Prime Mathematics 7

(d) A rectangular field has length 42.75m and breadth 32.24m. Find its area and
perimeter.

(e) A rectangle has length 16.6cm and area 229.08cm2. Find its breadth and
perimeter.

Unit Revision Test

1. Simplify:

a) 2 1 - 3 5 + 2 b) 4 1 ÷ 3 × 3
3 6 9 2 8 4

2. Simplify: 1- 1

1 + 1 1 1
+ 2

[ { ( )}]3. 1 1 3 1 1 1 - 1 1 1 - 1
Simplify: 3 - 4 ÷ 8 3 2 3 6

4. In a school there are 450 students. If 2 of the students are boys, find the number
3

of girls.

5. A can do 1 work and B can do 1 of the work in a day. How many days will they
10 15

take to complete the work working together?

6. Convert the following into fraction.

a) 0.016 b) 0.6

7. Simplify: a) 6.78 × 0.3.7

b) 0.2184 ÷ 0.28

8. Simplify: (216.35+111.25)÷ 3.2

9. A rectangular field has length 42.75 m and breadth 32.24 m. Find its area and

perimeter. 2 is divided by the sum of 1 and 1 ?
10. What is the result when 3 2

194 Fraction and Decimal

Answers

Unit:8 Fraction and Decimals
Exercise:8.1

1. a) 3 b) 1 c) - 3 d) 13 e) 22 f) 41 g) 13 h) 4
8 2 8 15 35 12 5

2. a) 1 b) 5 c) 2 d) 9 e) 2 f) -8 g) 2 h) 7
2 7 9 14 45 15 21 12

3. a) 1 27 b) -3 8 c) -1 11 d) 6 1 e) -2 7 f) 2 2 g) -1 3 h) 1 2
40 15 24 2 12 3 7 5

4. a) 1 1 b) 5 c) -1 5 d) -254 e) 2 7 g) 3 h) 38
20 12 18 5 f) 1 243 45

5. a) 3 1 b) 1 c) 1 5 d) 10 e) 1 1 f) 4 g)- 1
2 9 7 33 2 7 7

1. a) Rs 4200 Exercise:8.2 c) 72514 m d) girls 28, boys 20
2. a) Ritesh by 10 marks b) 1400 m c) d) Rs 8500.
b) daughter got more

3. a) 1 work b) 1 work c) 1
6 15 4
2 3
4. a) 3 b)6 c) -1 d) - 2

Exercise:8.3

1. a) non-termination and recurring b) terminating
c) terminating d) terminating e) non-terminating and recurring
f) non-terminating and recurring g) terminating h) terminating
i) non-terminating and recurring j) non-terminating and recurring
k) non-terminating & recurring l) terminating

2. a) 1 1 1 31
4 b) 5 c) 2 d) 250

e) 3 1 g)4 7 7
64 f) 625 20 h) 2 200
. b) 0.16. c) 0.0.9.
3. a) 0.3 ..
.. .. . d) 0.428571
e) 0.0714285 f) 0.076923 g) 0.5
.
h) 0.23

Prime Mathematics Book - 7 195

2 1 1 4
4. a) 9 b) 3 c) 9 d) 33

e) 35 f) 5 41 4
99 11 g) 330 h) 11

7 1 k) 3 d) 882.088
i)2 15 j)1 9 d) 11.458
d) 4.41
1. a) 438.981 b) 2.095 Exercise: 8.4 d) Rs. 164.25
2. a) 1.513 b) 5.82 c) 2114.807
3. a) 0.46 b) 0.62 c) 120.625
4. a) 4.6 b) 50 c) 14.29
c) 13.25m

Exercise: 8.5
1. a) 2.5068 b) 59.52 c) 68.4255 d) 9.688
2. a) 3.56 b) 9.6 c) 0.78 d) 8.9
3. a) 2.23 b) 4.096 c) 10.5 d) 18 e) 12
4. a) Rs.647.9 b) 50 c) 16.04m d) 1378.26m2 149.98m
e) 13.8cm, 60.8cm

196 Fraction and Decimal

9UNIT PERCENTAGE, RATIO
AND PROPORTION
Estimated periods 6
Objectives:

At the end of this unit students will be able to
● know the relation between fractions, decimals and percentages
● solve the simple problems on percentage.
● understand the simple problems on ratios and proportions.
● use the idea of proportion to solve problems related to unitary method.

Teaching Materials:
Grids, graph, menu, etc

Activities:
It is better to
● discuss about decimal fractions and percentages.
● give the concept of comparing quantities.
● discuss about ratios and proportions.
● discuss the method of unitary method and rule of variation (proportion).

9.1 Percentage

Percentage means per hundred or out of hundred, percentage is a fraction with
denominator 100.
1
Thus 1% = 1 out of 100 or 100

10% = 10 out of 100 = 10
100
35
35% = 35 out of 100 = 100

The symbol % is used for percentage. Thus % can be replaced by 1 .
100

We express quantities as

(i) Absolute quantities. (ii) Relative quantities.

Absolute quantities are complete and independent quantities. Such quantities
are expressed with units. e.g. 10kg, 106m, 10 hours e.t.c.

Relative quantities are the quantities compared to some standard quantities.
Such quantities have no units. Relative quantities may be expressed as

(a) Fraction having any denominator;

(b) Decimal fraction having denominator power of 10;

(c) Percentage with denominator 100.

Conversion of fraction or decimal into percentage.

Consider a fraction 3 which means
4
Out of 4 is 3
3
\ Out of 1 is 4 3 × 100 = 75%.
\ Out of 100 is 4

Thus we observed that (fraction × 100) becomes percentage.
Again
o34ut=
0.75, which means
of 1 is 0.75

\ Out of 100 is 0.75 x 100 = 75%

Thus (decimal × 100) gives percentage.

Conversion of percentage into fraction or decimal.
Consider 25%, which means
out of 100 is 25
1
\ Out of 1 is 25 = 4
100
i.e Out of 1 is 0.25
Thus, given number in percentage divided by 100 gives fraction or decimal
25
\ 25% = 25 = 1 or 25% = 100 = 0.25.
100 4

198 Percentage

To express a given quantity as the percentage of the given whole quantity.

Let’s express Rs 150 as percentage of Rs 600.

\ out of Rs 600 is Rs 150. Quantity as qpuearcnetinttyage = Given quantity
of other Total quantity
\ out of Rs 1 is Rs 150 × 100%
600

\ out of Rs 100 is Rs 150 × 100%.
600
150
= 600 × 100 %

= 25% Rs 150
Rs 600
Thus, Rs 150 as percentage of Rs 600 = × 100%

= 25%

Example 1. Express the following fraction or decimals into percentage.
3
Solution : (a) 8 (b) 0.125

(a) 3 = 3 × 25 = 75 % = 37.5%.
8 8 2 100% 2

(b) 0.125 = 0.125 × 100% = 12.5%

Example 2 . Express the following into fractions.

Solution: (a) 70% (b) 112%
(a) 70% =
1700071=0 7
10

\ 70% = 7
10
112 28
(b) 112% = 100 25 = 28 = 1 3
25 25

Example 3. What percentage is (a) 20 of 80 ? (b) 20 p of Re1?
Solution:
(a) 20 as percentage of 80 = 20 × 25 = 25 %
= Rs 1 80 4 100%

(b) Here, 20 p = Rs 20 5
100

Now 20p as percentage of Re 1. = 20p × 100%
Re 1
Rs 1/5
= Re 1 × 100%

= 1 × 20 = 20%
5 100%

Prime Mathematics Book - 7 199

Example 4. Find 75% of 80 full marks.
Solution:

Here 75% of 80
= 1710550×5804 = 60

\ 75% of 80 full marks = 60

Example 5. Find the sum whose 5% is Rs 80
Solution:

Here, 5% of the sum is Rs 80

\ 1% of the sum is Rs 80
5
80
\ 100 % of the sum is Rs 5 × 100 = Rs 1600

\ The sum is Rs 1600 whose 5% is Rs 80

Alternative method.

Let the sum be x, then

5% of x = Rs 80

or 5 ×x = Rs 80
100
20
or x = Rs 80 × 20
= Rs 1600

\ The sum is Rs 1600 whose 5% is Rs 80

Example 6. A man spends 65 % of his monthly income and saves Rs 9100. How much
does he spend every month?

Solution :

Let his monthly income is x.

his expenditure in percentage = 65%.

\ Saving in percentage = (100-65)% = 35%

But saving = Rs 9100

\ 35% of x = Rs 9100
35
\ 100 × x = Rs 9100

x = Rs 260 = 26000
1820
9100 × 100

35 7

200 Percentage

\ Income = Rs 26000

Now,
Expenditure = Income – Saving

= Rs 26000- Rs 9100
= Rs 16900
He spends Rs 16900 each month.

Exercise 9.1

1. Express the following fractions or decimals into percentage.
1 3 1
a) 4 b) 4 c) 10 d) 7
10

e) 0.152 f) 1.25 g) 0.72 h) 0.1234

2. Express the following percentage into fractions.
a) 85% b) 50% c) 42.5% d) 123%

3. What Percentage is? b) Rs 124 of Rs 400?
a) 27 of 60? d) 48p of Rs 4 ?
c) 704m of 2km?

4. Find the value.
a) 16 % of Rs 625 b) 35% of 60 students
c) 60% of 2km d) 80% of full mark 80
e) 25% of 30000000 people

5. Find the quantity whose b) 40% is 32 marks c) 65% is 780
a) 40% is 80 e) 0.1% is 4 f) 0.02% is 3.6 m
d) 90% is 540 students

3
6. a) What percentage is equivalent to 5 ?

