10)18 = > (10)21 / (10)18

=> (10)21-18 => (10)3 => 1000
Ans :( c )
(10) The value of (8-25-8-26) is

(a) 7* 8-25 (b) 7*8-26 (c ) 8* 8-26 (d) None
Solution: ( 8-25 - 8-26 )
=> 8-26 (8-1 )
=> 7* 8-26
Ans: (b)
Top
(11) 1 / (1+ an-m ) +1/ (1+am-n) = ?
(a) 0 (b) 1/2 (c ) 1 (d) an+m

Solution: 1/ (1+ an/am) + 1/ ( 1+ am/an)
=> am / (am+ an ) + an /(am +an )
=> (am +an ) /(am + an)
=> 1
Ans: ( c)
(12) 1/(1+xb-a+xc-a)+1/(1+xa-b+xc-b)+1/(1+xb-c+xa-c)=?

(a) 0 (b) 1 ( c ) xa-b-c (d) None of the above
Solution: 1/ (1+xb/xa+xc/xa) + 1/(1+xa/xb +xc/xb) +
1/(1+xb/xc +xa/xc)
=> xa /(xa +xb+xc) + xb/(xa +xb+xc) +xc/(xa +xb+xc)
=>(xa +xb+xc) /(xa +xb+xc)
=>1
Ans: (b)
(13) If x=3+2 âˆš2 then the value of (âˆšx â€“ 1/ âˆšx)
is [ âˆš=root]
(a) 1 (b) 2 (c ) 2âˆš2 ( d) 3âˆš3

Solution: (âˆšx-1/âˆšx)2 = x+ 1/x-2
=> 3+2âˆš2 + (1/3+2âˆš2 )-2
=> 3+2âˆš2 + 3-2âˆš2 -2
=> 6-2 = 4
(âˆšx-1/âˆšx)2 = 4
=>(âˆšx-1/âˆšx)2 = 22
(âˆšx-1/âˆšx) = 2.
Ans : (b)
(14) (xb/xc)b+c-a (xc/xa)c+a-b (xa/xb)b+a-c = ?
(a) xabc (b) 1 ( c) xab+bc+ca (d) xa+b+c

Solution: [xb-c]b+c-a [xc-a]c+a-b [xa-b]a+b-c

=>x(b-c)(b+c-a) x(c-a)(c+a-b) x(a-b)(a+b-c)
=>x(b2-c2-ab-ac) x(c2-a2-bc-ab) x(a2-b2-ac-bc)
=>x(b2-c2-ab-ac+c2-a2-bc-ab+a2-b2-ac-bc)
=> x0
=>1
Ans: (b)

(15) If 3x-y = 27 and 3x+y = 243 then x is equal to

(a) 0 (b) 2 (c ) 4 (d) 6
Solution: 3x-y = 27 => 3x-y = 33
x-y= 3
3x+y = 243 => 3x+y = 35
x+y = 5
From above two equations x = 4 , y=1
Ans: (c )
(16) If ax = by = cz and b2 = ac then â€˜yâ€™equals

(a)xz/x+z (b)xz/2(x-z) (c)xz/2(z-x) (d)2xz/x+z
Solution: Let us say ax = by = cz = k
ax =k => [ax]1/x = k1/x
=> a = k1/x
Simillarly b = k1/y
c = k1/z
b2 = ac
[k1/y]2=k1/xk1/z
=>k2/y = k1/x+1/z
=> 2/y = 1/x+1/y
=>y= 2xz/x+z
Ans: (d)
(17) ax = b,by = c ,cz = a then the value of
xyz is is

(a) 0 (b) 1 (c ) 1/abc (d) abc

Solution: ax = b
(cz)x = b [cz = a]
by)xz = b [by = c]
=>xyz =1
Ans: (b)
(18) If 2x = 4y =8z and (1/2x +1/4y +1/6z) =24/7 then
the value of 'z' is

(a) 7/16 (b) 7 / 32 (c ) 7/48 (d) 7/64

Solution: 2x = 4y=8z
2x = 22y = 23z
x= 2y = 3z
Multiply above equation with â€˜ 2â€™
2x = 4y= 6z
(1/2x+1/4y+1/6z) = 24/7
=>(1/6z+1/6z+1/6z) = 24/7
=> 3 / 6z = 24/7
=> z= 7/48
Ans: ( c)

Problems on Percentages
Simple problems:
1 . Express the following as a fraction.
a) 56%
SOLUTION:
56/100=14/25
b) 4%
SOLUTION:
4/100=1/25
c) 0.6%
SOLUTION:
0.6/100=6/1000=3/500

d) 0.08%
SOLUTION:
0.08/100=8/10000=1/1250
2. Express the following as decimals

a) 6%
SOLUTION:
6% = 6/100=0.06

b) 0.04%
SOLUTION:
0.04% = 0.04/100=0.0004
3 . Express the following as rate percent.
i).23/36

SOLUTION:
= (23/36*100) %
= 63 8/9%
ii).6 Â¾

SOLUTION:
6 Â¾ =27/4
(27/4 *100) % =675 %
4. Evaluate the following:
28% of 450 + 45% of 280 ?

SOLUTION:
=(28/100) *450 + (45/100) *280
= 28 * 45 / 5
= 252
5. 2 is what percent of 50?
SOLUTION:
Formula : (IS / OF ) *100 %
= 2/50 *100
= 4%
6. Â½ is what percent of 1/3?
SOLUTION:
=( Â½) / (1/3) *100 %
= 3/2 *100 %
= 150 %
7. What percent of 2 Metric tonnes is 40 Quintals?
SOLUTION:
1 metric tonne =10 Quintals
So required percentage=(40/(2*10))*100%
= 200%
8. Find the missing figure .
i) ? % of 25 = 2.125
SOLUTION :
Let x% of 25 = 2.125.then
(x/100) *25 =2.125
x = 2.125 * 4
= 8.5
ii) 9% of ? =6.3

SOLUTION:
Let 9 % of x = 6.3.
Then 9/100 of x= 6.3
so x = 6.3 *100/7
= 70.
9. Which is the greatest in 16 2/3 %, 2/15,0.17?
SOLUTION:
16 2/3 % = 50/3 %
=50/3 * 1/100
=1/6
= 0.166
2 / 15 =0.133
So 0.17 is greatest number in the given series.
10.If the sales tax be reduced from 3 Â½ % to 3 1/3 % ,
then what difference does it make to a person who
purchases an article with marked price of RS 8400?
SOLUTION:

Required difference = 3 Â½ % of 8400 â€“ 3 1/3 % of 8400
=(7/2-10/3)% of 8400
=1/6 % of 8400
= 1/6* 1/100* 8400
= Rs 14.
11. A rejects 0.08% of the meters as defective .How many
will he examine to reject 2?

SOLUTION:
Let the number of meters to be examined be x.
Then 0.08% of x=2.
0.08/100*x= 2
x= 2 * 100/0.08
=2 * 100 * 100/8
= 2500
12.65 % of a number is 21 less than 4/5 of that number.
What is the number?
SOLUTION: Let the number be x.
4/5 x- (65% of x) = 21
4/5x â€“ 65/100 x=21
15x=2100
x=140
13. Difference of two numbers is 1660.If 7.5 % of one number
is 12.5% of the other number. Find two numbers?

SOLUTION:
Let the two numbers be x and y.
7.5% of x=12.5% of y'
So 75x=125 y
3x=5y
x=5/3y.
Now x-y=1660
5/3y-y=1660
2/3y=1660
y=2490
So x= 2490+1660
=4150.
So the numbers are 4150 , 1660.
14. In expressing a length 81.472 KM as nearly as possible
with 3 significant digits ,Find the % error?
SOLUTION:
Error= 81.5-81.472=0.028
So the required percentage = 0.028/81.472*100%
= 0.034%
Top
15. In an election between two persons ,75% of the voters
cast their votes out of which 2% are invalid. A got
9261 which 75% of the total valid votes. Find total
number of votes?
SOLUTION:
Let x be the total votes.
valid votes are 98% of 75% of x.
So 75%(98%(75% of x))) = 9261
==> 75/100 *98 /100 * 75 100 *x = 9261
x= 1029 * 4 *100 *4 / 9
= 16800
So total no of votes = 16800
16 . A's maths test had 75 problems i.e 10 arithmetic, 30
algebra and 35 geometry problems.Although he answered
70% of arithmetic ,40% of algebra and 60 % of geometry
problems correctly he didn't pass the test because he
got less than 60% of the problems right. How many more
questions he would have needed to answer correctly to
get a 60% passing grade.

SOLUTION:
70% of 10 =70/100 * 10
=7
40% of 30 = 40 / 100 * 30
= 12
60 % of 35 = 60 / 100 *35
= 21
So correctly attempted questions = 7 + 12 + 21
=40.
Questions to be answered correctly for 60% grade
=60% of 75
= 60/100 *75
=45.
So required questions=45-40 = 5
17 . If 50% of (x â€“ y) = 30% of (x + y) then what percent
of x is y ?

SOLUTION:
50/100(x-y) =30/100(x+y)
Â½ (x-y)= 3/10(x+y)
5x-5y=3x+3y
x=4y
So Required percentage =y/x*100 %
=y/4y *100 %
= 25%.
18 .If the price of tea is increased by 20% ,find how much
percent must a householder reduce her consumption of tea
so as not to increase the expenditure?
SOLUTION:
Reduction in consumption= R/(100+R) *100%
=20/120 *100
= 16 2/3 %
19.The population of a town is 176400 . If it increases
at the rate of 5% per annum ,what will be the
population 2 years hence? What was it 2 years ago?

SOLUTION:
Population After 2 years = 176400[1+5/100]2
=176400 * 21/20 *21/20
=194481
Population 2 years ago = 176400/(1+5/100)2
= 176400 * 20/21 *20/ 21
=160000
20.1 liter of water is add to 5 liters of a 20 % solution
of alcohol in water . Find the strength of alcohol in
new solution?

SOLUTION:
Alcohol in 5 liters = 20% of 5
=1 liter
Alcohol in 6 liters of new mixture = 1liter
So % of alcohol is =1/6 *100=16 2/3%
21.If A earns 33 1/3 more than B .Then B earns less
than A by what percent?

SOLUTION:
33 1/3 =100 /
Required Percentage = (100/3)/(100 + (100/3)) *100 %
= 100/400 *100 = 25 %
22.A school has only three classes which contain
40,50,60 students respectively.The pass percent of
these classes are 10, 20 and 10 respectively . Then
find the pass percent in the school.

SOLUTION:
Number of passed candidates =
10/100*40+20/100 *50+10/100 * 60
=4+10+6
=20
Total students in school = 40+50+60 =150
So required percentage = 20/150 *100
= 40 /3
=13 1/3 %
23. There are 600 boys in a hostel . Each plays either
hockey or football or both .If 75% play hockey and
45 % play football ,Find how many play both?

SOLUTION:
n(A)=75/100 *600
=450
n(B) = 45/100 *600
= 270
n(A^B)=n(A) + n(B) â€“ n(AUB)
=450 + 270 -600
=120
So 120 boys play both the games.
24.A bag contains 600 coins of 25p denomination and
1200 coins of 50p denomination. If 12% of 25p coins
and 24 % of 50p coins are removed, Find the percentage
of money removed from the bag ?

SOLUTION:
Total money = (600 * 25/100 +1200 *50/100)
=Rs 750
25p coins removed = 12/100 *600
=72
50p coins removed = 24/100 *1200
=288
So money removed =72 *1/4 +288 *1/2
= Rs 162
So required percentage=162/750 *100
=21 .6%
25. P is six times as large as Q.Find the percent that
Q is less than P?

SOLUTION:
Given that P= 6Q
So Q is less than P by 5Q.
Required percentage= 5Q/P*100 %
=5/6 * 100 %
=83 1/3%
26.For a sphere of radius 10 cm ,the numerical value of
surface area is what percent of the numerical value
of its volume?
SOLUTION:
Surface area = 4 *22/7 *r2
= 3/r(4/3 * 22/7 * r3)
=3/r * VOLUME
Where r = 10 cm
So we have S= 3/10 V
=3/10 *100 % of V
= 30 % of V
So surface area is 30 % of Volume.
27. A reduction of 21 % in the price of wheat enables
a person to buy 10 .5 kg more for Rs 100.What is
the reduced price per kg.

SOLUTION:
Let the original price = Rs x/kg
Reduced price =79/100x /kg
==> 100/(79x/100)-100/x =10.5
==> 10000/79x-100/x=10.5
==> 10000-7900=10.5 * 79 x
==> x= 2100/10.5 *79
So required price = Rs (79/100 *2100/10.5 *79) /kg
= Rs 2 per kg.
28.The length of a rectangle is increased by 60 % .
By what percent would the width have to be decreased
to maintain the same area?
SOLUTION:
Let the length =l,Breadth= b.
Let the required decrease in breadth be x %
then 160/100 l *(100-x)/100 b=lb
160(100-x)=100 *100
or 100-x =10000/160
=125/2
so x = 100-125/2
=75/2=37.5

Profit and Loss
Important Facts:
Cost Price: The price at which an article is purchased,
is called its cost price,abbreviated as C.P.
Selling Price: The price at which an article is sold,
is called its selling price,abbreviated as S.P.
Profit or Gain: If S.P. Is greater than C.P. The seller
is said to have a profit or gain.
Loss:if S.P. Is less than C.P., the seller is said to
have incurred a loss.
Formulae

1.Gain=(S.P-C.P)
2.Loss=(C.P-S.P)
3.Loss or Gain is always reckoned on C.P.
4.Gain%=(gain*100)/C.P
5.Loss%=(loss*100)/C.P
6.S.P=[(100+gain%)/100]*C.P
7.S.P=[(100-loss%)/100]*C.P
8.C.P=(100*S.P)/(100+gain%)
9.C.P=(100*S.P)/(100-loss%)
10.If an article is sold at a gain of say,35%,then S.P=135% of C.P.
11.If an article is sold at a loss of say,35%,then S.P=65% of C.P.
12.When a person sells two similar items, one at a gain of say,
x%,and the other at a loss of x%,then the seller always incurs a
loss given by Loss%=[common loss and gain %/10]2=(x/10)2
13.If a trader professes to sell his goods at cost price,but uses
false weight,then Gain%=[(error/(true value-error))*100]%
14.Net selling price=Marked price-Discount
Top
Simple Problems
1.A man buys an article for Rs.27.50 and sells it for Rs.28.60
Find the gain percent.
Sol: C.P=Rs 27.50 S.P=Rs 28.60
then Gain=S.P-C.P=28.60-27.50=Rs 1.10
Gain%=(gain*100)/C.P%
=(1.10*100)/27.50%=4%
2.If a radio is purchased for Rs 490 and sold for Rs 465.50
Find the loss%?
Sol: C.P=Rs 490 S.P=Rs 465.50
Loss=C.P-S.P=490-465.50=Rs 24.50
Loss%=(loss*100)/C.P%
=(24.50*100)/490%=5%
3.Find S.P when C.P=Rs 56.25 and Gain=20%
Sol: S.P=[(100+gain%)/100]*C.P
S.P=[(100+20)/100]56.25=Rs 67.50
4.Find S.P when C.P=Rs 80.40,loss=5%
Sol: S.P=[(100-loss%)/100]*C.P
S.P=[(100-5)/100]*80.40=Rs 68.34
5.Find C.P when S.P=Rs 40.60,gain=16%?
Sol: C.P=(100*S.P)/(100+gain%)
C.P=(100*40.60)/(100+16)=Rs 35
6.Find C.P when S.P=Rs 51.70 ,loss=12%?
Sol: C.P=(100*S.P)/(100-loss%)
C.P=(100*51.70)/(100-12)=Rs 58.75

7.A person incurs 5% loss by selling a watch for Rs 1140 . At
what price should the watch be sold to earn 5% profit?
Sol: Let the new S.P be Rs x then,
(100-loss%):(1st S.P)=(100+gain%):(2nd S.P)
(100-5)/1140=(100+5)/x
x=(105*1140)/95=Rs 1260
8.If the cost price is 96% of the selling price,then what is
the profit percent?
Sol: let S.P=Rs 100 then C.P=Rs 96
profit=S.P-C.P=100-96=Rs 4
profit%=(profit*C.P)/100%
=(4*96)/100=4.17%
9.A discount dealer professes to sell his goods at cost price
but uses a weight of 960 gms for a Kg weight .Find his gain %?
Sol: Gain%=[(error*100)/(true value-error)]%
=[(40*100)/1000-40)]%=25/6%
10.A man sold two flats for Rs 675,958 each .On one he gains
16% while on the other he losses 16%.How much does he gain or
lose in the whole transaction?
Sol: loss%=[common loss or gain%/10]2=(16/10)2=2.56%
11.A man sold two cows at Rs 1995 each. On one he lost 10% and
on the other he gained 10%.what his gain or loss percent?
Sol: If loss% and gain% is equal to 10
then there is no loss or no gain.
12.The price of an article is reduced by 25% in order to
restore the must be increased by ?
Sol: [x/(100-x)]*100 =[25/(100-25)]*100
=(25/75)*100=100/3%
13.Two discounts of 40% and 20% equal to a single discount of?
Sol: {[(100-40)/100]*[(100-20)/100]}%=(60*80)/(100*100)%
=48%
single discount is equal to (100-48)%=52%
Top
Difficult Problems
1.The cost of an article including the sales tax is Rs 616.The
rate of sales tax is 10%,if the shopkeeper has made a profit
of 12%,then the cost price of the article is?
Sol: 110% of S.P=616
S.P=(616*100)/110=Rs 560
C.P=(100*S.P)/(100+gain%)
C.P. =(100*560)/(100+12)=Rs 500
2.Sam purchased 20 dozens of toys at the rate of 375 Rs per dozen.
He sold each one of then at the rate of Rs 33.What was his
percentage profit?
Sol: C.P of one toy=Rs 375/12=Rs 31.25
S.P of one toy=Rs 33
profit=S.P-C.P=33-31.25=Rs 1.75
profit %=(profit/C.P)*100
=(1.75/31.25)*100
profit% =5.6%
3.Two third of consignment was sold at a profit of 5% and the
remainder at a loss of 2%.If the total was Rs 400,the value of the
consignment was?
Sol: let the total value be Rs x
value of 2/3=2x/3, value of 1/3=x/3
total S.P value be Rs x
total S.P=[(105% of 2x/3)+(98% of x/3)]
=(105*2x)/(100*3)+(98/100)+x/3
=308x/300
(308x/300)-x=400
8x/300=400
x=(300*400)/8=Rs 15000
4.Kunal bought a suitcase with 15% discount on the labelled price.
He sold the suitcase for Rs 2880 with 20% profit on the labelled
price .At what price did he buy the suitcase?
Sol: let the labelled price be Rs x
then 120% of x=2880
x=(2880*100)/120=Rs 2400
C.P=85% of the 2400
(85*2400)/100=Rs 2040
5.A tradesman gives 4% discount on the marked price and gives
article free for buying every 15 articles and thus gains 35%. The
marked price is above the cost price by
Sol: let the C.P of each article be Rs 100
then C.P of 16 articles=Rs (100*16)=Rs 1600
S.P of 15 articles =1600*(135/100)=Rs 2160
S.P of each article =2160/15=Rs 144
If S.P is Rs 96, marked price =Rs 100
If S.P is Rs 144,marked price =(100/96)*144=Rs 15000
therefore marked price=50% above C.P
6.By selling 33m of cloth ,one gains the selling price of 11m.Find
the gain percent?
Sol: gain=S.P of 33m-C.P of 33m
=11m of S.P
S.P of 22m=C.P of 33m
let C.P of each meter be Rs 1,then C.P of 22m=Rs 22
S.P of 22m=Rs 33
gain=S.P-C.P=33-22=Rs 11
gain%=(gain/C.P)*100
=(11/22)*100=50%
Top
7.The price of a jewel, passing through three hands, rises on the
whole by 65%.if the first and second sellers earned 20% and 25%
profit respectively,find the percentage profit earned by the
third seller?
Sol: let the original price of the jewel be Rs P and
let the profit earned by the third seller be x%
then (100+x)% of 125% of P=165% of P
[(100+x)/100]*(125/100)*(120/100)*P=(165/100)*P
100+x=(165*100*100)/(125*120)
100+x=110
x=10%
8.When a producer allows 36% commission on the retail price of his
product ,he earns a profit of 8.8%.what would be his profit
percent if the commission is reduced by 24%
Sol: let retail price =Rs 100
commission=Rs 36
S.P=retail price-commission=100-36=Rs 64
But profit=8.8%
C.P=(100*C.P)/(gain+100)=(100*64)/(100+8.8)=Rs 1000/17
new commission=Rs 12
new S.P=100-12=Rs 88
gain=88-(1000/17)=Rs 496/17
gain%=gain*100/C.P
=(496*17*100)(17*1000)
gain%=49.6%
9.Vikas bought paper sheets for Rs 7200 and spent Rs 200 on
transport. Paying Rs 600,he had 330 boxes made,which he sold
at Rs 28 each. His profit percentage is
Sol: total investments=7200+200+600=Rs 8000
total receipt=330*28=Rs 9240
gain=S.P-C.P
=total receipt-total investments
gain=9240-8000=Rs 1240
gain% =gain*100/C.P=1240*100/8000=15.5%
10.A person earns 15% on investment but loses 10% on another
investment .If the ratio of the two investments be 3:5 ,what is the
gain or loss on the two investments taken together?
Sol: let the investments be 3x and 5x
then total investment=8x
total receipt=115% of 3x+90% of 5x
=115*3x/100+90*5x/100=7.95x
loss=C.P-S.P=8x-7.95x=0.05x
loss%=.05x*100/8x=0.625%
11.The profit earned by selling an article for Rs 900 is double the
loss incurred when the same article is sold for Rs 490 .At what
price should the article be sold to make 25% profit?
Sol: let C.P be Rs x
900-x=2(x-450)
3x=1800
x=Rs 600
C.P=Rs 600 , gain required=25%
S.P=(100+gain%)*C.P/100
S.P=(100+25)*600/100=Rs 750
12.If an article is sold at 5% gain instead of 5% loss,the seller
gets Rs 6.72 more. The C.P of the article is?
Sol: let C.P be Rs x
105% of x-95% of x=6.72
(105/100)*x-(95/100)*x=6.72
x/10=6.72
x=Rs 67.21.

