Numerical reasoning I

Problems on Numbers
Arithmetic Progression:
The nth term of A.P. is given by Tn = a + (n – 1)d
Sum of n terms of A.P S = n/2 *[2a+(n-1)d)]

Geometrical Progression:
Tn = arn – 1.
Sn = a(rn – 1) / (r-1)

Basic Formulae
1. ( a+b)2 = a2 + b2 + 2ab
2. (a-b)2 = a2 +b2 -2ab
3. ( a+b)2 - (a – b)2 = 4ab
4. (a+b)2 + (a – b)2 = 2 (a2 +b2)
5. (a2 – b2) = (a+b) (a-b)
6. (a+b+c)2 =a2 +b2 +c2 + 2(ab +bc+ca)
7. (a3 +b3) = ( a+b) (a2 –ab +b2)
8. (a3 –b3) = (a-b) (a2 +ab + b2)

Problem - 1
A 2 digit number is 3 times the sum of its digits
if 45 is added to the number. Its digits are
interchanged. The sum of digits of the number
is?

Solution
The number is 3 times the sum of its digits
45 is added = 4 +5 = 9
So, common numbers in 3 and 9th table.
9, 18, 27, 36, 45….
27 + 45 = 72
2 + 7 = 9 or 4 + 5 = 9

Problem - 2
A number when divided by 119 leaves a
remainder of 19. If it is divided by 17. It will
leave a remainder of?

Solution
= 19/17 = 2 remainder

Problem - 3
A boy was asked to find the value of 3/8 of sum of
money instead of multiplying the sum by 3/8 he
divided it by 8/3 and then his answer by Rs.55.
Find the correct answer?

Solution
8/3 – 3/8 = 55/24
= 55/55/24
= 24

Problem - 4
A man spends 2/5rd of his earning. 1/4th of the
expenditure goes to food, 1/5th on rent, 2/5th on
travel and rest on donations. If his total earning
is Rs.5000, find his expenditure on donations?

Solution
5000*2/5 = 2000
Remaining amount has given as donation
2000* (1/5 + 2/5 + ¼)
Total amount = 200*17/20 = 1700
2000 – 1700 = 300

Problem - 5
From a group of boys and girls 15 girls leave.
There are then left, 2 boys for each girl. After
this 45 boys leave, there are then left 5 girls for
each boy, find the number of girls in the
beginning?

Solution
15 girls leave = 2 boys for each girl
45 boys leave = 5 girls fro 1 boy
Let the boys be x; Girls = x/2 +15
After the boys have left,
No.of boys = x – 45 and girls = 5(x-45)
x/2 = 5(x-45)
X = 2(5x-22)
X = 10x – 450
X =50
50/2 +15 =40

Problem - 6
An organization purchased 80 chairs fro
Rs.9700. For chairs of better quality they paid
Rs.140 each and for each of the lower grade
chair they paid Rs.50 less. How many better
quality chairs did the organization buy?

Solution
Better quality chairs = x;
Lower quality = 80 –x
Price of better quality = Rs.140, Lower quality =
140-50 = 90
140*x + 90(80-x) = 9700
140x + 7200 – 90x = 9700
50x = 9700 – 7200;
50x = 2500
X = 50

Problem - 7
A labour is engaged for 30 days, on the condition
that Rs.50 will be paid for everyday he works
and Rs.15 will be deducted from his wages for
everyday he is absent from work. At the end of
30 days he received Rs.850 in all. For how many
days did he wanted?

Solution
Total wages = 30*50 = 1500 (without Absent)
Wages received in 30 days = 850 (with Absent)
Let the labourer work for x days
Absent = 30 – x
50x – (30-x)15 = 850
50x -450 +15x = 850
65x = 1300
X = 1300/65 = 20 days

Problem - 8
The rent is charged at Rs.50 per day for first 3
days Rs.100 per day next 5 days, and 300 per
day thereafter. Registration fee is 50 at the
beginning. If a person had paid Rs.1300 for his
stay how many days did he stay?

Solution
3 days = 150 + 50 = 200
5 days = 100*5 = 500
= 200 + 500 = 700
1300 – 700 = 600
2 days = 300*2 = 600
= 5 + 3 + 2 = 10 days

Problem - 9
In a school 20% of students are under the age of
8 years. The number of girls above the age of 8
years is 2/3 of the number of boys above the age
of 8 years and amount to 48. What is the total
number of students in the school?

Solution
Girls above 8 yrs = 48
Boys above 8 yrs = 48 / 2/3
80% of students above 8 yrs = 48 + 72 = 120
80 120
20 x
80x = 120*20
X = 120*20/80 = 30
Total No.of students = 120+30 = 150

Ratio and Proportion
Ratio : Relationship between two variables.
=a:b
Proportion : Relationship between two ratios.
=a:b::c:d
Proportion Calculation = a*d : b*c

Problem - 1
The ratio of number of boys to that of girls in a
school is 3:2. If 20% boys and 25% of girls are
scholarship holders, find the percentage of the
school students who are not scholarship holders?

Solution
Let the total number of students be 100
Boys = 100*3/5 = 60
Girls = 100*2/5 = 40
S. holders = 60*20/100 = 12, non S. holders = 60 -12 = 48
Girls s. holders = 40*25/100 = 10,
Non s. holders = 40 – 10 = 30
Students who do not have scholarship = 48 + 30 = 78
78/100*100 = 78%

Problem - 2
The cost of diamond varies as the square of its
weight. A diamond weighing 10 decigrams costs
Rs. 32000. Find the loss incurred when it breaks
into two pieces whose weights are in the ratio
2:3?

Solution
1st piece = 10*2/5 = 4
2nd piece = 10*3/5 = 6
Cost of the diamond varies as square of its weight
42 : 62 102 = 100k
16k : 36k
100k – 52 k = 48k(loss)
100k = 32000; k = 320
48*320 = 15360

Problem - 3
The ratio of the first and second class fares
between two railway stations 4 : 1 and the ratio
of the number of passengers traveling by first
and second class is 1:40. If the total of Rs.1100
is collected as fare from passengers of both
classes what was the amount collected from first
class passengers?

Solution
Fare = 4 : 1
Passengers traveling = 1 : 40
Amount = No. pas * fare = 4*1 :10*1 = 4 : 40
= 1:10
Total amount = 1100.
First class passengers‟ amount = 1*1100/11
= 100

Problem - 4
A vessel contains a mixture of water and milk in
the ratio 1:2 and another vessel contains the
mixture in the ratio 3:4. Taking 1 kg each from
both mixtures a new mixture is prepared. What
will be the ratio of water and milk in the new
mixture?

Solution
1st vessel = water = 1/3 , milk = 2/3
2nd vessel = water = 3/7, milk = 4/7
Water = 1/3 + 3/7 = 16/21
Milk = 2/3 + 4/7 = 26/21
16 : 26 = 8:13

Problem - 5
Ratio of the income of A, B, C last year 3 : 4 : 5.
The ratio of their individual incomes of last year
and this year are 4:5, 2:3 and 3:4 respectively. If
the sum of their present income is Rs.78,800.
Find the present individual income of A, B and
C.

Solution
A‟s Present Income = 5/4*3x = 15x/4
B‟s Present Income = 3/2*4x = 12x/2
C‟s Present Income = 4/7*5x = 20x/7
15x/4 + 6x+20x/3 = 78,800
197x/12 = 78,800
X = 945600/197
X = 4,800
A‟s Present income = 15x/4 = 15*4800/4 = 18,000
B‟s Present income = 6*x = 6*4,800 = 28,800
C‟s Present income = 20x/3 = 20*4800/3 = 32,000

Problem - 6
Of the three numbers, the ratio of the first and
the second is 8:9 and that of the second and third
is 3:4. If the product of the first and third
numbers is 2,400, then find the second number?

Solution
a:b=8:9
b:c=3:4

b : c = 3*3 : 4*3 = 9 : 12
a : b : c = 8 : 9 : 12
Product of first and third = 8k * 12k = 2400
96k2 = 2400; k2 = 2400/96 = 25
k=5
Second number = 9 * 5 = 45

Problem - 7
Annual income of A and B are in the ratio of 4 : 3
and their annual expenses are in the ratio 3 : 2. If
each of them saves Rs.600 at the end of the year,
what is the annual income of A?

Solution
Income = 4 : 3, Expenses = 3 : 2
Savings 600 each
A‟s income = 4x, expenses = 3x,
savings = x i.e 600

Income = 4*600 : 3*600
A : B = 2400 : 1800
A income = 2400

Problem - 8
The property of a man was divided among his
wife, son and daughter according to his will as
follows. Wife‟s hare is equal to 6/7th of son‟s
share and daughter share is equal of 4/7th of
Son‟s. If the son and daughter together receives
Rs.1,02,300. How much does his wife get?

Solution
Let the Son‟s share be x.
Daughter‟s share = x*4/7 = 4x/7
Wife‟s share = x* 6/7 = 6x/7
X + 4x/7 = 1,02,300
7x + 4x = 1,02,300
X = 1,02,300 /11 = 65,100
Wife Share = 65,100 *6/7 = Rs. 55, 800

Problem - 9
A pot containing 81 litres of pure milk of the
milk 1/3 is replaced by the same amount of
water. Again 1/3 of the mixture is replaced by
the same amount of water. Find the ratio of milk
to water in the new mixture?

Solution
Milk : Water
Initial = 81 : 0
1/3 removed = 54 : 27
1/3 mixture = 36 : 45

Ratio of Milk and Water = 4 : 5

Problem - 10
729 ml of mixture contains milk and water are in
the ratio 7 : 2. How much more water is to be
added to get a new mixture containing milk and
water in the ratio of 7 : 3.