7
b) Which one is greater 70% of 56 or 8 of 56?

c) Jangbu has Rs 5800, Sonam has only 70% of the sum that Jangbu has. How
much money does he have?

d) There are 45 students in a class. If 60% of them are girls, find the number of
girls and boys.

7. a) In a basket full of apples, 20% apples are rotten. If 712 apples are found good,
find the total number of apples in the basket.

b) A man saves Rs 5400 in a month which is 30% of his monthly income. How
much does he spend in a month?

c) Brass is an alloy of zinc and copper. If 30% of the content is zinc, find the
weight of zinc and copper in 300 grams of brass.

8. a) Last year, there were 480 students in a school and this year it has

Prime Mathematics Book - 7 201

become 600. Find the percentage increase in the number of students.
b) Price of petrol hiked from Rs 112.5 to Rs 117 per litre. Find the percentage

price hike.
c) Monthly salary of a teacher was Rs 12000 last month. If the salary increased

by 20% , find his salary this month?
d) Srishti obtained 92% out of 800 full marks. Krishti obtained 72% out of the

same full mark. By how much is Srishti’s marks more than Krishti’s?
9 . a) In a school of 800 students , 5/8 are boys and rest girls.

i) Find the number of boys and girls .
ii) Find the percentage of boys and girls.
b) A men earns Rs.50000 in a month. He spends 30% of his income on food,
2/5 on his child’s Education,0.25% on transportation and rest he saves.
i) Find the amount that he spends on each topic.
ii) Find his saving percentage.
10. a) The pass marks in an examination is 40%. Rajani secured 295 marks and
failed by 25 marks. Find the full marks of the examination.
b) In a exam, Biraj got 350 marks and failed by 20 marks .If the percentage
of pass marks is 40% , what is the full marks?
c) Out of 700 full marks, Rumba got 250 marks failed by 30 marks, what is
the pass percentage?
d) In an exam 90% students passed. If 15 Students failed , Find the number
of students appeared in the examination?
11. a) The population of a village was 30,000 in last year . Every year it
increases at the rate of 5%.
i) What is the population of that village this year?
ii) what will be its population next year?
b) Solman started his job with a basic salary of Rs 8000 per month in 2069.
An increment of 20% was made in 2070 and again increment of 15% in
2071. What was his salary in 2071?

202 Percentage

9.2 Ratios aknidndba(i,.ealsoofdesanmoteeduansitb),:atihsecnallbaed,

If a and b are two quantities of same also
denoted as a:b is called the ratio of a to b ratio

of b to a.

The symbol (:) read as ‘is to’ is used to denote ratio.
a and b are called the terms of the ratio.
While taking ratio, a and b should be in the same unit
Generally, ratio is expressed in its lowest term,
In the ratio a:b, a is called antecedent and b is called consequent.
Ratio is the way of comparing two terms of same unit showing how many times a
quantity is of the other.

Example 1: Find the ratio of 32 to 80.

Solution: = 32 2 (cancelled by 16)
Here, the ratio 32 to 80 80 5

\The ratio 32 to 80 = 2 or 2:5
5

Example 2: Find the ratio of 2Km to 800 m.

Solution.
Here, 2Km = 2 × 1000m = 2000m
2000m = 20 5 = 5:2
6 800m 82

Now , the ratio of 2 km to 800m =

Example 3: Express the following ratios in their simple forms.

(a) 0.75:3 b) 1 3 : 5 1
4 4
Solution

a) 0.75:3

= 0.75 = 75 31 = 1 = 1:4
3.00 300 12 4 4

\ 0.75:3 = 1:4

b) 1 3 : 5 1 Methods of comparison
4 4
143 1. By observing size
7
4 a>b or a<b or a=b
= = 21
5 1 2. Finding the difference.
4 If a-b = 5 then a is greater than b by 5 and b is smaller than a by 5
4
3. Taking ratio
7 1 4 a 2
4 21 3 b 1
= × =1:3 If = then a = 2b (a is twice of b)

\ 1 3 : 5 1 = 1:3 or b = 1 a (b is half of a)
4 4 2

Prime Mathematics Book - 7 203

Example 4: The age of a father and son are in the ratio 7:2. If father’s age is 42 years,
find the age of son.
Solution:
Let the son’s age be x years
Since ratio of father’s age to son’s age = 7:2

42 years = 7
2
x years

or, 7x = 42 × 2

\x= 42 × 2 = 12 \ Son’s age is 12 years.
7

To divide a given quantity in the given ratio.

Example 5: Divide 70 in the ratio 4:3

Solution :

To divide 70 in the ratio 4:3, let the two parts be 4x and 3x.

Then,

4x + 3x = 70

or, 7x=70 \ x=10

The two parts are 4x = 4 × 10 = 40

And 3x = 3 × 10 = 30

Alternate method.
Given quantity = 70
Given ratio = 4:3
Total parts = 4+3 = 7
4 4
Two parts are 4+3 of 70 = 7 × 70 = 40

And 3 of 70 = 3 × 70 = 30
4+3 7

Transforming the given ratio.

Example 6: Two numbers are in the ratio 5:7. When 3 is added to each, the ratio
becomes 3:4. Find the numbers.

Solution:

Let the numbers in the ratio 5:7 be 5x and 7x

According to the question,

5x +3 = 3
7x + 3 4

or 20x + 12 = 21x + 9

or, 20x + 21x = 9 - 12
or, -x = -3

204 Percentage

\x = 3 are 5x =
Now, the numbers and 7x 5 × 3 = 15
= 7 21
×3=

Exercise 9.2

1. Express each of the following ratios in their lowest terms :
(a) 3:9 (b) 15:25 (c) 12:20 (d) 48:56
(e) 96:48 (f) 80:120 (g) 78:65

2. Write the following ratios in the simple form.

a) Rs 64 to Rs 80 b) 12 kg to 84 kg c) 108 cm to 72 cm

d) 2 m to 125cm e) 5 kg to 750 gm f) 225 ml to 3l
g) Rs 6 to 75p h) 0.5 to 2.

3. Write the following ratios in their simple form.

a) 1.25 : 0.5 b) 3.5 : 0.125

c) 1 1 : 2 1 d) 5 3 : 3 1
3 2 4 15

e) 2 : 3 1 f) 2 2 :8
2 5

4. A bag contains 16 green pencils and 8 blue pencils. Find the ratio of the following
a) green pencils to blue pencils.
b) blue pencils to green pencils.
c) green pencils to total number of pencils.
d) blue pencils to total number of pencils.

5. In a class of 42 students, there are 28 girls and rest are boys. Find the ratio of
a) Girls to boys.
b) Boys to girls
c) Girls to the total number of students.
d) Boys to the total number of students.

6. a) The ages of a father and his son are in the ratio 8:3. If the age of the
father is 48 years, find the age of his son.

b) Brass is an alloy of copper and zinc mixed in the ratio 3:2. Find the
weight of copper and zinc in 120 gram of brass.

c) Angles of a triangle are in the ratio 2:3:4. Find the angles.
d) Divide Rs 84 in two parts in the ratio 5:7.
e) Mr. Sherchan divides Rs 60,000 to his two sons in the ratio 3:2. Find the share of each.

7. a) Two numbers are in the ratio 3:4. If 4 is added to each numbers the new ratio becomes
5:6, what are the numbers ?

b) Two numbers are in the ratio 4:5, if 6 is subtracted from each the new ratio becomes
5:7, what are the numbers ?

Prime Mathematics Book - 7 205

8. a) The ratio of two numbers is 5:7 and their sum is 36, what are the numbers ?
b) The ratio of two numbers is 5:3 and their difference is 18, what are the numbers.

9.3 Proportion

Let's consider the data. Class No. of boys No. of girls

VII A 15 20

VII B 18 24

In class VII A, ratio of number of boys to the number of girls = 15 = 3 =3:4
20 4

And in class VII B, ratio of number of boys to the number of girls = 18 = 3 =3:4
24 4

Here, the ratio of number of boys to the number of girls in two classes are same and
we say the number of boys and girls in two classes are proportional.

We write 15 : 20:: 18:24 (read as 15 is to 20 as 18 is to 24)
The equality of two ratio is called proportion .

If a : b = c : d, the elements a, b, c, and d are in proportion and we write a : b : : c : d.

The terms or elements a, b, c and d in the given proportion are also called proportional

Here, a is first proportional ,b second proportional, c third proportional and d fourth
proportional.

In the proportion a : b : : c : d, first proportional a and a:b::c:d
fourth proportional d are called extremes and second ExMtreeanms es

proportional b and third proportional c are called

means.

We get, product of extremes = product of the means
i.e. a × d = b × c
a c
which is simply the cross product in b = d

To check whether the four numbers are in proportion or not.

Method I.
Find the ratio of 1st two in lowest terms . Similarly find the ratio of next two terms in
their lowest terms. If two ratios are equal , the numbers are in proportion.

Method II

Suppose the numbers are in proportion
Find the products of extremes and means . If the product of extremes = the product
of means, the numbers are in proportion.

206 Percentage

Example1: To check whether 16 : 28 and 36 : 63 are in proportion.

Here , the first ratio = 16 = 4 = 4:7
28 7

The 2nd ratio = 36 = 4 =4:7
63 7

Since 16 = 36 = 4 \16 : 28 : : 36 : 63
28 63 7

Alternate Method

Here, 16 = 36
28 63

or, 16 × 63 = 28 × 36 \16:28::36:63
or, 1008=1008

Example 2: If 4, 6, 10 and x are in proportion, find the value of x.
Solution :

Since 4, 6, 10 and x are in proportion

so, 4 = 10
6
x
or, 4x = 60
60
or x = 4 \ x = 15

Types of proportions

Let a pen costs Rs. 180, then

No.of pens Rate(Rs) Cost(Rs) No of pens Rate(Rs) Cost(Rs)

2 180 360 2 180 360
5 180 900 8 180 1440

Here , two pens cost Rs.360 and Five pens cost Rs. 900

\ Ratio of no. of pens = 2 = 2:5
5

And ratio of their costs = Rs 360 = 2 = 2:5
Rs 900 5

\ Ratio of the numbers of pens = Ratio of the cost of the pens.