Ratio and Proportions
Important Facts:
Ratio: The ratio of two qualities a and b in the same units,
is the fraction a/b and we write it as a:b. In the ratio, a:b, we
call ‘a’ as the first term of antecedent and b, the second
term consequent.
Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9
Rule: The multiplication or division of each term of 9 ratio
by the same non-zero number does not affect the ratio.
Proportion: The equality of two ratios is called proportion.
If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in
proportion. Here a and b are called extremes, while b and c are
called mean terms.
Product of means=product of extremes
Thus, a:b::c:d => (b*c)=(a*d)

Fourth proportional: If a:b::c:d, then d is called the fourth
proportional to a,b and c.
Third proportional: If a:b::b:c, then c is called third
proportional to a and b.
Mean proportional: Mean proportional between a and b is SQRT(a*b).
Comparision of Ratios:
We say that (a:b)>(c:d) => (a/b)>(c/d)
Compounded ratio: The compounded ratio of the ratios (a:b),
(c:d),(e:f) is (ace:bdf).
Duplicate Ratio: If (a:b) is (a2: b2 )
Sub-duplicate ratio of (a:b) is (SQRT(a):SQRT(b))
Triplicate ratio of (a:b) is (a3: b3 )
Sub-triplicate ratio of (a:b) is (a1/3: b1/3 ).
If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d) (componend o and
dividend o)

Variation:
we say that x is directly proportional to y, if x=ky for some
constant k and we write.
We say that x is inversely proportional to y, if xy=k for some
constant and we write.
X is inversely proportional to y.
If a/b=c/d=e/f=g/h=k then
k=(a+c+e+g)/(b+d+f+h)
If a1/b1,a2/b2, a3/b3..............an/bn are unequal
fractions then the ratio.
a1+a2+a3+..........an/(b1+b2+b3+...............bn) lies between the
lowest & the highest of the three fractions.
Top
Simple Problems
1.If a:b =5:9 and b:c=4:7 Find a:b:c?
Sol: a:b=5:9 and b:c=4:7=4*9/4:9*4/9=9:63/9
a:b:c=5:9:63/9=20:36:63
2.Find the fourth proportion to 4,9,12
Sol: d is the fourth proportion to a,b,c
a:b=c:d
4:9=12:x
4x=9*12=>x=27
3.Find third proportion to 16,36
Sol: If a:b=b:c then c is the third proportion to a,b
16:36=36:x
16x=36*36
x=81
4.Find mean proportion between 0.08 and 0.18
Sol: mean proportion between a and b=square root of ab
mean proportion =square root of 0.08*0.18=0.12

5.If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d is
Sol: a:b=2:3 and b:c=4:5=4*3/4:5*3/4=3:15/4
c:d=6:7=6*15/24:7*15/24=15/4:35/8
a:b:c:d=2:3:15/4:35/8=16:24:30:35

6.2A=3B=4C then A:B:C?
Sol: let 2A=3B=4C=k then
A=k/2, B=k/3, C=k/4
A:B:C=k/2:k/3:k/4=6:4:3

7.15% of x=20% of y then x:y is
Sol: (15/100)*x=(20/100)*y
3x=4y
x:y=4:3

8.a/3=b/4=c/7 then (a+b+c)/c=
Sol: let a/3=b/4=c/7=k
(a+b+c)/c=(3k+4k+7k)/7k=2
9.Rs 3650 is divided among 4 engineers, 3 MBA’s and 5 CA’s
such that 3 CA’s get as much as 2 MBA’s and 3 Eng’s as
much as 2 CA’s .Find the share of an MBA.
Sol: 4E+3M+5C=3650
3C=2M, that is M=1.5C
3E=2C that is E=.66 C
Then, (4*0.66C)+(3*1.5C)+5C=3650
C=3650/12.166
C=300
M=1.5 and C=450.
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Difficule Problems
1.Three containers A,B and C are having mixtures of milk and
water in the ratio of 1:5 and 3:5 and 5:7 respectively. If
the capacities of the containers are in the ratio of all the
three containers are in the ratio 5:4:5, find the ratio of
milk to water, if the mixtures of all the three containers
are mixed together.
Sol: Assume that there are 500,400 and 500 liters respectively
in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of
the three containers.
Thus, the total milk is 441.66 liters. Hence, the amount of
water in the mixture is 1400-441.66=958.33liters.
Hence, the ratio of milk to water is
441.66:958.33 => 53:115(using division by .3333)
The calculation thought process should be
(441*2+2):(958*3+1)=1325:2875
Dividing by 25 => 53:115.
2.A certain number of one rupee,fifty parse and twenty five
paise coins are in the ratio of 2:5:3:4, add up to Rs 210.
How many 50 paise coins were there?
Sol: the ratio of 2.5:3:4 can be written as 5:6:8
let us assume that there are 5 one rupee coins,6 fifty
paise coins and 8 twenty-five paise coins in all.
their value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10
If the total is Rs 10,number of 50 paise coins are 6.
if the total is Rs 210, number of 50 paise coins would be
210*6/10=126.
3.The incomes of A and B are in the ratio of 4:3 and their
expenditure are in the ratio of 2:1 . if each one saves
Rs 1000,what are their incomes?
Sol: Ratio of incomes of A and B=4:3
Ratio of expenditures of A and B=2:1
Amount of money saved by A=Amount of money saved by B=Rs 1000
let the incomes of A and B be 4x and 3x respectively
let the expense of A and B be 2y and 1yrespectively
Amount of money saved by A=(income-expenditure)=4x-2y=Rs 1000
Amount of money saved by B=3x-y=Rs 1000
this can be even written as 6x-2y=Rs 2000
now solve 1 and 3 to get
x=Rs 500
therefore income of A=4x=4*500=Rs 2000
income of B=3x=3*500=Rs 1500

4.A sum of Rs 1162 is divided among A,B and C. Such that 4
times A's share share is equal to 5 times B's share and 7
times C's share . What is the share of C?
Sol: 4 times of A's share =5 times of B's share=7 times of
C's share=1
therefore , the ratio of their share =1/4:1/5:1/7
LCM of 4,5,7=140
therefore, ¼:1/5:1/7=35:28:20
the ratio now can be written as 35:28:20
therefore C's share=(20/83)*1162=20*14=Rs 280.

5.The ratio of the present ages of saritha and her mother is
2:9, mother's age at the time of saritha's birth was 28 years,
what is saritha's present age?
Sol: ratio of ages of saritha and her mother =2:9
let the present age of saritha be 2x years.
then the mother's present age would be 9x years
Difference in their ages =28 years
9x-2x=28 years
7x=28=>x=4
therefore saritha's age =2*4=8 years

Partnership
Important Facts:
Partnership:When two or more than two persons run a business
jointly, they are called partners and the deal is known as partnership.
Ratio of Division of Gains:
1.When the investments of all the partners are do the same time, the
gain or loss is distributed among the partners in the ratio of their
investments.
Suppose A and B invest Rs x and Rs y respectively for a year in a
business, then at the end of the year:
(A's share of profit):(B's share of profit)=x:y
2.When investments are for different time periods, then equivalent
capitals are calculated for a unit of time by taking (capital*number
of units of time). Now gain or loss is divided in the ratio of these
capitals.
Suppose A invests Rs x for p months and B invests Rs y for q months,
then (A's share of profit):(B's share of profit)=xp:yq
3.Working and sleeping partners:A partner who manages the business is
known as working partner and the one who simply invests the money is
a sleeping partner.
Formulae
1.When investments of A and B are Rs x and Rs y for a year in a
business ,then at the end of the year
(A's share of profit):(B's share of profit)=x:y
2.When A invests Rs x for p months and B invests Rs y for q months,
then A's share profit:B's share of profit=xp:yq
Short cuts:
1.In case of 3 A,B,C investments then individual share is to be found
then A=16000 , B=32,000 , C=40,000
Sol: A:B:C=16:32:40
=2:4:5`
then individual share can be easily known.
2.If business mans A contributes for 5 months and B contributes for 9
months then share of B in the total profit of Rs 26,8000 ,A = Rs 15000,
B =Rs 12000
Sol: 15000*5 : 12000*9
25 : 36
for 36 parts=268000*(36/61)
=Rs 158.16
Back Top
Difficult problems:
1.P and Q started a business investing Rs 85,000 and Rs 15,000
respectively. In what ratio the profit earned after 2 years be divided
between P and Q respectively?
Sol: 85,000*2 : 15,000*2
17*2 : 3*2=34:6
2.A,B and C started a business by investing Rs 1,20,000, Rs 1,35,000
and Rs 1,50,000.Find the share of each ,out of an annual profit of
Rs 56,700?
Sol: Ratio of shares of A,B and C=Ratio of their investments
120,000:135,000:150,000
=8:9:10
A's share=Rs 56,700*(8/27)=Rs 16,800
B's share =Rs 56,700*(9/27)=Rs 18,900
C's share =Rs 56,700*(10/27)=Rs 21,000
3.3 milkman A,B,C rented a pasture A grazed his 45 cows for 12 days
B grazed his 36 cows for 15 days and c 60 cows for 10 days.If b's
share of rent was Rs 540 What is the total rent?
Sol: 45*12:36*15:60*10
=9:9:10
9 parts is equal to Rs 540
then one part is equal to Rs 60
total rent=60*28=Rs 1680
4.Ramu and Krishna entered into a partnership with Rs 50,000 and
Rs 60,000, after 4 months Ramu invested Rs 25,000 more while Krishna
withdraw Rs 20,000 . Find the share of Ramu in the annual profit of
Rs 289,000.
Sol: Ramu : Krishna=50,000*4+75,000*8:60,000*4+40,000*8
=10:7
Ramu's annual profit=289000*(10/17)=Rs 170,000
5.A,B,C enter into partnership .A invests 3 times as much as B invests
and B invests two third of what C invests. At the end of the year ,the
profit earned is Rs 6600. what is the share of B?
Sol: let C's capital =Rs x
B's capital=Rs (2/3)*x
A's capital =3*(2/3)*x=Rs 2x
ratio of their capitals=2x:(2/3)*x:x
=6x:2x;3x
B's share =Rs 6600(2/11)=Rs 1200

6.A,B and C enter into a partnership by investing in the ratio of 3:2:4.
After one year ,B invests another Rs 2,70,000 and C,at the end of 2
years, also invests Rs 2,70,000.At the end of 3 years ,profit are shared
in the ratio of 3:4:5.Find the initial investment of each?
Sol: Initial investments of A,B,c be Rs 3x, Rs 2x, Rs 4x then for 3 years
(3x*36):[(2x*12)+(2x+270000)*24]:[(4x*24)+(4x+270000)*12]=3:4:5
108x:(72x+640,000):(144x+3240000)=3:4:5
108x:72x+6480000:144x+3240000=3:4:5
(108x)/(72x+6480000)=3/4
432x=216x+19440000
216x=19440000
x=Rs 90000
A's initial investment=3x=3*90,000=Rs 2,70,000
B's initial investment=2x=2*90,000=Rs 1,80,000
C's initial investment=4x=4*90,000=Rs 3,60,000

Chain Rule
Important Facts:
Direct Proportion: Two Quantities are said to be directly
proportional, if on the increase (or decrease) of th one, the
other increases(or decreases) to the same extent.
Ex:(i) Cost is directly proportional to the number of articles.
(More articles, More cost).
(ii) Work done is directly proportional to the number of men
working on it. (More men, more work).
Indirect Proportion: Two Quantities are said to be
indirectly proportional,if on the increase of the one , the other
decreases to the same extent and vice-versa.
Ex:(i) The time taken by a car covering a certain distance is
inversely proportional to th speed of the car.(More speed,
less is the time taken to cover the distance).
(ii) Time taken to finish a work is inversely proportional to
the number of persons working at it.
(More persons, less is the time taken to finish a job).
Nte: In solving Questions by chain rule, we compare every
item with the term to be found out.
Problems
1)If 15 toys cost Rs.234, what do 35 toys cost ?
Sol: Let the required cost be Rs. x then
more toys more cost(direct proportion)
15:35:: 234:x
(15*x)=(234*35)
x=(234*35) /(15)= 546 Rs
2)If 36 men can do a piece of work in 25hours, in how many hours
will 15men do it?
Sol: Let the required number of hours be x.
less men more hours(Indirect proportion).
15:36::25:x
(15*x)=(36*25)
x=(36*25) /15
x=60
For 15 men it takes 60 hours.
3)If 9 engines consume 24metric tonnes of coal, when each is working
8 hours a day, how much coal will be required for 8 engines, each
running 13 hours a day, it being given that 3 engines of former type
consume as much as 4 engines of latter type?
Sol: Let 3 engines of former type consume 1 unit in 1 hour.
4 engines of latter type consume 1 unit in 1 hour.
1 engine of former type consumes 1/3 unit in 1 hour.
1 engine of latter type consumes ¼ unit in 1 hour.
Let required consumption of coal be x units.
Less engines, less coal consumed.(direct)
More working hours, more coal consumed(direct)
Less rate of consumption, less coal consumed (direct)

9:8
8:13 :: 24:x
1/3:1/4
(9*8*(1/3)*x)=(8*13*(1/4)*24)
24x=624
x=26 metric tonnes.
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Complex Problems
1)A contract is to be completed in 46 days and 117 men were set to work,
each working 8 hours a day. After 33 days, 4/7 of the work is completed.
How many additional men may be employed so that the work may be
completed in time, each man now working 9 hours a day?
Sol: 4/7 of work is completed .
Remaining work=1- 4/7
=3/7
Remaining period= 46-33
=13 days
Less work, less men(direct proportion)
less days, more men(Indirect proportion)
More hours/day, less men(Indirect proportion)
work 4/7:3/7
Days 13:33 :: 117:x
hrs/day 9:8

(4/7)*13*9*x=(3/7)*33*8*117
x=(3*33*8*117) / (4*13*9)
x=198 men
So, additional men to be employed=198 -117=81
2)A garrison had provisions for a certain number of days. After 10 days,
1/5 of the men desert and it is found that the provisions will now last
just as long as before. How long was that?
Sol: Let initially there be x men having food for y days.
After, 10 days x men had food for ( y-10)days
Also, (x -x/5) men had food for y days.
x(y-10)=(4x/5)*y
=> (x*y) -50x=(4(x*y)/5)
5(x*y)-4(x*y)=50x
x*y=50x
y=50
3)A contractor undertook to do a certain piece of work in 40 days. He
engages 100 men at the beginning and 100 more after 35 days and completes
the work in stipulated time. If he had not engaged the additional men,
how many days behind schedule would it be finished?
Sol: 40 days- 35 days=5 days
=>(100*35)+(200*5) men can finish the work in 1 day.
4500 men can finish it in 4500/100= 45 days
This s 5 days behind the schedule.
4)12 men and 18 boys,working 7 ½ hors a day, can do a piece if work in
60 days. If a man works equal to 2 boys, then how many boys will be
required to help 21 men to do twice the work in 50 days, working
9 hours a day?
Sol: 1man =2 boys
12men+18boys=>(12*2+18)boys=42 boys
let the required number of boys=x
21 men+x boys
=>((21*2)+x) boys
=>(42+x) boys
less days, more boys(Indirect proportion)
more hours per day, less boys(Indirect proportion)
days 50:60
hrs/day 9:15/2 :: 42:(42+x)
work 1:2
(50*9*1*(42+x))=60*(15/2)*2*42
(42+x)= (60*15*42)/(50*9)= 84
x=84-42= 42
=42
42 days behind the schedule it will be finished.

Time and Work
Important Facts:
1.If A can do a piece of work in n days, then A's 1 day work=1/n
2.If A's 1 day's work=1/n, then A can finish the work in n days.
Ex: If A can do a piece of work in 4 days,then A's 1 day's work=1/4.
If A's 1 day’s work=1/5, then A can finish the work in 5 days
3.If A is thrice as good workman as B,then: Ratio of work done by
A and B =3:1. Ratio of time taken by A and B to finish a work=1:3
4.Definition of Variation: The change in two different variables
follow some definite rule. It said that the two variables vary
directly or inversely.Its notation is X/Y=k, where k is called
constant. This variation is called direct variation. XY=k. This
variation is called inverse variation.
5.Some Pairs of Variables:
i)Number of workers and their wages. If the number of workers
increases, their total wages increase. If the number of days
reduced, there will be less work. If the number of days is
increased, there will be more work. Therefore, here we have
direct proportion or direct variation.
ii)Number workers and days required to do a certain work is an
example of inverse variation. If more men are employed, they
will require fewer days and if there are less number of workers,
more days are required.
iii)There is an inverse proportion between the daily hours of a
work and the days required. If the number of hours is increased,
less number of days are required and if the number of hours is
reduced, more days are required.
6.Some Important Tips:
More Men -Less Days and Conversely More Day-Less Men.
More Men -More Work and Conversely More Work-More Men.
More Days-More Work and Conversely More Work-More Days.
Number of days required to complete the given work=Total work/One
day’s work.
Since the total work is assumed to be one(unit), the number of days
required to complete the given work would be the reciprocal of one
day’s work.
Sometimes, the problems on time and work can be solved using the
proportional rule ((man*days*hours)/work) in another situation.
7.If men is fixed,work is proportional to time. If work is fixed,
then time is inversely proportional to men therefore,
(M1*T1/W1)=(M2*T2/W2)
Problems
1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 men
working 8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.
How many more man should be engaged to finish the rest of the work in
6 days working 9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can
10 women and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls ---------------?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the
same job. How long it take both A & B, working together but independently,
to do the same job?
Sol: A's one hour work=1/8.
B's one hour work=1/10
(A+B)'s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the
time taken by A. Then, working together, what part of the same work they
can finish in a day?
Sol: Given that B alone can complete the same work in days=half the time
taken by A=9days
A's one day work=1/18
B's one day work=1/9
(A+B)'s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of
work in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How
many days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A's one hour's work=1/63 and
B's one hour work=1/42
(A+B)'s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?

Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
work then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z
can do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
Top
Complex Problems
1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in
6 days while B alone can do it in 8 days. With the help of C, they can finish
it in 3 days, Find the share of each?
Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24
Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1
A's share=Rs (600*4/8)=300
B's share= Rs (600*3/8)=225
C's share=Rs[600-(300+225)]=Rs 75
2)A can do a piece of work in 80 days. He works at it for 10 days & then B alone
finishes the remaining work in 42 days. In how much time will A and B, working
together, finish the work?
Sol: Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A's one day's work=1/80
B’s one day's work=1/48
(A+B)'s one day's work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
3)P,Q and R are three typists who working simultaneously can type 216 pages
in 4 hours In one hour , R can type as many pages more than Q as Q can type more
than P. During a period of five hours, R can type as many pages as P can
during seven hours. How many pages does each of them type per hour?
Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x+y+z=216/4=54 ---------------1
z-y=y-x => 2y=x+z -----------2
5z=7x => x=5x/7 ---------------3
Solving 1,2 and 3 we get x=15,y=18, and z=21
4)Ronald and Elan are working on an assignment. Ronald takes 6 hours to
type 32 pages on a computer, while Elan takes 5 hours to type 40 pages.
How much time will they take, working together on two different computers
to type an assignment of 110 pages?
Sol: Number of pages typed by Ronald in one hour=32/6=16/3
Number of pages typed by Elan in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours.
5)Two workers A and B are engaged to do a work. A working alone takes 8 hours
more to complete the job than if both working together. If B worked alone,
he would need 4 1/2 hours more to compete the job than they both working
together. What time would they take to do the work together.
Sol: (1/(x+8))+(1/(x+(9/2)))=1/x
=>(1/(x+8))+(2/(2x+9))=1/x
=> x(4x+25)=(x+8)(2x+9)
=> 2x2 =72
=> x2 = 36
=> x=6
Therefore, A and B together can do the work in 6 days.
6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.
If A,B and C work together, they will complete the work in how many days?
Sol: (A+B)'s one day's work=1/12;
(B+C)'s one day's work=1/15;
(A+C)'s one day's work=1/20;
Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5
(A+B+C)'s one day work=1/10
So, A,B,and C together can complete the work in 10 days.
7)A and B can do a work in 8 days, B and C can do the same wor in 12 days.
A,B and C together can finish it in 6 days. A and C together will do it in
how many days?
Sol: (A+B+C)'s one day's work=1/6;
(A+B)'s one day's work=1/8;
(B+C)'s one day's work=1/12;
(A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day
work+(B+C)'s one day work)
= (2/6)-(1/8+1/12)
=(1/3)- (5/24)
=3/24
=1/8
So, A and C together will do the work in 8 days.
8)A can do a certain work in the same time in which B and C together can do it.
If A and B together could do it in 10 days and C alone in 50 days, then B alone
could do it in how many days?
Sol: (A+B)'s one day's work=1/10;
C's one day's work=1/50
(A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25
Also, A's one day's work=(B+C)’s one day's work
From i and ii ,we get :2*(A's one day's work)=3/25
=> A's one day's work=3/50
B's one day’s work=(1/10-3/50)
=2/50
=1/25
B alone could complete the work in 25 days.
9) A is thrice as good a workman as B and therefore is able to finish a job
in 60 days less than B. Working together, they can do it in:
Sol: Ratio of times taken by A and B=1:3.
If difference of time is 2 days , B takes 3 days
If difference of time is 60 days, B takes (3*60/2)=90 days
So, A takes 30 days to do the work=1/90
A's one day's work=1/30;
B's one day's work=1/90;
(A+B)'s one day's work=1/30+1/90=4/90=2/45
Therefore, A&B together can do the work in 45/2days
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10) A can do a piece of work in 80 days. He works at it for 10 days and then
B alone finishes the remaining work in 42 days. In how much time will A&B,
working together, finish the work?
Sol: Work Done by A n 10 days =10/80=1/8
Remaining work =1-1/8=7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7= 48 days
=> A's one day's work =1/80 and
B's one day's work =1/48
(A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days.
11) 45 men can complete a work in 16 days. Six days after they started working,
so more men joined them. How many days will they now take to complete the
remaining work?
Sol: M1*D1/W1=M2*D2/W2
=>45*6/(6/16)=75*x/(1-(6/16))
=> x=6 days
12)A is 50% as efficient as B. C does half the work done by A&B together. If
C alone does the work n 40 days, then A,B and C together can do the work in:
Sol: A's one day's work:B's one days work=150:100 =3:2
Let A's &B's one day's work be 3x and 2x days respectively.
Then C's one day's work=5x/2
=> 5x/2=1/40
=> x=((1/40)*(2/5))=1/100
A's one day's work=3/100
B's one day's work=1/50
C's one day's work=1/40
So, A,B and C can do the work in 13 1/3 days.
13)A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work?
Sol: B's 10 day's work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the
work but are forced to leave after 3 days. The remaining work done by A in:
Sol: (B+C)'s one day's work=1/9+1/12=7/36
Work done by B & C in 3 days=3*7/36=7/12
Remaining work=1-(7/12)=5/12
Now , 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in 24*5/12=10 days
15)X and Y can do a piece of work n 20 days and 12 days respectively. X started
the work alone and then after 4 days Y joined him till the completion of work.
How long did the work last?
Sol: work done by X in 4 days =4/20 =1/5
Remaining work= 1-1/5 =4/5
(X+Y)'s one day's work =1/20+1/12 =8/60=2/15
Now, 2/15 work is done by X and Y in one day.
So, 4/5 work will be done by X and Y in 15/2*4/5=6 days
Hence Total time taken =(6+4) days = 10 days
16)A does 4/5 of work in 20 days. He then calls in B and they together finish
the remaining work in 3 days. How long B alone would take to do the whole work?
Sol: Whole work is done by A in 20*5/4=25 days
Now, (1-(4/5)) i.e 1/5 work is done by A& B in days.
Whole work will be done by A& B in 3*5=15 days
=>B's one day's work= 1/15-1/25=4/150=2/75
So, B alone would do the work in 75/2= 37 ½ days.
17) A and B can do a piece of work in 45 days and 40 days respectively. They
began to do the work together but A leaves after some days and then B completed
the remaining work n 23 days. The number of days after which A left the work was
Sol: (A+B)'s one day's work=1/45+1/40=17/360
Work done by B in 23 days=23/40
Remaining work=1-(23/40)=17/40
Now, 17/360 work was done by (A+B) in 1 day.
17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days
So, A left after 9 days.
18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days.
The rest of work finished by C in 2 days. If they get Rs 1500 for the whole
work, the daily wages of B and C are
Sol: Part of work done by A= 5/10=1/2
Part of work done by B=1/3
Part of work done by C=(1-(1/2+1/3))=1/6
A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1
A's share=(3/6)*1500=750
B's share=(2/6)*1500=500
C's share=(1/6)*1500=250
A's daily wages=750/5=150/-
B's daily wages=500/5=100/-
C's daily wages=250/2=125/-
Daily wages of B&C = 100+125=225/-
19)A alone can complete a work in 16 days and B alone can complete the same
in 12 days. Starting with A, they work on alternate days. The total work will
be completed in how many days?
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days
Sol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48
work done in 6 pairs of days =(7/48) * 6 = 7/8
remaining work = 1- 7/8 = 1/8
work done by A on 13th day = 1/16
remaining work = 1/8 – 1/16 = 1/16
on 14th day, it is B’s turn
1/12 work is done by B in 1 day.
1/16 work is done by B in ¾ day.
Total time taken= 13 ¾ days.
So, Answer is: D
Top
20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how
many days can A do the work if he is assisted by B and C on every third day?
Sol: A's two day's work=2/20=1/10
(A+B+C)'s one day's work=1/20+1/30+1/60=6/60=1/10
Work done in 3 days=(1/10+1/10)=1/5
Now, 1/5 work is done in 3 days
Therefore, Whole work will be done in (3*5)=15 days.
21)Seven men can complete a work in 12 days. They started the work and after
5 days, two men left. In how many days will the work be completed by the
remaining men?
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none
Sol: 7*12 men complete the work in 1 day.
Therefore, 1 man's 1 day's work=1/84
7 men's 5 days work = 5/12
=>remaining work = 1-5/12 = 7/12
5 men's 1 day's work = 5/84
5/84 work is don by them in 1 day
7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days.
Ans: E
22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more
men joined them. How many days will they take to complete the remaining work?
(a) 2 days (b) 3 days (c) 4 days (d) 5days
Sol : 1 man's 1 day work = 1/108
12 men's 6 days work = 6/9 = 2/3
remaining work = 1 – 2/3 = 1/3
18 men's 1 days work = 18/108 = 1/6
1/6 work is done by them in 1 day
therefore, 1/3 work is done by them in 6/3 = 2 days.
Ans : A
23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.
How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?
(a). 1 (b). 4 (c). 19 (d). 41
Sol : (1 man + 1 woman)'s 1 days work = 1/3+1/4=7/12
Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48
Remaining work= 1- 7/48= 41/48
Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48
Therefore, Number of boys required= ((41/48)*48)= 41 days
So,Answer: D
24)12 men can complete a piece of work in 4 days, while 15 women can complete
the same work in 4 days. 6 men start working on the job and after working for
2 days, all of them stopped working. How many women should be put on the job
to complete the remaining work, if it is to be completed in 3 days.
(A) 15 (B) 18 (C) 22 (D) data inadequate
Sol: one man's one day's work= 1/48
one woman's one day's work=1/60
6 men's 2 day's work=((6/48)*2)= ¼
Remaining work=3/4
Now, 1/60 work s done in 1 day by 1 woman.
So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman.
So, Answer: A
25)Twelve children take sixteen days to complete a work which can be completed
by 8 adults in 12 days. Sixteen adults left and four children joined them. How
many days will they take to complete the remaining work?
(A) 3 (B) 4 ( C) 6 (D) 8
Sol: one child's one day work= 1/192;
one adult's one day's work= 1/96;
work done in 3 days=((1/96)*16*3)= 1/2
Remaining work= 1 – ½=1/2
(6 adults+ 4 children)'s 1 day's work= 6/96+4/192= 1/12
1/12 work is done by them in 1 day.
½ work is done by them 12*(1/2)= 6 days
So, Answer= C
26)Sixteen men can complete a work in twelve days. Twenty four children can
complete the same work in 18 days. 12 men and 8 children started working and
after eight days three more children joined them. How many days will they now
take to complete the remaining work?
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days
ol: one man's one day's work= 1/192
one child's one day's work= 1/432
Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54
Remaining work= 1 -35/54= 19/54
(12 men+11 children)'s 1 day's work= 12/192 + 11/432 = 19/216
Now, 19/216 work is done by them in 1 day.
Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days
So,Answer: B
27)Twenty-four men can complete a work in 16 days. Thirty- two women can
complete the same work in twenty-four days. Sixteen men and sixteen women
started working and worked for 12 days. How many more men are to be added to
complete the remaining work in 2 days?
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men
Sol: one man's one day's work= 1/384
one woman's one day's work=1/768
Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4
Remaining work=1 – ¾=1/4
(16 men+16 women)'s two day's work =12*( 16/384+16/768)=2/16=1/8
Remaining work = 1/4-1/8 =1/8
1/384 work is done n 1 day by 1 man.
Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men
28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women
can complete it in 10 days. In how many days will 10 women complete it?
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days
Sol: Let 1 man's 1 day's work =x days and
1 woman's 1 day's work=y
Then, 4x+6y=1/8 and 3x+7y=1/10.
Solving these two equations, we get: x=11/400 and y= 1/400
Therefore, 1 woman's 1 day's work=1/400
=> 10 women will complete the work in 40 days.
Answer: B
29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boys
can do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do
it in?
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs
Sol: Let 1 man's 1 hour's work=x
1 woman's 1 hour's work=y
1 boy's 1 hour's work=z
Then, x+3y+4z=1/96 -----------(1)
2x+8z= 1/80 ----------(2)
adding (2) & (3) and subtracting (1)
3x+4z=1/96 ---------(4)
From (2) and (4), we get x=1/480
Substituting, we get : y=1/720 and z= 1/960
(5 men+ 12 boy)'s 1 hour's work=5/480+12/960 =1/96 + 1/80=11/480
Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours.
So,Answer: C