Solution
Water = 729 * 2/9 = 162

Ratio Water
2 162
3 x
2x = 3*162/2 = 243
243 – 161 = 81 ml water is to be added

Problem - 11
Price of a scooter and a television set are in the
ratio 3 : 2. If the scooter costs Rs.600 more than
the television set, then find the price of
television?

Solution
Diff. in ratio = 3 – 2 = 1
1 ratio is 600 means, the television cost is 2 ratio
so, cost of television = 1200

Problem - 12
The annual income and expenditure of man and
his wife are in the ratio of 5:3 and 3:1
respectively, if they decide to save equally and
find their balance is 4000. Find their income at
the end of the year?

Solution
Man and Wife income = 5 : 3 = 2 (diff)
Man and Wife Expenses = 3 : 1 = 2 (diff)
so, both of them are saving ratio of 2
Total saving of Man and Women = 4000,
individual saving 2000

So, Man income = 5000 and Women income =
3000

Problem - 13
In a class room, ¾ of the boys are above 160 cm
in height and they are in 18 number. Also out of
the total strength, the boys are only 2/3 and the
rest are girls. Find the total number of girls in a
class?

Solution
¾ of the boys in 18 numbers means, ¼ of the
boys = 6
Total number of boys = 18+6 = 24
Ratio Number
2/3 24
1/3 x
2/3*x = 24*1/3
x = 24/2 = 12 Girls

Problem - 14

Rs. 770 was divided among A, B and C such that

A receives 2/ 9th of what B and C together

receive. Find A‟s share?

Solution
A = 2/9 (B+C)
B+C =9A/2
A+B+C = 770
A + 9A/2 = 770
11A = 770*2
A = 140

Problem - 15
A sporting goods store ordered an equal
number of white and yellow balls. The tennis
ball company delivered 45 extra white balls
making the ratio of white balls to yellow balls
1/5 : 1/6. How many white tennis balls did the
store originally order for?

Solution
Let the number of yellow balls be x

(x + 45) : x = 1/5 : 1/6

Solving the above equation,

The number of white balls originally ordered

would be = 225 balls

Alligation and Mixture
Alligation : It is the rule that enables us to find the ratio
in which two or more ingredients at the given price
must be mixed to produce a mixture of a desired
price.
(Quantity of cheaper / Quantity of costlier)
(C.P. of costlier) – (Mean price)
= --------------------------------------
(Mean price) – (C.P. of cheaper)

Alligation or Mixture
Cost of Cheaper Cost of costlier
c d

Cost of Mixture
m

d-m m-c

(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)

Problem -1
Three glasses of size 3 lit, 4 lit and 5 lit contain
mixture of milk and water in the ratio of 2:3, 3:7
and 4:11 respectively. The content of all the
three glasses are poured into a single vessel.
Find the ratio of milk and water in the resulting
mixture.

Solution
1st Vessel
= Milk = 3*2/5 = 6/5
= Water = 3*3/5 = 9/5

2nd Vessel:
= Milk = 4*3/10 = 12/10
= Water = 4*7/10 = 28/10
3rd Vessel:
= Milk = 5*4/15 = 20/15
= Water = 5*11/15 = 55/15
Milk : Water = 6/5 +12/10 + 20/15 : 9/5 + 28/10 + 55/15
= 18/15 + 18/15 +20/15 : 27/15 + 42/15+55/15
= 56 : 124 (or) 14:31

Problem - 2
How many kg of tea worth Rs. 25 per kg must be
blended with 30 kg tea worth Rs. 30 per kg, so
that by selling the blended variety at Rs.30 per
kg there should be a gain of 10%?

Solution
30*100/110 = 300/11
25 30
300/11

30/11 25/11
30 : 25
6 : 5
36 : 30kg

Problem - 3
A man buys cows for Rs. 1350 and sells one so
as to lose 6% and the other so as to gain 7.5%
and on the whole he neither gains nor loses. How
much does each cow cost?

Solution
6 7.5
0
7.5 6
15 12
5 : 4
1350*5/9 = 750
1350 *4/9 = 600

Problem - 4
There are 65 students in a class, 39 rupees are
distributed among them so that each boy gets
80p and each girl gets 30p. Find the number of
boys and girls in a class.

Solution
Girls Boys
30 80
60
20 30
2 : 3
65*2/5 = 26
65*3/5 = 39

Problem - 5
A person covers a distance 100 kms in 10 hr
Partly by walking at 7 km per hour and rest by
running at 12 km per hour. Find the distance
covered in each part.

Solution
Speed = Distance / Time = 100 / 10 = 10
7 12
10
2 : 3
Time taken in 7 km/hr = 10 * 2/5 = 4
4*7 = 28 km
Time taken in 12 km/hours = 10*3/5 = 6
12*6 = 72 km

Problem - 6

A merchant has 100 kg of salt, part of which
he sells at 7% profit and the rest at 17% profit.
He gains 10% on the whole. Find the quantity
sold at 17% profit?

Solution

7 17

10

(17-10) (10-7)
7 : 3
The quantity of 2nd kind = 3/10 of 100kg
= 30kg

Problem - 7
In what ratio two varieties of tea one costing
Rs. 27 per kg and the other costing Rs. 32 per
kg should be blended to produce a blended
variety of tea worth Rs. 30 per kg. How much
should be the quantity of second variety of tea,
if the first variety is 60 kg?

Solution
27 32
30
2 3
Quantity of cheaper tea = 2
Quantity of superior tea 3

Quantity of cheaper tea =2*x/5 = 60 , x=150
Quantity of superior tea = 3 * 150/5 = 90 kg

Problem - 8
A 3-gallon mixture contains one part of S and

two parts of R. In order to change it to mixture

containing 25% S how much R should be

added?

Solution

R : S
2 : 1
75% : 25%
3 : 1
1 gallon of R should be added.

Problem - 9

Three types of tea A,B,C costs Rs. 95/kg, Rs.
100/kg. and Rs 70/kg respectively. How many
kg of each should be blended to produce 100 kg
of mixture worth Rs.90/kg given that the
quantities of B and C are equal?

Solution
B+C/2 A
85 95
90
5 5
Ratio is 1:1 so A = 50 , B + C = 50

The quantity would be 50 : 25 : 25

Problem - 10
In what proportion water must be added to

spirit to gain 20% by selling it at the cost price?

Solution
Profit%=20%
Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3

0 10
25/3
5/3 25/3

The ratio is 1: 5

Problem - 11

In an examination out of 480 students 85% of
the girls and 70% of the boys passed. How
many boys appeared in the examination if total
pass percentage was 75%

Solution
Solution:
70 85

75
10 5

Number of Boys = 480 * 10/15
Number of Boys = 320

Problem - 12
A painter mixes blue paint with white paint so
that the mixture contains 10% blue paint. In a
mixture of 40 litre paint how many litre of blue
paint should be added, so that the mixture
contains 20% of blue paint?

Solution
Quantity of blue paint in the mixture = 10% of
40
40*10/100 = 4
40 – 4 = 36 litre
Let x litre blur paint can be mixed
4+x/30 = 20/80 = 4+x = 9
x=5

Problem - 13
From a 100 litre mixture containing water and
milk equal proportion, 10 litres of mixture is
replaced by 10 litres of water in succession
twice. At the end, what is the ratio of milk and
water?

Solution
Milk Water
10 lit(1st) 50 : 50
45 : 45
45 : 55
2nd 10 lit 40.5 : 49.5
Add Water 40.5 : 59.5
81 : 119

Problem - 14
In a mixture of 400 gms, 80% is copper, sliver is
20%. How much copper is to be added, so that
the new mixture has 84% copper?

Solution
400*80/100 = 320 Copper
400*20/100 = 80 Sliver
Percen Mixture
80 320
84 x
= 320*84/80 = 336
(320+x) = (400+x) 84/100
320+x = 400+84/100 + 84x/100
16x/100 = 336 – 320; 16x/100 = 16; x = 100

Problem - 15
A jar full of whisky contains 50% alcohol. A part
of this whisky is replaced by another containing
30% alcohol and now the percentage of alcohol
was found to be 35%. Find the quantity of
whisky replaced?

Solution
50 30
35
5 : 15
5 : 15 = 1 : 3
Replaced = 3/4

Type - 1
A invest = 10000
B invest = 15000
Profit = 5000
Find their Individual Share ?
A : B = 10000 : 15000 = 2 : 3
A‟s Share = 5000*2/5 = 2000
B‟s Share = 5000*3/5 = 3000
This is a first and basic step for any Partnership
Problem.

Type - 2
A invest = 5000,
After 3 months B joined A, with an investment
of 3000
Profit at the end of the year = 3500
Find their Share ?
Any thing happen after a month, like a person
joining a business, or withdraw from
business or withdraw some amount means
given amount is for month.
Cont…

Type - 2

A : B = 5000 : 3000 = 5*12 : 3*9 = 60 : 27 = 20 : 9

A‟s share = 3500*20/29 = 2413.7

B‟s Share = 3500*9/29 = 1086.3

Type - 3
A invest 5000
B invest 6000
After 3 months A withdraw amount 1000, after 5
months a withdraw amount 1000 again.
Profit at the end of the Year = 5000
Find their Share ?
A = 5*3 + 4*5 + 3*4 = 15 +20 + 12 = 47
B = 6*12 = 72

Type - 3

A‟s share = 5000* 47/119 = 1974.8

B‟s share = 5000*72/119 = 3025.2

Type - 4
A invest twice as much as B, B invest 1/3rd of C. At the
end of the year their Profit is 6000. Find their Share?
A = 2B
B = 1/3C
C=x
A : B : C = 2x/3 : x/3 : x
A : B : C = 2x/3 : x/3 : 3x /3
A:B:C=3:2:6
A‟s Share = 6000*3/11 = 1636
B‟s Share = 6000*2/11 = 1091
C‟s Share = 6000*6/11 = 3273

Problem - 1
A, B and C started a business in partnership by
investing Rs.12000 each. After 6 months, C left
and after 4 months D joined with his capital of
Rs.24,000. At the end of a year, a profit of
Rs.8,500 shared among all the partners. Find B‟s
share?