Here, the number of pens and cost of the pens are in proportion.

If the number of pens is changed from 2 to 8 their costs change from Rs 360 to

Rs 1440. In this case also..

Ratio of number of pens = 2 = 1 =1:4
8 4

Ratio of costs of pens = Rs 360 = 1 =1:4
Rs 1440 4

Prime Mathematics Book - 7 207

Thus, ratio of the number of pens = Ratio of the costs of pens
If a quantity say number increase or decrease , other quantity, say cost increase or
decrease proportionally, the proportion is called direct proportion and the change or
variation is called direct variation.
Number of objects and their costs; number of objects and their weights ; number of
men and quantity of work; quantity of work and time taken etc vary directly.
Again let’s consider the data.
i)

No. of men No. of days to do work
1 10
5
ii) 2

No. of hours per day No. of days a gas lamp can be used

If a man takes 1042days to do work, 2 men take 5 days6300to do same work.
Here, ratio of number of 1
men = 2 = 1:2

Ratio of number of days taken = 10 = 2 =2:1
5 1
Ratios are equal but inversely

i.e. ratio of number of men = 1
ratio of number of days taken
Similarly, see the table (ii)

If burning 2 hours a day, a gas lamp can be used for 60 days, obviously burning 4 hours
a day it can be used only for 30 days.

Here, ratio of number of hour per day= 2 = 1 = 1:2
4 2
60
Ratio of number of days = 30 = 2:1

Here also, the ratios are equal but inversely.
1
i.e. ratio of number of hours of use = ratio of number of days of use

If the quantity increase or decrease other quantity decrease or increase, the
proportion is called indirect or inverse proportion and the variation is called indirect
variation.

Use of proportion or variation.

Idea of direct and indirect proportion is used in solving problems of unitary method.

Example 1: If the cost of 10 books is Rs. 1500, find the cost of 16 books.

208 Percentage

Solution: No. and cost vary directly.
Here, the cost of 10 books is Rs. 1500 So, less number, less cost.
So, we divide
\ The cost of 1 book is Rs 1500 = Rs 150
10

\ The cost of 16 books is Rs 150 × 16 = Rs 2400 No of books and cost vary directly.
The cost of 16 books is Rs 2400 So, more number more cost.
So, we multiply.

Alternate method (rule of variation)

No. of books Cost (Rs)
10 1500
16
x

Let the cost of 16 books is x, then since the number of books and their costs are

directly proportional (vary directly)

Ratio of the number of books = ratio of their costs
1500
Or, 10 =
16 x

Or, 10 x = 1500 × 16
1500 × 16
Or, x = 10

\x = Rs 2400

\16 books cost is Rs. 2400.

Example 2: If 8 men can complete a piece of work in 12 days, in how many days would
6 men complete the same work?
Solution:
\8 men can complete a piece of work in 12 days
\1 man can complete the same work in 12 × 8 days No of men and days vary inversely So
less no. of men more days it takes.
daySso we multiply
\6 men can complete the same work in 12 × 8 No. of men and days vary
6

= 2 × 8 days = 16 daysinversely, so more men less days
\ 6 men can complete the work in 16 days. it takes. So we divide.
Alternate method (rule of variation)

No. of men No. of days

8 12

6x

Let 6 men takes x days to complete the work. Since number of men and number of days
taken vary inversely (indirectly)

Prime Mathematics Book - 7 209

Ratio of no. of men = 1
ratio of no. of days

Or, 8 = 1
6 12
x

Or, 8 = x
6
12

Or, x = 8 ×12 2
6
\ x = 16

\6 men take 16 days to complete the work.

Exercise 9.3

1. Check whether the following ratios are in proportion or not.
a) 5:3 and 25:15 b) 6:8 and 48:64 c) 12:18 and 36:72
d) 11:9 and 121:99 e) 42:28 and 90:64 f) 30:90 and 150:450
2. Find the following quantities are in proportion or not.
a. 6,7,18 and 21 b. 3,4,6 and 7
c. 4,5,12 and 15 d. 4,5,6 and 7

3. If the following numbers are in proportion, find the value of x.
a) 3, 4 , x, 20 b) 16, x, 18, 27 c) 27, x, x, 3
d) 4, x, x, 16 e) a, 3a, 4. 5, x f) 15, x, 3k, 2k
4. Find the fourth proportion.
a. 3,4 and 9 b. 4,5 and 12
5. The following number are continued proportion . Find the value of x.
a) 2,X,8 b) 4,X,16 c) X,6 and 2 d) 4,6,X
6 Find a:c and a:b:c of the following.
a) a:b = 2:5,b:c = 10:15
b) a:b = 4:5,b:c = 3 : 5

7. a) Given Map is the plan of a house 16 cm long and 12cm broad. If the actual

length is 32 feet, find the actual breadth of the house.

b) The length and breadth of a rectangle are in the ratio 5:2. If its breadth is
16cm, find its length.

210 Percentage

8. a) The cost of 8 kg of sugar is Rs 512. What is the cost of 18 kgs of sugar?
b) A car consumes 8 litres of petrol to run 176 km. How much petrol is needed to
travel a distance of 1100 km by the car?
c) A dozen of copy cost Rs.288. How many copies can be bought with Rs 1200?
d) A man walks 24 km in 4 hours. How far will he walk in 11 hours?

9. a) 9 workers can accomplish a work in 24 days. How many workers are required to
accomplish the work in 18 days?

b) When the rate is Rs 210 per piece, a certain sum of money can buy 16 pieces of
T-shirts. If the rate increases to Rs. 280 per piece, how many pieces of T-shirts
can be bought with the sum?

c) Operating 4 hours a day a battery can be used for 18 days. For how many days
can the battery be used operating 6 hours a day?

d) A platoon of 60 soldiers has provision for a month. How long would the
same provision last for 72 soldiers?

e) A garrison of 200 men have provision for 45 days. For how many men would the
provision last for 30 days?

f) A piece of work can be completed in 30 days working 8 hours a day. In how days
the work will be completed working 6 hours a day?

Unit Revision Test:

1. What percentage is Rs 124 of Rs 400 ?

2. Find the quantity whose 0.1% is 4 kg.

3. Last year there were 500 students in a school This year it has become 625. Find

the percentage increase in the number of students.

4. Write the ratios in their simple form.1 1
(a) 5 kgs to 750 gm. (b) 1 3 : 23

5. The age of a father and his son are in the ratio 8:3. If the age of son is 21 year,

find the age of the father.

6. Divide Rs. 112 in the ratio 4:3.

7. Check whether the following ratios are proportional or not.

(a) 12:18 and 36:72 (b) 84:96 and 56:64

8. If 27, x, 3a, 2a are in proportion, find the value of x.

9. A motor bike consumes 8 litres of petrol to run 336 km. How much petrol is

needed to travel a distance of 672 km ?

10. 9 workers can do a piece of work in 24 days. How many workers are required to

do the work in 18 days?

Prime Mathematics Book - 7 211

Answers Unit:9 Percentage, Ratio and Proportion

b) 75% Exercise:9.1
1. a) 25% f) 125% c) 10% d) 70%
e) 15.2% g) 72% h) 12.34%

2. a) 17 1 17 23 e) 7500000 f) 18000
20 b) 2 c) 40 d) 1 100 e) 4000
b) 31% c) 35.2% d) 12%
3. a) 45% b) 21 c) 1.2 km d) 64
4. a) Rs. 100 b) 80 c) 1200 d) 600
5. a) 200

6. a) 60% b) 7 of 56 c) Rs. 4060 d) girls 27, boys 18
7. a) 890 b) R8s. 12600
d) 800 c) zinc 90gm, copper 210gm

8. a) 25% b) 4% c) Rs. 14400 d) 160
9) a) i) 500, 300 ii) 62.5%, 37.5% b) i) 15,000, 20,000, 125 ii) 29.75%
10) a) 800 b) 925 c) 40% d) 150
11) a) i)31,500 ii) 33,075 b) 11,040

Exercise:9.2

1. a) 1:3 b) 3:5 c) 3:5 d) 6:7
e) 2:1 f) 2:3 g) 6:5
2. a) 4:5 b) 1:7 c) 3:2 d) 8:5
e) 20:3 f) 3:40 g) 8:1 h) 1:4
3. a) 5:2 b) 28:1 c) 8:15 d) 15:8 e) 4:7 f) 3:10
4. a) 2:1 b) 1:2 c) 2:3 d) 1:3
5. a) 2:1 b) 1:2 c) 2:3 d) 1:3
6. a) 18 b) copper 72gm, zinc 48gm c) 40° 60° 80°
d) Rs. 35 and Rs. 49 e) Rs. 36000, 24000
7) a) 6 and 8 b) 16 and 20 8) a) 15 and 21 b) 45 and 27

Exercise:9.3

1. a) Yes b) Yes c) No d) Yes
e) No f) Yes d) No
b) No c) Yes d) 8 e) 13.5 f) 10
2. a) Yes b) 24 c) 9 d) 9
3. a) 15 b) 15
4. a) 12 b) 8 c) 18 d) 66 km
5. a) 4 b) 12:25, 12:15:25 d) 25 days
6. a)4:15, 4:10:15 b) 40cm
7. a) 24ft b) 50 Liters c) 50
8. a) Rs. 1152 b) 12 c) 12 days
9. a) 12 f) 40 days

e) 300

212 Percentage

10UNIT PROFIT AND LOSS

Estimated periods 5 Objectives:

At the end of this unit students will be able to

● find the profit or loss when selling price and cost price are given.
● find the selling price or cost price when profit or loss is given.
● find the profit or loss in percentage.
● solve the problems on profit and loss.