Pipes and Cisterns
Important Facts:
1.INLET:A pipe connected with a tank or cistern or a reservoir,
that fills it, it is known as Inlet.
OUTLET:A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as Outlet.
2. i) If a pipe can fill a tank in x hours, then :
part filled in 1 hour=1/x.
ii)If a pipe can empty a tank in y hours, then :
part emptied in 1 hour=1/y.
iii)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where y>x), then on
opening both the pipes, the net part filled in
1 hour=(1/x -1/y).
iv)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where x>y), then on opening
both the pipes, the net part filled in 1 hour=(1/y -1/x).
v) If two pipes A and B can fill a tank in x hours and y hours
respectively. If both the pipes are opened simultaneously, part
filled by A+B in 1 hour= 1/x +1/y.
Top
Simple Problems
1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.
If both the pipes are opened simultaneously, how much time will be
taken to fill the tank?
Sol: Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45;
Part filled by (A+B)'s in 1 hour=1/36 +1/45= 9/180 =1/20
Hence, both the pipes together will fill the tank in 20 hours.
2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While
3rd pipe empties the full tank n 20 hours. If all the three pipes
operate simultaneously,in how much time will the tank be filled?
Sol: Net part filled in 1 hour=1/10 +1/12 -1/20
=8/60=2/15
The tank be filled in 15/2hours= 7 hrs 30 min
3)A cistern can be filled by a tap in 4 hours while it can be emptied
by another tap in 9 hours. If both the taps are opened simultaneously,
then after how much time will the cistern get filled?
Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.
4)If two pipes function simultaneously, the reservoir will be filled in
12 days.One pipe fills the reservoir 10 hours faster than the other.
How many hours does it take the second pipe to fill the reservoir.
Sol: Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x=20
So, the second pipe will take 30 hours to fill the reservoir.
5)A cistern has two taps which fill it in 12 min and 15 min respectively.
There is also a waste pipe in the cistern. When all the three are opened,
the empty cistern is full in 20 min. How long will the waste pipe take to
empty the full cistern?
Sol: Work done by a waste pipe in 1 min
=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)
6)A tap can fill a tank in 6 hours. After half the tank is filled, three
more similar taps are opened. What is the total time taken to fill the
tank completely?
Sol: Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
Therefore, 2/3:1/2::1:x
or x=(1/2)*1*(3/2)=3/4 hours.
i.e 45 min
So, total time taken= 3hrs 45min.
7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B
can empty it in 6 min. If both pipes are open, how long will it take to
empty or fill the tank completely ?
Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied=2/5.
Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15
Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.
So, the tank be emptied in 6 min.
8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for
Bucket P to fill the empty drum. How many turns it will take for both the
buckets P&Q, having each turn together to fill the empty drum?
Sol: Let the capacity of P be x lit.
Then capacity of Q=x/3 lit
Capacity of the drum=60x lit
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45
Top
Complex Problems
1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The
pipes are opened simultaneously and it is found that due to leakage in the
bottom it took 32min more to fill the cistern. When the cistern is full,
in what time will the leak empty it?
Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.
2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After
7 min, C also opened. In how much time, the tank s full.
Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min
3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If
both the pipes are opened simultaneously, after how much time B should
be closed so that the tank is full in 18 min.
Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 =1
=> x=8
Hence B must be closed after 8 min.
4)Two pipes A& B together can fill a cistern in 4 hours. Had they been
opened separately, then B would have taken 6 hours more than A to fill
the cistern. How much time will be taken by A to fill the cistern
separately?
Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.
5)A tank is filled by 3 pipes with uniform flow. The first two pipes
operating simultaneously fill the tan in the same time during which
the tank is filled by the third pipe alone. The 2nd pipe fills the tank
5 hours faster than first pipe and 4 hours slower than third pipe. The
time required by first pipe is :
Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.
Top
6)A large tanker can be filled by two pipes A& B in 60min and 40 min
respectively. How many minutes will it take to fill the tanker from
empty state if B is used for half the time & A and B fill it together for
the other half?
Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.
7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.
If they are opened on alternate hours and if pipe A s opened first, in
how many hours, the tank shall be full.
Sol: (A+B)'s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)'s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank=(4+1)= 5 hours.
8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.
If A is open all the time and B and C are open for one hour each
alternatively, the tank will be full in.
Sol: (A+B)'s 1 hour's work=1/12+1/15=9/60=3/20
(A+C)'s 1 hour's work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.
Therefore, total time taken to fill the tank=6+1=7 hours
9)A Booster pump can be used for filling as well as for emptying a tank.
The capacity of the tank is 2400 m3. The emptying capacity of the tank is
10 m3 per minute higher than its filling capacity and the pump needs 8
minutes lesser to empty the tank than it needs to fill it. What is the
filling capacity of the pump?
Sol: Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump=(x+10) m3/min.
So,2400/x – 2400/(x+10) = 8
on solving x=50.
10)A leak in the bottom of a tank can empty the full tan in 8 hr. An inlet
pipe fills water at the rate of 6 lits a minute. When the tank is full,
the inlet is opened and due to the leak, the tank is empty in 12 hrs.
How many liters does the cistern hold?
Sol: Work done by the inlet in 1 hr= 1/8 -1/12=1/24
Work done by the inlet in i min= (1/24)*(1/60)=1/1440
Therefore, Volume of 1/1440 part=6 lit
Volume of whole=(1440*6) lit=8640 lit.
11)Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes
respectively. Both the pipes are opened. The cistern will be filled in
just half an hour, if the pipe B is turned off after:
sol: Let B be turned off after x min. Then,
Part filled by (A+B) in x min+ part filled by A in (30-x)min=1
Therefore, x(2/75+1/45)+(30-x)(2/75)=1
11x/225 + (60-2x)/75=1
11x+ 180-6x=225
x=9.
So, B must be turned off after 9 minutes.

Time and Distance
Formulae:
I)Speed = Distance/Time
II)Time = Distance/speed
III) Distance = speed*time
IV) 1km/hr = 5/18 m/s
V)1 m/s = 18/5 Km/hr
VI)If the ratio of the speed of A and B is a:b,then the ratio of
the time taken by them to cover the same distance is 1/a : 1/b
or b:a
VII) suppose a man covers a distance at x kmph and an equal
distance at y kmph.then the average speed during the whole
journey is (2xy/x+y)kmph
Problems
1)A person covers a certain distance at 7kmph .How many meters
does he cover in 2 minutes.
Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
2)If a man runs at 3m/s. How many km does he run in 1hr 40min
Solution::
speed of the man = 3*18/5 kmph
= 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
3)Walking at the rate of 4knph a man covers certain distance
in 2hr 45 min. Running at a speed of 16.5 kmph the man will
cover the same distance in.
Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
Top
Complex Problems
1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.
Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph
2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?
Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x --------------1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2
divide 1 & 2 equations
by solving we get x = 40
5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.
Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction
Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.
Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9
8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.
Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief
Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.
Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
11)A motorist covers a distance of 39km in 45min by moving at a
speed of xkmph for the first 15min.then moving at double the
speed for the next 20 min and then again moving at his original
speed for the rest of the journey .then x=?
Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a
uniform speed of 8kmph.his average speed during the whole
journey is.
Solution::
When same distance is covered with different speeds,then the
average speed = 2xy / x+y
=10.88kmph
13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.
Solution::
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
Top
14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.
Solution::
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its
speed to cover the remaining distance in the time left.
Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
16)The ratio between the speeds of the A& B is 2:3 an
therefore A takes 10 min more than the time taken by B to reach
the destination.If A had walked at double the speed ,he would
have covered the distance in ?
Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min
A's time=30 min
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed
doubles (i.e 2x)
17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?
Solution::
speed's ratio
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey.
Solution::
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another
Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
20)A man takes 5hr 45min in walking to certain place and riding
back. He would have gained 2hrs by riding both ways.The time he
would take to walk both ways is?
Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance
Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min

Trains
General Concept:
(1) Time taken by a train x mt long in passing a signal post
or a pole or a standing man = time taken by the train to cover x mt
(2) Time taken by a train x mt long in passing a stationary
object of length y mt = time taken by the train to cover x+y mt
(3) Suppose two trains or two bodies are moving in the same
direction at u kmph and v kmph such that u > v then their
relative speed is u-v kmph
(4)If two trains of length x km and y km are moving in opposite
diredtions at u kmph and vmph,then time taken by the train to
cross each other = (x+y)/(u+v) hr
(5) Suppose two trains or two bdies are moving in opposite direction
at u kmph and v kmph then,their relative speed = (u+v) kmph
(6)If two train start at the same time from 2 points A & B towards
each other and after crossing they take a & b hours in reaching B & A
respectively then A's speed : B's speed = (b^1/2 : a^1/2 )
Problems

(1)Find the time taken by a train 180m long,running at 72kmph in
crossing an electric pole
Solution:
Speed of the train =72*5/18m/s =20 m/s
Distance move din passing the pole = 180m
Requiredtime = 180/20 = 9 seconds
(2)A train 140 m long running at 60kmph.In how much time will it
pass a platform 260m long.
Solution:
Distance travelled =140 + 260 m =400 m,
speed = 60 * 5/18 = 50//3 m
time=400*3 / 50 = 24 Seconds
(3)A man is standing on a railway bridge which is 180 m.He finds
that a train crosses the bridge in 20 seconds but himself in
8 sec. Find the length of the train and its sppeed
Solution:
i)D=180+x
T = 20 seconds
S= 180+x / 20 ------------ 1
ii)D=x
T=8 seconds
D=ST
x=8S ------------- 2
Substitute 2 in 1
S=180 + 8 S / 20
S=15 m/s
Length of the train,x is 8 *15 = 120 m
(4)A train 150m long is running with a speed of 68 mphIn wht
time will it pass a man who is running at a speed of 8kmph in
the same direction in which the train is going
Solution:
Relative Speed = 68-8=60kmph*5/18 = 50/3 m/s
time= 150 * 3 /50 =9sec
5)A train 220m long is running with a speed of 59 k mph /..In
what time will it pass a man who is running at 7 kmph in the
direction opposite to that in which train is going.
Solution:
Relative Speed = 59+7=66kmph*5/18 = 55/3 m/s
time= 220/55 * 3 =12sec
Top
(6)Two trains 137m and 163m in length are running towards each
other on parallel lines,one at the rate of 42kmph & another at
48 mph.In wht time will they be clear of each other from the
moment they meet.
Solution:
Relative speed =42+48 = 90 *5/18 = 25m/s
time taken by the train to pass each other = time taken to cover
(137+163)m at 25 m/s
= 300 /25 s =12 s
(7)A train running at 54 kmph takes 20 sec to pass a platform.
Next it takes 12 sec to pass a man walking at 6kmph in the same
direction in which the train is going.Find length of the train
and length of platform
Solution:
Relative speed w.r.t man = 54-6=48kmph
the length of the train is 48 * 5/18 * 12 =160m
time taken to pass platform =20 sec
Speed of the train = 54 * 5/18 =15m/s
160+x =20 *15
x=140m
length of the platform is 140m
(8)A man sitting in a train which is travelling at 50mph observes
that a goods train travelling in opposite irection takes 9 sec
to pass him .If the goos train is 150m long fin its speed
Solution:
Relative speed =150/9 m/s =60 mph
speed of the train = 60-50 =10kmph
(9)Two trains are moving in the sam e direction at 65kmph and
45kmph. The faster train crosses a man in slower train in18sec.the
length of the faster train is
Solution:
Relative speed =65-45 kmph = 50/9 m/s
Distancce covered in18 s =50/9 * 18 = 100m
the length of the train is 100m
(10)Atrain overtakes two persons who are walking in the same
direction in which the train is going at the rate of 2kmph an
4kmph and passes them completely in 9 sec an 10 sec respectively.
The length of train is
Solution:
2kmph = 5/9 m/s
4 mph =10/9 m/s
Let the length of the trainbe x meters and its speed is y m/s
then x / (y- 5/9) = 9 and x / (y- 10/9) = 10
9y-5 =x and 10(9y-10)=9x
9y-x=5 and 90y-9x=100
on solving we get x=50,lenght of trains
(11) Two stations A & B are 110 km apart on a straight line.
One train starts from A at 7am and travels towards B at 20kmph.
Another train starts from B at 8am an travels toward A at a speed
of 25kmph.At what time will they meet
Solution:
Suppose the train meet x hr after 7am
Distance covered by A in x hr=20x km
20x+25(x-1) = 110
45x=135
x=3
So they meet at 10 am
(12)A traintravelling at 48kmph completely crosses another train
having half its length an travelling inopposite direction at 42kmph
in12 sec.It also passes a railway platform in 45sec.the length of
platform is
Solution:
Let the length of the first train be x mt
then,the length of second train is x/2 mt
relative speed = 48+42 kmph =90 * 5/18 m/s = 25m/s
(x+ x/2)/25 =12
x=200
Length of the train is 200m
Let the length of the platform be y mt
speed f the first train = 48*5/18 m/s = 40/3 m/s
200+y * 3/40 = 45
y=400m
Top
(13)The length of a running trsain in 30% more than the length of
another train B runnng in the opposite direction.To find out the
speed of trtain B,which of the following information given in the
statements P & Q is sufficient
P : The speed of train A is 80 kmph
Q : They too 90 sec to cross each other
(a) Either P & Q is sufficient
(b)Both P & Q are not sufficient
(c)only Q is sufficient
(d)Both P & Q are neeed
Ans: B
Solution:
Let the length of th e train A be x mt
Length of the train B = 130/100 x mt =13x/10 mt
Let the speed of B be y mph,speed of the train A=80mph
relative speed= y+80 * 5/18 m/s
time taken by the trains t cross each other is gven by
90 = (x + 13x/10)/ (5y+400 / 18)
to find y,clearly xis also needed
so,both P & Q are not sufficient
(14)The speed of a train A,100m long is 40% more than then the speed
of another train B,180m long running in opposite direction.To fin out
the speed of B,which of the information given in statements P & Q is
sufficient
P :The two trains crossed each other in 6 seconds
Q : The difference between the spee of the trains is 26kmph
(a)Only P is sufficient
(b)Only Q is sufficient
(c)Both P & Q are needed
(d)Both P & Q are not sufficient
Ans : A
Solution:
Let speed of B be x kmph
then,speed of A =140x/100 kmph =7x/5 mph
relative speed = x + 7x/5 =2x/3 m/s
time taken to cross each other = (100+180)*3/2x s =420/x s
now,420/x = 6
x=70 mph
thus,only P is sufficient
(15)The train running at certain speed crosses astationary enginein
20 seconds.to find out the sped of the train,which of the following
information is necessary
(a)Only the length of the train
(b)only the length of the engine
(c)Either the length of the train or length of engine
(d)Both the length of the train or length of engine
Ans : D
Solution:
Since the sum of lengths of the tran and the engine is needed,
so both the length must be known

Boats and Streams
Important facts:
1)In water, the direction along the stream is called down stream.
2)Direction against the stream is called upstream.
3)The speed of boat in still water is U km/hr and the speed of
stream is V km/hr then
speed down stream =U + V km/hr
speed up stream = U – V km/hr
Formulae:
If the speed down stream is A km/hr and the speed up stream is
B km/hr then speed in still water = ½(A+B) km/hr
rate of stream =1/2(A-B) km/hr
Problems:
1. In one hour a boat goes 11 km long the stream and 5 km
against the stream. The speed of the boat in still water is?
Sol:
Speed in still water = ½ ( 11+5) km/hr= 8 kmph
2.A man can row 18 kmph in still water. It takes him thrice
as long as row up as to row down the river. find the rate
of stream.
Sol:
Let man's rate up stream be xkmph
then, in still water =1/2[3x+x]=2x kmph
so, 2x= 18, x=9
rate upstream =9kmph
rate downstream =27 kmph
rate of stream = ½ [27-9]
= 9kmph
3.A man can row 71/2kmph in still watre . if in a river
running at 1.5 km an hour, if takes him 50 min to row to
place and back. how far off is the place?
Sol: speed down stream =7.5+1.5=9kmph
speed upstream =7.5-1.5=6kmph
let the required distence x km. then ,
x/9+x/6=50/60 = 2x+3x= 5/6*18
5x=15, x=3
Hence, the required distence is 3 km
4.A man can row 3 quarters of a km aganist the stream is
111/4 min. the speed of the man in still water is ?
Sol: rate upstream = 750/625 m/sec =10/9 m/sec
rate downstream =750/450 m/sec = 5/3 m/sec
rate in still water =1/2[10/9+5/3] = 25/18 m/sec
= 25/18*18/5=5 kmph
5.A boat can travel with a speed of 13 kmph in still water.
if the speed of stream is 4 kmph,find the time taken by
the boat to go 68 km downstream?
Sol: Speed down stream = 13+4= 17 kmph
time taken to travel 68km downstream =68/17 hrs
= 4 hrs
Top
6.A boat takes 90 min less to travel 36 miles downstream then
to travel the same distence upstream. if the speed of the
boat in still water is 10 mph . The speed of the stream is :
Sol: Let the speed of the stream be x mph .
then, speed downstream = [10+x]mph
speed upstream =[10-x] mph
36/[10+x] - 36/[10-x] = 90/60 =72x*60= 90[100-x2]
(x+50)(x-2) =0
x=2 kmph
7.At his usual rowing rate, Rahul 12 miles down stream in a
certain river in 6 hrs less than it takes him to travel the
same distence upstream. but if he could double his usual
rowing rate for his 24 miles roundthe down stream 12 miles
would then take only one hour less than the up stream 12 miles.
what is the speed of the current in miles per hours?
Sol: Let the speed in still water be x mph and the speed of
the curren be y mph.
then, speed upstream = (x-y)
speed downstream =(x+y)
12/(x-y) - 12/(x+y) = 6
6(x2 – y2) m= 2xy => x2 – y2 =4y -(1)
and 12/(2x-y) - 12/(2x+y) =1 => 4x2 – y2 = 24y
x2= ( 24y + y2)/4 -->(2)
from 1 and 2 we have
4y+ y2 =( 24y+y2)/4
y=8/3 mph
y= 22/3 mph