These are all the basic types remaining
we will see when we solve problems.

Solution
A : B : C : D
12000 : 12000 : 12000 : 24000
1 : 1 : 1 : 2
1*12 : 1*12 : 1*6 : 2*2
12 : 12: 6 : 4
6 : 6 : 3 : 2
B‟s share = 6/17*8500 = 3000

Problem - 2
A, B and C enter into partnership. A contributes
one third of the capital while B contributes as
much as A and C together contributed. If the
profit at the end of the year amounted to Rs.840.
What would be B‟s share?

Solution
A‟s share = 1/3 of the capital
A‟s share = 1/3*840 = 280
B‟s share = A + C = 280 + x
A + B + C = 840
280 + 280 + x + x = 840
560 + 2x = 840
2x = 840 – 560
X = 140
B‟s share = 280+140 = 420

Problem - 3
Akilesh and Jaga enter into a partnership.
Akilesh contributing Rs.8000 and Jaga
contributing Rs.10000. At the end of 6 months
they introduce Prakash, who contributes
Rs.6000. After the lapse of 3 years, they find that
he firm has made a profit of Rs.9660. Find
Prakash‟s share?

Solution
Akilesh : Jaga : Prakash
8 : 10 : 6
4 : 5 : 3
4*36 : 5836 : 3*30
144 : 180 : 90
8 : 10 : 5
Prakash‟s share = 9660*5/23 = 2100

Problem - 4
Priya and Vijay enter into partnership. Priya
supplies whole of the capital amounting to
Rs.45,000 with the conditions that the profit are
to be equally divided and that Vijay pays Priya
interest on half of the capital of 10% p.a. but
receives Rs.120 per month for carrying on the
concern. Find their total yearly profit. When
Vijay‟s income is one half of Priya‟s income?

Solution
45,000 *1/2 = 22,500
22,500 *10//100 = 2250 (interest p.a)
Vijay receives Rs.120 per month = 120*12 = 1440
Total profit be x
Ratio of Profit sharing = 1 : 1
Priya‟s income = x/2+2250
Vijay‟s income = x/2 – 2250 +1440
1 Priya = ½ Vijay
Priya Income = Twice of Vijay income
x/2 + 2250 = 2(x/2 – 2250 +1440)
X+4500/2 = 2(x/2 – 810)
X+4500/2 = x – 1620 = x +4500 = 2x – 3240
X = 7740
Total Profit of the year = 7740+1440 = 9,180

Problem - 5
Revathy and Shiva are partners sharing profits in
the ratio of 2:1. They admit Pooja into
partnership giving her 1/5th share in profits
which she acquires from Revathy and Shiva in
the ratio of 1:2. Calculate the new profit sharing
ratio?

Solution
Pooja gets her share of 1/5th of total share of
Profit from Revathy and Shiva in the ratio 1 : 2
From Revathy = 1/3*1/5 = 1/15
From Shiva = 2/3*1/5 = 2/15
Total Pooja share = 1/15+2/15 = 3/15 = 1/15
Revathy share = 2/3 – 1/15 = 9/15
Shiva share = 1/3 – 2/15 = 3/15
Shares = Revathy : Shiva : Pooja = 3 : 1 : 1

Problem - 6
A and B started a partnership business investing
some amount in the ratio of 3 : 5. C joined them
after six months with an amount equal to that of
B. In what proportion should the profit at the end
of 1 year be distributed among A, B and C?

Solution
Let the investment,
3 : 5 : 5
3*12: 5*12 : 5*6
36 : 60 : 30
6 : 10 : 5

Problem - 7
If 4(A‟s capital) = 6(B‟s capital) = 10 (C‟s
capital) then out of a profit of rs.4650. Find C‟s
share?

Solution
Let the unknown value be x
x/4 : x/6 : x/10
15x/60 : 10x/60 :6x/60
15 : 10 : 6
C‟s share = 6/31*4650 = Rs. 900

Problem - 8
A, B, C subscribe Rs.50,000 fro business. A
subscribes Rs.4000 more than B and B Rs.5000
more than C. Out of total profit of Rs.35,000.
Find A‟s share?

Solution
C = x,
B = x + 5000
A = x+5000+4000 = x + 9000
x +x+5000 +x+9000 = 50000
3x+14000 = 50000
3x = 50000 – 14000
3x = 36000,
x = 12000
C : B : A
12000 : 17000 : 21000
A = 35000*21/50 = 14,700

Problem - 9
A and B are partners in a business, A contributes
¼ of he capital for 15 months and B received 2/3
of the profit. For how long B‟s money was used?

Solution
B = 2/3
A = 1/3
A : B = 1/3 : 2/3 = 1 : 2
Investment
1/4x+15 : 3/4x*y
15x/4 : 3xy/4
15x/4 : 3xy/4 : : 1 : 2
30x/4 = 3xy/4
Y = 30x/4 * 4/3x = 10 months

Problem - 10
A, B and C invests Rs.4,000, Rs.5,000 and
Rs.6,000 respectively in a business and A gets
25% of profit for managing the business and the
rest of the profit is divided by A, B and C in
proportion to their investment. If in a year, A
gets Rs.200 less than B and C together, what was
the total profit for the year?

Solution
Total Profit = 100
25% for managing the business = 100 – 25 = 75%
A : B : C
4000 : 5000 : 6000
4 : 5 : 6
4x : 5x : 6x = 25x
100*15x/75 = 20x
A gets 4x + 25% of 20x
= 4x + 20x *25/100 = 9x
B = 5x, C = 6x
(5x + 6x) – 9x = 200
11x – 9x = 200
2x = 200; x = 100
Total Profit 20x = 20*100 = 2000

Problem - 11
A and B entered into partnership with capitals in
the ratio of 4 : 5. After 3 months, A withdraw ¼
of his capital and B withdraw 1/5 of his capital.
The gain at the end of 10 months was Rs.760.
Find the share of B?

Solution
A : B
4 : 5
4000 : 5000
A‟s share = 4000*1/4 = 4000 – 1000 = 3000
B‟s share = 5000*1/5 = 5000 – 1000 = 4000
A : B
3*4+3*7 : 5*3 +4*7
12 + 21 : 15+28
33 : 43
60*43/76 = 430

Problem - 12
Rs. 1290 is divided between A, B and C. So, that
A‟s share is 1 ½ times B‟s and B‟s share is 1 ¾
times C. What is C‟s share?

Solution
A : B = 1 ½ : 1 = 3/2 : 1 = 3 : 2
B : C = 1 ¾ : 1 = 7/4 : 1 = 7 : 4
A :B :C =3*7(A) : 2*7(B) : 7*2(B) : 4*2(C)
= 21 : 14 : 8
B = 1290*8/43 = Rs.240

Problem - 13
A man starts a business with a capital of
Rs.90000 and employs an assistant. From the
yearly profit he keeps an amount equal to 4 ½ of
his capital and pay 35% of the remainder of the
profits. Find how much the assistant receives in a
year, in which profit is Rs.30,000.

Solution
Investment = 90,000
4 ½ of investment = 9/2/100*90000 = Rs.4050
Profit = 30,000 – 4050 = 25,950
35/100*25,950 =9082.50

Problem - 14
A and B invest in a business in the ratio 3 : 2. If
5% of the total profit goes to charity and A‟s
share is Rs. 855, what is the total profit %?

Solution
Let the total profit be Rs. 100
After paying charity A‟s share = 3/5 *95 = 57
If A‟s share is Rs. 57, the total profit is 100
If A‟s share is Rs. 855, the total profit is
100 * 855/57 = Rs. 1500
The total profit = Rs. 1500

Problem - 15
A,B,C entered into a partnership by making an
investment in the ratio of 3 : 5 : 7. After a year
C invested another Rs. 337600 while A withdrew
Rs. 45600. The ratio of investments then
changed into 24: 59 : 167. How much did A
invest initially?

Solution
Solution:
Let the investments of A, B, and C be 3x, 5x, 7x
(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167
(3x – 45600)/5x = 24/59
x = 47200
Initial investment of A = 47200 * 3 = Rs. 141600

Problem - 1

The age of the Father is 4 times the age of his
Son. If 5 years ago, Father‟s age was 7 times the
age of his Son, what is the Father‟s present age?

Solution
F = 4S
F - 5 = 7(S - 5)
4S – 5 = 7S – 35
3S = 30
S = 10
Father‟s age = 4* 10 = 40 years

Problem - 2

The age of Mr. Gupta is four times the age of his
Son. After Ten years, the age of Mr. Gupta will
be only twice the age of his Son. Find the present
age of Mr. Gupta‟s Son.

Solution
G = 4S
G + 10 = 2 ( S + 10)
4S + 10 = 2S + 20
2S = 10
S=5
Son‟s Age = 5 years

Problem - 3

10 years ago Anu‟s mother was 4 times older
than her daughter. After 10 years, the mother
will be twice as old as her daughter. Find the
present age of Anu.

