Math 7

Example: 6 Example: 7

Simplify: Simplify: 5 1 ÷ [1 3 ÷{ 1 – 3 )}]
2 8 2 8
1 3 × 1175 4
7 2 ÷ 3 4 of 1 11 Solution:

Solution: = 5 1 ÷ [1 3 ÷( 1 – 3 )
2 8 2 8
1 3 7 1141
7 2 ÷ 3 4 × 1 15 of = 5 1 ÷ [1 3 ÷( 1 – 3 )
2 8 2 8
15 15 222 151
= 2 ÷ 4 × 15 × 11 = 5 1 ÷ [1 3 ÷1×48–3]
2 8
15 15
= 2 ÷ 4 ×2 = 5 1 ÷ [181 ÷ 1 ]
2 8
1 4 21
= 15 15 × = 11 ÷ [181 × 8] = 11 ÷11
= 2 × 2 2
1
4
= 121×111 = 1
2

Example: 8 3
4
Neema's monthly salary is Rs. 10400. He spends of his salary on household function.

Find how much money left with him?

Solution:

His monthly salary = Rs. 10,400

His expenditure = 3 of 10,400
4

= 3 × 10,400 = Rs. 3 × 2600 = Rs. 7800
4

Money left with him = Rs. 10,400 – Rs. 7,800 = Rs. 2600.

Exercise 14.4

1. Multiply together:

(a) 12 × 8 (b) 4 × 27 (c) 8 1 × 6
8 3 9 32 2 25

2. Find the quotient:

(a) 13 ÷ 3 (b) 14 ÷ 56
8 5 15

(c) 8 1 ÷ 4 1 (d) 2 3 ÷ 5 1
2 4 4 8

3. Simplify the following:

(a) 4 × 5 (b) 3 1 × 2 2
5 8 2 7

(c) 1 1 ×2 1 ×2 2 (d) 2 2 × 5 1 × 6 1
7 3 5 5 3 4

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4. Simplify the following:

(a) 1÷ 5 (b) 5 ÷1 (c) 16 2 ÷ 12 1 (d) 1 ÷ 3
7 7 3 2 2 4

5. Simplify the following:

(a) 2 2 (b) 10 (c) 8
5 2 5
6 1 3 4

(d) 1530 (e) 7 1 (f) 3 3
2 4
6. Find the value of : 5
8 1
1 2

(a) 3 of 50. (b) 5 of Rs. 64. (c) 1 of Rs. 1040
5 8 5

(d) 3 of 1 of Rs. 420 (e) 3 of 1 of 500 kg.
5 3 2 2

7. Simplify the following:

(a) 2 1 × 3 + 7 1 × 4 (b) 4 ÷ 1 – 1 × 3
2 5 2 15 3 2 3 4

(c) 1 × 2 of 1 – 3 (d) 4 ÷ 1 – ( 3 × 2 )
2 5 2 10 5 2 2 5

8. Simplify the following:

( )(a) 4− 3 1 − 3 ( )(b) 2 5 1 − 4 3
2 4 3 6 4

{ ( )}(c) 1 −  3 − 1 + 1 − 1  { ( )}(d) 3 2 1 ÷ 3 − 1 2 − 1 
 4 2 4 8  14 3 4 2 3 6 

{ ( ) } (e) 1 − 1 1 − 1 ÷ 1 + 1 + 1
3 3 3 3 2 3 2

9. (a) If the speed of the bus is 40 1 km per hour, find the distance covered by a bus in
2
2
10 3 hours.

(b) I purchased 15 3 metre of cloth costing Rs.20 4 per metre. How much do I have to
4 5

pay?

10. (a) If 7 of a number is 14, what is the number?
8

(b) If 1 parts of a sum is Rs. 125, find the sum.
5

(c) If 3 parts of a sum is Rs. 600, find 3 of the sum.
4 5

196 Oasis School Mathematics – 7

11. (a) In an exercise, there are 40 problems. A student has worked out 4 of them. How
many are unsolved? 5

(b) Ram's monthly income is Rs. 8400. If he saves 5 of his earning, find;
12

(i) his monthly saving (ii) his expenditure per month

12. (a) If the cost of 5 1 kg of apples is Rs. 110, what is the cost of 1 kg of apples?
2

(b) If the cost of 3 1 m cloths is Rs. 325, what is the cost of 1 m cloths?
4



Answer

1. (a) 4 (b) 3 (c) 2 1 2. (a) 8 (b) 3
8 25 9 4

(c) 2 (d) 22 3. (a) 1 (b) 8 (c) 6 2
41 2 5

(d) 80 4. (a) 152 (b) 5 (c) 1 1 (d) 2
7 3 3

5. (a) 2 (b) 6 (c) 2 (d) 1 (e) 12 (f) 221
5 (b) Rs. 40 5 6 (e) 375 kg

6. (a) 30 (c) Rs. 208 (d) Rs. 84

7. (a) 312 (b) 2152 (c) –1 (d) 1
8. (a) 141 5

9. (a) 432 km (b) 5 (c) 7 (d) 1 (e) 5
18 8 6
11. (a) 8
(b) Rs. 32753 10. (a) 16 (b) Rs. 625 (c) Rs. 480
(b) (i) Rs. 3500 (ii) Rs. 4900
12. (a) Rs. 20 (b) Rs.100

14.5 Decimals

Fractions with denominators 10, 100, 1000 etc. are called decimal fractions.

For examples: 8 , 14050 , 5 etc. are decimal fractions.
10 1000

Decimal fractions written in decimal form are called decimal numbers or simply

decimals. We use decimal point ( . ) to represent a decimal.

For examples: 0.5, 3.56, 0.345 etc. are decimals. A decimal consists of two parts
separated by decimal point. They are a whole number part or integral part and a
decimal part.

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Conversion of decimals into fractional form

A decimal number can be changed into fraction and vice – versa. See the following
examples.

(i) 0.25 = 25 = 1 Steps:
100 4
• Write the given number without decimal.
(ii) 2.4 = 1240 = 12 • Put 1 on the denominator.
5 • Put as many zeros as the number of digits after decimal point.
• Reduce the fraction into its lowers them.
(iii) 0.005 = 5 = 1 , etc.
1000 200

Addition and Subtraction of decimals

Let us consider two decimal numbers as 0.25 and 2.5 Here, in order to add or
subtract two decimal numbers, follow the following steps.

0.25 Steps:
2.50
2.75 • Keep two decimal numbers in such a way that two decimal
should lie on the same vertical line

• Add zeros at the last of the number having less digits after
decimal.

• Add two numbers.


Worked Out Examples

Example: 1 Example: 2

Add 0.25 and 2.5 Subtract 0.25 from 2.5

Solution: 0 .25 Solution: 2.50
+ 2.50 – 0.25
2.75
2.25

∴ 0.25 + 25 = 2.75

Example: 3 45.750
+ 6.375
Simplify: 15.3 – 45.75 + 60 – 6.375 52.125
75.300
Solution: –52.125
23.175
15.3 – 45.75 + 60 – 6.375 = 15.3 + 60 – 45.75 – 6.375

= 75.3 – 52.125 = 23.175

198 Oasis School Mathematics – 7

Example: 4

A T.V. set costs Rs. 17500; a calculator costs Rs. 945.45 and a fan costs Rs. 675.70. Find the
total cost of all the three things.

Solution:

Cost of a T.V. = Rs. 17500.00

Cost of a Calculator = Rs. 945.45

Cost of a Fan = Rs . 675.70
Total cost Rs. 19121.15

Hence, the total cost of all the three things is Rs. 19121.15

Example: 5

Senmikha bought an instrument box for Rs. 65.45 and paid a hundred rupee note. How

much did he get back?

Solution:

Cost of an instrument box = Rs. 65.45 100.00

Money given = Rs. 100.00 – 65.45

He got back = Rs. 100.00 – Rs. 65.45 5 34.55

= Rs. 34.55

Exercise 14.5

1. Add together:

(a) 2.453 + 3.657 (b) 4.005 + 3.125 (c) 12.8 + 9.258

(d) 0.0048 + 2.0215 + 5.5109 (e) 8.62 + 9.74 + 15.47

2. Subtract:

(a) 6.25 – 4.82 (b) 9.215 – 4.621 (c) 4.25 – 1.853

(d) 0.4132 – 0.0058 (e) 29.45 – 82.67 (f) 5 – 8.55

(g) 42.015 – 55.25

3. Simplify:

(a) 2.5 + 8.6 – 7.4 (b) 9.62 – 5.89 + 3.56

(c) 4.21 – 16.5 + 8.92 (d) 14.657 + 8.125 – 3.626

(e) 8.1 + 1.8 – 9.9 (f) 2.825 – 6.94 – 10.2 + 14.95

4. (a) Mohan obtained 75.65 marks in Maths, 65.5 in English and 70.45 in Science. Find
his total marks in three subjects.

Oasis School Mathematics – 7 199

(b) My father gave me Rs. 10.25 on my birthday. My brother gave me Rs. 52.75 and
my uncle gave me Rs. 31.25. Find the total amount that I had with me.

5. (a) Anil had Rs. 55 with him. He spent Rs. 35.35. How much money was left with him?

(b) The sum of the two numbers is 58.025. If one of them is 35.75, what is the other
number?

(c) What is to be added to 35.365 to get 150?

(d) A milkman had 45.5 litres of milk. He sold 25.75 litres of milk during the day.
How much milk is left with him?

6. The monthly expenditure of a family is as follows.

Food: Rs. 7500.75, Rent: Rs. 4000.25, Transport: Rs. 475.75 and school fees: Rs. 895.50.

If the monthly income is Rs. 15000, how much money does the family save every
month?

7. (a) Find the perimeter of the following figures.

(i) A (ii) A 5.8 cm

3.9 cm D
2.8 cm
4.5 cm 3.9cm

B C B 6.4 cm C

5.5 cm

(b) The perimeter of triangle is 25.6 cm. If the length of two sides of the triangles are
8.5 cm and 7.9 cm, find the length of the third side.