8.There is a road beside a river. two friends started from
a place A, moved to a temple situated at another place B
and then returned to A again. one of them moves on a cycle
at a speed of 12 kmph, while the other sails on a boat at a
speed of 10 kmph . if the river flows at the speedof 4 kmph,
which of the two friends will return to place A first ?
Sol: Clearly, The cyclist moves both ways at a speed of 12 kmph
so, average speed of the cyclist = 12 kmph
the boat sailor moves downstream = (10+4) = 14 kmph
upstream =(10-4) = 6 kmph
So, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph
=42/5 kmph =8.4 kmph
Since, the average speed of the cyclist is greater, he will
return to A first.
9.A boat takes 19 hrs for travelling downstream from point A to
point B. and coming back to a point C midway between A and B.
if the velocity of the sream is 4 kmph. and the speed of the
boat in still water is 14 kmph. what is the distence between
A and B?
Sol:
speed downstream =14+4 =18 kmph
speed upstream = 14 -4 = 10 kmph
let the distence between A and B be x km. then,
x/18 + (x/2)/10 = 19
x/18 + x/20 =19
19x/180 =19 =>x = 180km
Hence, the distence between A and B bw 180 km

Alligation or Mixtures
Important Facts and Formulae:
1.Allegation:It is the rule that enables us to find the
ratio in which two of more ingredients at the given price
must be mixed to produce a mixture of a desired price.
2.Mean Price:The cost price of a unit quantity of the
mixture is called the mean price.
3.Rule of Allegation:If two ingredients are mixed then
Quantity of Cheaper / Quantity of Dearer =
(C.P of Dearer – Mean Price) /(Mean Price–C.P of Cheaper).
C.P of a unit quantity of cheaper(c)
C.P of unit quantity of dearer(d)
Mean Price(m)
(d-m) (m-c)
Cheaper quantity:Dearer quantity = (d-m):(m-c)
4.Suppose a container contains x units of liquid from which
y units are taken out and replaced by water. After n
operations the quantity of pure liquid = x (1 – y/x)n units.
Simple Problems
1.In what ratio must rice at Rs 9.30 per Kg be mixed with rice
at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?
Solution:
C.P of 1 Kg rice of 1st kind 930 p
C.P of 1 Kg rice of 2n d kind 1080p
Mean Price 1000p
80 70
Required ratio=80:70 = 8:7
2.How much water must be added to 60 liters of milk at
11/2 liters for Rs 20 so as to have a mixture worth
Rs 10 2/3 a liter?
Solution: C.P of 1 lit of milk = 20*2/3 = 40/3
C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3
Mean Price 32/3
8/3 32/3
Ratio of water and milk =8/3 : 32/3 = 1:4
Quantity of water to be added to 60 lit of milk
=1/4*60=15 liters.
3.In what ratio must water to be mixed with milk to gain
20% by selling the mixture at cost price?
Solution:Let the C.P of milk be Re 1 per liter
Then S.P of 1 liter of mixture = Re.1
Gain obtained =20%.
Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6
C.P of 1 liter of water 0 C.P of 1 liter of milk1
Mean Price 5/6
1/6 5/6
Ratio of water and milk =1/6 : 5/6 = 1:5.
4.In what ratio must a grocer mix two varieties of pulses
costing Rs 15 and Rs 20 per Kg respectively so as to get
a mixture worth Rs 16.50 per Kg?
Solution:
Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of
2nd kind 20
Mean Price Rs 16.50
3.50 1.50
Required ratio =3.50 : 1.50 = 35:15 = 7:3.
5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice
at Rs 6 per Kg .Find the average price of the mixture?
Solution:
rice of 5 Rs per Kg rice of 6 Rs per Kg
Average price Aw
6-Aw Aw-5
(6-Aw)/(Aw-5) = 4/8 =1/2
12-2Aw = Aw-5
3Aw = 17
Aw = 5.66 per Kg.
Top
6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice to
get a mixture costing Rs 7 per Kg. Find the price
of the costlier rice?
Solution: Using the cross method:
rice at Rs 6 per Kg rice at Rs x per Kg
Mean price Rs 7 per Kg
5 4
x-7:1=5:4
4x-28 = 5
4x=33=>x=Rs 8.25.
Therefore price of costlier rice is Rs 8.25 per Kg
Medium Problems
1.A butler stole wine from a butt of sherry which contained
40% of spirit and he replaced,what he had stolen by wine
containing only 16% spirit. The butt was then of 24%
strength only. How much of the butt did he steal?
Solution:
Wine containing 40%spirit Wine containing 16% spirit
Wine containing 24% spirit
8 16
They must be mixed in the ratio of =1:2.
Thus 1/3 of the butt of sherry was left and hence the
butler drew out 2/3 of the butt.
2.The average weekly salary per head of the entire staff
of a factory consisting of supervisors and the laborers
is Rs 60.The average salary per head of the supervisors
is Rs 400 and that of the laborers is Rs 56.Given that
the number of supervisors is 12.Find the number of
laborers in the factory.
Solution:
Average salary of laborer Rs 56 Average salary of
supervisors Rs 400
Average salary of entire staff Rs 60
340 4
Number of laborer / Number of Supervisors = 340 / 4=85/1
Thus,if the number of supervisors is 1,number of
laborers =85.
Therefore if the number of supervisors is 12 number of
laborers 85*12=1020.
3.The cost of type 1 rice is Rs 15 per Kg and type 2 rice
is Rs 20 per Kg. If both type1 and type 2 are mixed in the
ratio of 2:3,then the price per Kg of the mixed variety
of rice is?
Solution:Let the price of the mixed variety be Rs x per Kg.
Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2
rice Rs 20
Mean Price Rs x
20-x x-15
(20-x) /( x-15) = 2/3
=> 60-3x = 2x-30
5x = 90=>x=18.
4.In what ratio must a grocer mix two varieties of tea worth
Rs 60 a Kg and Rs 65 a Kg so that by selling the mixture
at Rs 68.20 a Kg he may gain 10%?
Solution:S.P of 1 Kg of the mixture = Rs 68.20,gain =10%
S.P of 1 Kg of the mixture = Rs (100/110*68.20)=Rs 62.
Cost of 1 Kg tea of 1st kind 60 Cost of 1 Kg tea of 2nd
kind 65
Mean Price Rs 62
3 2
Required ratio =3:2.
5.A dishonest milkman professes to sell his milk at cost price
but he mixes t with water and there by gains 25% .The
percentage of water in the mixture is?
Solution:Let C. P of 1 liter milk be Re 1.
Then S.P of 1 liter mixture=Re 1. Gain=25%
C.P of 1 liter mixture =Re(100/125*1) = Re 4/5.
C.P of 1 liter milk Re 1 C.P of 1 liter of water 0
Mean Price 4/5
4/5 1/5
Ratio of milk to water =4/5 : 1/5 = 4:1
Hence percentage of water n the mixture=1/5*100=20%.
12.A merchant has 1000Kg of sugar,part of which he sells
at 8% profit and the rest at 18% profit. He gains 14% on the
whole. The quantity sold at 18% profit is?
Solution:
Profit on 1st part 8% Profit on 2nd part 18%
Mean Profit 14%
4 6
Ratio of 1st and 2nd parts =4:6 =2:3.
Quantity of 2nd ind =3/5*1000Kg =600 Kg.
6.A jar full of whiskey contains 40% alcohol. A part of
this whiskey is replaced by another containing 19% alcohol
and now the percentage of alcohol was found to be 26%.
The quantity of whiskey replaced is?
Solution:
Strength of first jar 40% Strength of 2nd jar 19%
Mean Strength 26%
7 14
So,ratio of 1st and 2nd quantities =7:14 =1:2
Therefore required quantity replaced =2/3.
7.A container contains 40lit of milk. From this container
4 lit of milk was taken out and replaced by water.
This process was repeated further two times.
How much milk is now contained by the container?
Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit
=(40*9/10*9/10*9/10)
= 29.16 lit
Top
Complex Problems
1.Tea worth Rs 126 per Kg are mixed with a third variety in
the ratio 1:1:2. If the mixture is worth Rs 153 per Kg ,
the price of the third variety per Kg will be?
Solution:
Since First and second varieties are mixed in equal proportions
so their average price =Rs (126+135)/2 = 130.50.
So the mixture is formed by mixing two varieties ,one at
Rs 130.50 per Kg and the other at say Rs x per Kg in the
ratio 2:2 i e,1:1 we have to find x.
Costof 1Kg tea of 1st kind RS 130.50 Costof 1Kg tea of 2n d
kind Rs x.
Mean Price Rs 153
x-153 22.50
(x=153)/22.5 = 1 =>x-153 = 22.5
x = 175.50.
Price of the third variety =Rs 175.50 per Kg.
2.The milk and water in two vessels A and B are in the ratio 4:3
and 2:3 respectively. In what ratio the liquids in both the
vessels be mixed to obtain a new mixture in vessel c
consisting half milk and half water?
Solution:Let the C.P of milk be Re 1 per liter.
Milk in 1 liter mixture of A = 4/7 liter.
Milk in 1 liter mixture of B = 2/5 liter.
Milk in 1 liter mixture of C = 1/2 liter.
C.P of 1 liter mixture in A=Re 4/7
C.P of 1 liter mixture in B=Re 2/5.
Mean Price = Re 1/2.
By rule of allegation we have:
C.P of 1 liter mixture in A C.P of 1 liter mixture in B
4/7 2/5
Mean Price ½
1/10 1/14
Required ratio = 1/10 : 1/14 = 7:5.
3.How many Kg s of wheat costing him Rs 1.20,Rs 1.44
and Rs 1.74 per Kg so that the mixture may be worth
Rs 1.41 per Kg?
Solution:
Step1:Mix wheat of first and third kind to get a mixture
worth Rs 1.41 per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of
3rd kind 174p
Mean Price 141p
33 21
They must be mixed in the ratio =33:21 = 11:7
Step2:Mix wheats of 1st and 2n d kind to obtain a mixture
worth of 1.41.per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 2n d
kind 144p
Mean Price 141p
3 21
They must be mixed in the ratio = 3:21=1:7.
Thus,Quantity of 2n d kind of wheat / Quantity of
3rd kind of wheat = 7/1*11/7= 11/1
Quantities of wheat of 1st :2n d:3rd = 11:77:7.
4.Two vessels A and B contain spirit and water mixed in
the ratio 5:2 and 7:6 respectively. Find the ratio n which
these mixture be mixed to obtain a new mixture in vessel
c containing spirit and water in the ratio 8:5?
Solution:Let the C.P of spirit be Re 1 per liter.
Spirit in 1 liter mix of A = 5/7 liter.
C.P of 1 liter mix in A =5/7.
Spirit in 1 liter mix of B = 7/13 liter.
C.P of 1 liter mix in B =7/13.
Spirit in 1 liter mix of C = 8/13 liter.
C.P of 1 liter mix in C =8/13.
C.P of 1 liter mixture in A 5/7 C.P of 1 liter mixture
in B 7/13
Mean Price 8/13
1/13 9/91
Therefore required ratio = 1/13 : 9/91 = 7:9.
Top
5.A milk vendor has 2 cans of milk .The first contains 5% water
and the rest milk. The second contains 50% water. How much milk
should he mix from each of the container so as to get 12 liters
of milk such that the ratio of water to milk is 3:5?
Solution:Let cost of 1 liter milk be Re 1.
Milk in 1 liter mixture in 1st can = 3/4 lit.
C.P of 1 liter mixture in 1st can =Re 3/4
Milk in 1 liter mixture in 2n d can = 1/2 lit.
C.P of 1 liter mixture in 2n d can =Re 1/2
Milk in 1 liter final mixture = 5/8 lit.
Mean Price = Re 5/8.
C.P of 1 lt mix in 1st Re3/4 C.P of 1 lt mix in 2nd Re1/2
Mean Price 5/8
1/8 1/8
There ratio of two mixtures =1/8 :1/8 = 1:1.
So,quantity of mixture taken from each can=1/2*12
= 6 liters.
6.One quantity of wheat at Rs 9.30 p
er Kg are mixed
with another quality at a certain rate in the ratio 8:7.
If the mixture so formed be worth Rs 10 per Kg ,what is
the rate per Kg of the second quality of wheat?
Solution:Let the rate of second quality be Rs x per Kg.
C.P of 1Kg wheat of 1st 980p C.P of 1 Kg wheat of 2nd 100x p
Mean Price 1000p
100x-1000p 70 p
(100x-1000) / 70 = 8/7
700x -7000 = 560
700x = 7560 =>x = Rs 10.80.
Therefore the rate of second quality is Rs10.80
7.8lit are drawn from a wine and is then filled with water.
This operation is performed three more times.The ratio of
the quantity of wine now left in cask to that of the water
is 16:81. How much wine did the cask hold originally?
Solution:
Let the quantity of the wine in the cask originally be
x liters.
Then quantity of wine left in cask after
4 operations = x(1- 8/x)4lit.
Therefore x((1-(8/x))4)/x = 16/81.
(1- 8/x)4=(2/3) 4
(x- 8)/x=2/3
3x-24 =2x
x=24.
8.A can contains a mixture of two liquids A and B in the
ratio 7:5 when 9 liters of mixture are drawn off and the
can is filled with B,the ratio of A and B becomes 7:9.
How many liters of liquid A was contained by the can initially?
Solution:
Suppose the can initially contains 7x and 5x liters
of mixtures A and B respectively .
Quantity of A in mixture left = (7x- (7/12)*9 )lit
= 7x - (21/4) liters.
Quantity of B in mixture left = 5x - 5/12*9
= 5x - (15/4) liters
Therefore (7x – 21/4)/ (5x – 15/4+9)=7/9
(28x-21)/(20x +21)= 7/9
(252x -189)= 140x +147
112x = 336
=> x=3.
So the can contains 21 liters of A.
9.A vessel is filled with liquid,3 parts of which are water
and 5 parts syrup. How much of the mixture must be drawn off
and replaced with water so that the mixture may be
half water and half syrup?
Solution:
Suppose the vessal initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water
then quantity of water in new mixture
= 3-(3x/8)+x liters.
Quantity of syrup in new mixture = 5 - 5x/8 liters.
Therefore 3 - 3x/8 +x = 5 - 5x/8
5x+24 = 40-5x
10x = 16.
x= 8/5.
So part of the mixture replaced = 8/5*1/8 =1/5.

Simple Interest
Important Facts and Formulae:
Principal or Sum:- The money borrowed or lent out for a
certain period is called Principal or the Sum.

Interest:- Extra money paid for using others money is
called Interest.

Simple Interest:- If the interest on a sum borrowed for
a certain period is reckoned uniformly,then it is called
Simple Interest.

Formulae:
Principal = P
Rate = R% per annum
Time = T years. Then,

(i)Simple Interest(S.I)= (P*T*R)/100

(ii) Principal(P) = (100*S.I)/(R*T)
Rate(R) = (100*S.I)/(P*T)
Time(T) = (100*S.I)/(P*R)
Simple Problems
1.Find S.I on Rs68000 at 16 2/3% per annum for 9months.
Sol:- P=68000
R=50/3% p.a
T=9/12 years=4/3 years
S.I=(P*R*T)/100
=(68000*(50/3)*(3/4)*(1/100))
=Rs 8500

Note:If months are given we have to converted into
years by dividing 12 ie., no.of months/12=years

2.Find S.I on Rs3000 at 18% per annum for the period from
4th Feb to 18th April 1995
Sol:- Time=(24+31+18)days
=73 days
=73/365=1/5 years
P= Rs 3000
R= 18% p.a
S.I = (P*R*T)/100
=(3000*18*1/5*1/100)
=Rs 108
Remark:- The day on which money is deposited is not
counted while the day on which money is withdrawn is
counted.

3. In how many years will a sum of money becomes triple
at 10% per annum.
Sol:- Let principal =P
S.I = 2P
S.I = (P*T*R)/100
2P = (P*T*10)/100
T = 20 years
Note:
(1) Total amount = Principal + S.I
(2) If sum of money becomes double means Total amount
or Sum
= Principal + S.I
= P + P = 2P
Top
Medium Problems
1.A sum at Simple interest at 13 1/2% per annum amounts
to Rs 2502.50 after 4 years.Find the sum.
Sol:- Let Sum be x. then,
S.I = (P*T*R)/100
= ((x*4*27)/(100*2))
= 27x/100
Amount = (x+(27x)/100)
= 77x/50
77x/50 = 2502.50
x = (2502.50*50)/77
= 1625
Sum = 1625

2. A some of money becomes double of itself in 4 years
in 12 years it will become how many times at the same
rate.
Sol:- 4 yrs - - - - - - - - - P
12 yrs - - - - - - - - - ?
(12/4)* P =3P
Amount or Sum = P+3P = 4 times

3. A Sum was put at S.I at a certain rate for 3 years.
Had it been put at 2% higher rate ,it would have
fetched Rs 360 more .Find the Sum.
Sol:- Let Sum =P
original rate = R
T = 3 years
If 2% is more than the original rate ,it would have
fetched 360 more ie., R+2
(P*(R+2)*3/100) - (P*R*3)/100 = 360
3PR+ 6P-3PR = 36000
6P = 36000
P = 6000
Sum = 6000.

4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest
rate is increased by 3%, it would amount to how much?
Sol:- S.I = 920 - 800 = 120
Rate = (100*120)/(800*3) = 5%
New Rate = 5 + 3 = 8%
Principal = 800
Time = 3 yrs
S.I = (800*8*3)/100 = 192
New Amount = 800 + 192
= 992

5. Prabhat took a certain amount as a loan from bank at
the rate of 8% p.a S.I and gave the same amount to Ashish
as a loan at the rate of 12% p.a . If at the end of 12 yrs,
he made a profit of Rs. 320 in the deal,What was the
original amount?
Sol:- Let the original amount be Rs x.
T = 12
R1 = 8%
R2 = 12%
Profit = 320
P = x
(P*T*R2)/100 - (P*T*R1)/100 =320
(x*12*12)/100 - (x*8*12)/100 = 320
x = 2000/3
x = Rs.666.67

6. Simple Interest on a certail sum at a certain rate is
9/16 of the sum . if the number representing rate percent
and time in years be equal ,then the rate is.
Sol:- Let Sum = x .Then,
S.I = 9x/16
Let time = n years & rate = n%
n = 100 * 9x/16 * 1/x * 1/n
n * n = 900/16
n = 30/4 = 7 1/2%
Complex Problems
1. A certain sum of money amounts t 1680 in 3yrs & it
becomes 1920 in 7 yrs .What is the sum.
Sol:- 3 yrs - - - - - - - - - - - - - 1680
7 yrs - - - - - - - - - - - - - 1920
then, 4 yrs - - - - - - - - - - - - - 240
1 yr - - - - - - - - - - - - - ?
(1/4) * 240 = 60
S.I in 3 yrs = 3*60 = 18012
Sum = Amount - S.I
= 1680 - 180
= 1500
we get the same amount if we take S.I in 7 yrs
I.e., 7*60 =420
Sum = Amount - S.I
= 1920 - 420
= 1500
2. A Person takes a loan of Rs 200 at 5% simple Interest.
He returns Rs.100 at the end of 1 yr. In order to clear
his dues at the end of 2yrs ,he would pay:
Sol:- Amount to be paid
= Rs(100 + (200*5*1)/100 + (100*5*1)/100)
= Rs 115

3. A Man borrowed Rs 24000 from two money lenders.For one
loan, he paid 15% per annum and for other 18% per annum.
At the end of one year,he paid Rs.4050.How much did he
borrowed at each rate?
Sol:- Let the Sum at 15% be Rs.x
& then at 18% be Rs (24000-x)
P1 = x R1 = 15
P2 = (24000-x) R2 = 18
At the end of ine year T = 1
(P1*T*R1)/100 + (P2*T*R2)/100 = 4050
(x*1*15)/100 + ((24000-x)*1*18)/100 = 4050
15x + 432000 - 18x = 405000
x = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000 - 9000)
= 15000
Top
4.What annual instalment will discharge a debt of Rs. 1092
due in 3 years at 12% Simple Interest ?
Sol:- Let each instalment be Rs x
(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092
28x/25 + 31x/25 + x =1092
(28x +31x + 25x) = (1092 * 25)
84x = 1092 * 25
x = (1092*25)/84 = 325
Each instalement = 325
5.If x,y,z are three sums of money such that y is the simple
interest on x,z is the simple interest on y for the same
time and at the same rate of interest ,then we have:
Sol:- y is simple interest on x, means
y = (x*R*T)/100
RT = 100y/x
z is simple interest on y,
z = (y*R*T)/100
RT = 100z/y
100y/x = 100z/y
y * y = xz

6.A Sum of Rs.1550 was lent partly at 5% and partly at 5%
and partly at 8% p.a Simple interest .The total interest
received after 3 years was Rs.300.The ratio of the money
lent at 5% to that lent at 8% is:
Sol:- Let the Sum at 5% be Rs x
at 8% be Rs(1550-x)
(x*5*3)/100 + ((1500-x)*8*3)/100 = 300
15x + 1500 * 24 - 24x = 30000
x = 800
Money at 5%/ Money at 8% = 800/(1550 - 800)
= 800/750 = 16/15

7. A Man invests a certain sum of money at 6% p.a Simple
interest and another sum at 7% p.a Simple interest. His
income from interest after 2 years was Rs 354 .one
fourth of the first sum is equal to one fifth of the
second sum.The total sum invested was:
Sol:- Let the sums be x & y
R1 = 6 R2 = 7
T = 2
(P1*R1*T)/100 + (P2*R2*T)/100 = 354
(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354
6x + 7y = 17700 â€”â€”â€”(1)
also one fourth of the first sum is equal to one
fifth of the second sum
x/4 = y/5 => 5x - 4y = 0 â€”â€” (2)
By solving 1 & 2 we get,
x = 1200 y = 1500
Total sum = 1200 +1500
= 2700

8. Rs 2189 are divided into three parts such that their
amounts after 1,2& 3 years respectively may be equal,
the rate of S.I being 4% p.a in all cases. The Smallest
part is:
Sol:- Let these parts be x,y and[2189-(x+y)] then,
(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100
4x/100 = 8y/100
x = 2y
By substituting values
(2y*1*4)/100 = (2189-3y)*3*4/100
44y = 2189 *12
y = 597
Smallest Part = 597

9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and
the remainder at 10%.If his annual income is Rs.561. The
capital is:
Sol:- Let the capital be Rs.x
Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)
+ (5x/12 * 10/100 * 1) = 561
7x/300 + x/50 + x/24 = 561
51x = 561 * 600
x = 6600