Solution
Ten years before:
M – 10 = 4(A – 10 )
M – 10 = 4A – 40
M = 4A – 40 + 10
M = 4A – 30
Ten Years After:
M + 10 = 2(A + 10)
M + 10 = 2A + 20
M = 2A + 20 – 10
M = 2A + 10
4A – 30 = 2A + 10
2A = 10 + 30
2A = 40: Anu‟s Age = 20

Problem – 4

The sum of the ages of A and B is 42 years. 3
years back, the ages of A was 5 times the age of
B. Find the difference between the present ages
of A and B?

Solution
A + B = 42
A = 42 – B
A – 3 = 5 ( B – 3)
A – 3 = 5B – 15
42 – B – 3 = 5B – 15
42 – 3 + 15 = 5B + B
54 = 6B
B = 54 /6 = 9
A = 42 – B; A = 42 – 9 = 33
Difference in their ages = 33 – 9 = 24 Years

Problem - 5

The sum of the ages of a son and father is 56
years. After 4 years, the age of the father will be
3 times that of the son. Find their respective
ages?

Solution
F + S= 56
S = 56 – F
F + 4 = 3 (S + 4)
F + 4 = 3 (56 – F + 4)
F + 4 = 168 – 3F + 12
4F = 168 + 12 – 4
4F = 176 ; F = 44
S = 56 – F ; S = 56 – 44 = 12
Father Age = 44; Son Age = 12

Problem – 6

The ratio of the ages of father and son at present
is 6:1. After 5 years, the ratio will become 7:2.
Find the Present age of the son.

Solution
6x + 5/x + 5 = 7/2
12x + 10 = 7x + 35
12x – 7x = 35 – 10
5x = 25
x = 25 / 5
x = 5 years
Son age = 1* 5 = 5 years

Problem - 7

The ages of Ram and Shyam differ by 16 years.
Six years ago, Shyam‟s age was thrice as that of
Ram‟s. Find their present ages?

Solution
S = R + 16
S – 6 = 3(R – 6)
S – 6 = 3R – 18
R + 16 – 6 = 3R – 18
R + 10 = 3R – 18
2R = 28 ; R = 14
Shyam‟s Age = 14 + 16 = 30.

Problem - 8

A man‟s age is 125% of what it was 10 years
ago, 83 1/3% of what it will be after 10 years.
What is his present age?

Solution
Let the age be x
125% of (x – 10) = 83 1/3 % of (x +10)
125/100 * x – 10 = 250/ 300 * x +10
5/4 x – 10 = 5/6 x – 10
5x / 4 – 5x / 6 = 50/6 + 50/4
5x /12 = 250/12
5x = 250 ; x = 50 years

Problem - 9

3 years ago, the average age of a family of 5
members was 17. A baby having born, the
average age of the family is the same today.
What is the age of the child?

Solution
Average age of 5 members = 17
Total age of 5 members = 17*5 = 85
3 years later, the age of 5 members will be
= 85 + 15 = 100
100 + x / 6 = 17
100 + x = 17*6
100 + x = 102
x = 102 – 100 = 2 years

Problem - 10
The sum of the age of father and his son is 100
years now. 5 years ago their ages were in the
ratio of 2 : 1. The ratio of the ages of father and
his son after 10 years will be?

Solution
F + S = 100
5 years ago 2 : 1
5 years ago
F + S = 100 – 10 = 90
90*2/3 = 60 : 30
Present age = 65 : 35
10 years ago = 75 : 45
=5:3

Problem - 11
Six years ago, Sushil‟s age was triple the age of
Snehal. Six years later, Sushil‟s age will be 5/3
of the age of Snehal. What is the present age of
Snehal?

Solution
Six years ago,
Snehal = x; Sushil = 3x
Six years later,
3x + 6+6 = 5/3(x+6+6)
9x +36 = 5x+60
4x = 60 – 36
X=6
Present Age of Snehal = 6+6 = 12 years

Problem - 12

Susan got married 6 years ago. Today her age is
1¼ times that at the time of her marriage. Her
son is 1/6 as old as she today. What is the age of
her son?

Solution
6 years ago Susan got married.
So her son‟s age will be less than 6 years.
Let as consider, her son‟s age is 5 years.
Susan‟s Age is 5*6 = 30 yrs, since the son is 1/6th of
Susan‟s age.
6 years ago her age must have been 24 yrs
24*1 ¼ = 24*5/4 = 30 yrs
As it satisfies the conditions her son‟s age is 5 years

Problem - 13
My brother is 3 years elder to me. My father was
28 years of age when my sister was born, while
my mother 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 of my
father and mother respectively when my brother
was born?

Solution
My brother was born 3 years before I was born
and 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

Problem - 14

If 6 years are subtracted from the present age of
Gagan and the reminder is divided by 18, then
the present age of his grandson Aunp is obtained.
If Anup is 2 years younger to Madan whose age
is 5 years, then what is Gagan‟s present age?

Solution
Anup‟s age = 5 – 2 = 3 years
Let Gagan‟s age be x
= x – 6 / 18 = 3
x – 6 = 3*18 ; x – 6 = 54
x = 54 + 6
Gagan‟s age = 60

Problem - 15

Ramu‟s grandfather says, “ Ram, I am now 30
years older than your father. 15 years ago, I was
2½ times as old as your father”. How old is the
grandfather now?

Solution
Let the father‟s age be x.
Grandfather‟s age will be 30 + x
15 years ago,
X + 30 – 15 = 5/2 (x – 15)
X + 15 = 5/2 (x – 15)
2x + 30 = 5x – 75
105 = 3x
X = 105 / 3 = 35
Grandfather‟s age = 35 + 30 = 65

Average
Average = Sum of Quantities
Number of Quantities
Sum of quantities
= Average*Number of Quantities.

Number of quantities
= Sum of Quantities
Average

Problem - 1

The average age of a class of 22 students is 21
years. The average increases by 1 when the
teacher‟s age is also included. What is the age of
the teacher?

Solution
Total age of the students be x
x/22 = 21; x = 21*22= 462
Teacher‟s age is also included
x/23 = 22; x = 22*23 = 506
Total age of 23 people – Total age of 22 people
will be the age of teacher
506 – 462 = 44 years
The age of teacher = 44

Problem - 2

The average of 7 numbers is 25. The average of
first three of them is 20 while the last three is 28.
What must be the remaining number?

Solution
Average of 7 numbers = 25,
Sum of 7 numbers = 25* 7 = 175
Avg. of first three numbers = 20, 20* 3 = 60
Avg. of last three numbers = 28, 28*3 = 84
The 4th number = 175 – (60+84) = 175 – 144
= 31

Problem - 3

The average age of a team of 10 people remains
the same as it was 3 years ago, when a young
person replaces one of the member. How much
younger was he than the person whose place he
took?

Solution
Let Average be x
10 members‟ Average = 10x
Average of 10 members (including new one) is
same as it was 3 yrs ago.
Now 10*3 = 30 years have increased, so a person
of 30 years should have replaced to keep the
average as same.

Problem - 4

The average age of a couple was 26 years at that
time of their marriage. After 11 years of marriage
the average age of the family with 3 children
become 19 years. What is the average age of the
Children?

Solution

Average of parents ages is 26, sum= 26*2 = 52
Parents age after 11 years = 52 +22 = 74
Average age of Family = 19, Sum = 19*5 = 95
Sum of family‟s age – Sum of parents‟ age
= 95 – 74 = 21
Sum of the ages of 3 children = 21,
Average Age = 21/3 = 7 yrs

Problem - 5

9 members went to a hotel for taking meals.
Eight of them spent Rs. 12 each on their meals
and the ninth person spent Rs. 8 more than the
average expenditure of all the nine. What was
the total money spent by them?

Solution
Average = x/9
Amount Spent by 8 members = 12 * 8 = 96
96 + x/9 + 8 = x
104 = x – x/9
104 = 8x/9
8x = 104 *9 = 936
x = 936/8 = 117

Problem - 7

A batsman makes a score of 87 runs in the 17th
inning and thus increases his average by 3. Find
his average after 17th innings?

Solution
17th innings avg. = x, Runs = 17x
16th innings avg. = x -3, Runs = 16 (x -3)
16 (x-3) + 87 = 17x
16x – 48 +87 = 17x
X = 39

Problem - 7
There are 24 students in a class. One of them,
who was 18 yrs old left the class and his place
was filled up by the newcomer. If the average of
the class thereby was lowered by one month,
what is the age of the newcomer?

Solution
Average reduced by 1 month,
24 * 1 = 2 years
So, the newcomer‟s age is 18 -2 = 16 years

Problem - 8
The average of marks in mathematics for 5
students was found to be 50. Later, it was
discovered that in the case of one student the
mark 48 was misread as 84. What is the correct
average?

Solution
Difference = 84 – 48 = 36
36 /5 = 7.2 (Increased)
The corrected average = 50 – 7.2 = 42.8

Problem - 9
The average salary of all the workers in a factory
is Rs. 8000. The average salary of 7 technicians
is Rs. 12000 and the average salary of the rest is
Rs. 6000. What is the total number of workers in
the factory?

Solution
Members Avg.
7 12000
X 6000
6x = 7*12
X = 7812/6 = 14
Total no. of workers = 7 + 14 = 21

Problem - 10
Average salary of all the 50 employees including
5 officers of the company is Rs. 850. If the
average salary of the officers is 2500, find the
average salary of the remaining staff of the
company.

Solution
x/50 = 850; x = 42,500
5 officers‟ salary = 2500*5 = 12500
50 – 5 members = 42500 – 12500
45 members = 30000
Avg. salary of 45 members = 30000/45
= 667(App)

Problem - 11

Find the average of 8 consecutive odd numbers
21,23,25,27,29,31,33,35

Solution

1st number + last Number /2
= 21 + 35 /2 = 28

Problem - 12

A train covers 50% of the journey at 30 km/hr,
25% of the journey at 25 km/hr, and the
remaining at 20 km/hr. Find the average speed of
the train during entire journey.

