Oasis Math 6

6. Solve the following problems:
(a) A rope of length 3.72 m is cut into 4 equal parts, find the length of each part.
(b) The cost of 5 pens is Rs 235.50, what is the cost of one pen?
(c) How many pieces of 0.75cm can be cut from the length of 3.75cm?

Answers (b) 0.3 (c) 0.154 (d) 3.24 (e) 42.64
(b) 0.43 (c) 0.182 (d) 3.048 (e) 6.3742
1. (a) 2.5 (b) 0.023 (c) 0.426 (d) 0.0182 (e) 0.3745
2. (a) 0.03 (b) 0.05 (c) 0.42 (d) 0.4 (e) 0.091
3. (a) 0.007 (b) 1.7 (c) 36 (d) 0.37 (e) 74
4. (a) 0.21
5. (a) 1.6 (b) Rs 47.10 (c) 5
6. (a) 0.93m

4.9 Rounding off Decimal Numbers

Rounding off decimal numbers
Rounding off means replacement of a number by another convenient number
which is easy to understand and is close to the original number.

We round off the decimal numbers nearest to ones, tenths, hundredths and so
on.

Rounding off to the nearest ones (whole number)
Lets be clear with the help of given example;

• See the digit in the tenths place.

• If it is less than 5 (1, 2, 3, 4) keep ones place same and make
tenth place '0'

• If it is more (or equal) to 5 (5, 6, 7, 8, 9). Add 1 on the ones place and
make tenth place 0.

Round off to the nearest ones:
13.6
Solution:

13.6
Digit in the tenths place = 6, which is greater than 5. So we have to add 1 in ones
place and replace the tenths place by 0.
\ 13.6 to the nearest ones = 14.

Oasis School Mathematics – 6 95

Again, look at one more example:
18.3
Digit in the tenths place = 3 which is less than 5. Digit in the ones place
remains same.
183. to the nearest ones = 18.
Rounding off to the nearest tenths (one decimal place)
For rounding off to the nearest tenths,

• See the digit in the hundredths place.

• If the digit in the hundredths place is less than 5 (1, 2, 3, 4), tenths
place remains same and hundredths place is replaced by ‘0’.

• If the digit in the hundredths place is 5 or more than 5, 1 is to be
added in the tenths place and hundredths place is replaced by ‘0’.

Round off to the nearest tenths (one decimal place).
4.62
Solution:
In 4.62, digit in the hundredths place = 2
which is less than 5
So we have to keep tenths place same, replace hundredths place by zero.
\ 4.62 to the nearest tenths = 4.6
Again, lets discuss one more example:
Round off 13.76 to the nearest tenths.
Solution: Here, digit in the hundredths place = 6 which is greater than 5.
So, we have to add 1 in tenths place and replace hundredths place yb 0.
Hence, 13.76 to its nearest tenths = 13.8.
Rounding off to the nearest hundredths (2 decimal place)
For rounding off to the nearest hundredths, we have to see the digit in thou-
sandths place.
Round off to the nearest hundredths (2 decimal places).
15.846

Solution: Here, in 15.846, digit in thousandths place = 6.
Which is greater than 5.
\ 15.846 to the nearest hundredths = 15.85
Again, look at one more example.

96 Oasis School Mathematics – 6

Round off 32.352 to its nearest hundredths.
Solution: Here, digit in the thousandths place = 2 which is less than 5.
Hence, hundredths place remains same and thousandths place is 0.
Hence, 32.352 to its nearest thousandths = 32.35.