Answer 

1. (a) 3 (b 4 1 (c) 1 (d) 1 3 (e) 5 4 2. (a) 0.25
25 4 200 4 5 (b) 7.13

(b) 3.33 (c) 6.25 (d) 1.25 (e) 5.25 3. (a) 6.11

(c) 22.058 (d) 7.5372 (e) 33.83 4. (a) 1.43 (b) 4.594 (c) 2.397

(d) 0.4074 (e) -53.22 (f) -3.55 (g) -13.235 5. (a) 3.7 (b) 7.29

(c) -3.37 (d) 19.156 (e) 0 (f) 0.635 6. (a) 211.6 (b) Rs. 94.25

7. (a) Rs. 19.65 (b) 22.275 (c) 114.635 (d) 19.75 litres 8. Rs. 2127.75

9. (a) (i) 12.2 cm (ii) 20.6 cm (b) 9.2 cm

14.6 Multiplication of Decimals

I. Multiplication of Decimal number by 10, 100, 1000, etc.
To multiply a decimal by 10, 100, 100 etc. move the decimal point in the

multiplicand by as many places to the right as there are zeros in the multiplier.

200 Oasis School Mathematics – 7

e.g. 0.456 × 10 = 4.56 (The decimal point is moved one place to the right)

0 .456 × 100 = 45.6 (the decimal point is moved two places to the right)

0.456 × 1000 = 456 (The decimal point is moved three places to the right)

0.456 × 10000 = 4560 (the decimal point is moved four places to the right)

II. Multiplication of Decimal by whole numbers

To multiply a decimal by a whole number, follow the following steps:

• Multiply the decimal without decimal point by the whole n umber.

• Count the number of digits after decimal point.

• Put the decimal on the product such that number of digits after decimal be same
as same as in the given decimal number.

L et us multiply 25.35 by 15.

25.35 → multiplicand has two decimal places.
× 15
12675
+25350
380.25 → product has two decimal places.

So, 25.35 × 15 = 380.25

III. Multiplication of decimal by decimal

To multiply a decimal by a decimal, follow these steps.

L et us multiply 5.653 by 4.5

• Multiply both decimals with out decimal points.

• Mark the decimal point in such a way that number of digits after decimal in
product is equal to the total number of digits after decimal in multiplier and
multiplicand.

5.653 → 3 decimals places
× 4.5 → 1 decimal place
.28265
+226120
25.4385 → 4 decimal places

Oasis School Mathematics – 7 201

14.7 Division of Decimals

I. Division of Decimal by the numbers like 10,100, 1000, etc.

To divide a decimal by 10, 100, 10000 etc. shift the decimal point in the dividend by
as many places to the left as there are zeros in the divisor.

For example: My quick idea !
basItnyehdp1a0v2l0eefsatttonepwdsshhsioillfenetftoddnewivc.hiimdilieanlgdpivboiyidnitn1g01
3.56 ÷ 10

= 356 ÷ 10
100

= 356 × 1 = 356
100 10 1000

= 0.356 (The decimal point is moved to 1 place to the left)

3.56 ÷ 100 = 356 ÷ 100 = 356 × 1
100 100 100

= 356 = 0.0356 (The decimal point is moved 2 places to the left.)
10000


II. To divide a decimal by a whole number, follow these steps.

• Perform the division, considering the dividend a whole number.

• After the completion of division of whole number part put decimal on the
quotient.

• Follow the same method to divide the decimal part also.

Let us divide 98.37 by 12.

12 98.3700 8.1975 Steps:
-96
23 • Convert divisor into the whole number.
-12 • follow the process of division of decimal
117
-108 number by whole number.
90
-84 75 1.575 0.021
60 -150
-60 75
–75
0 ×

(III) Division of Decimal by Decimal

To divide a decimal by another decimal .

Let us divide, 0.001575 by 0.075.

Now 0.001575 ÷ 0.075 = 0.001575 × 1000 = 1.575
0.075 × 1000 75

So, 0 .001575 ÷ 0.075 = 0.021

202 Oasis School Mathematics – 7

Worked Out Examples

Example: 1

If the cost of 1 metre cloth is Rs. 45.35, what is the cost of 3.5 metres of cloth?
Solution:

Cost of 1 metre cloth = Rs. 45.35

Cost of 3.5 metres of cloth = Rs. 45.35 × 3.5

Now, 45.35
× 3.5

22675
13605
158.725

Therefore, cost of 3.5 m cloth is Rs. 158.725

Example: 2

If the cost of a dozen note books is Rs. 225, what is the cost of one note book.

Solution:

Cost of 1 dozen (12) note books = Rs. 225 12 225 18.75
∴ Cost of 1 note book = Rs. 225 ÷ 12 -12
Now, 105
So, 225 ÷ 12 = 18.75 -96
90
Therefore, cost of one note book is Rs. 18.75 -84
60
-60
0

Note: Simplification of decimals is also performed according to the BODMAS rule.

Example: 3

Simplify: 2.6 ÷ [9.4 – {4.9 – (3.2 ÷ 1.6 – (4.8 – 3.5 )}]

Solution:
2.6 ÷ [9.4 – {4.9 – (3.2 ÷ 1.6 – 4.8 – 3.5 )}]

= 2.6 ÷ [9.4 – {4.9 – {3.2 ÷ 1.6 – 1.3)}]

= 2.6 ÷ [9.4 – {4.9 – (2 – 1.3)}]

= 2.6 ÷ [9.4 – {4.9 – 0.7}]

= 2.6 ÷ [9.4 – 4.2]

= 2.6 ÷ 5.2 = 2.6 = 26 = 1 = 0.5
5.2 52 2

Note: We can also simplify the above problem by first converting the decimals into
fractions and then proceeding with simplification of fractions.

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Exercise 14.6

1. Find the products in each of the following:

(a) (i) 5.864 × 10 (ii) 5.864 × 100 (iii) 5.864 × 1000 (iv) 5.864 × 10000

(b) (i) 4.0045 × 10 (ii) 4.0045 × 100 (iii) 4.0045 × 1000 (iv) 4.0045 × 10000

(c) (i) 0.9562 × 10 (ii) 0.9562 × 100 (iii) 0.9562 × 1000 (iv) 0.9562 × 10000

2. Find the products of the following.

(a) 4.5 × 9 (b) 15.65 × 12 (c) 25.05 × 25

(d) 8.36 × 2.5 (e) 0.03 × 0.9 (f) 0.025 × 0.4

3. Find the quotients of the following.

(a) (i) 345.6 ÷ 10 (ii) 345.6 ÷ 100 (iii) 345.6 ÷ 1000

(b) (i) 3.5 ÷ 10 (ii) 3.5 ÷ 100 (iii) 3.5 ÷ 1000

(c) (i) 0.5 ÷ 10 (ii) 0.5 ÷ 100 (iii) 0.5 ÷ 1000

4. Carry out the following (division).

(a) 0.8 ÷ 4 (b) 1.625 ÷ 5 (c) 4.230 ÷ 15

(d) 8.64 ÷ 0.24 (e) 3.48 ÷ 0.003 (f) 216 ÷ 0.6

5. Simplify the following:

(a) 3.5 × 4 + 0.65 (b) 16.6 – 4.5 × 3.5

(c) (2.5 ÷ 0.05) × 0.575 (d) {10 ÷ (3.7 – 1.4 + 0.7 )}

(e) 3.2 ÷ {1.8 + (3 ÷ 1.5 + 0.6 – 0.4 )}

6. (a) If rice costs Rs. 35.50 per kg, what is the cost of 15 kg rice?
(b) If the cost of one note book is Rs. 15.75, what is the cost of 12 such note books?

7. (a) 1 cm = 10 mm. How many millimetres are there in

(i) 4.5 cm (ii) 3.8 cm

(b) 10 mm = 1 cm. How many centimetres are there in

(i) 8 mm (ii) 5mm

8. (a) 1 m = 100 cm. How many centimetres are there in

(i) 4.5 m (ii) 3.57 m

(b) 100 cm = 1 m. How many metres are there in

(i) 8 cm (ii) 3.52 cm

204 Oasis School Mathematics – 7

9. (a) 1 km = 1000 m. How many metres are there in

(i) 1.5 km (ii) 3.52 km

(b) 1000 m = 1 km. How many kilometres are there in

(i) 5 m (ii) 45 m

10. (a) A ribbon 3.84 m long is cut into 4 equal pieces. What is the length of each piece?

(b) The cost of 27 kg of potatoes is Rs. 44.55. What is the cost of 1 kg of potatoes ?

11. If 15.6 km of road is graveled in 12 days. How much length of the road is graveled
in 1 day?

12. (a) A rectangular garden is 25.50 m long and 15.35 m wide.

(i) Find its perimeter (ii) Find its area

(b) The area of rectangular field is 625.25 m² and it is 41 m long.

(i) Find its breadth (ii) Find its perimeter.



Answer

1. (a) (i) 58.64 (ii) 586.4 (iii) 5864 (iv) 58640 (b) (i) 40.045 (ii) 400.45

(ii) 4004.5 (iv) 40045 (c) (i) 9.562 (ii) 95.62 (iii) 956.2 (iv) 9562

2. (a) 40.5 (b) 187.8 (c) 626.25 (d) 20.9 (e) 0.027 (f) 0.01

3. (a) (i) 34.56 (ii) 3.456 (iii) 0.3456 (b) (i) 0.35 (ii) 0.035 (iii) 0.0035

(c) (i) 0.05 (ii) 0.005 (iii) 0.0005 4. (a) 0.2 (b) 0.325 (c) 0.282

(d) 36 (e) 1160 (f) 360 5. (a) 14.65

(b) 0.85 (c) 28.75 (d) 6.25 (e) 0.8 6. (a) Rs. 532.5

(b) Rs. 189 7. (a) (i) 45 mm (ii) 38 mm (b) (i) 0.8 cm (ii) 0.5 cm

8. (a) (i) 450 cm (ii) 357 cm (b) (i) 0.08 m (ii) 0.0352 m 9. (a) (i) 1500 m (ii) 3520 m

(b) (i) 0.005 km (ii) 0.045 km 10. (a) 0.96 m (b) Rs. 1.65 11. 1.3 km

12. (a) (i) 81.7 m (ii) 391.425 m² (b) (i) 15.25 m (ii) 112.5 m

Oasis School Mathematics – 7 205

Objective Questions

Choose the correct alternatives.
1. A number multiplied by itself gives 121, then the number is:

(i) square of 121 (ii) square root of 121 (iii) cube of 121.