Solution
Total Journey = 100 km
S = Distance / Time = 100 / 5/3 + 1/1 + 5/4
= 100 * 12 /20+12+15
= 1200/47 = 25 25/47 km/hr

Problem - 13

The average of 10 numbers is 7. What will be the
new average if each number is multiplied by 8?

Solution
If numbers are multiplied by 8,
Average also to be multiplied by 8
= 7*8 = 56
{or}
x/10 = 7
x = 10*7 = 70
= 70* 5 = 560 /10 = 56

Problem - 14
The mean marks of 10 boys in a class is 70%
whereas the mean marks of 15 girls is 60%.
What is the mean marks of all 25 students?

Solution
Boys = x/10 = 70 = 700
Girls = x/15 = 60 = 900
10 + 15 = 700 + 900
25 = 1600
1600/25 = 64%

Problem - 15
Of the three numbers the first is twice the second
and the second is thrice the third. If the average
of the three numbers is 10, what are the
numbers?

Solution
A = 2x
B=x
C = x/3
2x + x + x/3/3 = 10
6x + 3x + x /9 = 10
6x + 3x + x = 90
10x = 90 ; x = 9.
A = 18, B = 9, C = 3

Percentage

• By a certain Percent, we mean that many

hundredths.

• Thus, x Percent means x hundredths, written

as x%

Percentage
•Finding out of Hundred.
If Length is increased by X% and Breadth is
decreased by Y% What is the percentage
Increase or Decrease in Area of the rectangle?
Formula: X+Y+ XY/100 %
Decrease 20% means -20

Problem -1

When 75% of the Number is added to 75%, the
result is the same number. What is the number?

Solution
Percentage Number
75 x+75
100 x
100x + 7500 = 75x
25x = 7500
x = 300

Problem - 2
A tank is full of milk. Half of the milk is sold
and the tank is filled with water. Again half of
the mixture is sold and the tank is filled with
water. This operation is repeated thrice. Find the
percentage of milk in the tank after the third
operation?

Solution
Milk Water
100 0
50 50(1st)
25 75 (2nd)
12.5 87.5 (3rd)
After 3 operation Milk 12.5%

Problem 3

A large water-melon weighs 20kg with 96% of
its weight being water. It is allowed to stand in
the sun and some of the water evaporates so that
now, only 95% of its weight is water. What will
be its reduced weight?

Solution
20 *96/100=19.2kg of water
Let the evaporated water be x
19.2-x=95%(20-x)
19.2-x=95(20-x)/100
1920-100x=1900-95x
5x=20 ;x=4
20-4=16kg.

Problem 4
The population of a city is 155625. For
every1000 men, there are 1075 women. If 40%
of men and 24% of women be literate, then what
is the percentage of literate people in the city?

Solution
Ratio of men and women=1000:1075=40:43
Number of men=40*155625/83=75000
Number of women=155625-7500=80625
Number of literate men=75000*40/100=3000
Number of literate women
=80625*24/100=19350
Literate people =30000+19350=49350
Percentage of literate people
=49350/155625*100=2632/83=31 59/83%

Problem 5

300 grams of sugar solution has 40% sugar in it.
How much sugar should be added to make it
50% in the solution?

Solution
Grams Sugar
300 40%
X 50%
50x = 40*300
x = 40*300/50 = 240
300 – 240 =60 Kg

Problem - 6

A man lost 12½% of his money and after
spending 70% of the remainder, he has Rs. 210
left. How much did the man have at first?

Solution
Let the amount be 100
Then, 100.00 – 12.50 = 87.50

70% of 87.50 = 87.50 *70/100 =61.25

The remaining amount will be Rs. 26.25
Initial Final
100 26.25
X 210
26.25x = 21000; x = 21000/26.25 = 800

Problem - 7
During one year the population of a town
increases by 10% and during next year it
diminished by 10%. If at the end of the second
year, the population was 89,100, what was the
Population at the beginning of first year?

Solution
Let the population be 100
1st Year = 100 + 10 = 110
2nd Year = 110 * 10/100 = 110 -11 = 99
Percentage Population
99 89100
100 x
99x = 89100*100;
x = 8910000/99 = 90000

Problem - 8

When a number is first increased by 20% and
then again 20% by what percent should the
increase number be reduced to get back the
original number?

Solution
Let the number be 100
20% increase = 100*20/100 = 20
New Value = 120
Again increase by 20% = 120*20/100 = 24
New value = 144
Increased amount = 44/144*100 = 30 5/9%

Problem - 9
The number of students studying Arts,
Commerce and Science in an institute were in
the ratio 6 : 5 : 3 respectively. If the number of
students in Arts, Commerce and science were
increased by 10%, 30% and 15% respectively,
what was the new ratio between number of
students in the three streams?

Solution
A:C:S
6:5:3
6x : 5x : 3x
6x*110/100 : 5x*130/100 : 3x*115/100
6x*110 : 5x*130 : 3x*115
660 : 650 : 345
132 : 130 : 69

Problem - 10

In measuring the sides of rectangle errors of 5%
and 3% in excess are made. What is the error
percent in the calculated area?

Solution

Area = xy
X = 5% Excess = 100* 5/100 = 105
Y = 3% Excess = 100*3/100 = 103
103*105/100 = 10815/100 = 108.15
Error – Actual = 108.15 – 100 = 8.15% Excess

Problem - 11

In a certain examination there were 2500
candidates. Of them 20% of them were girls and
rest were boys. If 5% of boys and 40% of girls
failed, what was the Percentage of candidates
passed?

Solution
Girls = 2500*20/100 = 500
Boys = 2500*80/100 = 2000
Students who failed were
Boys = 2000*5/100 = 100
Girls = 500*40/100 = 200
Total Failed Students = 300
Total Pass students = 2500 – 300 = 2200
Pass Percentage = 2200/2500*100 = 88%

Problem - 12

A person saves every year 20% of his income. If
his income increases every year by 10% then his
saving increases by?

Solution
Every year saving, if the income is Rs. 100
= 100 *20/100 =Rs. 20
Salary increases = 110*20/100 = 22
Percentage increase (Savings) = 2/20*100 = 10%

Problem - 13
On a test containing 150 questions carrying 1
mark each, meena answered 80% of the first
answers correctly. What percent of the other 75
questions does she need to answer correctly to
score 60% on the entire exam?

Solution
Required correct answer = 150*60/100 = 90
Questions need to be correct.
80% of 75 questions = 60 q answered correctly.
Remaining 30 questions need be correct out of 75
= 30/75*100 = 40

Problem - 14

A boy after giving away 80% of his pocket
money to one companion and 6% of the
remainder to another has 47 paise left with him.
How much pocket money did the boy have in the
beginning?

Solution
Let the amount be 100
To the first companion = 100*80/100 = 80
Remaining = 100 – 80 = 20
To the 2nd person = 20*6/100 = 1.20
The remaining = Rs.18.80 or 1880 paise
Initial Final
100 1880
X 47
1880x = 47*100
x = 4700/1880 = 2.5

Problem - 15
The length of a rectangle is increased by 10%
and breath decreased by 10%. Then the area of
the new rectangle?

Solution
I – D – I*D /100
10 -10 – 10*10/100
0 – 1 = -1
Decrease by 1%

Profit and Loss
• Gain =(S.P.)-(C.P.)
• Loss =(C.P.)-(S.P.)
• Loss or gain is always reckoned on C.P.
• Gain% = [(Gain*100)/C.P.]
• Loss% = [(Loss*100)/C.P.]
• S.P. = ((100 + Gain%)/100)C.P.
• S.P. = ((100 – Loss%)/100)C.P.

Problem - 1
A trade man allows two successive discount of
20% and 10%. If he gets Rs.108 for an article.
What was its marked price?

Solution
I1 + I2 – I1*I2/100
20 + 10 – 20*10 /100
= 28%
Discount = 28%, 72 Percent Cost is 108
Then 100percent cost = 72 108
100 x
100*108/72 = 150

Problem - 2
A trade man bought 500 metres of electric wire
at 75 paise per metre. He sold 60% of it at profit
of 8%. At what gain percent should he sell the
remainder so tas to gain 12% on the whole

Solution
500* 60/100 = 300
8 X
12
300 200
300 : 200 = 6 : 4
8 18
12
6 4
Remainder at 18% Profit

Problem - 3
A man purchased a box full of pencils at the rate
of 7 for Rs. 9 and sold all of them at the rate of 8
for Rs. 11. in this bargains he gains Rs. 10. How
many pencils did the box contains.

Solution
LCM = 7 and 8 = 56
56 pencil cost price = 8*9 = 72
56 Pencil selling price = 7*11
Profit = 77 – 72 = Rs. 5 for 56 pencil
Rs. 5 for 56 pencil means , for Rs. 10 the pencils
are 112

Problem - 4
A cloth merchant decides to sell his material at
the cost price, but measures 80cm for a metre.
His gain % is?

Solution
100 – 80 = 20 cm difference
Actual = 80
20/80*100 = 25% Gain

Problem - 5
Sales of a book decrease by 2.5% when its price
is hiked by 5%. What is the effect on sales?

Solution
Let the sales be 100 – 2.5 = 97.5
Profit = 100+5 = 105
Sales Profit
97.5 105
100 X
100x = 97.5*105
x = 97.5*105/100 = 102.375
100 – 102.375 = 2.375 = 2.4 profit (app)

Problem - 6
A dealer buys a table listed at Rs.1500 and gets
successive discount of 20% and 10%. He spends
Rs. 20 on transportation and sells it at a profit of
10%. Find the selling price of the table.