Exercise 4.9

1. Round off each of the following to the nearest ones. (whole number)

(a) 8.5 (b) 3.6 (c) 3.3 (d) 14.2 (e) 12.8

(f) 18.62 (g) 132.32 (h) 156.81

2. Round off each of the following to the nearest 1 decimal place (tenths).

(a) 3.57 (b) 3.85 (c) 0.33 (d) 4.09 (e) 6.72

(f) 86.12 (g) 932.65 (h) 57.18

3. Round off each of the following to the nearest 2 decimal places (hundredth).

(a) 5.583 (b) 3.956 (c) 6.769 (d) 35.102 (e) 47.008
(f) 245.386 (g) 36.8291 (h) 384.3721

4. Round off each of the following to the nearest 3 decimal places (thousandth).

(a) 4.5467 (b) 6.0072 (c) 3.3408 (d) 24.6702 (e) 3.4586
(f) 5.0055
(g) 19.62154 (h) 264.5732

5. Round off each of the following to the nearest decimal places given in brackets.

(a) 6.576 (1) (b) 7.954 (2) (c) 8.9517 (3) (d) 9.543 (1)
(e) 0.00582 (2) (h) 92.32145 (3)
(f) 6.324169 (3) (g) 0.05858 (3)

Answers

1. (a) 9 (b) 4 (c) 3 (d) 14 (e) 13 (f) 19 (g) 132 (h) 157

2. (a) 3.6 (b) 3.9 (c) 0.3 (d) 4.1 (e) 6.7 (f) 86.1 (g) 932.7 (h) 57.2
3. (a) 5.58 (b) 3.96 (c) 6.77 (d) 35.10 (e) 47.01 (f) 245.39 (g) 36.83 (h) 384.37

4. (a) 4.547 (b) 6.007 (c) 3.341 (d) 24.670 (e) 3.459 (f) 5.006 (g) 19.622 (h) 264.573

5. (a) 6.6 (b) 7.95 (c) 8.952 (d) 9.5 (e) 0.01 (f) 6.324 (g) 0.059 (h) 92.321

Oasis School Mathematics – 6 97

Objective Questions

1. If the numerator of two fractions are the same, then the fraction having
smaller denominator is :

(i) the greater one (ii) the smaller one

(iii) equal to the fraction with greater denominator

2. The value of 1 + 1 + 1 is equal to :
2 3

(i) 1 (ii) 11 (iii) 5
6 6 6

3. 0.0214 × 1000 is equal to :

(i) 2.14 (ii) 0.214 (iii) 21.4

4. Rounding off 13.642 to its nearest tenths is:

(i) 14 (ii) 13.64 (iii) 13.6
(iii) 3215.6
5. 32.156 ÷ 100 is equal to:

(i) 3.2156 (ii) 0.32156

6. Rounding off 15.83 to its nearest ones is :

(i) 15.8 (ii) 16 (iii) 15
(iii) 0.06
7. 6-hundredths is equal to:

(i) 0.6 (ii) 600

8. 0.1234 × 100 is equal to

(i) 1.234 (ii) 12.34 (iii) 123.4

9. ( 1 + 3 ) ÷ ( 1 – 1 ) is equal to
3 4 2 3

(i) 6 1 (ii) 1 (iii) 1 1
2 6 12
3
10. 5 ÷ 20 is equal to

(i) 3 (ii) 12 (iii) 3
100 10
6
11. 3 is equal to

5 2 2
5 5
(i) (ii) 10 (iii)

12. Rounding of 2.3756 to its nearest hundredth is

(i) 2.38 (ii) 2.4 (iii) 2.376

98 Oasis School Mathematics – 6

Assessment Test Paper Full marks: 30

Group "A" [5 × 1 = 5]

1. (a) Add : 3 + 2
7 7

(b) Multiply : 2 × 3
3 5

(b) Divide : 7 ÷ 1
10 5

2. (a) Convert 0.6 into fraction.

(b) Convert 53 into decimal.
100

(c) Multiply: 3.5 × 10

Group "B" [5 × 2 = 10]

3. (a) Subtract: 4 1 – 2 1
6 3

(b) Simplify: 1 ÷ 1 × 3
5 2 4

4. (a) Multiply: 0.256 × 10

(b) Divide : 3.476
1000

(c) Multiply: 0.45 × 0.6
Group "C" [5 × 3 = 12]

5. (a) Weight of 3 bags are 2 1 kg, 3 1 kg and 6 1 kg, what is the total weight
4 2 3

of 3 bags?

(b) A man bought 8 kg of rice at Rs.354 1 . Find the cost of 1 kg rice.
4
2 1
(c) Simplify: 4 – (2 3 –1 3 ).