Teaching Materials:

Chart of market where the people are buying the things, price tag, chart of

formula etc.

Activities:

It is better to

● demonstrate the activities of buying and selling the things in the
classroom to give the concept of cost price and selling price.

● display the chart of market where the people are buying and selling the
things to give the concept of cost price, selling price, profit and loss.

● derive the formula to find the profit and loss, selling price and cost price.
● derive the formula to find the profit and loss in percentage.
● say the students to drill the problems by using the formula.

● discuss about the process of solving the problems of profit and loss.

10.1 Profit and Loss

Review.

Suppose Mohan bought a watch for Rs. 1200 and sold it for Rs. 1350 in his business.
The money which is invested to buy the things in the business is called cost price. So,
in the above example, Rs. 1200 is the cost price. The cost price is also written as C.P.
Similarly, the money which is got by the shopkeeper or paid by the customer for the
things is called the selling price. So, in the above example, Rs. 1350 is the selling
price. The selling price is also written as S.P.
In the above example, the selling price is more than the cost price. So, Mohan is made
the profit.

here, S.P. > C.P, so that,

\profit = Rs. 1350 - Rs. 1200 = Rs. 150

\profit = selling price (S.P.) - cost price(C.P.)
But due to certain reasons sometimes the shopkeeper sells his goods at the price lower
than the cost price. In such cases, he makes the loss.

\ loss = cost price(C.P.) - selling price(S.P.)
From the formula, profit = S.P. - C.P. , we can write

S.P. = profit + C.P
C.P = S.P - profit
From the formula, Loss = C.P - S.P., we can write
C.P. = S.P + Loss
S.P. = C.P - loss
Example 1: A shopkeeper bought an article for Rs. 575 and sold it for Rs. 625. Find his
profit or loss.
Solution:
Here,
cost price(C.P.) = Rs. 575
selling price(S.P.) = Rs. 625
since, S.P. > C.P. so, there is profit.
profit = S.P. - C.P.
= Rs. 625 - Rs. 575
= Rs. 50
Example 2. A man buys a radio for Rs. 750. At what price should he sell it if he makes
a loss of Rs. 65 ?
Solution:
Here,
cost price(C.P.) = Rs. 750
loss = Rs. 65
selling price (S.P.) = ?
We know that,
S.P. = C.P — loss
= Rs. 750 — Rs. 65 = Rs. 685

214 Profit and Loss

Example 3. A fruit seller bought 8 dozen of bananas at Rs. 375 each and sold all
bananas at Rs. 4 each. Find his profit or loss in his business.
Solution:
Here,
8 dozen = 8 × 12 pieces [1 dozen = 12 pieces]
= 96 pieces
Cost price (C.P.) of 8 dozen bananas = Rs. 3.75 × 96
= Rs. 360
Selling price (S.P) of 8 dozen bananas = Rs. 4 × 96
= Rs. 384
Since, S.P. > C.P. So, there is profit.
Profit = S.P. - C.P.
= Rs. 384 - Rs. 360
= Rs. 24

Exercise 10.1

1. Find the profit or loss in each of the following cases.
a) C.P. = Rs. 620, S.P. = Rs. 760 b) C.P. = Rs. 4300, S.P. = Rs. 4500
c) C.P. = Rs. 273.50, S.P. = Rs. 260 d) S.P. = Rs. 1050, C.P. = Rs. 1100

2. Find the cost price(C.P.) or selling price(S.P) in each of the following cases.
a) C.P = Rs. 720, profit = Rs. 142 b) S.P. = Rs. 600, loss = Rs. 60
c) C.P. = Rs. 1205, loss = Rs. 120 d) S.P = Rs. 785, profit = Rs. 45

3. Mr. Khadka bought a watch for Rs. 375 and sold it for Rs. 405. Find his profit or loss.

4. A shopkeeper bought 1 dozen pens at Rs 207 and sold all pens at Rs. 18 each. Find
his profit or loss.

5. A fruit seller buys 60 apples at Rs. 4.50 per piece and sells all apples at Rs. 4.35
per piece. Find his profit or loss.

6. Rupak electronic buys a computer set for Rs. 21500 and sells it bearing a loss of
Rs. 750. Find the selling price of the computer set.

7. A shopkeeper sold a toy car for Rs. 450 making a profit of Rs. 95. At what price was
it buy?

8. A dealer buys an article for Rs. 3000. At what price should he sell it if he makes a
profit of Rs. 510 ?

9. a) A shopkeeper of Banepa bought 300 pieces of same type of the sari. He sold
70% of them at Rs. 1500 per piece and rest of them at Rs. 1350 per piece. At
what price did he purchese each sari if his profit is Rs. 42000 ?

b) A man bought some lemons at Rs. 5 each and sold 80 lemons at Rs. 640. Find
his profit or loss.

c) A man sold few lemons at Rs. 300 at the rate of Rs. 5 each. If he got a loss of
Rs. 75. Find the cost price of each lemon.

Prime Mathematics Book - 7 215

10.2 Problems involving percentage in profit and loss

Study and learn the activities:
Ramu bought 2 dozens bananas at Rs. 80. He could not sell 4 bananas due to rotten and
sold the rest bananas at Rs. 4.25 each. Find his profit or loss.
Here,

Total number of bananas = 2 dozen
= 2 × 12 = 24 pieces

Number of rotten bananas = 4

\Number of good bananas = 24 - 4 = 20

C.P. of 24 bananas = Rs. 80
S.P. of 20 bananas = Rs. 4.25 × 20

= Rs. 85.00 = Rs. 85
Rs. 85 > Rs. 80 means S.P. > C.P.

So, there is profit.

\Profit = S.P. - C.P.

= Rs. 85 - Rs. 80 = Rs. 5

In the above activity, Rs. 5 is profit on Rs. 80.

\ On Rs. 80, profit = Rs. 5
Rs. 5
On Re. 1, Profit = Rs. 80

On Rs. 100, Profit = 5 ×100 = 25 = 6.25
80 4

The result which So, 6.25 is called the
is calculated out profit percent. It is
of 100 is called written as 6.25%.

percentage.

\ Profit percentage = Profit × 100

C.P.
Similarly,

Loss percent = Loss × 100
C.P.

216 Profit and Loss

Example 1: Find the profit or loss percent when C.P = Rs. 150 and S.P = Rs. 140
Solution:
Here, C.P = Rs. 150
S.P = Rs. 140
S.P < C.P. So, there is loss
Loss = C.P. - S.P.
= Rs. 150 - Rs. 140 = Rs. 10

Now , Loss percent = loss × 100 = 10 × 100 = 20 = 6 2 %
C.P. 150 3 3

Example 2. A shopkeeper bought 160 apples for Rs. 640. 20 of them were rotten and
thrown away. He sold the good one at Rs. 4.25 each. Find his gain or loss percent.
Solution:
Here,
C.P. of 160 apples = Rs. 640
Number of rotten apples = 20
Number of good apples = 160- 20 = 140
S.P. of 1 apple = Rs. 4.25
S.P. of 140 apples = Rs. 4.25 × 140 = Rs. 595
C.P > S.P. So there is loss.
Loss = C.P. - S.P.
= Rs. 640 - Rs. 595 = Rs. 45
5
Now , Loss percent = loss × 100 = 45 × 100 = 225 = 7 1 %
C.P. 640 32 32

32

Exercise 10.2

1. Find the profit or loss percent in the following cases.

a. C.P. = Rs. 720, S.P. =Rs. 800 b. C.P = Rs 380, S.P = 304

c. C.P. = Rs. 400, S.P = 375 d. C.P = Rs. 1800, S.P. = Rs. 2000

2. Find the profit percentage.

a. c.p = Rs 400 b. s.p = Rs 780

profit (p) = Rs 200 profit = Rs 30

c. s.p. = Rs. 850

Profit (p) = Rs. 50

3) Find the loss percentage.

a. c.p = Rs 420 b. s.p = Rs 140 c. S.p = Rs 800

Loss (l) = 84 Loss = Rs 10 Loss(L) = Rs 80

4. A cap was purchased for Rs. 50 and sold it at Rs. 75. How much percent profit is

made? Find it.

Prime Mathematics Book - 7 217

5. A fruit seller bought a few oranges at Rs. 60 per dozen and sold them at Rs. 5.50
each. How much percent profit or loss did he make? Find it.

6. A shopkeeper bought 100 glasses at Rs. 16.50 each.5 glasses among them were
damaged. If he sold the rest glasses at Rs. 18 each, what is his gain or loss? Find it
in percentage.

7. A man buys 4 pens for Rs. 12 and sells each for Rs. 3.50. What percent did he make
profit or loss? Find it.

8. Nunulal Sahu bought 200 kg of rice at Rs. 40 per kg. He sold one half of them at
Rs. 45 per kg. He then found 12 kg of rice damaged and sold the rest at Rs. 38 per
kg. Find his profit or loss percent.

9. A fruit seller bought 150 apples for Rs. 975. 30 of them were rotten and thrown
away. He sold the good one at Rs. 7 each. Find his profit or loss percent.