2. L.C.M. of two prime numbers is :

(i) their product (ii) 1 (iii) their sum.

3. H.C.F. of two prime numbers is :

(i) their product (ii) their sum (iii) 1

4. 11112 is equal to (ii) 30 (iii) 8
(i) 15

5. Which one of the following may not be an integer:
(i) sum of two integers (ii) product of two integers (iii) quotient of two integers.

6. The least number which should be multiplied to 48 to make it a perfect square is:

(i) 4 (ii) 1 (iii) 3

7. 0.563 × 100 is equal to

(i) 56.9 (ii) 5.63 (iii) 0.00563

8. Which one of the following number is irrational?

(i) 4 (ii) 2 (iii) 9

9. Which one of the following statement is not true?
(i) All the rational numbers are integers.
(ii) All the integers are rational numbers.
(iii) All the natural numbers are whole numbers.
10. Which one of the given statements is not true?
(i) Quotient of two integers with the same sign is positive.
(ii) Quotient of two integers with the opposite sign is negative.
(iii) Product of two negative integers is negative.

11. H.C.F. and L.C.M. of two numbers are 6 and 120 respectively. If one number is 24,
then the another number is

(i) 30 (ii) 24 (iii) 60

12. The smallest number which should be multiplication to 27 to make a perfect
square is

(i) 9 (ii) 27 (iii) 3

206 Oasis School Mathematics – 7

Assessment Test Paper

Attempt all the questions. Full marks : 50

Group 'A' [12 × 1 = 12]

1. (a) Find the square of 15

(b) Find the cube root of 27

2. (a) Find the H.C.F. of 8 and 12.

(b) If 'a' and 'b' be any two prime numbers, what is their L.C.M.?

3. (a) Convert 1002 into decimal number.
(b) Convert 165 into decimal number.
4. (a) Write all the integers between –2 and + 5.

(b) Evaluate : (–3) × (+2) × (– 5)

5. (a) Reduce 72 into its lowest term. (b) Subtract : 5 – 2
90 7 7

6. (a) Divide : 72.3 ÷ 10 (b) Multiply : 2.8146 × 100

Group 'B' [13 × 2 = 26]

7. (a) Find the square root of 1225.
(b) By which number, 625 be divided to make it a perfect cube?
8. (a) Find the H.C.F. of : 48, 72 and 96
(b) Find the least number which is exactly divisible by 12, 18, 24.
9. (a) Convert 58 into binary number.

(b) Convert 168 into quinary number.

10. (a) Using number line, add (+6) + (– 3).

(b) Using number line, multiply (+3) × (+ 4)

11. (a) Simplify : 13 – [9 – {16 + 4 – 2}] (b) Add : 172 + 312 ÷ 65.

12. (a) Subtract : 613 – 521.

(b) Multiply : 0.89 × 14.6. (c) Divide : 312 – 121.
Group 'C' [3 × 4 = 12]

13. Simplify : 120 ÷ [17 – {15 – 3 (8 – 2)}]

14. Simplify: 5 + [112 of {2 + {5 –112 )}] 2
5
15. An exercise has 50 questions. A student solved of the questions. Find how many

questions are unsolved?

Oasis School Mathematics – 7 207

Unit Ratio and
15 Proportion

15.1 Ratio

L et us suppose that marks obtained by Lakpa in Mathematics is 80 and the marks
obtained by Dorje is 60.

Let's compare their marks,

Lakpa got (80-60) = 20 marks more than Dorje.

Let's compare their marks by division.

Marks obtained by Lakpa = 80 = 80 = 4 = 4:3
Marks obtained by Dorje 60 60 3

Hence the ratio is the comparison of two quantities having same unit by division.
We use the symbol : to express ratio.

Write Say
a:b 'a' is to 'b'

or

'a' to 'b'

Terms of ratio:

In the ratio 4 : 5

4 is antecedent and 5 is consequent

The first term of the ratio is antecedent and the second term is consequent.

Let us discuss more about the ratio from the following example.

Shila and Ramila are friends in class seven. Their details are given below:

Name Weight Height Ages Marks obtained in Maths
Shila 32 kg 1.4 m 13 years 70%
Ramila 36 kg 1.3 m 13 years 60%

The ratio of their weight = 32:36 = 8:9

The ratio of their height = 1.4:1.3 = 14:13

The ratio of their ages = 13:13 = 1:1

The ratio of their marks in maths = 70:60 = 7:6

208 Oasis School Mathematics – 7

But we cannot find the ratio of Shila's height and weight because they are different

in units of measurement.

So, if a and b are any two numbers in the same units, then the lowest form of a or a:b
b
is called ratio of a and b and the ratio does not have any unit. It is just comparison

like ratio of length and breadth of a play ground which is 2:1. It means that if the

length of the play ground is 20m then breadth will be 10m and if the length is 40m

then the breadth will be 20m and so on.

We can use the ratio to compare two numbers on the same units, to divide certain
sum for given group in the given ratio etc. These are clarified from the worked out
examples given below.

Worked Out Examples

Example: 1

The ratio of two numbers is 7:10. If the consequent is 40, find the antecedent.
Solution:

We have, CAonnteseceqdueenntt = 7
10

Antec4e0dent = 7
10

or, Antecedent = 7 × 40 = 28
10

∴ The antecedent is 28.

Example: 2

Find the ratio of height of a boy and the length of his foot, if the length of foot of 1.4 m
tall boy is 25 cm.
Solution:

Here, height of a boy 1.4 m and size of his foot 25 cm are in different units.

Height = 1.4 × 100 cm = 140 cm, Size of foot = 25 cm

So, the ratio of height: size of foot = 140 cm = 28 = 28:5
25 cm 5

Example: 3

In 200 m race the ratio of distance covered by Asha Kaji and Dawa is 9:10. When Dawa
completes the race, what is the distance covered by AshaKaji?

Solution:

The ratio of distance covered by Asha Kaji and Dawa is 9
10

Oasis School Mathematics – 7 209

Suppose Asha kaji covers x m when Dawa completes 200m then,

9 = x or, x = 9 × 200
10 200
10

∴ x = 180

∴ Asha kaji completes only 180 m when Dawa completes the race.

Example: 4

Divide Rs. 7200 between two workers A and B in the ratio of 7:8.

Solution:
Here, Rs. 7200 is to be divided in the ratio of 7:8, so let the two parts be 7x and 8x for a
constant number 'x' then we have,

7x + 8x = 7200

or, 15 x = 7200

or, x = 7200 [Dividing both sides by 15]
15

or, x = 480

∴ A receives Rs. 7x = 7 × 480 = Rs. 3360

∴ B receives Rs. 8x = 8 × 480 = Rs. 3840

Example: 5

Two numbers are in the ratio 1:3, when 2 is subtracted to each number. The new ratio
becomes 2:7. Find the numbers.

Solution:

Let, the required numbers be x and 3x.

If 2 is subtracted from each of them, they become x–2 and 3x–2.

From the given condition, 3xx––22 = 2
7

or, 7x – 14 = 6x – 4 [By cross multiplication]

or, 7x – 6x = 14 – 4

or, x = 10

∴ The numbers are x = 10

and 3x = 3 × 10 = 30

210 Oasis School Mathematics – 7

Exercise 15.1

1. Find the ratios in their lowest form:

(a) 20 kg and 60 kg (b) 80 cm and 70 cm

(c) 3.4 kg and 500 gm (d) 400 litres and 320 litres

2. (a) The antecedent of a ratio 2:3 is 21, find its consequent.

(b) The consequent of a ratio 5:8 is 24, find its antecedent.

3. (a) The ratio of ages of a brother and sister is 3:5, find the age of the sister if the
brother is 15 years.

(b) Jenusha and Subashna spent money in the ratio 3:4. If Jenusha spent Rs. 300,
how much did Subashna spend?

(c) Kastuv is 36 years old. The ratio of age of Kastuv and his daughter is 6:1. Find
the age of his daughter.

(d) A piece of land is 25 m long and its length and breadth are in the ratio of 5:3. Find
its breadth, perimeter and area.

4. (a) In a class there are 40 students out of which 25 are girls. Find the ratio of (i)
boys and girls (ii) girls and boys (iii) total students and boys (iv) girls and total
students.

(b) A room is 4m long, 5m high and 3m broad. Find the ratio of (i) length and breadth
of the room (ii) breadth and height of the room (iii) perimeter and height of the
room. (iv) can you find the ratio of area and perimeter of floor of the room?

5. (a) Divide 80 into two parts in the ratio of 3:5.

(b) Two brothers agree to divide Rs. 4000 in the ratio 3:1. Find their shares.

(c) Find two complementary angles in the ratio 2:3.

(d) The pair of supplementary angles are in the ratio 3:7. Find them.

6. (a) The ratio of two numbers are 3:4. If 5 is subtracted from each the ratio becomes
5:7. Find the numbers.

(b) The ratio of two numbers are 4:7. When 4 is added to each of term, it will be 5:8.
Find the numbers.

(c) The ratio of present age of Shyam and Bhagawat is 3:4. Five years hence, the ratio
of their ages will be 4:5. Find their present ages.

(d) The age of two sisters is 3:2. After 6 years their ages will be in the ratio 5:4. Find their
ages.