(d) Cost of 1 kg apple is Rs. 75.15. What is the cost of 9kg apples?

(e) Divide: 37.8 ÷ 0.18

Oasis School Mathematics – 6 99

Commercial Arithmetic

15Estimated Teaching Hours

Contents
• Percentage
• Profit and Loss
• Unitary Method
Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:
• To understand the relation among percentage, fraction

and decimal and to convert one from the other
• To convert one quantity as the percentage of the other
• To calculate profit, loss, profit percent and loss percent

from the given C.P. and S.P.
Teaching Materials
Graph sheet, A4 size paper, chart paper, coloured pencil, etc.

100 Oasis School Mathematics – 6

Unit

5 Percentage

5.1 Introduction Per Cent
For every Hundred
Percent is composed of two words per and cent. Per
means for every and cent means out of hundred. Therefore
percent means per hundred or out of hundred.

It is denoted by %.

I got I got
marks. marks.

Zenith got 90 marks Zenisha got 80 marks
out of every 100. out of every 100.

Relation between percent and fraction

Activity:
• Take rectangular sheet of paper

• Draw a square and divide it into 100 equal
parts.

• Then, discuss what part of fraction is shaded.
What decimal part of given figure is shaded.

• If 25 part out of 100 part is shaded, what is the
percentage shaded part?

• Find the relation of fraction and decimal with
percentage.

Method of converting percent into fraction

As we know that, 15% = 15 , 25% = 25
100 100

Oasis School Mathematics – 6 101

Hence, to covert percent into fraction
• Remove the symbol %
• Divide it by 100

• Simplify the fraction into its lowest term.

Conversion of percentage into decimals:
Hence, to convert percent into decimal,

• Convert percent into fraction

• Convert fraction into decimal.

Let's see the given example

85% = 85 Converting percentage into fraction
100

= 0.85 Converting fraction into decimal

Method of converting fraction into percentage

Let's see the given example:
iAT.ehs.iws12ie0s50kpno×osws1i0bt0lhea=to,n21l52y0%50wh=en251%205.0 is multiplied by 100.
Hence, to convert percent into decimal,

• Multiply given fraction by 100.

• Put the symbol %.

Method of converting decimal into percentage

Let's see the given example and get the idea of conversion of decimal into percentage.
56
0.56 = 100 = 56%

It is possible only when 0.56 is multiplied by 100.

Hence, to covert decimal into percentage

• Multiply the decimal by 100

• Put the symbol%

To find the value of percentage of given quantity:

Let's study the given example and get the idea to find 50% of Rs 200

Rs. 50% means 50 and 'of' means '×'
100

50% of 200 = 50 × 200 = 100
100

102 Oasis School Mathematics – 6

• Express the given percentage into the fraction.

• Multiply the fraction by the given quantity.
For example:

Find the value of 45% of 900.

Here, 45% of 900 = 45 × 900
100

= 45 × 9 = 405.

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 = 180
100

Number of girls = (300 – 180) = 120

Example: 2
A man earns Rs. 12,000 in a week. He spends 40% of his income on food and
30% on miscellaneous. Find

i) how much does he spend on food?

ii) how much does he spend on miscellaneous?

iii) how much does he save in a week?

Solution:

Weekly income of a man = Rs. 12,000

His expenditure on food = 40% of 12,000

= 30 × 12,000
100

= 4800

∴ His expenditure on food = Rs. 4800.

His expenditure on miscellaneous = 30% of 12,000

= 30 × 12,000 = 4600
100
His saving in a week = Rs. (12,000 – 4800 – 3600) = Rs. 3600.

Oasis School Mathematics – 6 103

Exercise 5.1

1. Find the fraction, percentage and decimal of shaded part of the given figures
having 100 squares.
(a) (b) (c)

2. (a) 56 students out of 100 are boys. Find the percentage of boys in the class.
(b) Sreeja scored 92 in Mathematics out of 100 full marks. What percentage did she
score?
(c) 64 students out of 100 are passed. Find the passed percentage and failed percentage
of the class.