Solution
Discount = 20+10 – 20*10/100 = 28%
Actual price = 100 – 28 = 72
100 1500
72 x
72*1500/100 = 1080
Transport = 1080 +20 = 1100
100 1100
110 x
1100*110/100 = 1210

Problem - 7
A fridge is listed at Rs. 4000. due to the off
season, a shopkeeper announces a discount of
5%. What is the S.P?

Problem - 8
If the cost price of 9 pens is equal to the S.P of
11 pens. What is the gain or loss?

Solution
= 11 – 9 = 2
= 2/11*100 = 18 2/11% loss

Problem - 9
A machine is sold for Rs.5060 at a gain of 10%
what would have been the gain or loss percent if
it had been sold Rs.4370?

Solution
S.P = Rs.5060 = Gain = 10%
C.P = 100/110*5060 = 4600
IF S.P = Rs.4370 and C.P = Rs.4600
Loss = 230
Loss % = 230/4600 * 100 = 5% loss

Problem - 10
A person purchased two washing machines each
for Rs.9000. he sold one at a loss of 10% and
other at a gain of 10%. What is his gain or loss?

Solution
Each Rs.9000. one is 10% profit and other is
10% loss. So No profit and No loss

Problem - 11
Four percent more is gained by selling an article
for Rs.180, then by selling if for Rs.175. then its
C.P is?

Solution
Let the cost price = Rs. X
4% of x = 180 – 175 = 4x/100 = 5
4x = 500; x = 500/4 = 125

Problem - 12
An article is sold at a profit of 20%. If it had
been sold at a profit of 25%. It would have
fetched Rs.35% more. The Cost Price of the
article is?

Solution
Let C.P = Rs. X
125% of x – 120% of x = 35
5% of x =Rs.35 = x = 35*100/5 = 700
C.P = Rs. 700

Problem - 13
A reduction of 20% in the price of orange
enables a man to buy 5 oranges more for Rs. 10.
The price of an orange before reduction was,

Solution
20% Rs. 10 = Rs.2
Reduced price of 5 oranges = Rs. 2
Reduced price of 1 oranges = 40 p
Original price = 40/ 1- 0.20 = 400/8 = 50 Paise

Problem - 14
A man sells two horses for Rs.1475. The cost
price of the first is equal to the S.P of the second.
If the first is sold at 20% loss and the second at
25% gain. What is his total gain or loss? ( in
rupees)

Solution
Let cost price of 1st horse = S.P of 2nd = x
C.P of 2nd = S.P of 2nd * 100/125 = x*100/125 =
4x/5
S.P of 1st = C.P of 1st *80/100 = x*80/100 = 4x/5
Neither loss nor gain

Problem - 15
Rekha sold a watch at a profit of 15%. Had he
bought it at 10% less and sold it for Rs. 28 less,
he would have gained 20%. Find the C.P of the
Watch.

Solution
C.P be Rs. X
First S.P = 115% of x = 23x/20 and second C.P =
90% x = 9x/10
Second S.P = 120% of 9x/10 = 120/100 * 9x/10
= 27x/25
Given 23x/20 – 27x/25 = 28 = 115x – 108x/100
= 28
7x/100 = 28 = x = 28*100/7 = 400
C.P = Rs.400

Probability
• Probability:
P(є) = n(є) / n(s)
• (Addition theorem on probability:
n(AUB) = n(A) + n(B) - n(A B)
• Mutually Exclusive:
P(AUB) = P(A) + P(B)
• Independent Events:
P(A B) = P(A) * P(B)

Problem - 1
Four cards are drawn at random from a pack of
52 playing cards. Find the probability of getting
all face cards?

Solution
n(E) = 52C4
n(S) = 12C4 = 12C4/52C4

Problem - 2
Four persons are to be chosen at random from a
group of 3 men, 2 women and 4 children. Find
the probability of selecting 1 man, 1 woman or 2
children?

Solution
Total 3 M + 2 W + 4 C = 9 C 4 = 126
n (E) = 3C1 * 2C1 * 4C2 = 36
36/126 = 2/7

Problem - 3
A word consists of 9 letters, 5 consonants and 4
vowels. Three letters are chosen at random.
What is the probability that more than one
vowels will be selected?

Solution
n(E) = 9C3 = 84
More than one Vowels. So,
2V +1C or 3 V
4C2 *5C1 + 4C3 = 34
= 34/84 = 17/42

Problem - 4
A bag contains 10 mangoes out of which 4 are
rotten. Two mangoes are taken out together. If
one of them was found to be good, then what is
the probability that the other one is also good?

Solution
10 mangoes – 4 are rotten = 6 good mangoes
Getting good mangoes = 6C1/10C1 = 6/10
Getting second mango to be good = 5/9
1st and 2nd mangoes
6/10 *5/9 = 1/3

Problem - 5
Out of 13 applicants for a job there are 5 women
and 8 men. It is desired to select 2 persons for
the job. What is the probability that at least one
of the selected person will be a woman?

Solution
n(E) = 13C2 = 78
n(S) = 1m and 1 w or 2 w
= 8C1*5C1 + 5C2 = 50
= 50/78 = 25/39

Problem - 6
Two cards are drawn at random from a pack of
52 cards. What is the probability that either both
are black or both are queen?

Solution
P(A) = Both are Black
P(B) = Both are Queen
P(AnB) = Both are queen and Black
P(A) = 26C2/52C2 = 325/1326
P(B) = 4C2 /52C2 = 6/1326
P(AnB) = 2C2 /52C2 = 1/1326
325/1326 + 6/1326 - 1/ 1326 = 55/221

Problem -7
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 probability
of wife‟s selection is 1/5. Find the probability
that only one of them is selected?

Solution
Husband‟s Selection = 1/7;
Not getting selected = 1 – 1/7 = 6/7
Wife‟s selection = 1/5;
Not getting selected = 1 – 1/5 = 4/5
Only one of them is selected =
(Husband‟s Selection + Wife Not getting selected) or
(Wife‟s selection + Husband‟s Not getting selected)
= (1/7*4/5) + 1/5*6/7) = 2/7

Problem - 8
Four persons are chosen at random from a group
of 3 men, 2 women and 4 children. What is the
chance that exactly 2 of them are children?

Solution
3 + 2 + 4 = 9C4 = 126
4 members 2(M and W) + 2(boy)
5C2 + 4C2 = 60
= 60 / 126 = 10/21

Problem - 9
Prakash can hit a target 3 times in 6 shots, Priya
can hit the target 2 times in 6 shots and Akhilesh
can hit the target 4 times in 4 shots. What is the
probability that at least 2 shots hit the target?

Solution
Prakash hitting = 3/6; not hitting = 3/6
Priya hitting = 2/6; not hitting = 4/6
Akilesh = 4/4 = 1
At least 2 shots hit target
= 3/6*4/6 + 3/6*2/6 = ½

Problem - 10
There are two boxes A and B. A contains 3 white
balls and 5 black balls and Box B contains 4
white balls and 6 black balls. One box is taken at
random and what is the probability that the ball
picked up may be a white one?

Solution
(Box A is selected and a ball is picked up ) or
(Box B is selected and a ball is picked up)
½*3/8 + ½*4/10 = 31/80

Problem - 11
A bag contains 6 white balls and 4 black balls.
Four balls are successively drawn without
replacement. What is the probability that they are
alternately of different colour?

Solution
Suppose the balls drawn are in the order white,
black, white, black…
= 6/10 *4/9*5/8*3/7 = 360/5040
Suppose the balls drawn are in the order black,
white, black, white…
= 4/10*6/9*3/8*5/7 = 360/5040
360/5040 +360/5040 = 1/7

Problem - 12
A problem in statistics is given to four students
A, B, C and D. Their chances of solving it are
1/3, ¼, 1/5 and 1/6 respectively. What is the
probability that the problem will be solved?

Solution
A is not solving problem = 2/3,
B is not solving problem = ¾
C not solving problem = 4/5
D not solving problem = 5/6
2/3*3/4*4/5*5/6 = 1/3
All together the probability of solving the
problem = 1 -1 /3 = 2/3

Problem - 13
There are 8 questions in an examination each
having only 2 answers choices „Yes‟ or „No‟. All
the questions carry equal marks. If a student
marks his answer randomly, what is the
probability of scoring exacting 50%?

Solution
Each questions having 2 ways of answering,
1 question = 2!........ 8 question = 2!
= 2!*2!*2!*2!*2!*2!*2!*2! = 256
To get 50%, 4 questions need to be correct,
8c4 = 8*7*6*5/1*2*3*4 = 70
= 70/256 = 35/128

Problem - 14

A group consists of equal number of men and
women. Of them 10% of men and 45% of
women are unemployed. If a person is randomly
selected from the group find the probability for
the selected person to be an employee.

Solution

Let the number of men is 100 and women be 100
Employed men and women = (100-10)+(100-45)
= 145
Probability = 145 / 200 = 29 / 40

Problem - 15
The probability of an event A occurring is 0.5
and that of B is 0.3. If A and B are mutually
exclusive events. Find the probability that
neither A nor B occurs?

Solution

It is Mutually exclusive events P(A n B)=0
Probability = 1 – ( P(A) + P (B) – P(A n B) )
= 1 – (0.5 + 0.3 – 0)
= 0.2

Permutation and Combination

Permutation means Arrangement

Combination means Selection

Permutation and Combination
• Permutations:
Each of the arrangements which can be made by
taking some (or) all of a number of items is called
permutations.
np = n(n-1)(n-2)…(n-r+1)=n!/(n-r)!
r
• Combinations:
Each of the groups or selections which can be made
by taking some or all of a number of items is called a
combination.
nC = n!/(r!)(n-r)!
r

Types
1. How many ways of Arrangement possible by
using word SOFTWARE?
SOFTWARE = 8!
2. How many ways of arrangement Possible by
using word SOFTWARE, vowels should come
together.
SFTWR (OAE) = 6! * 3!