Oasis School Mathematics – 7 211

Answer (b) 8:7 (c) 34:5 (d) 5:4 2. (a) 31.5 (b) 15
1. (a) 1:3

3. (a) 25 years (b) Rs. 400 (c) 6 years (d) 15m, 80m, 375m² 4. (a)(i) 3:5 (ii) 5:3

(iii) 8:3 (iv) 5:8 (b) (i) 4:3 (ii) 3:5 (iii) 14:5 (iv) No.

5. (a) 30 and 50 (b) Rs. 3000, Rs. 1000 (c) 36° and 54° (d) 54° and 126°

6. (a) 30, 40 (b) 16, 28 (c) 15, 20 (d) 9 and 6 years.

15.2 Proportion 10 1
20 2
Let us consider two ratios 10:20 and 4:8, the lowest form of 10:20 = =

and the lowest form of 4:8 = 4 = 1
8 2

So, the lowest form of the ratio 10:20 and 4:8 are the same i.e. 10:20 = 4:8 is called

proportion and 10, 20, 4 and 8 are called proportional numbers.

The equality of two ratios is called proportion. Let, a, b, c, and d are numbers such
that a:b = c:d then a, b, c and d are called proportional numbers.

b and c are called mean of proportions and a and d are called extremes of the
proportion.

We can write a:b = c:d as a:b::c:d Remember!

or, a = c If a, b, c, d are in proportion,
b d a = first proportional
b = second proportional
If a, b, c, d are proportional number then ad = cb c = third proportional
d = fourth proportional
(Product of extremes = Product of means)

because, a = c
b d

or, ad = bc

By the help of proportion we can solve the problem like cost of goods and respective
quantity working days and number of workers etc. We can be clear from the
following examples:

(a) Cost of 30 pencils is Rs. 120. Then, the cost of 40 pencils will be Rs. 160.

Here, the ratio of pencils = 30 : 40 = 3:4 and their respective cost is 120:160 = 3:4 so,

we can write 30 = 116200 .
40

Further more let us look at another example.

(b) A provision, completed by 30 people in 120 days, takes 40 days by 90 people.

Here the ratio of people = 30:90 = 1:3 and the ratio of days 120:40 = 3:1 so, 30 ≠ 14200
90
(not equal) but when we take reciprocal of one of the ratios.

212 Oasis School Mathematics – 7

i.e. 40:120 instead of 120:40 then 30 = 40 .
90 120

So, 30, 90, 40 and 120 are proportional numbers.

In the above example the number involving in 'a' are directly equal (proportional)
so, the example a is the example of direct proportion and the number involving in 'b'
are not directly equal but inversely equal so, it is the case of inversely proportional.

From above two examples, it is clear that these are two types of proportion: direct
and inverse.

Worked Out Examples

Example: 1

Check whether the ratios are in proportion or not

a) 2, 5, 8, 20 b) 4, 9, 25, 30

Solution:
(a) Here, product of extreme terms = 2 × 20 = 40
Production of mean terms = 5 × 8 = 40
Since the product of mean terms is equal to the product of extreme, 2, 5, 8, 20 are
in proportion.

(b) Here, product of extreme terms = 4 × 30 = 120

Product of mean terms = 9 × 25 = 225

Here, product of mean terms is not equal to the product of extreme terms.

4, 9, 25, 30 are not in proportion.

Example: 2

If 3, 6 and 9 are proportional numbers, find their fourth proportion.
Solution:

Let the fourth number be x then,

3, 6, 9 and x are in proportion

so, 3 = 9
6 x

or, 3x = 6 × 9 (by cross product)

or, x = 6×9 = 2 × 9 = 18
3

∴ The fourth proportional is 18.

Oasis School Mathematics – 7 213

Example: 3

If a, 4 and 9 are in proportion, find a.

Solution: a = 4
4 9

or, a = 4×4 = 16
9 9

Note: If a, b, c are three numbers in proportion then a = b where b is called mean
proportional between a and c. b c

Example: 4

If 3, x, 8, 12 are in proportion, find the value of x.

Solution:

3, x, 8, 12 are in proportion,

So, 3 = 8
x 12

or, 8x = 3 × 12
or,
∴ 3 × 12 3 9
82 2
x = =
= 92

Example: 5

If the cost of 4 kg of apples is Rs. 400,find the cost of 9 kg of apples.

Solution:

To solve these types of problems by proportion method, first we have to identify the
variables, they are amount of apples and their cost.

Quantity of apples Cost
4 kg Rs. 400

9 kg x

Let us suppose the cost of 9 kg of apples by Rs. x. Here amount of apples and their cost

are in direct variation so we can write

4 = 400
or, 9 x

4x = 400 × 9

or, x = 400 × 9 = 100 × 9 = 900
4

∴ The cost of 9 kg of apples is Rs. 900.

214 Oasis School Mathematics – 7

Example: 6

20 people have provision for 30 days. For how many days is it enough for 15 people?

Solution:

Here the variables are days and people, so,

People Days
20 30

15 x

Let us suppose 15 people have provision for x days then people and days are in

inverse proportion. So, 20 = x
15 30

or, 20 × 30 = 15 × x

or, 600 = x
or, ∴ 15
x = 40

∴ 15 people have the provision for 40 days.

Exercise 15.2

1. Identify whether the following numbers are in proportion or not.

(a) 4, 8, 5, 10 (b) 7, 5, 35, 25

(c) 1, 8, 2, 10 (d) 12, 6, 18, 12

2. Find the value of 'a' from the following proportions.

(a) 6, 9, 4, a (b) 36, a, 16, 20 (c) a, 18, 50,100

(d) 15, 75, a,155 (e) a, 9, 30, 27

3. The following terms are in proportion, find the first proportional.

(a) 3, 8, 12 (b) 2, 8, 16 (c) 9, 16, 48 (d) 5, 15, 30

4. The following terms are in proportion, find the second proportional.

(a) 4, 8, 12 (b) 5, 80, 48 (c) 11, 7, 28 (d) 2, 9, 6

5. Find the third proportional number of the following terms

(a) 6, 14, 35 (b) 5, 9, 45 (c) 16, 8, 12

6. Find the fourth proportional in each of the following cases

(a) 6, 8, 9 (b) 25, 50, 7 (c) 40, 60, 30

Oasis School Mathematics – 7 215

7. Identify which of the following quantities are in direct proportion and which are
indirect proportion?

(a) Quantity of apples and their corresponding cost.
(b) Rate of a book and total cost of books.
(c) Number of teachers and their total salary.
(d) Number of workers and working days.
(e) Provision and days.
(f) Speed of a boy and time taken to cover certain distance.

8. Find the value of x in each of the following cases:

(a) M en Days (b) Quantity of Rice (kg) Cost (Rs)
20 30 50 10000
40 x 30 x

(c) Dis tance (k m) Time (hrs) (d) Provision (in kg) Days
80 4
50 5
120 x 30 x

9. (a) If the cost of 8 plates of Mo:Mo is Rs. 400, what is the total cost of 19 plates of
Mo:Mo?

(b) Cost of 20 pens is Rs. 700. What is the cost of 8 pens.

(c) Rs. 1125 is the cost of 25 kg of rice. What is the cost of 40 kg of rice?

(d) Cost of 10 packets of tea is Rs. 1600. How many packets can be bought for Rs. 3200?

10. (a) 20 students of a hostel have provision for 30 days. For how many days will
(b) 30 students use that provision?

16 people can finish a work in 32 days. When will 24 people complete it?

(c) A piece of work can be finished in 15 days working 8 hours a day. In how
many days will the work have been finished working 6 hours a day?

(d) 16 boys can do as much work as 20 men in the same time. How many boys
can do as much as 60 men?

(e) A garrison of 200 men have provisions for 40 days. For how many men will

the provisions last for 50 days?

Answer

1. a, b 2. (a) 6 (b) 45 (c) 9 (d) 31 (e) 10
5
3. (a) 2 (b) 1 (c) 3 (d) 2 4. (a) 6 (b) 3 (c) 44
4 5. (a) 15 (b) 25 6. (a) 12 (b) 14 (c) 45
(d) 3 (c) direct (c) 24 (f) inverse
(b) direct
7. (a) direct (d) inverse (e) direct

8. (a) 15 (b) Rs. 6,000 (c) 6 hr. (d) 3 days 9. (a) Rs. 950 (b) Rs. 280 (c) Rs. 1800
(d) 20 1 (e) 160 men.
10. (a) 20 days (b) 21 3 days (c) 20 days (d) 48 boys

216 Oasis School Mathematics – 7

Unit

16 Percentage

Introduction

Percent is composed by two words per and cent. Per means out of and cent means
hundred. Therefore percent means per hundred or out of hundred.

It is denoted by % .

10% means 10 out of hundred.

60% students passed in an examination means 60 students in every hundred passed
in the examination. Some operations related to percentage.

Conversion of percentage into fraction or decimal: To convert the percentage into
fraction remove the symbol % and divide by 100 and simplify the fraction into its
lowest term.

Percentage Fraction Decimal

30% 30 0.3
100

= 3
10

Conversation of fraction or decimals into percentage:

To convert fraction or decimal into percentage multiply it by 100 and put the symbol
of percentage.

Fraction Percentage Decimal Percentage

2 2 × 100% 0.3 0.3 × 100%

5 5 = 30%
= 40%

Calculation of the value given percentage of any number:

Let a number be x. If we have to find the value of 5% of x.

then, 5% of x = 5 × x.
100

25% of 1000 = 25 × 1000 = 250
100

i.e. if we multiply the fraction of given percentage with the given number, then we
will get the required value.

Oasis School Mathematics – 7 217

Worked Out Examples

Example: 1

(a) Convert 43 and 0.17 into percentage.
Solution:
Here, 34 = 34 × 100%
= 75%

Again, 0.17 = 0.17 × 100% = 17%

Example: 2

(b) Convert 6 1 % into fraction. Calculate the value of 20% of 1700.
4
Solution:

(a) Here, 6 1 % = 25 % (b) Here, 20% of 1700
4 4
20
25 = 100 × 1700
= 340
= 4
100

= 25 × 1 = 1 .
4 100 16

Example: 3

In a class, 1 of the total students failed in Mathematics. Find the failed and passed
4

percentage.