Compound Interest
Important Facts and Formulae:
Compound Interest:
Sometimes it so happens that the borrower and the lender
gree to fix up a certain unit of time ,say yearly or
half-yearly or quarterly to settle the previous account.
In such cases ,the amount after the first unit of time
becomes the principal for the 2nd unit ,the amount after
second unit becomes the principal for the 3rd unit and so
on. After a specified period ,the difference between the
amount and the money borrowed is called Compound Interest
for that period.
Formulae:
Let principal=p,Rate=R% per annum Time=nyears
1.When interest is compounded Annually,
Amount=P[1+(R/100)]n
2.When interest is compounded Halfyearly,
Amount=P[1+((R/2)100)]2n
3.When interest is compounded Quaterly,
Amount=P[1+((R/4)100)]4n
4.When interest is compounded Annually,but time in fractions
say 3 2/5 yrs Amount=P[1+(R/100)]3[1+((2R/5)/100)]
5.When rates are different for different years R1%,R2%,R3%
for 1st ,2nd ,3rd yrs respectively
Amount=P[1+(R1/100)][1+(R2/100)][1+(R3/100)]
6.Present Worth of Rs.X due n years hence is given by
Present Worth=X/[1+(R/100)]n
Simple Problems
1.Find CI on Rs.6250 at 16% per annum for 2yrs ,compounded
annually.
Sol: Rate R=16,n=2,Principle=Rs.6250
Method1:
Amount=P[1+(R/100)]n
=6250[1+(16/100)]2
=Rs.8410
C.I=Amount-P
=8410-6250
=Rs.2160
Method2:
Iyear------------------6250+1000
\\Interest for 1st yr on 6250
II yr---------------6250+1000+160
\\Interest for I1yr on 1000
C.I.=1000+1000+160
=Rs.2160
2.Find C.I on Rs.16000 at 20% per annum for 9 months
compounded quaterly
Sol:
MethodI:
R=20%
12months------------------------20%
=> 3 months------------------------5%
For 9 months,there are '3' 3months
--------16000+800
--------16000+800+40
--------16000+800+40+10+2
=>Rs.2522
MethodII: Amount=P[1+(R/100)]n
=16000[1+(5/100)]3
=Rs.18522
C.I=18522-16000
=Rs.2522
Top
Complex Problems
1.The difference between C.I and S.I. on a certain sum
at 10% per annum for 2 yrs is Rs.631.find the sum
Sol:
MethodI:
NOTE:
a) For 2 yrs -------->sum=(1002D/R2)
b) For 3 yrs -------->sum=(1003D/R2(300+R))
Sum=1002*631/102
=Rs.63100
MethodII:
Let the sum be Rs.X,Then
C.I.=X[1+(10/100)]2-X
S.I=(X*10*2)/100
=X/5
C.I-S.I.=21X/100-X/5
=X/100
X/100=631
X=Rs.63100

2.If C.I on a certain sum for 2 yrs at 12% per annum is
Rs.1590. What would be S.I?
sol:
C.I.=Amount-Principle
Let P be X
C.I=X[1+(12/100)]2-X
=>784X/625-X=1590
=>X=Rs.6250
S.I=(6250*12*2)/100=Rs.1500

3.A sum of money amounts to Rs.6690 after 3 yrs and to
Rs.10035b after 6 yrs on C.I .find the sum
sol:
For 3 yrs,
Amount=P[1+(R/100)]3=6690-----------------------(1)
For 6 yrs,
Amount=P[1+(R/100)]6=10035----------------------(2)
(1)/(2)------------[1+(R/100)]3=10035/6690
=3/2
[1+(R/100)]3=3/2-----------------(3)
substitue (3) in (1)
p*(3/2)=6690
=>p=Rs.4460
sum=Rs.4460
4.A sum of money doubles itself at C.I in 15yrs.In how many
yrs will it become 8 times?

sol: Compound Interest for 15yrs p[1+(R/100)]15
p[1+(R/100)]15=2P
=>p[1+(R/100)]n=8P
=>[1+(R/100)]n=8
=>[1+(R/100)]n=23
=>[1+(R/100)]n=[1+(R/100)]15*3
since [1+(R/100)] =2
n=45yrs

5.The amount of Rs.7500 at C.I at 4% per annum for 2yrs is
sol:
Iyear------------------7500+300(300------Interest on 7500)
IIyear ----------------7500+300+12(12------------4% interest
on 300)
Amount=7500+300+300+12
=Rs.8112

6.The difference between C.I and S.I on a sum of money for
2 yrs at 121/2% per annum is Rs.150.the sum is
sol:
Sum=1002D/R2=( 1002*150) /(25/2)2=Rs.9600

7.If the S.I on sum of money at 15% per annum for 3yrs is Rs.1200,
the C.I on the same sum for the same period at same rate is------
sol:
S.I=1200
P*T*R/100=1200
P*3*5/100=1200
=>P=Rs.8000
C.I for Rs.8000 at 5% for 3 yrs is-------------8000+400
-----8000+400+20
-------------8000+400+20+20+1
C.I =400+400+20+400+20+20+1
=Rs.1261

Areas
Important Facts and Formulae:
Results On Triangle
1.Sum of the angles of a triangle is 180 degrees.
2.The sum of any two sides of a triangle is greater
than third side.
3.Pythagoras Theorem:
In a right angled triangle (Hypotenuse)2 = (Base)2 +(Height)2
4.The line joining the mid point of a side of a triangle
to the opposite vertex is called the MEDIAN.
5.The point where the three medians of a triangle meet,
is called CENTROID. The centroid divides each of the
medians in the ratio 2:1
6.In an isosceles triangle, the altitude from the
vertex bisects the base
7.The median of a triangle divides it into two triangles
of the same area.
8.The area of the triangle formed by joining the mid points
of the sides of a given triangle is one-fourth of the area
of the given triangle.
Results On Quadrilaterals
1.The diagonals of a Parallelogram bisect each other.
2.Each diagonal of a Parallelogram divides it into two
triangles of the same area.
3.The diagonals of a Rectangle are equal and bisect
each other
4.The diagonals of a Square are equal and bisect each
other at right angles.
5.The diagonals of a Rhombus are unequal and bisect
each other at right angles.
6.A Parallelogram and a Rectangle on the same base
and between the same parallels are equal in area.
7.Of all he parallelogram of given sides the parallelogram
which is a rectangle has the greatest area.
Top
Formulae
1.Area of a RECTANGLE = length * breadth
Length = (Area/Breadth) and Breadth = (Area/Length)
2.Perimeter of a RECTANGLE = 2(Length + Breadth)
3.Area of a SQUARE = (side)2 = Â½ ( Diagonal)2
4.Area of four walls of a room = 2(length + breadth) * height
5.Area of a TRIANGLE = Â½ * base * height
6.Area of a TRIANGLE = âˆš[s * (s-a) * (s-b) * (s-c)],
where a,b,c are the sides of the triangle and s = 1/2(a+b+c)
7.Area of EQUILATERAL TRIANGLE = âˆš(3/4)* (side)2
8.Radius of in circle of an EQUILATERAL TRIANGLE of
side a = r / 2âˆš3
9.Radius of circumcircle of an EQUILATERAL TRIANGLE
of side a = r / âˆš3
10.Radius of incircle of a triangle of area âˆ† and
semi perimeter S = âˆ† / s
11.Area of a PARALLELOGRAM = (base * height)
12.Area of RHOMBUS = 1/2 (product of diagonals)
13.Area of TRAPEZIUM =
=1/2 * (sum of parallel sides)* (distance between them)
14.Area of a CIRCLE = ï›‡ r2 where r is the radius
15.Circumference of a CIRCLE = 2ï›‡r
16.Length of an arc = 2 ï›‡rÃ¸ / 360, where Ã¸ is central angle
17.Area of a SECTOR = Â½ (arc * r) = ï›‡r2Ã¸ / 360
18.Area of a SEMICIRCLE = ï›‡r2 / 2
19.Circumference of a SEMICIRCLE = ï›‡r
Top
Simple Problems
1.One side of a rectangular field is 15m and one of its diagonal
is 17m. Find the area of field?
Sol: Other side = âˆš[(17*17) â€“ (15*15)] = âˆš(289-225) = 8m
Area = 15 * 8 =120 sq. m
2.A lawn is in the form of a rectangle having its sides in the
ratio 2:3 The area of the lawn is 1/6 hectares. Find the length
and breadth of the lawn.
Sol: let length = 2x meters and breadth = 3x mt
Now area = (1/6 * 1000)sq m = 5000/3 sq m
2x * 3x = 5000/3 =>x * x =2500 / 9
x = 50/3
length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m
3.Find the cost of carpeting a room 13m long and 9m broad with
a carpet 75cm wide at the rate of Rs 12.40 per sq meter
Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m
length of the carpet = (Area/width) = 117 * (4/3) = 156 m
Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40
4.The length of a rectangle is twice its breadth if its length
is decreased by 5cm and breadth is increased by 5cm, the area
of the rectangle is increased by 75 sq cm. Find the length of
the rectangle.
Sol: let length = 2x and breadth = x then
(2x-5) (x+5) â€“ (2x*x)=75
5x-25 = 75 => x=20
length of the rectangle = 40 cm
5.In measuring the sides of a rectangle, one side is taken 5%
in excess and the other 4% in deficit. Find the error percent
in the area, calculate from the those measurements.
Sol: let x and y be the sides of the rectangle then
correct area = (105/100 * x) * (96 / 100 *y)
=(504/500 xy) â€“ xy = 4/500 xy
Error% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%
6.A room is half as long again as it is broad. The cost of
carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost of
papering the four walls at Rs 10 per sq m is Rs 1720. If a door
and 2 windows occupy 8 sq cm. Find the dimensions of the room?
Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt
Area of the floor = (total cost of carpeting /rate)
= 270/5 sq m = 54 sq m
x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m
so breadth = 6m and length=3/2*6 = 9m
now papered area = 1720 /10 = 172 sq m
Area of one door and 2 windows =8 sq m
total area of 4 walls = 172+8 = 180 sq m
2(9+6)*h = 180 => h=180/30 = 6m
7.The altitude drawn to the base of an isosceles triangle is 8cm
and the perimeter is 32cm. Find the area of the triangle?
Sol: let ABC be the isosceles triangle, the AD be the altitude
let AB = AC=x then BC= 32-2x
since in an isoceles triange the altitude bisects the base so
BD=DC=16-x
in âˆ†ADC,(AC) 2 = (AD) 2 + (DC) 2
x*x=(8*8) + (16-x)*(16-x)
32x =320 => x = 10
BC = 32-2x = 32-20 = 12 cm
Hence, required area = Â½ * BC * AD
= Â½ * 12 * 10 = 60 sq cm
8.If each side of a square is increased by 25%, find the
percentage change in its area?
Sol: let each side of the square be a , then area = a * a
New side = 125a / 100 = 5a / 4
New area =(5a * 5a)/(4*4) = (25aÂ²/16) â€“ aÂ²
= 9aÂ²/16
Increase %= 9aÂ²/16 * 1/aÂ² * 100%
= 56.25%
9.Find the area of a Rhombus one side of which measures 20cm
and one diagonal 24cm.

Sol: Let other diagonal = 2x cm
since diagonals of a rhombus bisect each other at right angles,
we have
20Â² = 12Â² + xÂ² => x = âˆš[20Â² -12Â²]= âˆš256 = 16cm
so the diagonal = 32 cm
Area of rhombus = Â½ * product of diagonals
= Â½ * 24 * 32
= 384 sq cm
10. The area of a circular field is 13.86 hectares. Find the cost
of fencing it at the rate of Rs. 4.40 per meter.

Sol: Area = 13.86 * 10000 sq m = 138600 sq m
ï›‡rÂ²= 138600 => rÂ² = 138600 * 7/22 => 210 m
circumference = 2ï›‡r = 2 * 22/7 * 210m = 1320 m
cost of fencing = Rs 1320 * 4.40 = Rs. 5808
Top
Medium Problems:
11.Find the ratio of the areas of the incircle and circumcircle of
a square.
Sol: let the side of the square be x, then its diagonal = âˆš2 x
radius of incircle = x/2 and
radius of circmcircle =âˆš2 x /2 = x/âˆš2
required ratio = ï›‡xÂ²/4 : ï›‡xÂ²/2 = Â¼ : Â½ = 1:2
12.If the radius of a circle is decreased by 50% , find the
percentage decrease in its area.
Sol: let original radius = r and new radius = 50/100 r = r/2
original area = ï›‡rÂ² and new area = ï›‡(r/2)Â²
decrease in area = 3 ï›‡rÂ²/4 * 1/ ï›‡rÂ² * 100 = 75%
13.Two concentric circles form a ring. The inner and outer
circumference of the ring are 352/7 m and 528/7m respectively.
Find the width of the ring.
sol: let the inner and outer radii be r and R meters
then, 2ï›‡r = 352/7 => r = 352/7 * 7/22 * Â½ = 8m
2ï›‡R = 528/7 => R= 528/7 * 7/22 * Â½ = 12m
width of the ring = R-r = 12-8 = 4m
14.If the diagonal of a rectangle is 17cm long and its perimeter
is 46 cm. Find the area of the rectangle.
sol: let length = x and breadth = y then
2(x+y) = 46 => x+y = 23
xÂ²+yÂ² = 17Â² = 289
now (x+y)Â² = 23Â² =>xÂ²+yÂ²+2xy= 529
289+ 2xy = 529 => xy = 120
area =xy=120 sq. cm
15.A rectangular grassy plot 110m by 65cm has a gravel path .5cm
wide all round it on the inside. Find the cost of gravelling the
path at 80 paise per sq.mt
sol: area of theplot = 110 * 65 = 7150 sq m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq m
area of the path = 7150- 6300 =850 sq m
cost of gravelling the path = 850 * 80/100 = 680 Rs
16. The perimeters of ttwo squares are 40cm and 32 cm. Find the
perimeter of a third square whose area is equal to the difference
of the areas of the two squares.
sol: side of first square = 40/4 =10cm
side of second square = 32/4 = 8cm
area of third squre = 10Â² â€“ 8Â² = 36 sq cm
side of third square = âˆš36 = 6 cm
required perimeter = 6*4 = 24cm
Top
17. A room 5m 44cm long and 3m 74cm broad is to be paved with squre
tiles. Find the least number of squre tiles required to cover the
floor.
sol: area of the room = 544 * 374 sq cm
size of largest square tile = H.C.F of 544cm and 374cm= 34cm
area of 1 tile = 34*34 sq cm
no. of tiles required = (544*374) / (34 * 34) = 176
18. The diagonals of two squares are in the ratio of 2:5. Find
the ratio of their areas.
sol: let the diagonals of the squares be 2x and 5x respectively
ratio of their areas = Â½ * (2x)Â² : Â½*(5x)Â² = 4:25
19.If each side of a square is increased by 25%. Find the percentage
change in its area.
sol: let each side of the square be a then area = a Â²
new side = 125a/100 = 5a/4
new area = (5a/4)Â² = 25/16 aÂ²
increase in area = (25/16)aÂ² - aÂ² = (9/16)aÂ²
increase % = (9/16)aÂ² * (1/aÂ²) * 100 = 56.25%
20.The base of triangular field os three times its altitude. If the
cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18.
Find its base and height.
sol:
area of the field = total cost/ rate = 333.18 /24.68 = 13.5 hectares
=> = 13.5 * 10000 = 135000 sq m
let the altitude = x mt and base = 3x mt
then Â½ *3x * x = 135000 => xÂ² = 90000 => x = 300
base= 900m and altitude = 300m
21.In two triangles the ratio of the areas is 4:3 and the ratio of
their heights is 3:4. Find the ratio of their bases?
Sol:
let the bases of the two triangles be x &y and their heights
be 3h and 4h respectively.
(1/2*x*3h)/(1/2*y*4h) =4/3 => x/y = 4/3 *4/3 = 16/9
22.Find the length of a rope by which a cow must be tethered in order
that it may be able to graze an area of 9856 sq meters.
Sol:
clearly the cow will graze a circular field of area 9856 sq m and
radius equal to the length of the rope.
Let the length of the rope be r mts
then ï›‡rÂ²=9856 => rÂ²=9856*7/22 = 3136 => r=56m
23.The diameter of the driving wheel of a bus is 140cm. How many
revolutions per minute must the wheel make inorder to keep a speed of
66 kmph?
Sol: Distance to be covered in 1min = (66*1000)/60 m =1100m
diameter = 140cm => radius = r =0.7m
circumference of the wheel = 2*22/7*0.7 = 4.4m
no of revolutions per minute = 1100/4.4 = 250
24.The inner circumference of a circular race track, 14m wide is 440m.
Find the radius of the outer circle.
Sol: let inner radius be r meters.
Then 2ï›‡r =440 => r=440*7/22*1/2 = 70m
radius of outer circle = 70+4 =84m
25.A sector of 120 degrees, cut out from a circle, has an area of
66/7 sq cm. Find the radius of the circle.
Sol: let the radius of the circle be r cm. Then
ï›‡rÂ²Ã¸/360 =66/7=> 22/7*rÂ²*120/360 = 66/7 =>rÂ² = 66/7 *7/22*3 =9
radius = 3cm
26.The length of the room is 5.5m and width is 3.75m. Find the cost
of paving the floor by slabs at the rate of Rs.800 per sq meter.
Sol: l=5.5m w=3.75m
area of the floor = 5.5 * 3.75 = 20.625 sq m
cost of paving = 800 *20.625 =Rs. 16500
Top
27.A rectangular plot measuring 90 meters by 50 meters is to be
enclosed by wire fencing. If the poles of the fence are kept 5 meters
apart. How many poles will be needed?
Sol: perimeter of the plot = 2(90+50) = 280m
no of poles =280/5 =56m
28.The length of a rectangular plot is 20 meters more than its breadth.
If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is
the length of the plot in meter?
Sol: let breadth =x then length = x+20
perimeter = 5300/26.50 =200m
2(x+20+x) =200 => 4x+40 =200
x = 40 and length = 40+20 = 60m
29.A rectangular field is to be fenced on three sides leaving a side of
20 feet uncovered. If the area of the field is 680 sq feet, how many
feet of fencing will be required?
Sol: l=20feet and l*b=680 => b= 680/20 = 34feet
length of fencing = l+2b = 20+68 =88 feet
30.A rectangular paper when folded into two congruent parts had a
perimeter of 34cm foer each part folded along one set of sides and
the same is 38cm. When folded along the other set of sides. What is
the area of the paper?
Sol: when folded along the breadth
we have 2(l/2 +b) = 34 or l+2b = 34...........(1)
when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)
from 1 &2 we get l=14 and b=10
Area of the paper = 14*10 = 140 sq cm
31.A took 15 seconds to cross a rectangular field diagonally walking at
the rate of 52 m/min and B took the same time to cross the same field
along its sides walking at the rate of 68m/min. The area of the field is?
Sol: length of the diagonal = 52*15/60 =13m
sum of length and breadth = 68*15/60 = 17m
âˆš(lÂ²+bÂ²)=13 or l+b = 17
area =lb = Â½ (2lb) = Â½[(l+b)Â² â€“ (lÂ²+bÂ²)] = Â½[17Â² -169]
=1/2*120 = 60 sq meter
32 . A rectangular lawn 55m by 35m has two roads each 4m wide running in
the middle of it. One parallel to the length and the other parallel to
breadth. The cost of graveling the roads at 75 paise per sq meter is
sol: area of cross roads = 55*4 +35*4-4*4 = 344sq m
cost of graveling = 344 *75/100 =Rs. 258
33.The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.
How much will it cost to lay a three meter wide pavement along the
fencing inside the field @ Rs. 50 per sq m
sol: perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126-6 = 120m
area of the pavement = (126*126)-(120*120)=246â€*6 sq m
cost of pavement = 246*6*50 = Rs. 73800
34.Amanwalked diagonally across a square plot. Approximately what was
the percent saved by not walking along the edges?
Sol: let the side of the square be x meters
length of two sides = 2x meters
diagonal = âˆš2 x = 1.414x m
saving on 2x meters = .59x m
saving % = 0.59x /2x *100%
= 30% (approx)
36.A man walking at the speed of 4 kmph crosses a square field
diagonally in 3 meters.The area of the field is
sol: speed of the man = 4*5/18 m/sec = 10/9 m/sec
time taken = 3*60 sec = 180 sec
length of diagonal = speed * time = 10/9 * 180 = 200m
Area of the field = Â½ *(dioagonal)Â²
= Â½ * 200*200 sq m = 20000sq m
37.A square and rectangle have equal areas. If their perimeters
are p and q respectively. Then
sol: A square and a rectangle with equal areas will satisfy the
relation p < href="file:///G:/Aptitude/areas.html">
38.If the perimeters of a square and a rectangle are the same,
then the area a & b enclosed by them would satisfy the condition:
sol: Take a square of side 4cm and a rectangle having l=6cm and
b=2cm
then perimeter of square = perimeter of rectangle
area of square = 16 sq cm
area of rectangle = 12 sq cm
Hence a >b
39.An error of 2% in excess is made while measuring the side of a
square. The percentage of error in the calculated area of the
square is
sol: 100cm is read as 102 cm
a = 100*100 sq cm and b = 102 *102 sq cm
then a-b = 404 sq cm
percentage error = 404/(100*100) = 4.04%
40.A tank is 25m long 12m wide and 6m deep. The cost of plastering
its walls and bottom at 75 paise per sq m is
sol: area to be plastered = [2(l+b)*h]+(l*b)
= 2(25+12)*6 + (25*12)= 744 sq m
cost of plastering = Rs . 744*75/100 = Rs. 5581
41.The dimensions of a room are 10m*7m*5m. There are 2 doors and 3
windows in the room. The dimensions of the doors are 1m*3m. One
window is of size 2m*1.5m and the other 2 windows are of size 1m*1.5m.
The cost of painting the walls at Rs. 3 per sq m is
sol: Area of 4 walls = 2(l+b)*h
=2(10+7)*5 = 170 sq m
Area of 2 doors and 3 windows = 2(1*3)+(2*1.5)+2(1*1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m
cost of painting = Rs. 158 *3 = Rs. 474
42.The base of a triangle of 15cm and height is 12cm. The height of
another triangle of double the area having the base 20cm is
sol: a = Â½ *15*12 = 90 sq cm
b = 2a = 2 * 90 = Â½ * 20 *h => h= 18cm
43.The sides of a triangle are in the ratio of Â½:1/3:1/4. If the
perimeter is 52cm, then the length of the smallest side is
sol: ratio of sides = Â½ :1/3 :1/4 = 6:4:3
perimeter = 52 cm, so sides are 52*6/13 =24cm
52*4/13 = 16cm
52 *3/13 = 12cm
length of smallest side = 12cm
44.The height of an equilateral triangle is 10cm. Its area is
sol: aÂ² = (a/2)Â² +(10)Â²
aÂ² â€“ aÂ²/4 = 100 =>3aÂ² = 100*4
area = âˆš3/4 *aÂ² = âˆš3/4*400/3 = 100/âˆš3 sq cm
45.From a point in the interior of an equilateral triangle, the
perpendicular distance of the sides are âˆš3 cm, 2âˆš3cm and
5âˆš3cm. The perimeter of the triangle is
sol: let each side of the triangle be â€˜aâ€™ cm
then area(AOB) +area(BOC)+area(AOC) = area(ABC)
Â½ * a *âˆš3 +1/2 *a *2âˆš3 +1/2 * a*5âˆš3 = âˆš3/4 a Â²
a/2âˆš3(1+2+5) = âˆš3/4 a Â² => a=16
perimeter = 3*16 = 48cm
Top
Complex Probems:
1.If the area of a square with side a s equal to the area of a
triangle with base a, then the altitude of the triangle is
sol: area of a square with side a = a Â² sq unts
area of a triangle with base a = Â½ * a*h sq unts
a Â² =1/2 *a *h => h = 2a
altitude of the triangle is 2a
2.An equilateral triangle is described on the diagonal of a
square. What is the ratio of the area of the triangle to that of
the square?
Sol: area of a square = a Â² sq cm
length of the diagonal = âˆš2a cm
area of equilateral triangle with side âˆš2a
= âˆš3/4 * (âˆš2a) Â²
required ratio = âˆš3aÂ² : a Â² = âˆš3 : 2
3.The ratio of bases of two triangles is x:y and that of their
areas is a:b. Then the ratio of their corresponding altitudes
wll be
sol: a/b =(Â½ * x*H) /(1/2 * y * h)
bxH = ayh =>H/h =ay/bx
Hence H:h = ay:bx
4 .A parallelogram has sides 30m and 14m and one of its diagonals
is 40m long. Then its area is
sol: let ABCD be the given parallelogram
area of parallelogram ABCD = 2* (area of triangle ABC)
now a = 30m, b = 14m and c = 40m
s = Â½(30+14+40) = 42m
Area of triangle ABC = âˆš[ s(s-a)(s-b)(s-c)
= âˆš(42*12*28*2 = 168sq m
area of parallelogram ABCD = 2 *168 =336 sq m
5.If a parallelogram with area p, a triangle with area R and a
triangle with area T are all constructed on the same base and all
have the same altitude, then which of the following statements
is false?
Sol: let each have base = b and height = h
then p = b*h, R = b*h and T = Â½ * b*h
so P = R, P = 2T and T = Â½ R are all correct statements
6.If the diagonals of a rhombus are 24cm and 10cm the area
and the perimeter of the rhombus are respectively.
Sol: area = Â½*diagonal 1 *diagonal 2= Â½ * 24 * 10= 120 sq cm
Â½ * diagonal 1 = Â½ * 24 = 12cm
Â½ * diagonal 2 = Â½ *10 =5 cm
side of a rhombus = (12) Â² + (5) Â² = 169 => AB = 13cm
Top
7.If a square and a rhombus stand on the same base, then the ratio
of the areas of the square and the rhombus is:
sol: A square and a rhombus on the same base are equal in area
8.The area of a field in the shape of a trapezium measures
1440sq m. The perpendicular distance between its parallel sides
is 24cm. If the ratio of the sides is 5:3, the length of the
longer parallel side is:
sol: area of field =1/2 *(5x+3x) *24 = 96x sq m
96x = 1440 => x = 1440 /96 = 15
hence, the length of longer parallel side = 5x = 75m
9.The area of a circle of radius 5 is numerically what percent its
circumference?
Sol: required percentage = ï›‡(5)Â²/(2ï›‡*5) *100 = 250%
10.A man runs round a circular field of radius 50m at the speed of
12m/hr. What is the time taken by the man to take twenty rounds of
the field?
Sol: speed = 12 k/h = 12 * 5/18 = 10/3 m/s
distance covered = 20 * 2*22/7*50 = 44000/7m
time taken = distance /speed = 44000/7 * 3/10 = 220/7min
11.A cow s tethered in the middle of a field with a 14feet long
rope.If the cow grazes 100 sq feet per day, then approximately
what time will be taken by the cow to graze the whole field?
Sol: area of the field grazed = 22/7 * 14 * 14 = 616 sq feet
12.A wire can be bent in the form of a circle of radius 56cm.
If it is bent in the form of a square, then its area will be
sol: length of wire = 2ï›‡ r = 2 *22/7 *56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88*88 = 7744sq cm
13.The no of revolutions a wheel of diameter 40cm makes in
traveling a distance of 176m is
sol:
distance covered in 1 revolution = 2ï›‡ r = 2 *22/7 *20
= 880/7 cm
required no of revolutions = 17600 *7/880 = 140
Top
14.The wheel of a motorcycle 70cm in diameter makes 40
revolutions in every 10sec.What is the speed of motorcycle
n km/hr?
Sol: distance covered in 10sec = 2 *22/7 *35/100 *40 =88m
distance covered in 1 sec =88/10m = 8.8m
speed =8.8m/s = 8.8 * 18/5 *k/h = 31.68 k/h
15.Wheels of diameters 7cm and 14cm start rolling simultaneously
from x & y which are 1980 cm apart towards each other in opposite
directions. Both of them make the same number of revolutions per
second. If both of them meet after 10seconds.The speed of the
smaller wheel is
sol: let each wheel make x revolutions per sec. Then
(2ï›‡ *7/2 *x)+(2ï›‡ * 7*x)*10 = 1980
(22/7 *7 * x) + (2 * 22/7 *7 *x) = 198
66x = 198 => x = 3
distance moved by smaller wheel in 3 revolutions
= 2 *22/7 *7/2 *3 = 66cm
speed of smaller wheel = 66/3 m/s = 22m/s
16.A circular swimming pool is surrounded by a concrete wall
4ft wide. If the area of the concrete wall surrounding the pool
is 11/25 that of the pool, then the radius of the pool is?
Sol: let the radius of the pool be R ft
radius of the pool including the wall = (R+4)ft
area of the concrete wall = ï›‡ [(R+4)2 - R2 ]
=> = ï›‡[R+4+R][R+4-R]
= 8ï›‡(R+2) sq feet
8ï›‡(R+2) = 11/25 ï›‡ R2 => 11 R2 = 200 (R+2)
Radius of the pool R = 20ft
17.A semicircular shaped window has diameter of 63cm. Its
perimeter equals
sol: perimeter of window = ï›‡r +2r = [22/7 * 63/2 +63] = 99+63
= 162 cm
18.Three circles of radius 3.5cm are placed in such a way that
each circle touches the other two. The area of the portion
enclosed by the circles is
sol:
required area = (area of an equilateral triangle of side 7 cm)
- (3 * area of sector with Ã˜ = 6o degrees and r = 3.5cm)
= ( âˆš Â¾ * 7 * 7) â€“ (3* 22/7 *3.5 *3.5*60/360 ) sq cm
= 49âˆš3/4 â€“ 11*0.5*3.5 sq cm = 1.967 sq cm
19. Four circular cardboard pieces, each of radius 7cm are placed
in such a way that each piece touches two other pieces. The area
of the space encosed by the four pieces is
sol: required area = 14*14 â€“ (4 * Â¼ * 22/7 * 7 *7) sq cm
= 196 â€“ 154 = 42 sq cm