Types
3. How many ways of Arrangement Possible by
using word SOFTWARE, vowels should not
come together?
SFTWR ( ARE)
Not together
= Total arrangement – Vowels together
= 8! – (5! * 3!)

Types
4. How many ways of arrangement possible by using
word MACHINE, so that vowels occupy only ODD
places.
- - - - - - - (7 places)
MCHN (AIE) 4 Consonant and 3 vowels.
7 places = 4 ODD places, 3 EVEN places
Vowels = 4P3 = 4!
Consonant = 4P4 = 4!
Total Number of arrangement = 4!*4!

Types
5. How many ways of arrangement possible by
using word ARRANGEMENT
Letter‟s Repetition = 2(A) 2(R) 2 (E) 2 (N)
= 11!/2!*2!*2!*2!
In a given problem, any letter is repeated more
than once that should be divided with total
number.

Problem - 1

A committee of 5 is to be formed out of 6 gents
and 4 ladies. In how many ways this can be
done, when at least 2 ladies are included?

Solution
a. 2 ladies * 3 Gents
4C2 * 6 C3 = 120
b. 3 ladies * 2 Gents
4C3 * 6C2 = 60
c. 4 ladies * 1 Gent
4C4 *6C1 = 1*6 = 6
Total ways = 120 +60 +6 = 186

Problem - 2

It is required to seat 5 men and 4 women in a
row so that the women occupy the even places.
How many such arrangements are possible?

Solution
Total places = 9
Odd places = 5
Even places = 4
4 even places occupied by 4 women
= 4P4 = 4! = 24
5 odd places occupied by 5 men
= 5P4 = 5! = 120
Total ways = 120*24 = 2880 ways

Problem - 3

A set of 7 parallel lines is intersected by another
set of 5 parallel lines. How many parallelograms
are formed by this process?

Solution

Two parallel lines from the first set and any two
from the second set will from a parallelogram.
7C2 *5C2 = 21 * 10 = 210

Problem - 4

There are n teams participating in a football
championship. Every two teams played one
match with each other. There were 171 matches
on the whole. What is the value of n?

Solution
Total number of matches played = nC2
nC2 = 171
n(n-1)/2= 171
n2 – n – 342 = 0
(n+18) (n-19) = 0
n = 19

Problem - 5

In an examination, a candidate has to pass in
each of the 6 subjects. In how many ways can he
fail?

Solution

6C1 + 6C2 + 6C3 + 6C4+6C5+6C6
1 + 6 + 15 + 20 + 15 + 6 = 63 ways

Problem - 6

In how many ways can a pack of 52 cards be
distributed to 4 players, 17 cards to each of 3 and
one card to the fourth player?

Solution
17 cards can be given to 1st player = 52 C17
2nd player = 35C17
3rd player = 18C17
4th player = 1
= 52C17*35C17*18C17
= 52!/17!35! * 35!/17!*18! * 18!/17!*1!
= 52!/(17!)3

Problem - 7
A foot race will be held on Saturday. How many

different arrangements of medal winners are

possible if medals will be for first, second and

third place, if there are 10 runners in the race …

Solution
n = 10
r=3
n P r = n!/(n-r)!
= 10! / (10-3)!
= 10! / 7!
= 8*9*10
= 720
Number of ways is 720.

Problem - 8

To fill a number of vacancies, an employer must
hire 3 programmers from among 6
applicants, and two managers from 4 applicants.
What is total number of ways in which she can
make her selection ?

Solution
It is selection so use combination formula

Programmers and managers = 6C3 * 4C2

= 20 * 6 = 120

Total number of ways = 120 ways.

Problem - 9

A man has 7 friends. In how many ways can

he invite one or more of them to a party?

Solution
In this problem, the person is going to select his
friends for party, he can select one or more
person, so addition
= 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7
= 127
Number of ways is 127

Problem - 9

Find the number of different 8 letter words

formed from the letters of the word EQUATION

if each word is to start with a vowel

Solution
For the words beginning with a vowel, the first
letter can be any one of the 5 vowels, the
remaining 7 places can be filled by
7P7 = 5040
The number of words = 5 * 5040 = 25200

Problem - 10

In how many different ways can the letters of the

word TRAINER be arranged so that the vowels

always come together?

Solution
A,I,E can be arranged in 3! Ways

(5! * 3!) / 2! = 360 ways

Problem - 11
In how many different ways can the letters of the
Word DETAIL be arranged so that the vowels
may occupy only the odd positions?

Solution
___ ___ ___ ___ ___ ___
3P3 = 3! = 1*2*3 = 6
3P3 = 3! = 1*2*3 = 6
= 6*6 = 36

Problem - 12
There are 5 red, 4 white and 3 blue marbles in a
bag. They are taken out one by one and arranged
in a row. Assuming that all the 12 marbles are
drawn, find the number of different
arrangements?

Solution
Total number of balls = 12
Of these 5 balls are of 1st type (red), 4 balls are
the 2nd type and 3 balls are the 3rd type.
Required number of arrangements =
12!/5!*4!*3!
= 27720

Problem - 13
5 men and 5 women sit around a circular table,
the en and women alternatively. In how many
different ways can the seating arrangements be
made?

Solution
5 men can be arranged in a circular table in 4
ways = 24 ways
There are 5 seats available for 5 women they can
be arranged in 5 ways
No. of ways = 5!*4! = 2880 ways

Problem - 14

In a chess board there are 9 vertical and 9
horizontal lines. Find the number of rectangles
formed in the chess board.

Solution
Solution:

9C2 * 9C2 = 1296

Problem - 15
In how many ways can a cricket team of 11
players be selected out of 16 players, If one
particular player is to be excluded?

Solution
Solution:
If one particular player is to be excluded, then
selection is to be made of 11 players out of 15.
15C11= 15!/( 11!*4!)=1365 ways

Area and Volume
Cube:

• Let each edge of the cube be of length a. then,

• Volume = a3cubic units

• Surface area= 6a2 sq.units.

• Diagonal = √3 a units.

Area and Volume
Cylinder:

• Let each of base = r and height ( or length) = h.

• Volume = πr2h

• Surface area = 2 πr h sq. units

• Total Surface Area = 2 πr ( h+ r) units.

Area and Volume
Cone:
• Let radius of base = r and height=h, then
• Slant height, l = √h2 +r2 units
• Volume = 1/3 πr2h cubic units
• Curved surface area = πr l sq.units
• Total surface area = πr (l +r)

Area and Volume
Sphere:

• Let the radius of the sphere be r. then,

• Volume = 4/3 πr3

• Surface area = 4 π r2sq.units

Area and Volume
Circle: A= π r 2
Circumference = 2 π r
Square: A= a 2
Perimeter = 4a
Rectangle: A= l x b
Perimeter= 2( l + b)

Area and Volume
Triangle:

A = 1/2*base*height

Equilateral = √3/4*(side)2

Area of the Scalene Triangle

S = (a+b+c)/ 2

A = √ s*(s-a) * (s-b)* (s-c)

Problem - 1
A rectangular sheet of size 88 cm * 35 cm is bent
to form a cylindrical shape with height 35 cm.
What is the area of the base of the cylindrical
shape?

Solution
The circumference of the circular region = 88 cm
2 r = 88
r = 88*7/22*2 = 14 cm
Area of the base = r2 = 22/7*14*14 v= 616 cm2

Problem - 2
The radius of the base of a conical tent is 7
metres. If the slant height of the tent is 15
metres, what is the area of the canvas required to
make the tent?

Solution
R=7m
L = 15 m
Area of Canvas required = Curved Surface Area
of cone
rl = 22/7*7*15 = 330 sq.m

Problem - 3
Three spherical balls of radius 1 cm, 2 cm and 3
cm are melted to form a single spherical ball. In
the process, the material loss was 25%. What
would be the radius of the new ball?

Solution
Vol. of sphere = 4/3 r3
Vol. of 3 small spherical balls = 4/3 ( 13+23+33)
= 4/3 (1+8+27) = 4 /3 (36) = 48
Material loss = 25%
Vol. of the single spherical ball = 48 *75/100
= 48 * ¾ = 36
V = 4/3 r3 = 36
r3 = 36*3/4 = 27
r = 3 cm

Problem - 4

A rectangular room of size 5m(l)*4m(w)*3m(h)
is to be painted. If the unit of painting is Rs. 10
per sq.m, what is the total cost of painting?

Solution
Area of 4 walls = 2h(l+b)
The area to be painted includes the 4 walls and
the top ceiling.
Area to be painted = 2h (l+b) +lb
= 2*3 (5+4) + 5*4
= 54+20 = 74 sq.m.
Total cost of painting = 74*10 = Rs.740

Problem - 5
The radius of a sphere is r units. Each of the
radius of the base and the height of a right
circular cylinder is also r units. What is the ratio
of the volume of the sphere to that of the
cylinder?

Solution
Vol. of sphere = 4/3 r3 and Vol. of Cylinder =
r2h = r3

Required Ratio = 4/3 r3: r3 = 4/3 : 1
=4:3

Problem - 6
A measuring jar of internal diameter 10 cm is
partially filled with water. Four equal spherical
balls of diameter 2 cm each, are dropped in it
and they sink down in the water completely.
What will be the increase in the level of water in
the jar?