Solution: 1
Here, 14 4
= ×100%

= 25 %

∴ 25% students failed in the examination.

Now, passed percentage = (100 – 25)%

= 75 %

Example: 4

In an election, 60% people in the voter list cast their votes, find the fraction of people
who didn't cast their votes.

Solution:

Percentage of people who cast their votes = 60%

= 60 = 3
100 5

( ) 3 2
Fraction of people who didn't cast their votes = 1– 5 = 5

218 Oasis School Mathematics – 7

Exercise 16.1

1. Convert the following fractions and decimals into percentage.

(a) 35 (b) 7 (c) 6 (d) 0.37 (e) 0.125 (f) 0.63
10 25
2. Express the following percentage into fraction and reduce them into lowest term.

(a) 65% (b) 80% (c) 3% (d) 0.5% (e) 5.5%

3. Find the value of each of the following.

(a) 20% of 1000. (b) 3% of 3000 (c) 1.5% of 4000

(d) 80% of 60 marks (e) 20% of Rs. 850 (f) 2% of 2 m.

4. (a) What percent of 60 is 12? (b) Express 8 as the percentage of 80.

(c) What percent of 80 marks is 60? (d) Express Rs. 18 as the percentage of Rs. 360.

(e) What percent of 950 is 190? (f) What percent of 5m is 30 cm?

5. Find the value whose:

(a) 12% is 60 (b) 20% is 320 (c) 80% is 480 (d) 1.5% is 10 cm.

6. (a) Find the value of Rs. 300 after the increment of 20%.

(b) What is the value of 10m after the decrement by 20%?

(c) What is the value of Rs. 9000 after the increment by 15%?

7. (a) 52 of the total students in a school are boys. Find the percentage of boys and girls.

(b) 1 part of a road is graveled and the rest is pitched. Find the percentage of pitched
4
road.

(c) A man spends 7 part of his income. What is his percentage of saving?
25
8. (a) Age of 40% people in a village is more than 40 years. What fraction of people are
less than 40 years?

(b) In a school, 55% students passed their examination. What fraction of students failed
in the examination?

Answer

1. (a) 60% (b) 70% (c) 24% (d) 37% (e) 12.5% (f) 63%

2. (a) 13 (b) 4 (c) 3 (d) 1 (e) 11 (f) 4 cm
20 5 100 200 200 (f) 6%

3. (a) 200 (b) 90 (c) 60 (d) 48 mark (e) Rs. 170

4. (a) 20% (b) 10% (c) 75% (d) 5% (e) 20%

5. (a) 500 (b) 1600 (c) 600 (d) 666.67 cm

6. (a) Rs. 360 (b) 8 m (c) Rs. 10,350 3 9
5 20
7. (a) Boys → 40%, Girls → 60% (b) 75 % (c) 72% 8. (a) (b)

Oasis School Mathematics – 7 219

Conversion of one quantity as the percentage of others:

If the units of two quantities are the same, then one quantity can be expressed as the
percentage of others.

If the first quantity is to be expressed as the percentage of second, the ratio of first
to the second quantity is to be multiplied by hundred.

If x is to be expressed as the percentage of y.

then required percent is x ×100% = 100x %
y y

e.g. what percent of 80 is 16?

Here, required percentage = 16 ×100% = 20%
80

Calculation of a quantity whose value of certain percentage is given:

To calculate this, we have to assume whole quantity as 'x'.

For example:

If 20% of a number is 35, find the number.

Let the number be x.

then, 20% of x = 35

20 × x = 35
100 x = 175

∴ Required number is 175.

Worked Out Examples

Example: 1

There are 300 students in a school. If 60% of them are boys, find the number of boys
and girls in the school.

Solution:

Here, Total students = 300

Number of boys = 60% of 300 .

= 60 × 300
100

= 180

Number of girls = (300 – 180) = 120

Example: 2

Cost of sugar increased from Rs. 40 per kg to Rs. 50 per kg. Find the percentage of
increment.

Solution:

220 Oasis School Mathematics – 7

Here, Original cost = Rs. 40 per kg.

Cost after increment = Rs. 50 per kg

Increment in the cost = Rs. (50–40) = Rs. 10

We have, = Increment in the cost × 100% = 10 ×100% = 25%
Percentage of increment Original cost 40

Example: 3

In a school, there are 175 girls and 65% boys. Find the total number of students and the
number of boys.

Solution:

Here, Number of girls = 175.

Let, The total number of students = x.

From the given condition,

Number of boys = 65% of x.

then, Number of girls = (100 – 65)% of x.

= 35% of x.

= 35 ×x.
100

∴ = 7x .
or, 20
or,
or, 7x = 175.
20

7x = 175×20

x = 175 × 20
7

x = 500

Total students = 500.

Number of boys = (500 – 175) = 325



Exercise 16.2

1. (a) There are 640 students in a school. If 40% students are girls, find the number of
boys.

(b) In a class of 80 students, 20% students are absent. Find the number of present and
absent students.

(c) Among 180 students appeared in an examination, 90% students passed. Find the
number of failed students.

2. (a) The cost of sugar increased from Rs. 50 per kg to Rs. 60 per kg. Find the percentage
increase in the price.

Oasis School Mathematics – 7 221

(b) A man's salary has increased from Rs. 4000 to Rs. 4400 per month. Find the
percentage of increment in the salary.

(c) The price of an article decreased from Rs. 200 to Rs. 150. Find the percentage
decrease in the price.

3. (a) Monthly income of a man is Rs. 32000. He spends 20% of his income on food, 30%
on cloths and 10% on miscellaneous. Find his expenses in each item and his saving
per month.

(b) A mixture contains 50% gold, 20% copper and remaining silver. Find the mass of
each in 1000 gm of mixture.

4. (a) After spending 80% of his income, a man has Rs. 1200 left with him. Find his
monthly income.

(b) Zenith spends 60% of his income and saves Rs. 1600. Find his income and his
expenditure.

(c) 60% of the total students in a school are girls. If there are 112 boys, find

(i) total students (ii) number of girls.

Answer (b) Present - 64, absent - 16 (c) 18 2. (a) 20% (b) 10% (c) 25%
1. (a) 384

3. (a) Food - Rs. 6400, cloths - Rs. 9600, Miscellaneous - Rs. 3200, Saving - Rs. 12,800

(b) Gold - 500 gm, copper - 200 gm, silver - 300 gm

4. (a) Rs. 6000 (b) Rs. 4000 and Rs. 2400 (c) (i) 280 (ii) 168

Self Practice Material

Interpret parts of a whole as fraction, decimal and percentage
Objective : To interpret parts of a whole as fraction, decimal and percentage
Material : Paper, pencil, Scale
Procedure : (i) Make the following figure on the paper and colour the shaded partion.
(ii) Write value of shaded parts in the form of fraction, decimal and
percentage.

(i) (ii) (iii)

Figure
(i) Fraction Decimal Percentage
(ii) …… 0. …… …… %
(iii) …… 0. …… …… %
…… 0. …… …… %



222 Oasis School Mathematics – 7

Unit

17 Unitary Method

Introduction

Cost of 10 pens is Rs. 150. Cost of 1 pen is less than

What is the cost of 1 pen ? the cost of 10 pens.
150
Cost of 1 pens is Rs. 10 = Rs. 15.

If the cost of 1 pen is Rs. 15 what is the cost of 18 pens? Cost of 18 pens is more

Cost of 18 pens = Rs. 18 × 15 = Rs. 270. than the cost of 1 pen.

Hence, the method of finding unit value from the value of given quantities and
finding the value of given quantities with the help of unit value is the unitary
method.

Look at the following examples:

(i) If the cost of 10 pens is Rs. 50. What is the cost of 6 such pens?

(ii) 8 men can do a piece of work in 5 days. When will 6 men finish the work?

In above two examples, clearly we see two variations.

In example (i) the cost of pens is less when less number of pens are there, so it is a
direct variation.

In direct variation, unit values are obtained by division.

In example (ii) when the number of men increase then the time gets less. So, it is
indirect variation.

In indirect variation
• Unit value is obtained by multiplication.

Worked Out Examples

Example: 1

The cost of 1 pen is Rs. 35, what is the cost of 12 pens.
Solution:

Cost of 1 pen is Rs. 35

Cost of 12 pens is Rs. 35 × 12

= Rs. 420

Oasis School Mathematics – 7 223

Example: 2

The cost of 10 m of clothes is Rs. 500. Find the cost of 8 m of clothes.

Solution:

The cost of 10m of clothes = Rs. 500

The cost of 1m of cloth = Rs 500 = Rs. 50
10

∴ The cost of 8m of clothes = Rs. 50 × 8 = Rs. 400

Example: 3

12 man can a piece of work in 30 days. In how many days 18 men can do the same work?

Solution:

12 men can do a work in 30 days.

1 man can do a work in 30 × 12 days.

18 men can do a work idnay3s0=1×812301×d8a1y2s. = 20 days.
∴ Required number of

Exercise 17.1

1. (a) Cost of a copy is Rs. 50. What is the cost of 12 such copies?
(b) The cost of 1 m cloth is Rs. 250. What is the cost of 8 m cltohs?

2. (a) The cost of 20 pens is Rs. 1200, what is the cost of one pen?
(b) 30 kg of rice costs Rs. 1200, what is the cost of 1 kg rice?
3. (a) 20 men can do a work in 5 days, in how many days can 1 man finish the same

work?
(b) 20 students of a hostel had provisions for 12 days, how long would the provi-

sion last if there was 1 student?
4. (a) The cost of 15 copies is Rs. 840.
(i) Find the cost of 1 copy.
(ii) Find the cost of 10 copies.
(b) 10 m of cloth costs Rs. 800. Find the cost of 12m of the same cloth.
(c) The cost of 20 pens is Rs. 1000. How many pens can be bought at Rs. 1200?
5. (a) A piece of work can be done by 24 men in 15 days. In how many days can 18

men do the same work?