Races and Games of Skill
Important Facts and Formulae:
Races:
A contest of speed in running ,riding,driving,sailing or
rowing is called a Race
Race Course:
The ground or path on which contests are made is called
a race course
Starting Point:
The point from which a race begins is called starting point.
Winning point or goal:
The point set to bound a race is called a winning point.
Dead Heat Race:
If all the persons contesting a race reach the goal exactly
at the same time,then the race is called a dead heat race.
Start:
Suppose A and B are two contestants in a race .If before the
start of the race,A is at the satrtint point and B is ahead
of A by 12 metres. Then we say that "A gives B a start 12 metres.
->To cover a race of 100metres in this case,A will have to cover
100m while B will have to cover 88m=(100-12)
->In a100m race 'A can give B 12m' or 'A can give B a start of
12m' or 'A beats B by 12m'means that while A runs 100m B runs 88m.
Games:
A game of 100m,means that the person among the contestants who
scores 100 points first is the winner.
If A scores 100 points while B scores only 80 points then we say
that 'A can give B 20 points'.
Top
Problems:
1) In a 1 km race,A beats B by 28 m or 7sec.Find A's time over the
course?
Sol: B covers 28 m in 7sec so,B's time over the course
= 7/28 *1000 =250 sec
A's time over the course =250-7 =243 sec = 4min ,3 sec.
2) A runs 1 3/4 times as fast as B.If A gives B a start of 84 m,how
far must be winning post be so that A and B might reach it at the
same time ?
Sol: Ratio of rates of A:B =7/4 :1 =7 :4 In a game of
7 m A gains 3m over B. 3m are gained by A in a race of 7m 84 m are
gained by A in a race of 7*84/3 =196 m Winning post must be 196m
away from the starting point.
3)A can run 1km in 3 min ,10sec and B can cover same distance in
3 min 20 sec. By what distance A beat B?
Sol: clearly A beats B by
10 sec. Distance covered by B in 10 sec =1000/200 *10 =50 m A beats
B by 50 metres.
4) In a 100m race,A runs at 8km per hour.If A gives b a start of
4 m and still beats him by 15 sec,what is the speed of B?
Sol: 8000 m -------60*60 sec 100m ------- 60*60*1000/8000 =45 sec.
Time taken by A to cover 100m =45 sec.
B covers 100-4 m =96 min 45 sec =60 sec
B's speed =96 *60*60/60*1000 =5.76 km/hr
5) A and B take part in 100m race .A runs at 5km per hour. A gives
B a start of 8 m and still beats him by 8 sec. What is the
speed of B?
Sol : A'speed = 5km/hr =5*5/18 =25/18 m/s Time taken by A to
cover 100m =100*18/25 =7.2 sec
Time taken by A to cover 92m = 72+8 =80 sec
B's speed =92*18/80*5 = 4.14 kmph.
6) A runs 1 2/3 times as fast as B.If A gives B a start of 80 m,
how far must the winning post be so that A and B might reach if
at the same time?
Sol: Ratio of the speed of A and B =5/3 :1 Thus
in a race of 5m ,A gains 2m over B 2m are gained by in a race of
5m 80 m will be gained by A in a race of 5/2* 80 =200 m
7) A,B and C are three contestans in a kmrace .If A can give B a
start of 40 m and A can give C a start of 64 m,how amny metres
start can b give C?
Sol: A covers 100m,B covers (1000-40) =960 m C covers 1000-64 m
or 936m when B covers 960 m,C covers 936 m when B covers 1000 m,
C covers 936*1000/960 m =975 m B can give C a start of 1000-975
or 25 m.
8) In a 100m race,A covers the distance in 36 sec and B in
45 sec.In this race A beats B by?
Sol: Distance covered by B in 9secs =100*9/45 =20 m A beats
B by 20m
9) In a 200 m race A beats B by 35m or 7 sec.What is the A's time
over the course?
Sol: B runs 35 m in 7sec. B covers 200m in =7*200/35 = 40 sec.
B's time over the course =40 sec
A's time over the course =40-7 =33 sec.
10) In a 300 m race A beats B by 22.5 m or 6 sec.What is the
B's time over the course?
Sol: B runs 22.5 m in 6sec.
B runs 300m in =6*300*2/45 =80 sec.
B's time over the course =80 sec.
Top
11) A can run 22.5 m while B runs 25 m.In a kilometre race B
beats A by?
Sol: B runs 25 m ,A runs 45/2 m B runs 1000 A runs
= 1000*45/2*25 =900m B beats A by 100m
12)In a 500 m race, the ratio of the speeds of two contestants
A and B is 3:4.A has a start of 140 m.Then ,A win by B?
Sol: The speeds of A and B =3:4 To reach the winning post A will
have to cover a distance of 500-140 m i.e 360 m while A covers 3m,
b covers 4m A covers 360m B covers 4/3*360=480m Thus when A
reaches the winning post, b covers 480m and therefore remains
20m behind. A wins by 20m.
13) In a 100m race,A can beat B by 25 m and B can beat c by 4m.
In the same race A can beat C by/
Sol: If A:B =100 :75 B :C=100 :96 then A :C =A/B*B/C=100/75* 100/96
= 100/72 A beats C by 100 -72 m=28 m
14) In a 100 race,A can give B 10 mand C 28 m.In the same race B
can give C?
Sol: A:B =100 :90 A :C=100 :72 B:C =B/A *A/C =90/100*100/72
=90/72
When B runs 90 C runs 72 when B runs 100 C runs
=72*100/90 B beats C by 20m
15) In a 100m race ,A beats B by 10 m and C by 13 m.In the race
of 180m. B will beat c by?
Sol :A : B =100 :90 A/B =100/90 A/C =100/87 B/C =B/A *A/C =90 /87
When B runs 90m C runs 87 When B runs 180 m then C runs
=87 *180/90 B beats C by 180-174= 6m
16) In a race of 200 m, A can beat B by 31 m and C by 18m.In arace
of 350 m, C will beat B by?
Sol : In a race of 200 m A :B =200:169 A :C =200 :182 C/B =C/A*A/B
=182/200*200/169 When C runs 182 m B runs 169
when C runs 350 m B runs =350*169/182 =325m
17) In agame of 100 points A can give B 20 points and C 28 points
then B can give C?
Sol: In a game of 100 points.A :B =100 :80 A :C =100 :72 B/C=B/A*A/C
=80/100*100/72 = 80/72 when B runs 80m C runs 72 when B runs
100m C runs =100*72/80 =90 B can give C 10 points in agame of 100.
18)At a game of billiards,A can give B 15 points in 60 and A can
give C 20 points in60.How amny points can B give C in a game
of 90?
Sol: A:B =60:45 A:C =60:40 B/C =B/A*A/C =45/60*60/40 =90/80
B can give C 10 points in agame of 90.
19) in agame of 80 points,A can give B 5 points and C 15 points.
Then how many points B can give C in agame of 60?
Sol: A :B =80 :75 A :C =80:65 B/C=B/A*A/C = 75/80*80/65
=15/13 B:C =60:52

Calenders
Important Facts and Formulae:
a) Odd Days : The number of days more than the complete number
of weeks in a given period is number of odd days during that
period.
b) Leap year : Every year which is divisible by 4 is called a
leap year. Thus each one of he year 1992,1996,2004,2008,..etc,
is a eap year.
Every 4th century is a leap year but no other century is a
leap year thus each one of 400,800,1200,1600,2000,etc is a
leap year. None of the 1900,2010,2020,2100,etc is a leap
year. An year which is not a leap year is called Ordinary year.
c)An ordinary year has 365 days.
d) A leap year has 366 days.
e)Counting of odd days :
i)1 ordinary year = 365 days =52 weeks+1 day,Therefore An
ordinary year has 1 Odd day.
ii)One leap year = 366 days =52 weeks+2 days, Therefore a leap
year has 2 Odd days.
iii) 100 years = 76 ordinary years+ 24 leap years
= [(76*52) weeks+76 days]+[(24*52)weeks+48 days]
= 5200 weeks+124days=[5217 weeks+5 days]
therefore 100 years contain 5 odd days
iv)200 years contain 10(1week+3days), i.e 3 odd days
v)300 years contain 15(2 weeks+1 day), i.e 1 odd day
vi)400 years contain (20+1), i.e 3 weeks,so 0 Odd days
similarly each one of 800,1200,1600,etc contains 0 odd days.
Note:(7n+m) odd days , where m less than or equal to 7
is equivalent to m odd days ,thus ,8 odd days = 1 odd day etc.
f) Some codes o remember the months and weeks:
a) Week
Sunday - 1
Monday - 2
Tuesday - 3
Wednesday - 4
Thursday - 5
Friday - 6
Saturday - 0
b) Month
jan - 1 july - 0
feb - 4 Aug - 3
Mar - 4 Sep - 6
Apr - 0 Oct - 1
May - 2 Nov - 4
june - 5 Dec - 6
Top
Simple problems:
Shortcuts : This shortcut must be applied only starting
with 19 series.
Example:
What day of the week on 17th june , 1998?
Solution : 5 -> the given month code(august)
17 -> the given date
98->(19 th century after years)
24-> ((47/4) = 11 i.e how many leap years
--------
total = 144 ((144/7) = 20 and the remainder is 4)
therefore in the above week table the no 4 code
represents wednesday
so the required day is wednesday.
Problem 1:
The first republic day of the India was celebrated on 26th
January,1950. It was
Solution : 01
26
50
12
----------
total = 89 ((89/7) = 12 and the remainder is 5)
therefore in the above week table represents the number 5
as thursday, so the required day was Thursday.
Problem 2:
Find on which day 15th august1947 ?
Solution :
03
15
47
11
----------
total = 76
Then (76)/7 = 6 odd days
6 indicates friday in the above week table.
Therefore required day is friday.
Problem 3:
Find on which day jan 26th 1956 ?
Solution :
01
26
56
14

-1 (-1 indicates leapyear(i.e 1956),so 1 reduce from the total)
---------
total = 96
Then (96)/7 = 5 odd days
5 indicates thursday in the above week table
Therefore our required day is Thursday.
Problem 4:
Today is friday after 62 days,it will be :
Solution : Each day of the week is repeated after 7 days.
so, after 63 days,it will be friday. Hence ,after 62 days,
it will be thursday.
Therefore the required day is thursday.
Problem 5:
Find the day of the week on 25th december,1995?
Solution :
06
25
95
23
---------
total = 149
Then (149)/7=(23)=2 odd days
Therefore the required day is "Monday".
Top
Medium Problems
Problem 1:
jan 1, 1995 was a sunday.what day of the week lies on
jan 1,1996?
Solution :
01
01
96
24
-1(since 1996 was leap year)
---------
total = 121
Then (121)/7 = (17) = 2 odd days
Therefore our required day wasMonday.
Problem 2:
On 8th feb,1995 it wednesday. The day of the week on
8th feb,1994 was?
Solution :
04
08
94
23
---------
total = 129
Then (129)/7 = (18) = 3 odd days
Therefore the required day is Tuesday.
Problem 3:
may 6,1993 was thursday.what day of the week was on
may 6,1992 ?
Solution :
02
06
92
23
-1
----------
total = 122
Then (122)/7 = (17) = 3 odd days
Therefore the required day is Tuesday
Problem 4:
jan 1, 1992 was wednesday. What day of the week was
on jan 1,1993 ?
Solution :
01
01
93
23
----------
total = 118
Then (118)/7 = (16) = 6 odd days
Therefore the required day is Friday.
Problem 5:
January 1,2004 was a thursday,what day of the week lies
on jan ,2005?
solution :
The year 2004 being a leap year, it has 2 odd days. so,
first day of the 2005 will be 2 days beyond thursday and
so it will be saturday
therefore the required day is Thursday.
Problem 6:
On 8th march,2005,wednesday falls what day of the week was
it on 8th march,2004?
Solution : the year 2004 being a leap year,it has 2 odd days.
so, the day on8th march,2005 will be two days beyond the day
on 8th march,2004.but 8th march,2005 is wednesday. so,
8th march,2004 is monday.
Therefore the required day is Monday.
Problem 7:
what was the day of the week on 19th september ,1986 ?
Solution :
06
19
86
21
---------
total = 132
Then ((132/7 = 18 and the remainder is 6)
In the above week table represents the number 6 is friday.
Therefore the required day is Friday.
Top
Typical problems
Problem 1:
On what dates of october,1994 did monday fall ?
Solution : 01
01
94
23
-------
total = 119
Then (119)/7 = (17) = 0 odd days
so the day is saturday
Therefore in october first the day is saturday.so,
the monday fell on 3rd october 1994.During october 1994,
monday fell on 3rd ,10th,17th and 24th.
Problem 2:
How many days are there from 2nd january 1995 to
15 th march,1995 ?

Solution : Jan + Feb + March
30 + 28 + 15 = 73 days
Problem 3:
The year next to 1996 having the same calendar as that
of 1996 is ?
Solution : Starting with 1996 , we go on countig the
number of odd days till the sum is divisible by 7.
Year 1996 1997 1998 1999 2000
odd days 2 1 1 1 2
2 + 1 + 1 + 1 + 2 = 7 odd days i.e odd day.
Therefore calendar for 2001 will be the same as
that of 1995.

Problem 4:
The calendar for 1990 is same as for :
Solution:
count the number of days 1990 onwards to get
0 odd day.
Year 1990 1991 1992 1993 1994 1995
oddd days 1 1 2 1 1 1
1 + 1 + 2 + 1 + 1 + 1 = 7 or 0 odd days
Therefore calendar for 1990 is the same as for the
year 1996.
Problem 5:
The day on 5th march of year is the same day on what
date of the same year?
Solution:
In the given monthly code table represents the march
code and november code both are same.that means any
date in march is the same day of week as the
corresponding date in november of that year, so the
same day falls on 5th november.