10. a An old motorbike is bought at Rs 58000 and its repair was done for Rs 2000.
Then bike was sold at Rs 80000, find profit or loss percentage?

b. An old mobile set is bought at 10000 and its repair was done for Rs 350 then the
mobile was sold Rs 12830, find profit or loss percentage.

10.2.1 To calculate the cost price or selling price from the given
profit or loss percentage

Study and learn the following examples.

Example 1. If C.P. = Rs. 40 and profit = 20%, find the S.P.
Solution: Here,
C.P. = Rs. 40
Profit = 20%
Actual profit = profit percent of C.P.
= 20% of Rs. 40
20
= 100 × Rs. 40 = Rs. 8

Now, S.P. = C.P. + Actual profit
= Rs. 40 + Rs. 8 = Rs. 48.

Example 2. By selling an article for Rs. 4000, a shopkeeper makes a profit of 25%. Find
the cost price of the article.

Solution :
Here,
S.P = Rs 4000 and profit = 25%

Let cost price of the article be Rs.x

Then, profit = 25% of C.P.

218 Profit and Loss

= 25 × x = x
100 4

Now, S.P = C.P+Profit
or, x
or, 4000 = x + 4

or, 4000 = 4x+x
4
5x
or, 4000 = 4

800 = x \ x = 3200
4000 × 4

5

Therefore, the cost price of the article is Rs. 3200.

Example. 3. By selling an electric iron for Rs. 900, a shopkeeper makes a loss of 10%.
At what price should he sell it to gain 5% ?

Solution:

Here,

S.P.= Rs 900 and loss = 10%
Let C.P be Rs. x

Then, loss = 10% of C.P.

= 10 × x = x
100 10

Now, S.P. = C.P. - Loss
x
or, 900 = x - 10 100

or, 900 = 10x-x or, 900 = 9x \ x = 900 × 10 = 1000
10 10 9

Therefore, C.P. of the iron is Rs. 1000.
Again, C.P. = Rs.1000
Profit = 5%
Actual profit = 5% of C.P = 5 × Rs. 1000 = Rs. 50
100

Now, S.P. = C.P + Actual profit
= Rs 1000 + Rs 50
= Rs. 1050

Thus, he should sell the iron for Rs.1050 to make a profit of 5%.

Prime Mathematics Book - 7 219

Exercise 10.2.1

1. Find the selling price (S.P.) in the following cases.

a) C.P. = Rs. 75, gain = 20% b) C.P. = Rs. 900, profit = 12%

c) C.P. = Rs. 740, loss = 15% d) C.P. = Rs. 2500, loss = 8%

2. A man makes a profit of 5% by selling a book. If his actual profit is
Rs. 125, at what price did he purchase the book? Find it.

3. Mr. Malla bought an article for Rs. 1260 and sold it at a loss of 8%. Find the selling
price of the article.

4. A fruit seller bought oranges at Rs. 60 per dozen and sold them at a profit of 12%.
How much will the customer pay for 30 oranges?

5. Find the cost price (C.P.) in the following cases.
a) S.P. = Rs. 4000, Profit = 25%
b) S.P. = Rs. 540, Profit = 20%
c) S.P. = Rs. 600, Loss = 6%

6. By selling an article for Rs. 2300, a shopkeeper makes a profit of 15%. Find the cost
price of the article.

7. A man sold a watch for Rs. 3200 and there by gained 20%. What is the cost price
of the watch ?

8. By selling a book for Rs. 500, a shopkeeper lost 10%. Find the cost price of the
book.

9. Mr. Chaudhary makes a profit of 20% by selling a watch for Rs. 1260.
a) what is the cost price of the watch?
b) At what price should he sell it to make a profit of 30%?

10. A radio was sold for Rs. 924 at a loss of 19%. For what should it have been sold to
gain 10%?

11. Biraj bought a book for Rs. 200 and sold it to Manish at a profit of 20%. Manish sold
that book to Prakash at a loss of 10%. At what price Prakash should sell the book
to receive 5% profit ?

12. a. Pemba sold a pair of shoes at Rs. 1600 at a loss of 20%. At what price should he
sell it in order to make profit of 30% ?

b. Birman sold a book at Rs.400 at a profit of 10%. At what price should be sell it
in order to make profit of 15% ?

220 Profit and Loss

Unit Revision Test:

1. Find the profit or loss b) S.P. = Rs. 800, C.P. = Rs.850
a) S.P. = Rs. 420, C.P. = Rs.310

2. A man bought two dozens pencils at Rs. 5 each and sold them at Rs. 5.50 each.
Find the profit or loss amount.

3. A dealer buys an article for Rs. 4000. At what price should he sell it if he makes a
profit of Rs. 310 ?

4. Find the profit or loss percent in the following cases.
a. C.P. = Rs. 500, S.P. =Rs. 620 b. C.P = Rs 285, S.P = 250

5. A radio was purchased for Rs. 3050 and sold it at Rs. 3250. How much percent
profit is made? Find it.

6. Prabhulal Sahu bought 200 kg of sugar at Rs. 65 per kg. He sold one half of them
at Rs. 70 per kg. He then found 10 kg of sugar lost and sold the rest at Rs. 80 per
kg. Find his profit or loss percent.

7. Find the selling price (S.P.) in the following cases.

a) C.P. = Rs. 175, gain = 15% b) C.P. = Rs. 800, loss = 12%

8. Mr. Malla bought an article for Rs. 1460 and sold it at a loss of 6%. Find the selling
price of the article.

9. Find the cost price (C.P.) in the following cases.
a) S.P. = Rs. 5000, Profit = 20%
b) S.P = Rs. 400, Loss = 8%

10. A man sold a second hand mobile for Rs. 3200 and there by losing 20%. What was
the cost price of the mobile ?

11. A cap was sold for Rs. 324 at a loss of 18%. For what price should it have been sold
to gain 10% ?

Prime Mathematics Book - 7 221

Answers

Unit:10 Profit and Loss
Exercise: 10.1

1. a) profit = Rs. 140 b) profit = Rs. 200 c) loss = Rs. 13.50 d) loss = Rs. 50
2. a) S.P. = Rs. 862 b) C.P. = Rs. 660 c) S.P. = Rs. 1085 d) C.P. = Rs. 740
3. a) Profit = Rs. 30 4. Profit = Rs. 9 5. Loss = Rs. 9 6. Rs. 20750
7. Rs. 355 8. Rs. 3510
9. a) Rs. 1315 b) 60% c) Rs. 6.25

Exercise: 10.2

1. a) Profit = 11 1 % b) loss = 20% c) loss = 6 1 % d) profit = 11 1 %
9 4 9

2. a) 50% b) 4% c) 6.25%

3. a) 20% b) 6.67% c) 9.09% 31317%1

4. 50% 5. 10% 6. gain = % 7. profit = 16 2 %
10. a) 33 3
8. loss = 1.95% 9. loss = 13 11 % b) 23.96%
13

Exercise:10.2.1

1. a) Rs. 90 b) Rs. 1008 c) Rs. 629 d) Rs. 2300
2. Rs. 2500 3. Rs. 1159.20 4. Rs.168 11) Rs. 226.8
5 a. Rs. 3200 b) Rs. 450 c) Rs. 638.30
6. Rs. 2000 7. Rs. 2666.67 8 Rs. 555.56
9. a) Rs. 1050 b) Rs. 1365 10 Rs. 1254.81
12. a) 2600 b) 418.18

222 Profit and Loss

11UNIT UNITARY METHOD

Estimated periods 5 Objectives:

At the end of this unit students will be able to

● use the idea of direct proportion and indirect proportion.
● find the value of certain number of object when the value or cost of

one object is known.
● find the value of cost of one object when the value or cost of certain

number of objects is known.

=?

= Rs. 900

Teaching Materials:

Menu, price list, etc.

Activities:

It is better to

● discuss about use of direct proportion and indirect proportion for
solving unitary methods problems.

● let the students to solve the related problems by group discussion.

11.1 Direct Variation

Let’s suppose that cost of two pens is Rs 5 and Rs 7. The ratio of their cost is 5:7. We
5
compare quantities of same kind. This ratio can also be written as 7 in a fraction.
are equal, they
We know, 1 = 2 , then 1,3,2 and 6 form proportion. Thus, if two ratios
3 6

form proportion, where 1 and 6 are called extremes and 2 and 3 are called means.

Lets observe the table

No. of pencils 12 3 45
Costs (in Rs.) 10 20 30 40 50

From the above table, it is clear that if the number of pencils increases (or
decreases), the cost of the pencils also increases (or decreases) in the same ratio. The
variation between such two quantities is called a direct variation.

Example 1: If the price of 5 exercise books is Rs 100, then what is the cost of 8 exercise

books ?

Solution:
Using unitary method.
The price of 5 exercise books is Rs. 100
100
The price of 1 exercise book is Rs 5 = Rs 20

The price of 8 exercise books = Rs 20 × 8 = Rs 160
In the above example, first the price of unit exercise book is determined. With the
help of unit exercise book, the price of 8 exercise books is determined. This method
is called unitary method.

Using method of proportion. Cost (Rs)
No. of exercise books.

5 100

8x

Let, Rs x be the cost of 8 exercise books, the cost of exercise books varies directly with

the number of exercise books.

\ 5 = 100
8
x

or 5x = 100 × 8

20

or x = 100 × 8
5

or x = Rs 160.

Hence the cost of 8 exercise books = 160.

224 Unitary Method

Example 2: If Raj runs 16 km in 4 hours, then find the distance covered by him in 6 hours ?
Solution.
Here, In 4 hours, Raj runs 16 km
In 1 hour, Raj runs 16 ÷ 4 = 4 km

In 6 hours, Raj runs 4 × 6 = 24 km.
Hence Raj runs 24 km in 6 hours.