224 Oasis School Mathematics – 7

(b) 24 masons can build a wall in 20 days. How many masons should be added to
build the wall in 12 days?

(c) 40 workers can build a wall in 60 days. How many workers can build the same
wall in 20 days?

6. (a) A group of 80 people had provisions for 30 days. If 20 more people joined the
group, how long would the provisions last?

(b) 50 students had provisions for 60 days. If there were 10 students less, how long
would the provisions last?

(c) 75 students of a hostel had provisions for 80 days. How many students should
leave the hostel so that the provisions would be enough for 100 days?

Answer (b) Rs. 2000 2. (a) Rs. 60 (b) Rs. 40


1. (a) Rs. 600

3. (a) 100 days (b) 240 days 4. (a) (i) Rs. 56 (ii) Rs. 560 (b) Rs. 960

(c) 24 pens 5. (a) 20 days (b) 16 masons (c) 120 workers

6. (a) 24 days (b) 75 days (c) 15 students

Oasis School Mathematics – 7 225

Unit

18 Simple Interest

Introduction

When money is lent, the lender charges extra money to the borrower for the benefit
of using that money.

The amount of money lent is called the principal. Charge imposed on it is called the
interest. The sum of the principal and the interest is called amount. Interest calcu-
lated on Rs. 100 for one year is the rate percent per annum.

Calculation of simple interest:

Let Principal = P

Time = T years.

Rate = R% per annum.

Interest = I Remember !

Amount = A If the interest of Rs. 100
in 1 year is Rs. 15 then
Then, by the definition of rate percent per annum, the rate percent is 15%
p.a.
Rate is the interest on Rs. 100 in 1 year.

i.e. Interest on Rs. 100 in 1 year = Rs. R.

Interest on Re. 1 in 1 year = Rs. R
100

I nterest on Rs. P in 1 year = Rs. 1R00 × P

Interest on Rs. P in T years = Rs. 1R00 × P × T

∴ Interest (I) = P ×T× R .............. (i)
100

From equation (i),

P × T × R = 100 × I

P = 100 × I .............. (ii)
T×R

R = 1P00××TI .............. (iii)

T = 1P00××RI .............. (iv)

226 Oasis School Mathematics – 7

Again, amount (A) = P + I

or, P = A–I

or, I = A–P

Note: If the time is given in month or days, we have to convert it in year, using the

relation. 1 day = 1 year. 1 month= 1 year.
365 12

Worked Out Examples

Example: 1

Find the simple interest on:

(a) Rs. 2000 at the rate of 5% p.a. for 2 years.

(b) Rs. 5000 at the rate of 10% p.a. for 73 days.

(c) Rs. 4500 at the rate of 12% p.a. for 6 months.
Solution:

(a) Here, Principal (P) = Rs. 2000
Rate (R) = 5% p.a.
Time (T) = 2 years.

Simple interest (I) = ?

We have, I = P ×T× R
100

= Rs. 2000×2×5
100

= Rs. 200

(b) Here, Principal (P) = Rs. 5000

Rate (R) = 10% p.a.

Time (T) = 73 days.

= 73 year = 1 year
365 5

Interest (I) = ?
We have,
I = P ×T× R
100
1
Rs. 5000 × 5 × 10
100 = Rs.100.
=

Oasis School Mathematics – 7 227

(c) Principal (P) = Rs. 4500

Rate (R) = 12% p.a.
61

Time (T) = 6 months = 12 year = 2 year.
Interest (I) = ?

We have,

I = P×T×R
=
100 × 1 × 12
4500 2
Example: 2
Rs. 100 = Rs. 270

Find the simple interest and amount on Rs. 6000 for 3 years at the rate of 15% p.a.
Solution:

Here, Principal (P) = Rs. 6000

Time (T) = 3 years

Rate (R) = 15% p.a.

Interest (I) = ?

Amount (A) = ?

We have, I = P×T×R
100

Again, = Rs. 6000 ×3 × 15 = Rs. 2700
100

Amount (A) = P + I

= Rs. 6000 + Rs. 2700

= Rs. 8700

Example: 3

At what rate of interest Rs. 5000 amounts to Rs. 6500 in 3 years?
Solution:

Here, Principal (P) = Rs. 5000

Amount (A) = Rs. 6500

Time (T) = 3 years

Rate (R) = ?

We have, I = A–P

= Rs. 6500 – Rs. 5000 = Rs. 1500

228 Oasis School Mathematics – 7

Again, we have, R = 100 × I
P×T

= 100 × 1500 = 10 %
Example: 4 5000 × 3

At what time will the interest on Rs. 500 is Rs. 75 at the rate of 5% p.a.

Solution:

Here, Principal (P) = Rs. 500

Interest (I) = Rs. 75

Rate (R) = 5% p.a.

Time (T) = ?

We have, T = 100 × I
P×R

= 100 × 75 = 3 years.
Example: 5 500 × 5

Find the sum that will produce an interest Rs. 630 in 3 years at the rate of 10% p.a.

Solution:

Here, Principal (P) = ?

Interest (I) = Rs. 630

Time (T) = 3 years.

Rate (R) = 10% p.a.

We have, P = 100 × I
T×R

= 100 × 16030
3×
Exercise 18.1
= Rs. 2100

1. Find the simple interest and amount on:

(a) P = Rs. 800, T = 2 years, R = 5% p.a.

(b) P = Rs. 1260, T = 6 months, R = 10% p.a.

(c) P = Rs. 200, T = 73 days, R = 12 % p.a.

(d) P = Rs. 3000, T = 1 years 6 months, R = 10% p.a.

(e) P = Rs. 2500, T = 3 years, R = 12% p.a.

Oasis School Mathematics – 7 229

2. Find the principal in each of the following cases.
(a) T = 2 years, R = 10% p.a., I = Rs. 450
(b) T = 3 years, R = 10% p.a., I = Rs. 630
(c) T = 4 years, R = 12% p.a., I = Rs. 480
(d) T = 5 years, R = 7% p.a., I = Rs. 875

3. Find the rate of interest in the following cases.
(a) P = Rs. 200, I = Rs. 48, T = 8 years
(b) P = Rs. 5000, I = Rs. 1500, T = 3 years
(c) P = Rs. 6000, I = Rs. 1200, T = 2 years
4. Find the time in each of the following cases.
(a) P = Rs. 500, R = 5% p.a., I = Rs. 50
(b) P = Rs. 1500, R = 10% p.a., I = Rs. 450
(c) P = Rs. 2000, R = 12% p.a., I = Rs. 480
5. (a) At what rate of simple interest, Rs. 3600 amounts to Rs. 5328 in 4 years?
(b) At what rate of simple interest, Rs. 3000 amounts to Rs. 3360 in 3 years?
6. (a) In how many years, Rs. 800 amounts to Rs. 1200 at 5% p.a. simple interest?
(b) In how many years, Rs. 1000 amounts to Rs. 1300 at 10%p.a. simple interest?
7. (a) A man deposited Rs. 4200 in a bank at the rate of 12% p.a., find the interest and

the amount at the end of 3 years.
(b) Himanka deposited a sum of money in a bank at the rate of 12% p.a. simple

interest. At the end of 2 years, he received an interest Rs. 540. Find the
sum.
(c) Elvis lent Rs. 1250 to Hrit. At the end of 1 year Hrit paid Rs. 75 as simple interest.
Find the rate of interest.
(d) A man deposited Rs. 1200 in a bank at the rate of 5% p.a.. After how many years
his sum amounts to Rs. 1500?



Answer

1. (a) I = Rs. 80, A = Rs. 880 (b) Rs. 63, Rs. 1323 (c) Rs. 4.80, Rs. 204.80

(d) Rs. 450, Rs. 3450 2. (a) Rs. 2250 (b) Rs. 2100

(c) Rs. 1000 (d) Rs. 2500 3. (a) 3 % (b) 10% (c) 10% 4. (a) 2 yrs.

(b) 3 yrs. (c) 2 yrs. 5. (a) 12% (b) 4% 6. (a) 10 yrs. (b) 3 yrs.

7. (a) Rs. 1512, Rs. 5712 (b) Rs. 2250 (c) 6 % (d) 5 yrs.

230 Oasis School Mathematics – 7

Unit

19 Profit and Loss

Introduction

We have already learnt about profit and loss in previous class. Some basic terms
which are used in profit and loss are cost price, selling price, profit, loss etc. These
terms are used while buying and selling goods. Cost price is the price for which an
article is purchased. It is denoted by C.P. Selling price is the price at which an article
is sold. It is denoted by S.P.

If an article costing Rs. 1500 is sold for Rs. 1800 there is the profit of Rs. (1800–1500) = Rs. 300

If the same article is sold for Rs. 1300 there is loss of Rs. (1500 – 1300) = Rs. 200

From these examples, it is clear that, if S.P. > C.P. there is profit.

∴ Profit = S.P. – C.P.

C.P. = S.P. – Profit.

S.P. = C.P + Profit.

Again, if C.P > S.P., there is loss.

∴ Loss = C.P. – S.P.

C.P. = S.P. + loss

S.P = C.P. – loss.

19.1 Profit and Loss Percent

Profit percent means profit per–hundred on the investment. If C.P. = Rs. 100 and
S.P. = Rs. 102, there is a profit of Rs. 2 on investing Rs. 100.

∴ There is a profit of 2%.

Loss percent means loss per hundred on the investment.

If C.P. = Rs. 100, S.P. = Rs. 99.

Here is a loss of Re. 1 on investing Rs. 100.

∴ There is loss of 1%.

Since the invested sum is cost price, profit and loss percent is always calculated on
the cost price.

Oasis School Mathematics – 7 231

Look at one more example,

If an article costing Rs. 400 is sold at a loss of Rs. 20. Find loss percent.