Clocks
General Concepts:
The face or dial of a watch is a circle whose circumference is
divided into 60 equal parts,called minute spaces.
A clock has two hands, the Smaller one is called the hour hand
or short hand while the larger one is called the minute hand or
long hand.
Important points:
a) In every 60 minutes, the minute hand gains 55 minutes on the
hour hand
b)In every hour, both the hands coincide once ,i.e 0 degrees.
c)the hands are in the same straight line when they are coincident
or opposite to each other. i.e 0 degrees or 180 degrees.
d)when the two hands are at right angles, they are 15 minute spaces
apart,i.e 90 degrees.
e)when the hands are in the opposite directions,they are 30 minute
spaces apart,i.e 180 degrees.
f)Angle traced by hour hand in 12hrs = 360 degrees.
g)Angle traced by minute hand in 60 min = 360 degrees. If a watch
or a clock indicated 8.15,when the correct time is 8, it is said
to be 15 minutes too fast. On the other hand, if it indicates 7.45,
when the correct time is 8,it is said to be 15 minutes slow.
h)60 min --> 360 degrees
1 min --> 60
i)the hands of a clock coincide in a day or 24 hours is 22 times,
in 12hours 11minutes.
j)the hands of clock are straight in a day is 44 times .
k)the hands of a clock at right angle in a day is 44 times .
l)the hands of a clock in straight line but opposite in direction is
22 times per day
Top
Simple Problems:
Type1:
Find the angle between the hour hand and the minute hand
of a clock when the time is 3.25
solution : In this type of problems the formulae is as follows
30*[hrs-(min/5)]+(min/2)
In the above problem the given data is time is 3.25. that is
applied in the
formulae
30*[3-(25/5)]+(25/2)30*(15-25)/5+25/2
= 30*(-10/5)+25/2
= -300/5+25/2
= -600+(25/2)=-475/10=-47.5
i.e 47 1/20
therefore the required angle is 47 1/20.
Note:The -sign must be neglected.
Another shortcut for type1 is :
The formulae is
6*x-(hrs*60+X)/2
Here x is the given minutes,
so in the given problem the minutes is 25 minutes,
that is applied in the given formulae
6*25-(3*60+25)/2
150-205/2
(300-205)/2=95/2
=47 1/20.
therefore the required angle is 47 1/20.
Type2:
At what time between 2 and 3 o' clock will be the hands of a
clock be together?
Solution : In this type of problems the formulae is
5*x*(12/11)
Here x is replaced by the first interval of given time.
here i.e 2. In the above problem the given data is between
2 and 3 o' clock
5*2*12/11 =10*12/11=120/11=10 10/11min.
Therefore the hands will coincide at 10 10/11 min.past2.
Another shortcut for type2 is:
Here the clocks be together but not opposite
to each other so the angle is 0 degrees. so the formulae is
6*x-(2*60+x)/2=06*x-(120+x)/2=012x-120-x=0
11x=120
x=120/11=10 10/11
therefore the hands will be coincide at 10 10/11 min.past2.
Top
Medium Problems
Type3:
At what time between 4 and 5 o'clock will the hands of a clock
be at rightangle?
Solution : In this type of problems the formulae is
(5*x + or -15)*(12/11)
Here x is replaced by the first interval of given time
here i.e 4
Case 1 : (5*x + 15)*(12/11)
(5*4 +15)*(12/11)
(20+15)*(12/11)
35*12/11=420/11=38 2/11 min.
Therefore they are right angles at 38 2/11 min .past4
Case 2 : (5*x-15)*(12/11)
(5*4-15)*(12/11)
(20-15)*(12/11)
5*12/11=60/11 min=5 5/11min
Therefore they are right angles at 5 5/11 min.past4.
Another shortcut for type 3 is:
Here the given angle is right angle i.e 900.
Case 1 : The formulae is 6*x-(hrs*60+x)/2=Given angle
6*x-(4*60+x)/2=90
6*x-(240+x)/2=90
12x-240-x=180
11x=180+240
11x=420
x=420/11= 38 2/11 min
Therefore they are at right angles at 38 2/11 min. past4.
Case 2 : The formulae is (hrs*60+x)/2-(6*x)=Given angle
(4*60+x)/2-(6*x)=90
(240+x)/2-(6*x)=90
240+x-12x=180
-11x+240=180
240-180=11x
x=60/11= 5 5/11 min
Therefore they art right angles at 5 5/11 min past4.
Type 4:
Find at what time between 8 and 9 o'clock will the hands of a
clock be in the same straight line but not together ?
Solution : In this type of problems the formulae is
(5*x-30)*12/11
x is replaced by the first interval of given time Here i.e 8
(5*8-30)*12/11
(40-30)*12/11
10*12/11=120/11 min=10 10/11 min.
Therefore the hands will be in the same straight line but not
together at 10 10/11 min.past 8.
Another shortcut for type 4 is:
Here the hands of a clock be in the same
straight line but not together the angle is 180 degrees.
The formulae is (hrs*60+x)/2-(6*x)=Given angle
(8*60+x)/2-6*x=180
(480+x)/2-(6*x)=180
480+x-12*x=360
11x=480-360
x=120/11=10 10/11 min.
therefore the hands will be in the same straight line but not
together at 10 10/11 min. past8.
Top
Type 5:
At what time between 5 and 6 oâ€™ clock are the hands of a
3 minutes apart ?
Solution : In this type of problems the formuae is
(5*x+ or - t)*12/11
Here x is replaced by the first interval of given time here xis 5.
t is spaces apart
Case 1 : (5*x+t)*12/11
(5*5+3)*12/11
28*12/11 = 336/11=31 5/11 min
therefore the hands will be 3 min .apart at 31 5/11 min.past5.
Case 2 : (5*x-t)*12/11
(5*5-3)*12/11
(25-3)*12/11=24 min
therefore the hands wi be 3 in apart at 24 min past 5.
Typicalproblems
problems:
A watch which gains uniformly ,is 5 min,slow at 8 o'clock in
the morning on sunday and it is 5 min.48 sec.fast at 8 p.m on
following sunday. when was it correct?
Solution :
Time from 8 am on sunday to 8 p.m on following sunday = 7 days
12 hours = 180 hours
the watch gains (5+(5 4/5))min .or 54/5 min. in 180 hours
Now 54/5 minare gained in 180 hours.
Therefore 5 minutes are gained in(180*5/54*5)hours=83 hours20 min.
=3 days11hrs20min.
therefore watch is correct at 3 days 11 hours 20 minutes after 8 a.m
of sunday
therefore it wil be correct at 20 min.past 7 p.m on wednesday

True Discount
Formulae:
Let rate=R%per annum, Time= T years
1.Present worth (PW) = (100*Amount)/(100+(R*T))
= (100* TrueDiscount)/(R*T)
2.TrueDiscount (TD) = (P.W*R*T)/100
= (Amount*R*T)/(100+(R*T))
3.Sum =(SimpeInterest*TrueDiscount)/(SimpleInterest-TrueDiscount)
4.SimpleInterest-TrueDiscount=SimpeInterest on TrueDiscount
5.When the sum is put at CompoundInterest,then
PresentWorth=Amount/(1+(R/100))^T
GeneralConcept:
Suppose a man has to pay Rs.156 after 4 years
and the rate of interest is 14%per annum
Clearly ,Rs.100 at 14% will amount to Rs156 in 4 years
So,the payment of Rs.100 now wil cear off the debt
of Rs156 due 4 years hence
We say that
Sum due = Rs156 due 4 years hence
Present Worth =Rs100
True Discount=Rs.156-Rs100=Rs56
=Sumdue-PW
We define
TrueDiscount= Interest on Present Worth(PW)
Amount = PresentWorth+TrueDiscount
1.Find the present worth of Rs.930 due 3 years hence at 8% per
annum.Aso find the discount?
Sol: Amount=RS.930,Time=3years,Rate=8%
TrueDiscount = (Presentworth*Time*Rate)/100
Presentworth = (Amount*100) /(100+(R*T))
Presentworth = (930*100)/(100+(8*3)
=Rs.750
TrueDiscount = (930*3*8)/(100+(8*3)) =Rs.180 (or)
TrueDiscount = Amount-Presentworth
=Rs.930-Rs.750
=Rs.180
2.The truediscount on a bill due 9 months hence at 12% per
annum is Rs540.Find the amount of the bill and its presentworth?
Sol: Time=9months=9/12years=3/4years
Rate=12%
TrueDiscount=Rs540
TrueDiscount=(Amount*100)/(100+(R*T))
=>Amount=(TrueDiscount*(100+(R*T))) /100
=(540*(100+12*(3/4))/100
=Rs.6540
Presentworth=Amount-TrueDiscount
=Rs.6540-Rs540
=Rs.6000
3.The TrueDiscount on a certain sum of money due 3 years hence
is Rs.250 and SimpeInterest on the same sum for same time and
same rate is Rs375 find sum and rate%?
Sol: Time=3 years
Truediscount=Rs.250
SimpeInterest=Rs375
Sum = (SimpeInterest*TrueDiscount)/(SimpeInterest-TrueDiscount)
= (375*250)/(375-250)
=Rs.750
SimpeInterest=(Principle*Time*Rate)/100
375 = ( 750*3R)/100
R=50/3%
=> Rate=162/3%
Top
4.The difference between SimpeInterest and TrueDiscount on a
certain sum of money for 6 months at 121/2% per annum is Rs25.
Find the sum.
Sol: Let amount be Rs.x
SimpleInterest=(Amount*T*R)/100
TrueDiscount=(Amount*T*R)/(100+(R*T))
SI-T.D=Rs25
=>((x*6/12*25/2)/100)-((x*6/12*25/2)/(100+(6/12*25/2))=25
=>x=Rs.6800
5.The Present worth of Rs.2310 due 21/2yrs henceThe interest
being 15% per annum is
sol: Amount=Rs2310,Time=21/2yrs,Rate=15%
Presentworth = (Amount*100) /(100+(R*T))
=2310*100/(100+15*5/2)
=Rs1680.
Medium Problems
1.A bill falls due in 1 year.The creditor agrees to accept
immediate payment of the half and to defer the payment of
the other half for 2 years .By this arrangement he gains
Rs.40.what is the amount of bill,if the money be worth 121/2%?
sol: Let the amount be Rs.x
If the bill falls due for 1 yr,
Truediscount=(Amount*100)/(100+R*T)
=(x*100)/(100+(25/2*1)
If Accepting immediate payment of the half and to
defer the payment of the other half for 2 years,
Truediscount=(Amount*100)/(100+R*T)
=[x/2+((x/2*100)/(100+(25/2*2))]
x/2 for immediate payment,
((x/2*100)/(100+(25/2*2)) for paying after 2 yrs
He gains Rs40
=>[x/2+((x/2*100)/(100+(25/2*2))] â€“ (x*100)/(100+(25/2*1) = Rs40
=> x/2+2x/5-8x/9=40
=>x=Rs.3600
Amount of the bill=Rs.3600
2.If the truediscount on a sum due 2yrs hence at 14% per annum
be Rs168.The sum due is?
Sol: Time=2yrs,
Rate=14%
Truediscount=Rs168,Amount=x
TrueDiscount=(Amount*R*T)/(100+R*T)
168 =(x*14*2)/(100+(14*2))
=>x=Rs.768
3.The truediscount on Rs.2562 due 4 months hence is Rs.122.The
rate % is?
Sol: Amount=Rs.2562
Time=4/12yrs
TrueDiscount=Rs122
Rate=X%
TrueDiscount=(Amount*R*T)/(100+R*T)
=>122=(2562*x*(4/12))/(100+(x*(4/12))
=>36600+122x=2562x
=>Rate=15%
4.The Truediscount on Rs1760 due after a certain time at
12% per annum is Rs160The time after which it is due is:
Sol: Amount=Rs.1760
Rate=12%
Truediscount=Rs160
Time=x
TrueDiscount=(Amount*R*T)/(100+R*T)
=>x*11*12=100+12x
=>x=5/6yr
Time=10months
5.The interest on Rs.760 for 2yrs is the same as the
Truediscount on Rs960 due 2yrs hence .If the rate of
interest is same in both cases,it is
sol:
Principal=Rs.760
Time=2yrs
amount=Rs960
P*T*R/100=(Amount*R*T)/(100+R*T)
=>(750*2*r)/100=(960*r*2)/(100+2r)
=>150r=2100
rate=14%
6.The SimpleInterest & TrueDiscount on a certain sum of
money for a given time & at a given rate are Rs85
& Rs.80.The sum is.............
sol:
S.I= Rs.85
TrueDiscount=Rs80
Sum=(S.I.*TD)/(SI-T.D)
=Rs.1360
Top
7.A trader owes a maerchant Rs10028 due 1yr hence The trader
wants to settle the account after 3 months .If the rate of
interest is 12%per annum How much cash should he pay?
Sol:
Time=1yr but he settles account after 3months
so time = 9months
Cash to pay=(Amount*R*T)/(100+R*T)
=(10028*100)/(100+12*(9/12))
=Rs.9200
8.A man buys a watch for Rs1950 in cash and sells it
for Rs2200 at a credit of 1yrThe rate of interest is 10%
per annum Then he gains / loose---------------amount?
Sol:
If he sells it for Rs2200 at a credit of 1yr then the
present worth of that amount is
Presentworth = (Amount*100) /(100+(R*T))
=(2200*100)/(100+10*1)
=2000/-
So he gains 2000-1950=Rs.50
9.A owes Rs.1573 payable 11/2yrs hence ,Aso B owes A Rs.144450
payabe 6months hence If they want to settle the accunt forth
with ,keeping 14% as the rate of interest ,then who should pay
& how much?
Sol:
B->A-----------Amount=Rs.1573
Rate=14%
Time=3/2
Presentworth = (Amount*100) /(100+(R*T))
=(1573*100)/(100+(3/2)*14)
=Rs1300
A->B------------Amount=Rs.1444.50
Rate=14%
Time=1/2
Presentworth = (Amount*100) /(100+(R*T))
=(1444.50*100)/(100+(1/2)*14)
=Rs1350
so B should pay Rs.1350
10.If Rs20 is the TrueDiscount on Rs260due after a certain time.
What wil be trueDiscount on same sum due after Â½ of former time,
the rate of interest beng the same?
Sol:
Simple Interest on (260-20) for a gven time=Rs20
Simple Interest on (260-20) for a half time=Rs20*(1/2)
=Rs10
True Discount on Rs.250=Rs10
True Discount on Rs.260=Rs10*260/250
=Rs10.40
11.A has to pay Rs.220 to B after 1yr.B asks A to pay Rs.110
in cash and defer the payment of Rs,.110 for 2 yrs.A agrees
to it.If the rate of interest be 10% per annum in this mode
of payment
sol:
A has to pay =Present worth of Rs220 due 1yr hence
=Rs.(220*100)/(100+(10*1))
=Rs.200 A actually pays=Rs.110+Presentworth of Rs110 due 2 yrs hence
=[110+ (Amount*100) /(100+(R*T))]
=[110+((110*100)/(100+(10*2)))]
=Rs192.66
So A gains Rs(200-192.66)=Rs.7.34
Top
Complex Problems
1.If Rs.10 be allowed as truedscount on a bill of Rs.110 at the
end of a certaimn time ,then discount allowed on the same sum
due at the end of double the time..
sol:
Amount=Rs110
TrueDiscount=Rs10
Present worth=Amount-TrueDiscount
=Rs110-10
=Rs.100
SI on Rs.100 for a certain time =Rs.10
SI on Rs.100 for doube the time =Rs.20
TrueDiscount on (100+20)=120-100
=Rs20
TrueDiscount on Rs.110 =(110*20)/120
=Rs18.33
2.A man wants to se hs scooter .There are two offers one at
Rs12000 cash and other at acredit of Rs12880 to be paid
after 8 months ,money being at 18% per annum which is better offer?
Sol:
offer1=Rs12000
offer2 Present worth= (Amount*100) /(100+(R*T))
=Rs.12880*100/(100+(18*(8/12))
=Rs11,500
The first offer is better as if he gains 500/-
if he sells at Rs.12000 is better
3.Goods were bought for Rs.600 and sold the same day for
Rs.688.80 at a credit of 9 months and thus gaining 2%
rate of interest per annum is
sol:
MethodI:
Amount=Rs.688.50
Time=9.12
Gaining 2%----------->Presentworth=102%of 600=(102*600)/100
=Rs.612
Presentworth=(Amount*100)/(100+(T*R))
612=(688.80*100)/(100+((9/2)*R))
=>R=162/3%
MethodII:
TrueDiscount=688.50-612
=Rs.76.50
Rate=TrueDiscount*100/(P.W*T)
=(76.50*100)/(612*(9/12))
=162/3%
4.The present worth of Rs.1404 due in 2 equal halfyearly instalments
at 8% per annu8m S.I is
SOl:
Presentworth=(Amount*100)/(100+(T*R))
PresentWorth=Presentworth of Rs.702 6 months hence +
Presentworth of Rs.702 1yr hence
= (702*100)/(100+(1/2*8)) + (702t*100)/(100+(1*8))
=Rs.1325

Bankers Discount
Formulae:
i. Suppose a merchant A buys goods worth Rs.10000 from another
merchant B at a credit of say 5 months
ii.Then,B prepares a bill , called the bill of exchange
iii. A signs this bill & allows B to withdraw the amount from his
bank account after exactly 5 months,the date exactly after
5 months is called Nominally due date
iv. Three days (grace days) are added to it get a date known as
legally due date
v.Suppose B wants to have money before legally due date then he
can have the money from banker or a broker who deducts S.I on the
face value (i.e., 10000) for the period from the date on which
the bill was discounted (i.e paied by the banker) & the legally
due date this amount is known as Bankers Discount
vi.Thus , B.D is the S.I on the face for the period from the date
on which the bill was discounted and the legally due date
vii.Bankers Gain (B.G) = (B.D) â€“ (T.D) for the unexpired time
Note:
When the date of the bill is not given,grace days are not to be added
Formulae:
(1)B.D = S.I on bill for unexpired time
(2)B.G = (B.D) â€“ (T.D) = S.I on T.D = (T.D)^2 /P.W
(3)T.D = sqrt(P.W * B.G)
(4)B.D = (Amount * Rate * Time)/100
(5)T.D = (Amount * Rate * Time)/(100+(Rate * time)
(6)Amount = (B.D * T.D)/(B.D â€“ T.D)
(7)T.D = (B.G * 100)/(Rate * Time)
Top
Simple Problems
1.If the true discount on a certain sum due 6 months hence at
15% is Rs 120.What is the bankers discount on the same for same
time and the same rate.
Sol:- B.G = S.I on T.D
= RS (120 * 15 * Â½ * 1/100)
= 9
(B.D) â€“ (T.D) = 9
B.D = 120 + 9 =129
2.The bankers discount on Rs 1800 at 12 % per annum is equal to the
true discount on Rs 1872 for the same time at the same rate .Find
the time.
Sol:- S.I on Rs 1800 = T.D on Rs 1872
P.W of Rs 1872 is Rs 1800
Rs 72 is S.I on Rs 1800 at 12%
Time = (100 * 72)/(12 * 1800)
= 1/3 years = 4 months
3.The bankers discount and true discount on a sum of money due 8 months
hence are Rs.120 & Rs.110 resp. Find the sum & the rate per cent
Sol:- Sum = (B.D * T.D) / (B.D) â€“ (T.D)
= (120 * 110) / (120 â€“ 110)
= 1320
Since B.D is S.I on sum due, so S.I on Rs 1320 for 8 months is
Rs 120
Rate = (100 * 120) / (1320 * 2/3)
= 13 7/11%
Medium Problems
1.The Bankers discount on Rs 1650 due a certain time hence is Rs 165.
find the true discount and the bankers gain.
Sol :- Sum = (B.D * T.D) / (B.D â€“ T.D)
= (B.D * T.D) / B.G
T.D/B.G = Sum/B.D =650/165 =10/1
Thus if B.G is Rs 1 ,T.D = Rs 10
if B.D is Rs 11 ,T.D = Rs 10
if B.D is Rs 165,T.D = Rs (10/11 * 165) = 150
B.G =Rs(165 â€“ 150) = Rs 15
2.The Present worth of a bill due something hence is Rs 1110 and the
true discount on the bill is Rs.110 . Find the bankes discount & the
bankers gain.
Sol:- T.D = sqrt (P.W * B.G)
B.G = (T.D)^2 /P.W
= (110 * 110) / 1110
= 11
B.D = (T.D + B.G)
= (110 + 11)
= Rs 121
3.What rate percent does a man get for his money when in discounting
Sol:- Let the amount of the bill =100
Money deducted = 10
Money received by the holder of the bill = (100 â€“ 10)
= 90
S.I on Rs 90 for 10 months = 10
Rate = (100 * 10) / (90 * 10/12)
= 13 1/3%
Top
Complex Problems
1.A bill for Rs.6000 is drawn on July 14 at 5 months . It is
discounted on 5th October at 10%.Find the bankers discount
true discount, bankers gain and the money that the holder of
the bill receives.
Sol:-
Face value of the bill = Rs.6000
date on which the bill was drawn = July 14 at 5 months
nominally due date = December 14
legally due date = December 17
Date on which the bill was discounted = October 5
Unexpired time : Oct Nov Dec
26 + 30 + 17 = 73days =1/5 years
B.D = S.I on Rs 6000 for 1/5 year
= Rs (6000 * 10 * 1/5 * 1/100) = Rs 120
T.D = Rs(6000 * 10 *1/5)/(100 + (10 * 1/5))
= Rs. 117.64
B.G = (B.D) â€“ (T.D)
= Rs(120 -117.64)
= Rs 2.36
Money received by the holder of the bill
= Rs(6000 â€“ 120) = 5880

2. The bankers gain on a certain sum due 1 Â½ year hence
is 3/25 of the bankers discount .The rate percent is
Sol:- Let B.D = 1 then B.G = 3/25
T.D = (B.D â€“ B.G)
= (1 â€“ 3/25)
= 22/25
sum = (1 * 22/25) / (1 â€“ 22/25)
= 22/3
S.I on Rs 22/3 for 1 Â½ year is 1.
Rate = (100 * 1) / (22/3 * 3/2)
= 9 1/9%
3. The bankers gain of a certain sum due 2 years hence
at 10% per annum is Rs 24 .The percent worth is
Sol:-
T.D = (B.G * 100) / (Rate * Time)
00) / (10 * 2)
= 120.
P.W = (100 *T.D) / (Rate * Time)
= (100 * 120) /(10 * 2)
= 600

Oddman Out and Series
Introduction:
In any type of problems,a set of numbers is given in such a way
that each one except one satiesfies a particular definite
property.The one which does not satisfy that characteristic is
to be taken out. Some important properties of numbers are
given below :
1.Prime Number Series
Example:
2,3,5,7,11,..............
2.Even Number Series
Example:
2,4,6,8,10,12,...........
3.Odd Number Series:
Example:
1,3,5,7,9,11,...........
4.Perfect Squares:
Example:
1,4,9,16,25,............
5.Perfect Cubes:
Example:
1,8,27,64,125,.................
6.Multiples of Number Series:
Example:
3,6,9,12,15,..............are multiples of 3
7.Numbers in Arthimetic Progression(A.P):
Example:
13,11,9,7................
8.Numbers in G.P:
Example:
48,12,3,.....
Some More Properties:
1. If any series starts with 0,3,.....,generally the relation
will be (n2-1).
2. If any series starts with 0,2,.....,generally the relation
will be (n2-n).
3. If any series starts with 0,6,.....,generally the relation
will be (n3-n).
4. If 36 is found in the series then the series will be in n2
relation.
5. If 35 is found in the series then the series will be in
n2-1 relation.
6. If 37 is found in the series then the series will be in n2+1
relation.
7. If 125 is found in the series then the series will be in n3
relation.
8. If 124 is found in the series then the series will be in n3-1
relation.
9. If 126 is found in the series then the series will be in n3+1
relation.
10. If 20,30 found in the series then the series will be in n2-n
relation.
11. If 60,120,210,........... is found as series then the series
will be in n3-n relation.
12. If 222,............ is found then relation is n3+n
13. If 21,31,.......... is series then the relation is n2-n+1.
14. If 19,29,.......... is series then the relation is n2-n-1.
15. If series starts with 0,3,............ the series will be on
n2-1 relation.
Top
Problems
1.Find the odd one out.3,5,7,12,17,19
SOLUTION:
The above series except 12 all elements are odd
numbers.so 12 is the odd one.
2.Find the odd one out.1,4,9,16,23,25,36
SOLUTION:
In the above series all elements except 23 are perfect sqares.
so 23 is odd one.
3.Find the odd one out.41,43,47,53,61,71,73,81
SOLUTION:
In the above series all elements except 81 are prime numbers.
so 81 is odd one.
4.Find the odd one out.1,4,9,16,20,36,49

SOLUTION:
In the above series all elements except 20 are perfect squares.
So 20 is odd one.
5.Find the odd one out.8,27,64,100,125,216,343
SOLUTION:
In the above series all elements except 100 are perfect cubes.
so 100 is odd one.
6. Find the odd one out.1,5,14,30,50,55,99
SOLUTION:
In the above series all elements in the pattern like 12,
12+22,12+22+32,................. But 50 is not in this
pattern,so odd one.

7.Find the odd one out.835,734,642,751,853,981,532
SOLUTION:
In the above series,the difference between third and first digit
of each element is equal to its middle digit.But 751 is not in
this pattern,so odd one.

8.Find the odd one out.385,4462,572,396,427,672,264
SOLUTION:
In the above series,the sum of first and third digit of each
element is equal to its middle digit.But 427is not in this
pattern,so odd one.