By the method of proportion Time (in hours)
Distance (in km)

16 4

x6
Let, x km be the distance run by Raj in 6 hours.

Distance varies direclty with the time.

\ 16 = 4
6
x
or 16 × 6 = 4 × x

16 × 6

or 4 = x

\ x = 24 km

Hence, Raj covers 24 km distance in 6 hours.

Exercise 11.1

1. Solve the following problems by method of proportion
a. Jay earns Rs 2500 in 5 days. Find the sum earned by him in 8 days.
b. A car needs 12 liters of petrol to travel 168 km. How many litres of petrol is
required to travel 350 km?
c. If 5m of cloth costs Rs 1150, then what is the cost of 8 m of cloth ?
d. In a map 3 cm represents 15 km, then how many kilometers is represented by 5 cm?
e. If 4 kg of sugar cost Rs 500, how many kg of sugar cost Rs 750?

2. Solve the following problems by using unitary method.
a. If 6 pens cost Rs 120 what is the cost of 8 pens ?
b. Anu cycles 15km in 3 hours .What distance does she cover in 5 hours?
c. 5 kg of sugar cost Rs 400. What is the cost of 7 kg of sugar ?
d. Chunu earns Rs 1000 in 4 days. In how many days will she earn Rs 2250?
e. A car covers a distance of 60 km in 45 minutes. Find the distance covered by
the car in 30 minutes.

3. Copy and complete the following table if 3 pencils cost Rs. 24.

No. of pencils 3 5 12 20

Cost (Rs.) 24 80 120

Prime Mathematics Book - 7 225

11.2 Indirect Variation

Let’s observe the table

No. of worker 1 2 5 10
Days 50 25 10 5

From the above table, it is clear that if the number of worker increases (or decreases),
the days to complete the one work is decreased (or increased) in the same ratio. The
variation between such two quantities is called a indirect variation.

Example 1: 12 men can do a work in 4 hours, how long would 16 men take to do the

same work ?

Solution:

Using unitary method.

12 men can do a work in 4 hours

1 man can do the work in 4×12 hours

16 men can do the work in 4×12 hours
16

= 3 hours

In the above example, first the time taken by one man is determined. With the help
of the work of 1 man the time taken to complete the work by 16 men is determined .
So, this method is called unitary method.

Using method of indirect proportion. hours
No. of men. 4
12
16 x

Let, x be the time taken by 16 men, the time varies indirectly with the number of men.

\ 12 = x
16
4

or 16x = 12 × 4

or x = 123× 4
16 4

or x = 3.

Hence 16 men can do the work in 3 hours.

226 Unitary Method

Example 2: The hostel has food enough for 50 students for a month. For how many days
would the food last if 25 more students join them?

Solution:
The hostel has food enough for 50 students for 30 days [1 month = 30 days]

\ The hostel has food enough for 1 student for 30 × 50 days.
\ The hostel has food enough for (50 + 25 = 75) students for 30 × 50 days.

75

= 20 days.

Exercise 11.2
1. Solve the following problems by method of proportion.

a. 15 men can do a work in 3 hours, how long would 20 men take to do the same work?
b. If 16 workers can build a wall in 7.5 days, how many workers can build it in 8 days?
c. A garrison had food enough for 120 soldiers for 40 days. How many more days

would the food last if after 10 days 20 soldeers had to leave?
d. A bus takes 8 hours to do a journey if it runs at 60 km per hour. What should be

its speed to do the journey in 6 hours ?
e. 30 men can do a piece of work in 21 days. How long will it take to finished the

work with the help of 5 additional men?
2. Solve the following problems by using unitary method.

a. 12 men can do a work in 20 days, how many men would do it in 15 days ?
b. A school hostel had food enough for 40 students for a month. For how many

days would the food last for 60 students.
c. A car takes 12 hours to cover the journey if it runs at 60 km per hour. What

should be its speed to cover the journey in 10 hours?
d. A school hostel having 60 students can feed for a month with its provision.

How long will the provision last if 12 more students join them?
3. a 8 workers can do a piece of work in 15 days. How many workers should be

added, to complete the work in 12 days?
b. In a hostel 50 students have food enough for 32 days . How many students

should be leave the hostel so that the food is enough for 40 days.

Prime Mathematics Book - 7 227

Unit Revision Test

1. How do the following vary:
a) Number of objects and their cost.
b) Number of workers and number of days of work.

2. In a map 15 km is represented by 3 cm. How many km is represented by 8cm.

3. If 2 men can do a work in 4 days, how long would 16 men take to do the same
work?

4. A car consumes 12 litres of petorl to travel 168 km. How many litres of petrol is
needed to travel 350 km ? (Solve by using method of proportion)

5. If 18 men can complete a piece of work in 12 days, in how many days would 6 men
complete the same work ? (Solve by using method of proportion)

6. If 5 kg of sugar cost Rs 625, how much sugar con be bought for Rs. 1124 ?

7. Operating 4 hours a day, battery can be used for 9 days. For how many days can
the battery be used operating 4 hours a day ?

8. Complete the given table.

a) 3 5?
No. of pens 240 ? 640
Cost (Rs)

b) 5 20 ?
12 ?4
No. of men
No. of days

Answers

Unit:11 Unitary Method
Exercise:11.1
1. a) Rs. 4000 b) 25 l c) Rs. 1840 d) 25 km. e) 6 kg
2. a) Rs. 160 b) 25 km. c) Rs. 560 d) 9 days e) 40 km.
3. 5 pencils = Rs. 40, Rs. 80 = 10 pencils, 12 pencils = Rs.96, Rs.120 = 15 pencils,
20 pencils = Rs 160

1. a) 2.25 hours b) 15 Exercise:11.2 e) 18 days
2. a) 16 b) 20 days c) 6 d) 80 km/hr
3. a) 2 b) 10 c) 72 km/hr d) 25 days.

228 Simple Interest

12UNIT SIMPLE INTEREST

Estimated periods 5 Objectives:

At the end of this unit students will be able to

● understand what is interest, principal, amount and rate of interest
● calculate simple interest, amount, time, rate of interest and

principal by using formula.

Teaching Materials:
The chart of current interest rates of different banks and finance companies.
Sample of loan agreements .
The chart of formula to calculate interest, principal, amount, time and rate of interest.

Activities:

It is better to

● discuss about interest, process of providing loan.

● show, how the formula I = PTR is obtained.
100

● show, how the formula T = I × 100 , R = I × 100 , P = I × 100 and P = A × 100
P×R P ×T T×R 100 + T × R

is obtained.

● discuss about the use of formula to calculate interest, time, rate of interest and
principal

12.1 Introduction

Suppose a man deposits Rs. 500 in a bank. After 2 years he get back Rs. 540 with Rs.
40 extra from the bank.

Here,
Rs. 500 which is deposited by the man in the bank is called principal. The principal is
denoted by ‘P’.

Rs. 40 which is got by the man as extra payment is called interest. The interest is
denoted by ‘I’.

Rs. 540 which is total sum of money got back by the man at the end of the period is
called Amount. The amount is denoted by ‘A’.
The bank paid Rs. 40 as extra money after 2 years. So 2 years is called Time. It is
denoted by ‘T’.
( )The extra money is paid by the bank in percentage =
40 × 100 % = 8%
500

8% which is paid by the bank on Rs. 100 for every year. so, 8% is called the rate of
interest The rate of interest is denoted by R.

What is the rate of interest on Rs. P for T years at the rate of interest R% per annum?

We know that the rate R% per annum means the interest on Rs. 100 for 1 year is Rs. R.

So, the interest on Rs. 100 for 1 year is Rs. R.
R
the interest on Re. 1 for 1 year is Rs. 100

the interest on Rs. P for 1 year is Rs. R × P
100
R
the interest on Rs. P for T years is Rs. 100 × P × T

PTR

\ Interest (I) = 100

And Amount (A) = P+I

Example 1. Find the interest and amount on Rs. 600 for 3 years at the rate of interest
6% per annum.
Solution :
Here

Principal (P)= Rs. 600, Time(T) =3years

Rate (R)=6%, Interest (I)=? and Amount (A)=?

By using formula, = PTR
Interest (I) 100

= Rs. 600×3×6 = Rs. 108
100

230 Simple Interest

Example 2. A man borrowed Rs. 2400 at 4% per annum from a bank. What was the
amount paid by the man to the bank at the end of 3 years 6 months?

Solution:
Here,

Principal(P) = Rs. 2400,
Rate (R) = 4% per annum
Time(T) = 3 years 6 months

( )=3years + 6 years. [Time should be always in years].
12

= 3 + 1 years = 7 years
2 2

Amount(A) = ?

By using formula,

Interest (I) = PTR = Rs. 2400×7×4 = Rs. 336
100 100×2

Now, Amount (A) = P + I = Rs. 2400 + Rs. 336 = Rs. 2736
Hence the man paid Rs. 2736 as amount to clear the loan to the bank at the end of
the given time.