If C.P. is 400, loss is Rs. 20.

If C.P. is 1, loss is Rs. 20 .
400

If C.P. is 100, loss is Rs. 20 × 100
400

= Rs. 5

∴ Loss percent = 5%

We can use following formula to find profit or loss percent

Profit percent = Profit × 100% = S.P. – C.P. × 100%
C.P. C.P.

Loss percent = Loss × 100% = C.P. – S.P. × 100%
C.P. C.P.

Worked Out Examples

Example: 1

Find the profit or loss percent in the following cases.

(a) Cost price (C.P.) = Rs. 540, Profit = Rs. 27

(b) Cost price (C.P.) = Rs. 630, Selling price (S.P.) = Rs. 600.
Solution:

(a) Cost price (C.P.) = Rs. 540

Profit = Rs. 27

We have, Profit percent = Profit × 100%
C.P.
∴
= 27 × 100%.
540

Profit percent = 5%

(b) Here, Cost price (C.P.) = Rs. 630

Selling price (S.P.) = Rs. 600

Since, C.P.> S.P. there is loss.

We have, Loss = C.P. – S.P.

= Rs. 630 – Rs. 600

= Rs. 30

232 Oasis School Mathematics – 7

We have, Loss percent = Loss × 100%
C.P.

= 30 × 100%
630

= 4.76%

Example: 2

An article is bought at Rs. 950 and sold at Rs. 1140. Find the profit or loss percent.
Solution:

Here, Cost price (C.P.) = Rs. 950

Selling price (S.P.) = Rs. 1140

We have, Profit = S.P. – C.P.

= Rs. 1140 – Rs. 950 = Rs. 190

Again we have, Profit percent = Profit × 100%
C.P.

= 190 × 100% = 20%
Example: 3 950

A fruit seller bought 100 oranges for Rs. 250 . 20 of them were rotten and he sold the
remaining at the rate of Rs. 3 per orange. Find the profit or loss percent.

Solution:

Here, C.P. of 100 oranges = Rs. 250

Number of rotten oranges = 20

Remaining oranges = (100 – 20 )

= 80

S.P. of 80 oranges = Rs. 80 × 3

= Rs. 240

Here, C. P. > S. P.

∴ Loss = C.P. – S.P.

= 250 – 240

= Rs. 10
We have,
Loss percent = Loss × 100%
C.P.

= 10 × 100 %
250

= 4%

Oasis School Mathematics – 7 233

Exercise 19.1

1. Find profit or loss in the following cases:

(a) Cost price = Rs. 500, Selling price = Rs. 550

(b) Cost price = Rs. 800, Selling price = Rs. 640

2. (a) If Cost price = Rs. 640, Profit = Rs. 80, find S.P.

(b) If Cost price = Rs. 300, Loss = Rs. 40, find S.P.

(c) If S.P. = Rs. 1020, Loss = Rs. 40, find C.P.

(d) If S.P. = Rs. 860, Profit = Rs. 120, find C.P.

3. (a) If C.P. = Rs. 1200 and Profit = Rs. 24, find profit percent.

(b) If C.P. = Rs. 480 and Loss = 120, find loss percent.

(c) If S.P. = Rs. 330 and profit = Rs. 30, find profit percent.

(d) If S.P. = Rs. 630 and loss = Rs. 70, find loss percent.

4. Find profit or loss percent in each of the following cases:

(a) C.P. = Rs. 500, S.P. = Rs. 450 (b) C.P. = Rs. 550, S.P. = Rs. 440

(c) C.P. = Rs. 360, S.P. = Rs. 432 (d) C.P. = Rs. 1450, S.P. = Rs. 1595.

5. (a) An article is purchased at Rs. 350 and sold for Rs. 420. Find the profit
percent.

(b) A chair is bought for Rs. 600 and sold for Rs. 500. Find the loss percent.

6. (a) Semima bought 5 dozens of oranges for Rs. 300 and sold them at Rs. 3.50 each.
Find profit or loss.

(b) A shopkeeper bought 90 eggs for Rs. 450. 10 of them were broken. He sold the
remaining at the rate of Rs. 6 each. Find gain or loss.

7. (a) A man bought 20 mangoes for Rs. 90 and sold them at the profit of Rs. 45. At
what rate did he sell the mangoes?

(b) A shopkeeper bought 100 eggs and 10 of them were rotten. He sold the remaining
eggs at Rs. 2.50 each and made a profit of Rs. 25. At what rate did he buy them?

8. (a) Samyog bought 5 dozens of pen for Rs. 240 and sold them at Rs. 5 per piece, find
his profit percent.

(b) A man bought 30 soaps for Rs. 600 and sold all of them for Rs. 75 for 3 soaps.
Find his profit percent.

(c) A man bought 5 tables and 5 chairs altogether for Rs. 1500. He sold the table at
the rate of Rs. 150 per piece and chair at the rate of Rs. 100 per piece. Find his
profit or loss percent.

234 Oasis School Mathematics – 7

9. (a) A fruit seller bought 300 apples for Rs. 2 each. He found 50 of them were rotten and
sold the remaining apples at the rate of Rs. 3 each. Find his profit or loss percent.

(b) Bidhya bought 200 pencils for Rs. 6 each and 20 of them were stolen. If she sells
remaining pencils for Rs. 6.50 each. Find her profit or loss percent.

(c) A man bought 6 dozens of eggs at Rs. 5 each, 1 dozen of them were damaged and
he sold the remaining at the rate of Rs. 5.50 each. Find his profit or loss percent.

(d) A milkman bought 50 litres of milk for Rs. 30 per litre. 10 litres of milk was lost

by leakage and sold the remaining quantity for Rs. 40 per litre. Find his profit or

loss percent.

Answer 

1. (a) Profit Rs. 50 (b) Loss Rs. 160 2. (a) Rs.720 (b) Rs. 260 (c) Rs. 1060

(d) Rs.740 3. (a) 2% (b) 25 % (c) 10% (d) 10%

4. (a) Loss = 10% (b) Loss = 20% (c) Profit = 20% (d) Profit = 10%

5. (a) 20% (b) 16.67 % 6. (a) loss = Rs. 90 (b) gain: Rs. 30

7. (a) Rs. 6.75 each (b) Rs. 2 each 8. (a) 25% (b) 25%(c) Loss 16.67 %

9. (a) Profit: 25% (b) Loss 2.5 % (c) Loss 8.33% (d) Profit 6.67%

(i) S.P. when C.P. and profit or loss percent are given

As we know that profit or loss percent are always calculated on C.P. Therefore S.P.
is calculated by calculating profit or loss and adding or subtracting from C.P.

If profit percent is given:

Profit = Profit percent of C.P.

then, S.P. = C.P. + Profit.

∴ S.P. = C.P. + Profit percent of C.P.

If loss percent is given.

Loss = Loss percent of C.P.

then, S.P. = C.P. – Loss

∴ S.P. = C.P. – Loss percent of C.P.

Alternatively,

S.P. can be calculated assuming C.P. as 100.

If profit percent is given,

C.P. = 100, S.P. = 100 + P%

If loss percent is given,

C.P. = 100, S.P. = 100 – L%

Oasis School Mathematics – 7 235

For example,

If C.P. = Rs. 450, Profit = 20%, S.P. = ?

In this case,

When C.P. is Rs. 100, S.P. is Rs. 120 [∵ If C.P. = 100, S.P. = 100 + P%]

When C.P. is Re. 1, S.P. is Rs. 120
100

When C.P. is Rs. 450, S.P. is Rs. 120 × 450
100

∴ S.P. = Rs. 120 × 450 = Rs. 540
100

S.P. can also be calculated by using formula,

S.P. = C.P.× (100 + P%)
100

Where, P % = profit percent.

If loss percent is given,

S.P. = C.P. × (100 – L%)
100

(ii) To find out C.P., when S.P. and profit or loss percent are given:

As profit or loss percent is always calculated on C.P.

Therefore, assume C.P. as x.

If profit percent is given,
Profit = Profit percent of C.P.
= Profit percent of x.

then, S.P. = C.P. + Profit

and solve for x.
Again, if loss percent is given,

then, Loss = Loss percent of C.P.

= Loss percent of x.

Then, use S.P. = C.P. –Loss

and solve for x.

Alternative method:

C.P. can also be calculated by assuming CP = 100

If Profit percent is given.

C.P. = 100, S.P. = 100 + P%

If loss percent is given.

C.P. = 100, S.P. = 100 – L%
And use unitary method to find C.P.

236 Oasis School Mathematics – 7

For example:

If S.P. = Rs. 270, loss = 10%, C.P.= ?

If S.P. is Rs.90,C.P is Rs. 100 [ ∵ If C.P. = 100, S.P. = 100 – L%]

If S.P. is 1, C.P. is Rs. 100 . Remember !
90
100
If S.P. is Rs. 270, C.P. is Rs. 90 × 270 If profit percent is given.

= Rs. 300 C.P. = 100 × S.P.
100 + P%

C.P. can also be calculated by using the formulae. If loss percent is given,

C.P. = 100 × S.P.
100 – L%
Worked Out Examples

Example: 1

Find the selling price of an article whose C.P. is Rs. 1200 and profit is 30%.
Solution:

Here, Cost price (C.P.) = Rs. 1200.

Profit percent = 30%

Selling price = ?

We have, S.P. = C.P. + Profit.

= C.P. + Profit percent of C.P.

= Rs. 1200 + 30% of Rs. 1200

= Rs. 1200 + Rs. 30 × 1200
100

= Rs. 1200 + Rs. 360 = Rs. 1560

Alternative Method: I Alternative method: II

Here, C.P. = Rs. 1200 Here, C.P. = Rs. 1200.

Profit Percent = 30% Profit percent = 30%

S.P. = ? S.P. = ?