9.Find the odd one out.331,482,551,263,383,242,111

SOLUTION:
In the above series,the product of first and third digit of
each element is equal to its middle digit. But 383 is not in this
pattern,so odd one.
10. Find the odd one out.2,5,10,17,26,37,50,64
SOLUTION:
In the above series,the elements are in the pattern of x2+1,
Where x is 1,2,3,4,5,6,7.but 82+1 is not equal to 64.It is
65.64 is odd one.
11.Find the odd one out. 19,28,39,52,67,84,102

SOLUTION:
In the above series,the elements are in the pattern of x2+3,
Where x is 4,5,6,7,8,9,10.but 102+3 is not equal to 102.
It is 103.so 102 is odd one.
Top
12.Find the odd one out.253,136,352,460,324,613,244

SOLUTION:
In the above series,the elements are in the pattern of x2+3,
Where x is 4,5,6,7,8,9,10.but 102+3 is not equal to 102.It is
103.so 102 is odd one.
13.Find the odd one out. 2,5,10,50,500,5000
SOLUTION:
In the above series,the pattern as follows:
1st term * 2nd term = 3rd term
2nd term * 3rd term = 4th term
3rd term * 4th term = 5th term
But 50*500=25000 which is not equal to 5000.
so 5000 is odd one.
14.Find the odd one out. 582,605,588,611,634,617,600

SOLUTION:
In the above series, alternatively 23 is added and 17 is
subtracted from the terms. So 634 is odd one.

15.Find the odd one out.46080,3840,384,48,24,2,1

SOLUTION:
In the above series,the terms are successively divided by
12,10,8,6,..... so 24 is not in this pattern.
so 24 is odd one.
16.Find the odd one out.5,16,6,16,7,16,9
SOLUTION:
In the above series,the terms at odd places are 5,6,7,8.......
and at even places is 16. So 9 is odd one.
17.Find the odd one out.6,13,18,25,30,37,40
SOLUTION:
In the above series,the difference between two successive terms
from the beginning are 7,5,7,5......... so 40 is odd one.
18.Find the odd one out.56,72,90,110,132,150

SOLUTION:
The above series as follows:
7*8,8*9,9*10,10*11,11*12,12*13.
So it will be 56,72,90,110,132,156 so 150 is wrong.
19.Find the odd one out. 1,2,6,15,31,56,91
SOLUTION:
Add 1square ,2square ,....,6square to the terms.
so 91 is wrong.
20.Find the odd one out.105,85,60,30,0,-45,-90
SOLUTION:
Subtract 20,25,30,35,40,45 from the terms.
So 0 is odd one.
21.Find out the odd one out.3,10,21,36,55,70,105
SOLUTION:
The pattern in the series is
1*3, 2*5, 3*7, 4*9, 5*11, 6*13, 7*15.
So the series will be 3,10,21,36,55,78,105.
So 70 is wrong term in the series.
Top
22.Find out the odd one out. 4,9,19,39,79,160,319
SOLUTION:
Double the number and add 1 to it.
So the series will be 4,9,39,79,159,319. So 160 is wrong.
23.Find out the odd one out.10,14,28,32,64,68,132.
SOLUTION:
Alternatively add 4 and double the next term.
So 132 is wrong.
COmplex Problems
1.Find the missing term in the series:
4,-8,16,-32,64,( )
SOLUTION:
The terms are doubled and change the sign.
So the next term is -128
2.16,33,65,131,261,( )

SOUTION:
The terms are doubled and 1 is added.
So 261*1+1=522+1=523
So the missing term is 523.

3.2,6,12,20,30,42,56,( )

SOLUTION:
The pattern is
1 * 2 , 2 * 3 , 3 * 4 , 4 * 5 , 5 * 6 , 6 * 7 ,
7 * 8 , 8 * 9 .
So the series is 2,6,12,20,30,42,56,72.
So 72 is the missing term.

4. 8,24,12,36,18,54,( )

SOLUTION:
Numbers are alternatively multiplied by 3 and
divided by 2. So the next term is 54 / 2 = 27.
5. 165,195,255,285,345,( )

SOLUTION:
Each number is 15 multiplied by a prime number.
i.e the series is 15*11,15*13,15*17,15*19,15*23,15*29.
So series is 165,195,255,285,345,435.
So 435 is the missing term.
Top
6. 7,16,63,124,215,342,( ).

SOLUTION:
Numbers are 23 -1,33-1,43-1,....................so 83-1=511.
So 511 is the missing term.
7.2,4,12,48,240,( )

SOLUTION:
Go on multiplying by 2,3,4,5,6.
So the last term in the series is 240*6=1440.
8.8,7,11,12,14,17,17,22,( )
SOLUTION:
There are two series 8,11,14,17,20 and 7,12,17,22
So increasing by 3 and 5.So 20 the missing term.
9.71,76,69,74,67,72,( )

SOLUTION:
Alternately add 5 and subtract 7.
so the series is 71+5=76
10.10.2,5,9,19,37
SOLUTION:
Second number is one more than twice the first,Third
number is one less than twice the second,Forth is one
more than twice the third and so on.
So the next number is 2 * 37 + 1 = 74+1 = 75.

11. Find the wrong number in the given series.
3,8,15,24,34,48,63
SOLUTION:
The difference between consecutive terms are
respectively 5,7,9,11,13.
So 34 is the wrong number in the series.
12. 125,106,88,76,65,58,53

SOLUTION:
Subtract 24,21,18,15,12,9 from the numbers to get the
next number.
So 128 is wrong.
13. 1,1,2,6,24,96,720

SOLUTION:
Multiply with 1,2,3,4,5,6 to get the next number.
So 96 is wrong.
14 . 32,36,41,61,86,122,171,235
SOLUTION:
Second term = First term + 22
Third term = Second term + 32
Fourth term = Third term + 42
Fifth term = Forth term + 52
Sixth term = Fifth term + 62
Seventh term = Sixth term + 72
So the third term should be 45 instead of 41.

15 . 15,16,34,105,424,2124,12576

SOLUTION:
Second term = First term * 1 + 1 = 16
Third term = Second term * 2 + 2 = 34
Forth term = Third Term * 3 + 3 = 105
Fifth term = Forth term * 4 + 4 = 424
Sixth term = Fifth term * 5 + 5 =2125
Seventh term = Sixth term * 6 + 6 = 12576.
So 2124 is wrong.
16 . 40960,10240,2560,640,200,40,10

SOLUTION:
Go on dividing by 4 ,the series will be
40960,10240,2560,640,160,40,10.
So 200 is wrong.
17.7,8,18,57,228,1165,6996

SOLUTION:
Let the numbers be A,B,C,D,E,F,G then
A,A*1+1,B*2+2,C*3+3,............
so 288 is wrong.
18. 19,26,33,46,59,74,91

SOLUTION:
Go on adding 7,9,11,13,15,17.
So 33 is wrong.
19 . 10,26,74,218,654,1946,5834.

SOLUTION:
Second term = first term * 3 â€“ 4 = 26.
Third term = Second term * 3 â€“ 4 =74
Forth term = Third term * 3 â€“ 4 =218
Fifth term = Forth term * 3 â€“ 4 =650
So 654 is wrong .

Probability
Introduction:
Experiment:
An operation which can produce some well-defined outcome is
called an experiment.
Random Experiment:
An experiment in which all possible out comes are known and
the exact output cannot be predicted in advance is called a
random experiment.
EX:
1) Rolling an unbiased dice.
2) Tossing a fair coin.
3) Drawing a card from a pack of well-shuffled cards .
4)Picking up a ball of certain colour from a bag containing
balls of different colours.
Details:
1) When we thrown a coin ,then either a Head(H) or a Tail(T)
appears.
2)A dice is a solid cube ,having 6 faces,marked 1,2,3,4,5,6
respectively. When we throw a die ,the outcome is the number
that appears on its upper face.
3)A pack of cards has 52 cards.
It has 13 cards of each suit,namely spades,clubs,hearts and
diamonds. Cards of spades and clubs are balck cards.
Cards of hearts and diamonds are red cards.
There are four honours of each suit.
These are Aces,Kings,queens and Jacks.
These are called Face cards.
Sample Space:
When we perform an experiment ,then the set of S of all
possible outcomes is called the Sample space .
EX:
1)In tossing a coin S= {H,T}.
2)If two coins are tossed then S= {HH,HT,TH,TT}.
3)In rolling a dice ,we have S={1,2,3,4,5,6}.
Event:Any subset of a sample space is called an Event.
Probability of occurrence of an Event:
Let S be the sample space.
Let E be the Event.
Then E cS i.e E is subset of S then
probability of E p(E) =n(E)/n(S).
Reults on Probability:
1)P(S) =1.
2)0 Top
Problems:
1)An biased die is tossed.Find the probability of getting a
multiple of 3?
Sol: Here we have sample space S={1,2,3,4,5,6}.
Let E be the event of getting a multiple of 3.
Then E={3,6}.
P(E) =n(E)/n(S).
n(E) =2,
n(S) =6.
P(E) =2/6
P(E) =1/3.
2)In a simultaneous throw of a pair of dice,find the
probability of getting a total more than 7?
Sol: Here we have sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7.
={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),
(5,3),(6,2),(4,5),(5,4),
(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S)
= 15/36.
P(E) = 5/12.
3)A bag contains 6 white and 4 black balls .Two balls are
drawn at random .Find the probability that they are of the
same colour?
Sol: Let S be the sample space.
Number of ways for drawing two balls out of 6 white and
4 red balls = 10C2
=10!/(8!*2!)
= 45.
n(S) =45.
Let E =event of getting both balls of the same colour.
Then
n(E) =number of ways of drawing ( 2balls out of 6) or
(2 balls out of 4).
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2 +4 *3/2
=15+6 =21.
P(E) =n(E)/n(S) =21/45 =7/45.
4)Two dice are thrown together .What is the probability that
the sum of the number on
the two faces is divisible by 4 or 6?
Sol: n(S) = 6*6 =36.
E be the event for getting the sum of the number on the two
faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)
(5,1)(5,3)(6,2)(6,6)}
n(E) =14.
Hence P(E) =n(E)/n(S)
= 14/36.
P(E) = 7/18
5)Two cards are drawn at random from a pack of 52 cards What
is the probability that either both are black or both are
queens?
Sol: total number of ways for choosing 2 cards from
52 cards is =52C2 =52 !/(50!*2!)
= 1326.
Let A= event of getting bothe black cards.
Let B= event of getting bothe queens
AnB=Event of getting queens of black cards
n(A) =26C2.
We have 26 black cards from that we have to choose 2 cards.
n(A) =26C2=26!/(24!*2!)
= 26*25/2=325
from 52 cards we have 4 queens.
n(B) = 4C2
= 4!/(2!* 2!) =6
n(AnB) =2C2. =1
P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A n B) = n(A n B)/n(S) = 1/1326
P(A u B) = P(A) +P(B) -P(AnB)
= 325/1326 + 6/1326 -1/1326
= 330/1326
P(AuB) = 55/221
6)Two diced are tossed the probability that the total score
is a prime number?
Number of total ways n(S) =6 * 6 =36
E =event that the sum is a prime number.
Then E={(1,1)(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(4,1)
(4,3)(5,2)(5,6)(6,1)(6,5)}
n(E) =15
P(E) =n(E)/n(S)
= 15/36
P(E) = 5/12
Top
7)Two dice are thrown simultaneously .what is the probability
of getting two numbers whose product is even?
Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6
= 36
E=Event of getting two numbers whose product is even
E={(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)
(3,4)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,4)(5,6)(6,1)
(6,2)(6,3)(6,4)(6,5)(6,6)}
n(E) = 27
P(E) = n(E)/n(S)
= 27 /36
P(E) =3/4
probability of getting two numbers whose product is even is
equals to 3/4.
8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is
drawn at random. what is the probability of getting a prize ?
Sol: By drawing lottery at random ,we have n(S) =10C1+25C1
= 10+25
= 35.
E =event of getting a prize.
n(E) =10C1 =10
out of 10 prozes we have to get into one prize .The number of
ways 10C1.
n(E) =10
n(S) =35
P(E) =n(E)/n(S)
=10/35
= 2/7
Probability is 2/7.
9)In a class ,30 % of the students offered English,20 % offered
Hindi and 10 %offered Both.If a student is offered at random,
what is the probability that he has offered English or Hindi?
Sol:English offered students =30 %.
Hindi offered students =20%
Both offered students =10 %
Then only english offered students E =30 -10
=20 %
only Hindi offered students S =20 -10 %
= 10 %
All the students =100% =E +S +E or S
100 =20 +10 + E or S +E and S
Hindi or English offered students =100 -20-10-10
=60 %
Probability that he has offered English or Hindi =60/100= 2/5
10) A box contains 20 electricbulbs ,out of which 4 are defective,
two bulbs are chosen at random from this box.What is the
probability that at least one of these is defective ?
Sol: out of 20 bulbs ,4 bulbs are defective.
16 bulbs are favourable bulbs.
E = event for getting no bulb is defective.
n(E) =16 C 2
out of 16 bulbs we have to choose 2 bulbs randomly .so the number
of ways =16 C 2
n(E) =16 C2
n(S) =20 C 2
P(E) =16 C2/20C2
= 12/19
probability of at least one is defective + probability of one
is non defective =1
P(E) + P(E) =1
12/19 +P(E) =1
P(Eâ€™) =7/19
11)A box contains 10 block and 10 white balls.What is the
probability of drawing two balls of the same colour?
Sol: Total number of balls =10 +10
=20 balls
Let S be the sample space.
n(S) =number of ways drawing 2 balls out of 20
= 20 C2
= 20 !/(18! *2!)
= 190.
Let E =event of drawing 2 balls of the same colour.
n(E) =10C2+ 10C2
= 2(10 C2)
= 90
P(E) =n(E)/n(S)
P(E) =90/190
= 9/19
12) A bag contains 4 white balls ,5 red and 6 blue balls .Three
balls are drawn at random from the bag.What is the probability
that all of them are red ?
Sol: Let S be the sample space.
Then n(S) =number of ways drawing 3 balls out of 15.
=15 C3.
=455
Let E =event of getting all the 3 red balls.
n(E) = 5 C3 =5C2
= 10
P(E) =n(E) /n(S) =10/455 =2/91.
Top
13)From a pack of 52 cards,one card is drawn at random.What is the
probability that the card is a 10 or a spade?
Sol: Total no of cards are 52.
These are 13 spades including tne and there are 3 more tens.
n(E) =13+3
= 16
P(E) =n(E)/n(S).
=16/52
P(E) =4/13.
14) A man and his wife appear in an interview for two vacancies in
the same post.The probability of husband's selection is 1/7 and the
probabililty of wife's selection is 1/5.What is the probabililty
that only one of them is selected?
Sol: let A =event that the husband is selected.
B = event that the wife is selected.
E = Event for only one of them is selected.
P(A) =1/7
and
p(B) =1/5.
P(A') =Probability of husband is not selected is =1-1/7=6/7
P(B') =Probaility of wife is not selected =1-1/5=4/7
P(E) =P[(A and B') or (B and A')]
= P(A and B') +P(B and A')
= P(A)P(B') + P(B)P(A')
= 1/7*4/5 + 1/5 *6/7
P(E) =4/35 +6/35=10/35 =2/7
15)one card is drawn at random from a pack of 52 cards.What is the
probability that the card drawn is a face card?
Sol: There are 52 cards,out of which there 16 face cards.
P(getting a face card) =16/52
= 4/13
16) The probability that a card drawn from a pack of 52 cards will
be a diamond or a king?
Sol: In 52 cards 13 cards are diamond including one king there are
3 more kings. E event of getting a diamond or a king.
n(E) =13 +3
= 16
P(E) =n(E) /n(S) =16/52
=4/13
17) Two cards are drawn together from apack of 52 cards.What is the
probability that one is a spade and one is a heart ?
Sol: S be the sample space the n (S) =52C2 =52*51/2
=1326
let E =event of getting 2 kings out of 4 kings
n(E) =4C2
= 6
P(E) =n(E)/n(S)
=6/1326
=1/221
18) Two cards are drawn together from a pack of 52 cards.What is the
probability that one is a spade and one is a heart?
Sol: Let S be the sample space then
n(S) =52C2
=1326
E = Event of getting 1 spade and 1 heart.
n(E) =number of ways of choosing 1 spade out of 13 and 1 heart out
of 13.
= 13C1*13C1 =169
P(E)= n(E)/n(S)
=169/1326 =13/102.
19) Two cards are drawn from a pack of 52 cards .What is the
probability that either both are Red or both are Kings?
Sol: S be the sample space.
n(S) =The number of ways for drawing 2 cards from 52 cards.
n(S) =52C2
=1326
E1 be the event of getting bothe red cards.
E2 be the event of getting both are kings.
E1nE2 =Event of getting 2 kings of red cards.
We have 26 red balls.From 26 balls we have to choose 2 balls.
n(E1) =26C2
= 26*25/2
=325
We have 4 kings .out of 4 kings,we have to choosed 2 balls.
n(E2) =4C2
=6
n(E1nE2) =2C2 =1
P(E1) = n(E1)/n(S)
=325/1326
P(E2) =n(E2)/n(S)
=6/1326
P(E1nE2) =n(E1nE2)/n(S) =1/1326
P(both red or both kings) = P(E1UE2)
= P(E1) +P(E2)-P(E1nE2)
=325/1326 +6/1326 -1/1326
=330/1326 =55/221

Permutations and Combinations
Formulae:

Factorial Notation:
Let n be positive integer.Then ,factorial n dentoed by n!
is defined as n! = n(n-1)(n-2). . . . . . . .3.2.1
eg:- 5! = (5 * 4* 3 * 2 * 1)
= 120
0! = 1
Permutations:
The different arrangements of a given number of things by
taking some or all at a time,are called permutations.
eg:- All permutations( or arrangements)made with the letters
a,b,c by taking two at a time are (ab,ba,ac,ca,bc,cb)
Numbers of permutations:
Number of all permutations of n things, taken r at a time is
given by nPr = n(n-1)(n-2). . .. . . (n-r+1)
= n! / (n-r)!
An Important Result:
If there are n objects of which p1 are alike of one kind;
p2 are alike of another kind ; p3 are alike of third kind and
so on and pr are alike of rth kind, such that
(p1+p2+. . . . . . . . pr) = n
Then,number of permutations of these n objects is:
n! / (p1!).(p2!). . . . .(pr!)
Combinations:
Each of different groups or selections which can be formed by
taking some or all of a number of objects,is called a combination.
eg:- Suppose we want to select two out of three boys A,B,C .
then ,possible selection are AB,BC & CA.
Note that AB and BA represent the same selection.
Number of Combination:
The number of all combination of n things taken r at atime is:
nCr = n! / (r!)(n-r)!
= n(n-1)(n-2). . . . . . . tor factors / r!
Note: nCn = 1 and nC0 =1
An Important Result:
nCr = nC(n-r)
Top
Problems
1.Evaluate 30!/28!
Sol:- 30!/28! = 30 * 29 * (28!) / (28!)
= 30 * 29 =870
2.Find the value of 60P3
Sol:- 60P3 = 60! / (60 â€“ 3)! = 60! / 57!
= (60 * 59 *58 * (57!) )/ 57!
= 60 * 59 *58
= 205320
3. Find the value of 100C98,50C 50
Sol:- 100C98 = 100C100-98)
= 100 * 99 / 2 *1
= 4950
50C50 = 1
4.How many words can be formed by using all the letters of the
word â€œDAUGHTRâ€ so that vowels always come together &
vowels are never together?
Sol:-
(i) Given word contains 8 different letters
When the vowels AUE are always together we may suppose
them to form an entity ,treated as one letter
then the letter to be arranged are DAHTR(AUE)
these 6 letters can be arranged in 6p6 = 6!
= 720 ways
The vowels in the group (AUE) may be arranged in 3! = 6 ways
Required number of words = 760 * 6 =4320
(ii)Total number of words formed by using all the letters of
the given words

8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 40320
Number of words each having vowels together
= 760 * 6
= 4320
Number of words each having vowels never together
= 40320 â€“ 4320
= 36000
5.In how many ways can a cricket eleven be chosen out of a batch
of 15 players.
Sol:- Required number of ways
= 15C 11 = 15C (15-11)
= 15 C 4
15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1
= 1365
6.In how many a committee of 5 members can be selected from 6men
5 ladies consisting of 3 men and 2 ladies
Sol:- (3 men out of 6) and (2 ladies out of 5) are to be chosen
Required number of ways
=(6C3 * 5C2)
= 200
7.How many 4-letter word with or without meaning can be formed out
of the letters of the word 'LOGARITHMS' if repetition of letters is
not allowed
Sol:- 'LOGARITHMS' contains 10 different letters
Required number of words
= Number of arrangements of 100 letters taking
4 at a time
= 10P4
= 10 * 9 * 8 * 7
= 5040
8.In how many ways can the letter of word 'LEADER' be arranged
Sol:- The word 'LEADER' contains 6 letters namely
1L,2E,1A,1D and 1R
Required number of ways
= 6! / (1!)(2!)(1!)(1!)(1!)
= 6 * 5 * 4 * 3 * 2 *1 / 2 * 1
=360
Top
9.How many arrangements can be made out of the letters of the word
'MATHEMATICS' be arranged so that the vowels always come
together
Sol:- In the word ' MATHEMATICS' we treat vowels
AEAI as one letter thus we have MTHMTCS(AEAI)
now we have to arrange 8 letters out of which M occurs
twice ,T occurs twice & the rest are different
Number of ways of arranging these letters
= 8! / (2!)(2!)
= 10080
now AEAI has 4 letters in which A occurs 2 times and the rest
are different
Number of ways of arranging these letters
= 4! / 2! = 12
Required number of words = (10080 * 12)
= 120960
10.In how many different ways can the letter of the word 'DETAIL' be
arranged in such a way that the vowels occupy only the odd positions
Sol:- These are 6 letters in the given word , out of which
there are 3 vowels and 3 consonants
Let us mark these positions as under
(1)(2) (3) (4)(5)(6)
now 3 vowels can be placed at any of the three places out of 4
marked 1,3,5
Number of ways of arranging the vowels = 3P3 = 3! =6
Also,the 3 consonants can be arranged at the remaining 3 positions
Number of arrangements = 3P3 = 6
Total number of ways = (6 * 6) =36
11.How many 3 digit numbers can be formed from the digits 2,3,5,6,7
and 9 which are divisible by 5 and none of the digits is repeated?
Sol:- Since each desired number is divisible by 5,
so we much have 5 at the unit place. The hundreds place
can now be filled by any of the remaining 4 digits .so, there
4 ways of filling it.
Required number of numbers = (1 * 5 * 4)
= 20
12.In how many ways can 21 books on English and 19 books on Hindi
be placed in a row on a self so that two books on Hindi may not
be together?
Sol:- In order that two books on Hindi are never together,
we must place all these books as under:
X E X E X . . . . . . . . . . X E X
Where E denotes the position of an English and X that of
a Hindi book.
Since there are 21 books on English,the number of places
marked X are therefore 22.
Now, 19 places out of 22 can be chosen in
22 C 19 = 22 C 3 =22 * 21 * 20 / 3 * 2 *1
Hence the required number of ways = 1540
13.Out of 7 constants and 4 vowels how many words of 3 consonants
and 2 vowels can be formed?
Sol:- Number of ways of selecting (3 consonants out of 7) and
(2 vowels out of 4)
= 7C3 * 4C2
= 210
Number of groups each having 3 consonants and 2 vowels = 210
Each group contains 5 letters
Number of ways of arranging 5 letters among themselves
= 5! = (5 * 4 * 3 * 2 * 1)
= 210
Required number of words = (210 * 210)
= 25200