3. Express the following fractions into percentage.

(a) 1 (b) 7 (c) 1 (d) 23 (e) 12
20 10 4 50 25
4. Express the following decimals into percentage.

(a) 0.2 (b) 0.05 (c) 0.68 (d) 0.315 (e) 0.826

5. Express the following percentage into fractions.

(a) 10% (b) 25% (c) 12.5% (d) 85 % (e) 2.5%

6. Express the following percentage into decimals.

(a) 20% (b) 5% (c) 75% (d) 64% (e) 92%

7. Find the value of :

(a) 25% of Rs. 1500 (b) 50 % of 800 (c) 10% of 240
1
(d) 12 2 % of Rs 600 (e) 12% of 900 students (f) 60% of 170

8. (a) A man spends 80% of his income. If his monthly income is Rs 50,000, find his

monthly expenditure.

(b) There are 50 students in a class. If 30% are boys, find the number of boys.

(c) In a school, there are 600 students. If 10% students are absent, find the number of

absent students.

9. (a) There are 800 students in a class. If 45% are boys, find the number of boys and
girls.

(b) A man earns Rs. 1000 in a month. If he spends 80% of his income, find, how

104 Oasis School Mathematics – 6

much does he save in a month?
(c) Out of 400 students in a school, 20% are bus students and rest are the walkers,

find how many students use the school bus and how many are walkers?.
(d) In a school of 200 students, 10% are absent. How many students are present?

10. (a) Monthly income of a man is Rs.6000. He spends 60% of his income. Find:
(i) his monthly expenditure. (ii) his monthly saving
(iii) his yearly saving.
(b) A man earns Rs.18000 per month and he spends 20% of his income on food and
30% on education. Find;
(i) how much does he spend on food?
(ii) how much does he spend on education?
(iii) how much does he save in a month?

õõõ

Answers

1. Consult your teacher 2. (a) 56% (b) 92%

(c) Passed percent = 64%, Failed percent = 36% 3. (a) 5% (b) 70%

(c) 25% (d) 46% (e) 48% 4. (a) 20% (b) 5%

(c) 68% (d) 31.5% (e) 82.6%

5. (a) 110 (b) 41 (c) 1 (d) 1207 (e) 1
8 40

6. (a) 0.2 (b) 0.05 (c) 0.75 (d) 0.64 (e) 0.92

7. (a) Rs. 375 (b) 400 (c) 24m (d) Rs 75 (e) 108 (f) 102

8. (a) 40,000 (b) 15 (c) Rs. 60 9. (a) Boys – 360,Girls – 450

(b) Rs. 200 (c) Bus students – 80,Walkers–320 (d) 180 10. (a) (i) Rs. 3600

(ii) Rs. 2400 (iii) Rs. 28800 (b) (i) Rs. 3600 (ii) Rs. 5400 (iii) Rs 9000

5.2 Calculation of Percentage

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

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

For example:

What percent of 30 is 6?

Here,

Required percent is 6 × 100% = 20%
30
20% of 30 is 6.

Oasis School Mathematics – 6 105

Worked Out Examples

Example 1

What percent of 10 is 2?

Solution:

Required percentage = 2 × 100% = 20%
10

Example 2

Out of 250 students in a school, 150 are girls. Find the percentage of girls.

Solution:

Here,

Total number students = 250

Number of girls = 150

Percentage of girls = 150 × 100%
250

= 15 × 100% = 15 × 100%20 4
25 25 5

= 15 × 4% = 15 × 4%

= 60%

Example 3

Monthly income of a man is Rs. 5000 and spends Rs. 2000. What percent
does he save?

Solution:

Here,

Monthly income of a man = Rs. 5000.

His monthly expense = Rs. 2000.

His monthly saving = Rs. 5000 – Rs. 2000

= Rs. 3000

Percentage of saving = 3000 × 100%
5000

= 3 × 100%
5

= 60%

106 Oasis School Mathematics – 6

Exercise 5.2

1. What percent of: (b) 600 is 240 ? (c) 3 is 1 ?
(a) 20 is 10 ? (e) Rs 5 is 20 paisa ? (f) 2m is 30cm ?
(d) 1 hour to 15 minutes ?