W hen C.P. is Rs. 100, S.P. is Rs. 130. We have, S.P. = Rs. C.P. (100 + P%)
100
130
When C.P. is Rs.1, SP. is Rs. 100 = Rs. 1200(100 + 30)
100
130
W hen C.P. is Rs. 1200, S.P. is Rs. 100 × 120 0. = Rs. 1560

∴ Selling price (S.P.) = Rs. 130 × 1200
100

= Rs. 1560

Oasis School Mathematics – 7 237

Example: 2

A shopkeeper sold a watch at Rs. 900 at a loss of 20%. Find the cost price of the watch.
Solution:

Here, Selling price (S.P) = Rs. 900.

Loss Percent = 20% Alternative method: III
Here, S. P. = Rs. 900
Let, Cost price (C.P.) = ? L% = 20%
Then, C. P. = ?
C.P. = x We have,
S.P. = C.P. – L% of C.P.
loss = 20% of C.P. or, 900 = C.P. – 20% of C.P.

We have, = 20 × x
100

= x
5

S.P. = C.P. – Loss

or, 900 = x – x or, 900 = C. P. × 20 C.P.
5 100
or,
or, 900 = 4x or, 900 = C.P. – C.P.
5 5

900 × 5 = 4x or, 900 = 4C.P.
5
900 × 5
or, x = 4 or, 900 × 5 = 4 C.P.

= Rs. 225 × 5 or, C. P. = 900×5
4
= Rs. 1125
∴ C.P. = Rs. 1125.
∴ Cost price (C.P.) = Rs. 1125

Alternative method: I Alternative method: II

Here, S.P. = Rs. 900 Here, S.P. = Rs. 900,
Loss percent = 20% Loss percent = 20%,
C.P. = ? C.P. = ?

We have, C .P. = Rs.110000 × S.P When S.P. is Rs. 80,
– L%

100 × 900 C .P. is Rs. 100 [If C.P. = 100, S.P. = 100 – L%)
100 – 20
= Rs. 100
80
100 × 900 When S.P. is Re. 1, C.P. is Rs. .
80
= Rs. W hen S.P. is Rs. 900, C.P. is Rs. 100 × 900
80
= Rs. 1125
= Rs. 1125

238 Oasis School Mathematics – 7

Example: 3

A man bought a second hand motorbike for Rs. 50,000 and spend Rs. 6,000 for its main-
tenance. At what price should it be sold to make the profit of 20%?

Solution:

Cost price = Rs. 50,000

Maintenance cost = Rs. 6000

Net C.P. = Rs. 50,000 + Rs. 6000

= Rs. 56,000

Profit % = 20%

We have, S.P. = C.P. + P% of C.P.

= Rs. 56,000 + 20 × 56,000
100

= Rs. 56,000 + 1 × 56,000
5

= Rs. 56,000 + 11,200

= Rs. 67,200.

Exercise 19.2

1. Find S.P. in each of the following cases.

(a) C.P. = Rs. 650, Profit percent = 20%.

(b) C.P. = Rs. 800, Loss percent = 12%

(c) C.P. = Rs. 350, Loss percent = 2%

(d) C.P. = Rs. 760, Profit percent = 5%

2. Find C.P. in each of the following cases.

(a) S.P. = Rs. 2310, Gain percent = 5%. (b) S.P. = Rs. 2610, Loss percent = 10%.

(c) S.P. = Rs. 384, Loss percent = 4%. (d) S.P. = Rs. 110, Profit percent = 10%.

3. (a) A Yangchhen bought a mobile set at Rs. 960 and sold it at 15% profit. Find

(i) profit (ii) selling price.

(b) Anasuya bought a camera at Rs. 6300 and sold it at a loss of 9%, find.

(i) amount of loss. (ii) selling price of the camera.

Oasis School Mathematics – 7 239

4. (a) Aryan purchased a mobile set for Rs. 4500 and spent Rs. 500 to repair it. If he
sold at 20% loss, find its selling price.

(b) A television was bought for Rs. 13,500 and spent Rs. 1500 to repair it. At what
price should it be sold in order to gain 20%?

5. (a) Kritika sold a watch for Rs. 1504 at a loss of 6%. Find its cost price.

(b) A man sold a radio at Rs. 4200 at a profit of 5%. Find its cost price.

(c) A Prastuti sold a calculator at Rs. 360 at a gain of 20%. At what price did he
purchase the calculator?



Answer

1. (a) Rs. 780 (b) Rs. 704 (c) Rs. 343 (d) Rs. 798 2. (a) Rs. 2200

(b) Rs. 2900 (c) Rs. 400 (d) Rs. 100 3. (a) Rs. 144, Rs. 1104

(b) Rs. 567, Rs. 5733 4. (a) Rs. 4000 (b) Rs. 18000

5. (a) Rs. 1600 (b)Rs. 4000 (c) Rs. 300

Objective Questions

Choose the correct alternatives.

1. 60% of x = 12, then the value of x is

(i) 20 (ii) 15 (iii) 18

2. If S.P. is less than C.P. by Rs. 20, then

(i) cost price is Rs. 20

(ii) profit is Rs. 20

(iii) loss is Rs. 20

3. If the interest of Rs. 100 in 1 year is Rs. 10 then

(i) the rate of interest is 10% p.a.

(ii) the principal is Rs. 10.

(iii) the total interest is Rs. 10.

240 Oasis School Mathematics – 7

4. Fourth proportional of 12, 16, 21, is

(i) 28 (ii) 14 (iii) 8

5. 3 students of a school are girls. Then the percentage of boys in the school is
5

(i) 60% (ii) 40% (iii) 50%

6. The cost of 10m cloths is Rs. 2000 then the cost of 2 m cloths is

(i) Rs. 200 (ii) Rs. 600 (iii) Rs. 400

7. If 'P', 'T', 'R' and 'I' be the principal, time rate and interest then which of the follow-

ing relation is not true?

(i) I = PTR (ii) T = P×I×R (iii) T = 100 I
100 100 PR

8. Which one of the following relation is not true?

(i) CP = 100×SP (ii) CP = 100×SP (iii) CP = 100×SP
100+P% 100–L% 100+L%

Assessment Test Paper

Attempt all the questions. Full marks : 20

Group 'A' [8 × 2 = 16]

1. (a) Convert 15% into decimal.

(b) If a:b = 4:7, b:c = 14:17, find a : c.

2. (a) The cost of 12 articles is Rs. 720, what is the cost of an article?

(b) If C.P. = Rs. 500, profit = Rs. 45, find S.P.

3. (a) In a class of 60 students, 60% are girls, find the number of girls.

(b) The cost of 15 apples is Rs. 300. Find the cost of 5 apples.

4. (a) If C.P. = Rs. 400, loss % = 10%, find S.P.

(b) Find the simple interest on Rs. 400 for 2 years at the rate of 10% per annum.

Group 'B' [1 × 4 = 4]

5. An article is sold at Rs. 400 at the loss of 20%, find its cost price. At which price
should it be sold to gain 20%?

Oasis School Mathematics – 7 241

Statistics

10Estimated Teaching Hours

Contents

• Collection of Data
• Frequency Distribution Table
• Bar Graph
• Arithmetic Mean
• Median
• Mode

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:
• Prepare the frequency distribution table from the

given data
• Prepare the cumulative frequency distribution table

from the given data
• Draw the simple bar graph and multiple bar graph

from the given data
• Calculate the arithmetic mean from the individual

and discrete data

Teaching Materials

• Graph sheet

242 Oasis School Mathematics – 7

Unit

20 Statistics

20.1 Review

Definition of Statistical Data
Statistics is the branch of mathematics in which information in terms of numerical

values is collected for the purpose of investigation. The numerical values, so
collected are called statistical data. The collected data are used to compute for
various statistical purposes by the statisticians.

Data Collection
The teacher decided to take class VII students in an educational tour to Bhaktapur

Durbar square. He told the students to find out the following things.

• The number of temples.
• The average number of flow of visitors in a day.
• The number of statues.
The students collected the information and handed over to the teacher. The
teacher finally said that the information that they had gathered was simply
a data.

Frequency
If the marks obtained by 10 students are given as:
1, 5, 2, 3, 6, 2, 4, 10, 3, 2
The number of times a variable occurs is called the frequency of the variable

denoted by 'f'.
The frequency of the marks 2 is 3, since there are three students who obtained 2

marks. In the same way we can find the frequency of 1 is 1, frequency of 4 is 1, etc.

Frequency Distribution
Presentation of data in the form of frequency table along with the respective

variables is called frequency distribution. There are mainly two ways to construct
frequency distribution table which are.

Oasis School Mathematics – 7 243

(i) Discrete (Ungrouped) frequency distribution table.
(ii) Continuous (grouped) frequency distribution table.

Discrete (Ungrouped) Frequency Distribution Table
Let us consider the weight in kg of 20 students of class VII as follows.
40, 39, 35, 42, 45, 43, 42, 41, 40, 45, 35, 38, 42, 43, 39, 39, 38, 37, 40, 40

Note: Arranging the given data in ascending or descending order is called an array.
Now, in order to represent the above data in the table we use tally marks (/)

Weight (in Tally marks Frequency Remember ! 1
kg) 2
// 2 I represents 3
35 /1 II represents 4
// 2 III represents 5
37 /// 3 IIII represents 10
IIII represents
38 IIII IIII represents

39

40 //// 4

41 / 1

42 /// 3

43 // 2

45 // 2

Total 20

Such representation as shown above is called discrete (Ungrouped) frequency
distribution.

Continuous (grouped) Frequency Distribution

If the collected numbers of data are enough and the range of data is also too high,
then we prefer to dense them by presenting into different groups. This way of
presentation is shown as below.

Suppose the marks of 20 students of class VII are as follows.

1, 19, 14, 3, 4, 20, 8, 12, 3, 9, 8, 17, 8, 1, 5, 13, 15, 11, 2, 14

Now changing above all data of marks in the class difference of 4, we can select the
classes as 0 – 4, 4 – 8, 8 – 12, 12 – 16 and 16 – 20.

244 Oasis School Mathematics – 7