Solution
Radius of each ball = 1 cm
Vol. of 4 balls = 4* 4/3 (r)3 = 16/3 cm3
Vol. of water raised in the Jar = Vol. of 4 balls
Let h be the rise in water level, then
Area of the base *h = 16/3
*5*5*h = 16/3
H = 16/3*25 = 16/75 cm

Problem - 7

What is the cost of planting the field in the form

of the triangle whose base is 2.8 m and height

3.2 m at the rate of Rs.100 / m2

Solution
Area of triangular field = ½ * 3.2 * 2.8 m2
= 4.48 m2
Cost = Rs.100 * 4.48
= Rs.448..

Problem - 8
Find the length of the longest pole that can be
placed in a room 14 m long, 12 m broad, and 8 m
high.

Solution
Length of the longest pole = Length of the
diagonal of the room
= √(142 + 122 + 82)
= √ 404 = 20.09 m

Problem - 9

Area of a rhombus is 850 cm2. If one of its

diagonal is 34 cm. Find the length of the other

diagonal.

Solution
850 = ½ * d1 * d2
= ½ * 34 * d2
= 17 d2
d2 = 850 / 17
= 50 cm
Second diagonal = 50cm

Problem - 10
A grocer is storing small cereal boxes in large
cartons that measure 25 inches by 42 inches by 60
inches. If the measurement of each small cereal
box is 7 inches by 6 inches by 5 inches then what is
maximum number of small cereal boxes that can be
placed in each large carton ?

Solution

No. of Boxes = 25*42*60 / 7*6*5 = 300

300 boxes of cereal box can be placed.

Problem - 11

If the radius of a circle is diminished by 10%,
what is the change in area in percentage?

Solution
= x + y + xy/100
= -10 - 10 + 10*10/100
= -19%
Diminished area = 19%.

Problem - 12

A circular wire of radius 42 cm is bent in the

form of a rectangle whose sides are in the ratio

of 6:5. Find the smaller side of the rectangle?

Solution
length of wire = 2 πr = (22/7*14*14)cm
= 264cm
Perimeter of Rectangle = 2(6x+5x) cm
= 22xcm
22x =264 x = 12 cm
Smaller side = (5*12) cm = 60 cm

Problem - 13
A beam 9m long, 40cm wide and 20cm deep is
made up of iron which weights 50 kg per cubic
metre. Find the weight of the Beam.

Solution

Vol. of the Beam = lbh = 9*40/100*10/100
= 72 m3
Weight of the iron beam is given as lm3 = 50 kg

72/100 m3 = 72/100*50 = 36 kg

Problem - 14

If the length of a rectangle is reduced by 20%
and breadth is increased by 20%. What is the
percentage change in the area?

Solution
x + y + (xy/100)%
= - 20 + 20 – 400/100
= -4
The area would decrease by 4%

Problem - 15
Find the number of bricks measuring 25 cm in
length, 5 cm is breadth and 10 cm in height for a
wall 40 m long, 75 cm broad and 5 metres in
height?

Solution
Vol. of the wall = 40*72/100*5 = 150 m3
Vol. of 1 bricks = 25/100*5/100*10/100
= 1/80 m3
Number of bricks required = 150/1/800
= 150*800
= 120000

Calendar
Odd days:
0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday

Calendar
Month code: Ordinary year
J=0 F=3
M=3 A=6
M=1 J=4
J=6 A=2
S=5 O=0
N=3 D=5
Month code for leap year after Feb. add 1.

Calendar
Ordinary year = (A + B + C + D )-2
-----------------------take remainder
7

Leap year = (A + B + C + D) – 3
------------------------- take remainder
7

Problem - 1
11th January 1997 was a Sunday. What day of
the week on 7th January 2000?

Solution
11th Jan 1997 = Sunday
11th Jan 1998 = Monday
11th Jan 1999 = Tuesday
11th Jan 2000 = Wednesday
7th Jan 2000 is on Saturday

Problem - 2
What day of the week was on 5th June 1999?

Solution
A+B+C+D – 2 / 7
A = 1999/7 = 4
B = 1999/4 = 499/7 = 2
C = June = 4
D = 5/7 = 5
= 4+2+4+5 – 2/7 = 13/7 = 5 = Saturday

Problem - 3
On what dates of August 1988 did Friday fall?

Solution
A = 1988 / 7 = 0
B = 1988/4 = 497/7 = 0
C=3
D=x
0+0+3+x+3/7 = x/7 = 5(Friday)
Friday falls on = 5,12,19,26

Problem - 4
India got independence on 15 August 1947.
What was the day of the week?

Solution
A = 1947/7 = 1
B = 1947/4 = 486/7 = 3
C = 15/7 = 1
D=2
1+3+1+2 – 2 /7 = 5/7 = Friday

Problem - 5
7th January 1992 was Tuesday. Find the day of
the week on the same date after 5 years. i.e on
7th January 1997.

Solution
7th January 1992 = Tuesday
7th January 1993 = Thursday (Leap)
7th January 1994 = Friday
7th January 1995 = Saturday
7th January 1996 = Monday ( Leap)
7th January 1997 = Tuesday

Problem - 6
The first Republic day of India was celebrated on
26th January 1950. What was the day of the
week on that date?

Solution
A = 1950/7 = 4
B = 1950/4 = 487/7 = 4
C=0
D = 26/7 = 5
4+4+0+5 – 2/7 = 11/7 = 4 = Thursday

Problem - 7
Find the Number of times 29th day of the month
occurs in 400 consecutive year?

Solution
1 year = 1 (Ordinary Year)
1 year = 12 (Leap Year)
400 years = 97 leap year
97 * 12 = 1164
303*11 = 3333
= 1164+3333 = 4497 times

Problem - 8
If 2nd March 1994 was on Wednesday, 25 Jan
1994 was on,

Solution
A = 1994/7 = 6
B = 1994/4 = 498/7 = 1
C=0
D = 25/7 = 4
= 6 + 1 + 0 + 4 – 2 / 7 = 3 = Tuesday

Problem - 9
Calendar for 2000 will serve also?

Solution
= 2000 + 2001 + 2002 + 2003 + 2004
= 2 + 1 + 1 + 1 + 2 = 7 (Complete Week)
2005

Problem - 10
If Pinky‟s 1st birthday fell in Jan 1988 on one of
the Monday‟s, the day on which are was born is,

Solution
Jan = 1988 = Monday
Jan = 1987 = Sunday

Problem - 11
Akshaya celebrated her 60th birthday on Feb 24,
2000. What was the day?

Solution
A = 2000 /7 = 7
B = 2000/4 = 500/7 = 3
C=3
D = 24/7 = 0
= 7+3+3+0-3/7 = 10/7 = 3 = Wednesday

Problem - 12
On what dates of April 2008 did Sunday Fall?

Solution
Calculate for 1st April 2008
A = 2008/7 = 6
B = 2008/4 = 502/7 = 5
C = 1/7 = 1
D=0
= 6+5+1+0 – 3/ 7 = 2 = Tuesday
1st April on Tuesday, then 1st Sunday fall on 6.
Sunday falls on 6, 13, 20, 27.

Problem - 13
Today is Friday. After 62 days it will be,

Solution
62 / 7 = 6 days after Friday then it will be
Tuesday

Problem - 14
What will be the day of the week on 1st Jan
2010?

Solution
A=1
B=5
C=0
D=1
= 1+5+0+1 – 2/ 7 = 5/7 = 5 = Friday

Problem - 15

What is the day of the week on 30/09/2007?

Calendar
Solution:
A = 2007 / 7 = 5
B = 2007 / 4 = 501 / 7 = 4
C = 30 / 7 = 2
D=5
( A + B + C + D )-2
= -----------------------
7
= ( 5 + 4 + 2 + 5) -2
----------------------- = 14/7 = 0 = Sunday
7

Clocks
Clock:

Angle between hour hand and minute hand
= (11m/2) – 30h

Angle between minute hand and hour hand
=30h – (11m/2)

Problem - 1

What is the angle between the minute hand and
hour hand when the time is 2.15?

Solution
= 11 m/2 – 30(h)
= 11 15/2 – 30(2)
= 11(7.5) – 60
= 82.5 – 60 = 22 1/2

Problem - 2

At what time between 5 and 6 o‟clock the hands
of a clock coincide?

Solution
Coinciding Angle = 0
Min. hand to hour hand = 25 min apart
60/55*25 = 12/11 * 25 = 300/11
= 27 3/11min past 5

Problem - 3

At what time between 12 and 1 o‟clock both the
hands will be at right angles?

Solution
Right angle = 90 degrees
= 30(h) – 11 m/2
90 = 30(12) – 11 m/2
180 = 360 – 11m
11m = 360 – 180
M = 180/11
16 4/11 past 12

Problem - 4

Find at what time between 7 and 8 o‟clock will
the hands of a clock be in the same straight line
but not together?

Solution

Minute hand to hour hand = 35 min apart
Straight line not together = 30 min apart
Difference = 35 – 30 = 5 min
= 60/55*5 = 12/11*5 = 60/11
= 55 5 / 11 past 7

Problem - 5

At what time between 5 and 6 are the hands of
the clock 7 minutes apart?

Solution
7 min space behind the hour hand:
25 min – 7 min = 18 min
60/55 *18 = 216/11 = 19 7/11 min past 5
7 Min space ahead the hour hand
25 min + 7 min = 32 min
60/55*32 = 12/11*32 = 384/11
= 34 10/11 min past 5

Problem - 6

A clock strikes 4 and takes 9 seconds. In order to
strike 12 at the same rate what will be the time
taken?

Solution
Strike Sec
3 (interval) 9
11 x
3x = 11*9
X = 11*9/3 = 33 Sec

Numerical reasoning I

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