Exercise 12.1

1. Find the simple interest on
a) Rs. 400 for 2 years at 7% per annum.
b) Rs. 6000 for 3 years at 12% per annum.
c) Rs. 6500 for 2 years 3 months at 6% per annum.
14calcyuelaatrinagt 8th12e%inpt.ear.est
2. d) Rs 5000 for 7 on
Find the amount
a) Rs. 300 for 2 years at 5% per annum.
b) Rs. 8000 for 4 years at 2% per annum.
dcA))mRRass.n61d050e00p0ofsfooirtre73d12yaeysaeuramsrs6oafmtRo1sn2.t.4h50s%0apti.na4a.% per annum.
3. bank. The rate of interest paid by the bank
is 7% per annum. Find the interest and the amount which the man will receive at
the end of 2 years 3 months.
4. A farmer borrowed Rs. 8000 from A.D. Bank to buy the seeds of rice at 12% simple
interest. At the end of the first year, he paid back Rs. 4260 to the bank. How much
he pay to clear the debt at the end of the second year?
5. A money lender lent a shopkeeper a sum of Rs. 1560 at the rate of 18% per annum.
What amount will he get from the shopkeeper at the end of 1 year 6 months?
6. Nuni Lal borrowed Rs. 2400 as a loan for 5 years at the rate of 6% per annum. Then
immediately he lent the same amount to Briju Lal for 4 years at the rate of 10% per
annum. How much profit or loss did Nuni Lal make in this transaction? Find it.

Prime Mathematics Book - 7 231

12.2 Finding Time (T), Rate of Interest (R) and Principal (P)

In the previous exercise we discuss of the formula to calculate simple interest (I)

so, Interest (I) = PTR
100
or, PTR = I × 100

\T = I × 100 [ Calculation for Time]
PR
Again, R I × 100 [Calculation for Rate]
and P = PT [Calculation for Principal]

= I × 100
TR

We know that,

A =P+I

or, A =P+ PTR
100

or, A = 100P + PTR
100

or, P(100 + TR) = A × 100

\ P = A × 100 (Calculating for principal if amount is given)
100 + TR

Example 1. Find the time in which the interest on Rs.600 at 25 % per annum is Rs. 150
Solution. 3

Here,

Principal (P) = Rs 600, Rate (R) = 25 %, p.a. Interest (I) = Rs.150, Time (T) = ?
3

Now, Time (T) = I × 100
PR

= Rs. 150 × 100

Rs. 600 × 25
2 3

30 6 3
150
= 2 × 25

5

= 3 years.

232 Simple Interest

Example 2. In how many years will Rs. 650 amount to Rs. 910 at 12% per annum simple
interest ?
Solution:
Here,
Principal (p) = Rs. 650,
Amount (A) = Rs. 910,
Rate (R) = 12 % per annum

\ Interest (I) = A - P = Rs. 910 - Rs. 650 = Rs. 260

Time (T) = ?
We know that,

Time (T) = I × 100 20 10
PR 2
Rs. 260 × 100
=
Rs 650 × 12
13 63

= 10 = 3 1 = 3 years 1 4
3 3 3 × 12 months

= 3 years 4 months.

Example 3. At what rate percentage will the interest on Rs. 1500 for 2 years be Rs. 150 ?
Solution:
Here, Principal (P) = Rs 1500,

Time (T) = 2 years.
Interest (I) = Rs 150
Rate (R) = ?

Now, Rate (R) = I × 100
PT

= Rs 150 × 100 = 5% p.a.
Rs 1500 × 2

Example 4. At what rate percentage will a sum of money treble itself in 10 years ?
Solution:

Here, Suppose the sum of money be Rs. x.

Then it is trebled itself in 10 years, which is the amount

\ Principal (P) = x, Amount (A) = 3x, Time (T) = 10 years.

Rate (R) = ?

Now, Interest (I) = A - P = 3x - x = 2x

Rate (R) = I × 100 = 2x × 100 = 20% p.a.
PT x × 10

Prime Mathematics Book - 7 233

Example 5. The interest on the certain sum of money at 8% per annum is Rs. 130 for 2
years 6 months. What is the sum ?

Solution:
Here, Interest, (I) = Rs 130, Rate (R) = 8%

Time (T) = 2 years 6 months

= 2 years + 6 1 years.
12 2

( )= 4+1 years = 5 years.
2 2

Sum of money (P) = ?

Now, P= I × 100
TR

= Rs.130 × 100

5 × 84
2

= Rs. 13000 = Rs. 650
20

Example 6. A man borrowed a sum of money for 3 years from a money lender at the
rate of 12% per annum. The man paid Rs. 1360 at the end of 3 years to the
money lender to clear the debt. How much money had he borrowed from
the money lender ?

Solution:

Here,

Time (T) = 3 years,

Rate (R) = 12% p.a.

Amount (A) = Rs.1360,

Principal (P) = ?

Now, A × 100
Principal (P) = 100 + TR

Rs. 1360 × 100
= 100 + 3 × 12

= Rs. 136000
136

= Rs.1000

234 Simple Interest

Exercise 12.2
1. Find the time in which the interest on

a) Rs. 2500 at the rate of 10% per annum is Rs.500.
b) Rs. 2200 at 4% per annum is Rs.220.
c) Rs. 5000 at the rate fo 12% per annum is Rs.240.

2. Find the time in which the amount on
1
a) Rs. 4500 at 3 3 % per annum is Rs.4700.

b) Rs. 3000 at 15% per annum is Rs.3200.

3. Find the rate percent per annum if
a) the interest on Rs. 600 for 5 years is Rs. 90.
b) the interest on Rs. 2000 for 9 months is Rs. 180

4. In what time will Rs. 2000 yield an interest of Rs.600 at the rate of 10% per annum ?
5. At what rate percent will the amount on Rs.650 for 3 years 4 months be Rs. 910?
6. In how many years will Rs.6250 amount to Rs.7500 at 10% per annum simple interest ?
7. At what rate percentage will a sum of money double itself in 10 years ?
8. In what time will a sum of money treble itself at the rate of 16% per annum?

9. Find the principal which yields an interest of
a) Rs. 468 in 4 years at 6% per annum.
b) Rs. 56 in 2 years 4 months at 8% per annum.
c) Rs. 405 in 2 years at 15% per annum.

10. Find the principal if the amount is
a) Rs.1755 in 2 years at the rate of 15% per annum.
b) Rs.13890 in 2 years 3 months at the rate of 7% per annum.
c) Rs. 480 in 288 days at the rate of 6% per annum.

11. Mr. Karki deposits a sum of money in a bank at the rate of 12.5% per annum.
After 3 years he receives Rs. 1375 as simple interest. How much does he deposit
in the bank ?

12. Anima deposited a sum of money for 3 years 6 months in a bank at the rate of 9%
per annum. She received Rs. 6575 at the end of 3 years 6 months from the bank.
How much money did she deposite in the bank ?

13. The difference between the simple interest on a sum of money in 5 years and 7
years at the same rate of 6% per annum is Rs. 1260. Find the sum.

Prime Mathematics Book - 7 235

Unit Revision Test

1. Find the simple interest and amount on
a) Rs. 5000 for 3 years and 8% per annum.
b) Rs. 4500 for 2 years 6 months at 10% per annum.

2. A man deposited a sum of Rs. 200 in a bank. The rate of interest paid by the bank
is 12% per annum. Find the interest and amount which the man will receive at the
end of 1 year 6 months.
froarte2 fyoe3ars31
3. Ramu lent Hira a sum of Rs. 3500 as a loan for 3 years at the % per
annum. Then immediately Hira lent the same amount to Kabir at the
rate of 6% per annum. How much profit or loss did Hira make in this transaction ?
Find it ?

4. In how many years will Rs.3000 amount to Rs.3900 at 15% per annum simple
interest ?

5. At what rate percent will the amount on Rs.1200 for 2 years 6 months be
Rs 1500 ?

6. In what time will a sum of money treble itself at the rate of 10% per annum ?

7. Find the principal which yields an interest of Rs.360 in 2 years at 9% per annum.

8. Aadarsh deposited a sum of money for 2 years 6 months in a finance company at
the rate of 12% per annum. He received Rs.19500 at the end of time from the
company. How much money did he deposit in the company ?

Answers

Unit:12 Simple Interest
Exercise:12.1
1. a) Rs. 56. b) Rs. 2160 c) Rs. 877.50 d) Rs. 3081.25
2. a) Rs. 330 b) Rs. 8640 c) Rs. 1710 d) Rs. 11,625
3. Rs. 63, Rs.463 4. Rs.5264 5. Rs.1981.20 6. Rs. 240
Exercise:12.2
1. a) 2 years b) 2 years 6 months c) 4 months 24 days.
2. a) 1 year 4 months b) 5 months 10 days

3. a) 3% b) 12 % 6. 2 years 7.10%
4. 3 years 5. 12% c) Rs. 1350
8. 12 years 6 months b) Rs. 300 c) Rs. 458.30
9. a) Rs.1950 b) Rs.12000 13. Rs. 10,500
10. a) Rs.1350 12. Rs. 5000
11. Rs. 3666.67

236 Statistics

13UNIT Statistics

Estimated periods 10 Objectives:

At the end of this unit students will be able to

● prepare frequency table and cumulative frequency table from the given
raw data.

● present the data in multiple bar diagram of ungrouped and grouped
data.

● find the mean (arithmetic mean) of the individual and discrete data.

No. of students in different Classes of a School

Classes VI VII VIII IX X
No. of boys 20 18 15 18 26
No. of girls 24 22 25 18 12

Teaching Materials:

Square papers, stat paper, attractive diagrams (simple bars, multiple bars,
subdivided bars).

Activities:

It is better to

● discuss about different methods of presentation of data.
● clarify the meaning of central tendency.
● discuss about arithmetic mean.

13.1 Introduction

The word statistics seems to have been derived from the Latin word 'status' which
means a 'political state'. Originally statistics was simply the collection of numerical
data on some aspects of life of the people useful to the government. However with the
passage of time, its scope has been broadened. Today, Statistics includes the collection
and analysis of numerical data concerning almost every aspect of our life.

Statistics includes the following steps.

i. Collection of data
ii. Presentation of data
iii. Analysis of data
iv. Interpretation of data
The facts or figures which are numerical or otherwise collected with definite purpose
are called data. Data is the plural form of the Latin word datum.
The quantity which is being measured in an experiment is called a variable.
e.g. height, age, weight, number of members etc.

Types of Variables.

(i) Continuous variable: A variable which can take any value between two given values
is called a continuous variable. e.g. age, height, weight etc.
(ii) Discontinuous (or discrete) variable: A variable which can not take all the possible
values between two given values is called a discontinuous (or discrete) variable.
e.g. number of workers.
The particular value of variable is called a variate (observation)

The number which tells us how many times a variate appears in the given data is the
frequency of the variate.
Presentation of data

The data primarily collected is called raw data which is simply a collection of haphazardly
arranged numbers which need to be presented in a form which is meaningful, easily
understood and gives its main features at a glance. Presentation of data includes
arranging in order, tabulation and presenting in diagram graphs, charts etc.
1. Tabulation of raw data: Out of 50 marks, the marks obtained by 32 students of class
VII are as follows.

21, 25, 28, 12, 40, 28, 21, 36, 15, 28, 49, 28, 36, 15, 25, 21,
15, 28, 25, 36, 28, 21, 40, 25, 36, 12, 28, 25, 36, 36, 44, 28.
The data is ungrouped or unclassified. Such data is called raw data. From a raw data
we cannot get much information.

Let us arranged the above data in ascending or descending order.
12, 12, 15, 15, 15, 21, 21, 21, 21, 25, 25, 25, 25, 25, 28, 28,
28, 28, 28, 28, 28, 28, 36, 36, 36, 36, 36, 36, 40, 40, 44, 49,

238 Statistics

The data presented in this form gives better information. The data arranged in this
form is called arrayed data. However it is also tedious and time consuming when the
number of observations is large.

To make it easily understandable we present it in a form of table called frequency
distribution table

A tabular arrangement of given numerical data showing the frequency of variables is
called frequency distribution and the table itself is called a frequency distribution
table.

To prepare a frequency distribution table, we take the observation taken in ascending
order in a column, we take each observation from the data, one at a time and mark
a stroke ( ) called tally in the next column opposite to the variates (we write tally
marks in bunches of five, the fifth one crossing the four diagonally). The number of
tally marks is called frequency and is written in the next column opposite to the
corresponding tally marks of the variates. Note that the sum of the frequencies is
equal to the number of observation in the given data.

The frequency distribution table of the above ungrouped data is given below.

Marks obtained Tally marks Freq uency
12 2
15 3
21 4
25 5
28 8
36 6
40 2
44 1
49 1

Total = 32

Grouped frequency distributions: When the number of observation is very large,
presenting the data in array or simple frequency distribution table is tedious.

Lets consider the data

15, 25, 30, 28, 13, 18, 8, 9, 17, 20, 25, 25, 28, 27, 32, 39, 16, 19, 9, 8, 4, 38, 35,
23, 20, 37, 25, 35, 7, 4, 12, 15, 3, 30, 27, 23, 18, 8, 5, 12

Arranging in order

3, 4, 4, 5, 7, 8, 8, 8, 9, 9, 12, 12, 13, 15, 15, 16, 17, 18, 18, 19, 20, 20, 23, 23, 25,
25, 25, 25, 27, 27, 28, 28, 30, 30, 32, 35, 35, 37, 38, 39

Prime Mathematics Book - 7 239

Marks Tally Frequency Marks Tally Frequency
marks marks
3 1 19 1
4 2 20 2
5 1 23 2
7 1 25 4
8 3 27 2
9 2 28 2
12 2 30 2
13 1 32 1
15 2 35 2
16 1 37 1
17 1 38 1
18 2 39 1
total = 40

In such case, the table can be made short and attractive by grouping in classes of
suitable class width.

Inclusive frequency distribution:

Class Tally marks Frequency

0-9 10

10-19 10

20-29 12

30-39 8

N = 40

Exclusive frequency distribution:

Class Tally marks Frequency
0-10 10

10-20 10

20-30 12

30-40 8

N = 40

While preparing a grouped frequency distribution table upper limit of a class and lower
limit of next class are made same and the observation corresponding to the upper limit
is not counted in that class but counted as lower limit of next class.

240 Statistics

Cumulative frequency distribution table: Observe the given data. How many students
got less than 20 marks? To answer this question, of course you will add frequencies of
the class 0-10 and 10-20 i.e. 2+5 = 7

Class No of student

0-10 2

10-20 5

20-30 12

30-40 10

40-50 3

N = 32
Some times we need to know how many observation are there less than or greater
than certain observation or class. In such case we prepare a cumulative frequency
distribution table.

Marks obtained by 32 students of class VII

Marks Marks less than (X) Frequency (f) Cumulative Frequency (c.f)

0-10 10 22
10-20 20 5 2+5=7
20-30 30 12 7 + 12 = 19
30-40 40 10 19 + 10 = 29
40-50 50 3 29 + 3 = 32

N = 32

The sum of the frequency of a particular class and the frequencies of the classes
preceding that class is called cumulative frequency of that class. And the cumulative
frequency distribution is a table showing the number of observation that are less than
specified value.

Marks Marks greater than f c.f

0-10 0 2 30 + 2 = 32

10-20 10 5 25 + 5 = 30

20-30 20 12 13 + 12 = 25

30-40 40 10 3 + 10 = 13

40-50 50 33

N = 32

Similarly greater than frequency distribution table can be prepared.

Note: - Take less than upper limit, greater than lower limit.
- For preparing greater than cumulative distribution table proceed from the
last class.

Prime Mathematics Book - 7 241

Exercise 13.1
1. The following data represents the marks obtained by 30 students of class VII of a

school (out of 20 marks ) in mathematics, construct a frequency distribution table
with tally marks.
12, 15, 14, 16, 10, 14, 15, 14, 18, 15, 15, 10, 12, 15, 18, 16, 12, 12, 15, 16, 14,
16, 14, 15, 16, 10, 15, 14, 18, 16
2. Daily wages (in Rs.) of 25 workers of a factory are given below. Construct a
frequency distribution table representing the data.
380, 360, 375, 370, 385, 380, 370, 380, 380, 375, 370, 390, 380, 390, 375, 380,
385, 375, 380, 370, 375, 385, 380, 385, 370

3. Pocket money (in Rs.) of 50 students of a school are as follows.
22, 55, 49, 27, 30, 27, 30, 27, 25, 42, 40, 13, 24, 38, 10, 24, 30, 33, 27, 29, 10,
50, 18, 34, 13, 40, 15, 32, 36, 32, 27, 35, 17, 36, 18, 41, 20, 41, 35, 51, 29, 27,
44, 43, 15, 32, 29, 54, 45, 14
Construct a frequency distribution table with one of the class 10-20 exclusive
form.

4. Taking class intervals 0-4, 5-9, 10-14, etc, construct the frequency distribution of
the data.
18, 12, 8, 1, 18, 21, 20, 12, 7, 5, 19, 11, 13, 17, 13, 9, 16, 14, 2, 8, 15, 11, 5, 10, 6

5. Marks obtained by 50 students in a test out of 100 marks are given below.
32, 84, 32, 92, 35, 95, 34, 92, 35, 5, 21, 37, 82, 31, 84, 31, 82, 31, 72, 29, 25, 54,
25, 64, 25, 70, 28, 66, 58, 25, 20, 36, 22, 38, 24, 50, 22, 39, 74, 31, 18, 35, 6, 35,
5, 36, 12, 36, 85, 32
Taking class interval of size 10 exclusive prepare a frequency as well as cumulative
frequency table.

6. Marks secured by 35 students in a test are given below.
33, 48, 44, 34, 24, 14, 4, 30, 38, 29, 24, 25, 18, 5, 0, 42, 47, 11, 21, 35, 32, 41,
36, 38, 41, 46, 8, 34, 39, 11, 13, 27, 26, 3, 43
Taking a class internal 0-9 inclusive, construct a frequency as well as cumulative
frequency table.

13.2 Graphical representation of statistical data:

Pictures or graphs are good visual aids and leave a deeper and more lasting impression
on the mind of an observer. The information contained in a numerical data can be
easily understood if we represent it in a form of graphs or diagrams.

242 Statistics

There are various ways of representing numerical data graphically.
i. Bar graphs
ii. Pie chart
iii. Line graph
iv. Histogram
v. Frequency polygon etc.

In this class we will learn about bar graphs.
i. Bar graphs (charts): In bar charts (bar graphs), bars (rectangle) of equal widths are
used to represent the numerical data under consideration.
“The statistical data are represented by graph, diagram or chart. Thus, graph is a
pictorical statement of statistical data. The significance of statistical data is understood
easily and quickly if they are represented in a graph or diagram or a chart.”
Bars can be of any width but width of all the bars must be same.

Spaces between the consecutive bars must be same.
Heights of different bars are proportional to the corresponding frequencies.

Simple bar diagram: This is the simple method to represent the data of single variable
(all classes of same character). It contains a set of rectangles of equal width raised at
equal distances. The heights of the rectangles are proportional to their frequencies.
Generally classes or subjects are taken along X-axis and frequencies along Y-axis

Example1: The data given below shows the numbers of students in class 7 of a school
in five years. Represent the data in a bar-chart.

Years(B.S.) 2066 2067 2068 2069 2070

No. of Students 25 40 38 48 55

Solution:

Years

Note: Since each bar represent same variable, all the bars should be given same shadow.

Multiple bar diagrams: The statistical data with two or more variables (more than
one character in a class) are represented in a multiple bar diagram. It consists a set of
adjacent rectangles of equal width raised on a class or group. Heights of the rectangle
are proportional to their respective frequencies.

Prime Mathematics Book - 7 243