2. (a) Out of 20 students in a class, 16 were present. What percentage of students were
present?

(b) In a class, there are 20 girls and 10 boys. What is the percentage of the boys?

3. (a) A man earns Rs. 6000 and spends Rs. 3000. What percentage does he save?

(b) Out of 40 students of a class, 10 were absent. What is the percentage of present
students?

4. (a) The number of students in a school increased from 800 to 1000. Find the
percentage of increment.

(b) Rohan's monthly salary was Rs. 15,000. It increased by Rs. 2400. What percent
of salary is increased?

5. (a) In a village, there are 500 men, 300 women and 200 children. What percentage of
the population are (i) men (ii) women (iii) children.

(b) Out of monthly income Rs 24000 a household spends Rs 6000 on food and Rs
8000 on cloths. Find the percentage of expenditure on

(i) food (ii) cloths.

Answers

1. (a) 50% (b) 40% (c) 33 1 % (d) 25% (e) 4% (f) 15% 2. (a) 80% (b) 33 1 %
3 3
3. (a) 50% (b) 75% 4. (a) 25% (b) 16% 5. (a) Men = 50% Women = 30% Children

= 20% (b) Food = 25% Cloths = 33 1 %
3

Project Work

Collect the data of the number of boys and girls of different classes of your
school. Convert that data into percentage and present that in your class.

Oasis School Mathematics – 6 107

Unit

6 Profit and Loss

6.1 Profit and Loss

I bought an article for Rs. 900.
\ Cost price of the article (C.P.) = Rs. 900.
I sold it for Rs. 1000.
\ Selling price of the article (S.P.) = Rs. 1000.
I paid Rs. 900 for an article and I got Rs. 1000 by selling it.

I got the profit of Rs 1000 – Rs 900 = Rs 100

S.P. C.P. Profit
\ Profit = Selling price – Cost price

Profit = S.P. – C.P.

If I had sold it for Rs. 800, i.e.
There would be the loss of Rs 900 – Rs 800 = Rs 100

C.P. S.P. Loss
\ Loss = Cost Price – Selling price

Loss = C.P. – S.P.

Cost price: The amount of money which we pay to buy an article.
Selling price:
Profit : The amount of money at which an article is sold.

Loss : The difference between the selling price and cost price.
\ Profit = S.P – C.P.

T he difference between the cost price and selling price.
\ Loss = C.P. – S.P.

If S.P. > C.P., there is profit. If C.P. > S.P., there is Loss.

Remember !

• Profit = S.P. – C.P.
• C.P. = S.P. – Profit
• S.P. = C.P. + Profit
• Loss = C.P. – S.P.
• C.P. = Loss + S.P.
• S.P = C.P – Loss.

108 Oasis School Mathematics – 6

Worked Out Examples

Example: 1

A shopkeeper bought a camera for Rs. 15,200 and sold it for Rs. 16,000.
Find his profit.

Solution:

Here, Cost price of camera(C.P.) = Rs. 15,200

Selling price of camera (S.P.) = Rs. 16,000

Since, S.P.> C.P. there is a profit.

We have, Profit = SP – CP

= Rs 16000 – 15,200

= Rs 800.

Hence, the required profit = Rs 800.

Example: 2

A man bought a mobile set for Rs. 25,000 and sold it for a profit of Rs. 3700.
Find it SP.

Solution:

Here, Cost price of a mobile set (C.P.)= Rs. 25000

Profit = Rs. 3700

Selling price (S.P.) = ?

We have, Profit = S.P. – C.P.

S.P. = C.P. + Profit

∴ SP = 25000 + 3700 = Rs. 29,200

Hence,t he selling price of the mobile set = Rs. 17500

Example: 3

A shopkeeper sold an article for Rs 5000 at the loss of Rs 550, find the cost
price of the article.

Solution:

Here, Selling price of the article (SP) = 5000

Loss = Rs 550

Cost price of the article (CP) = ?

We have, Loss = CP – SP

550 = CP – 5000

or, CP = 550 + 5000

= Rs 5550.

Oasis School Mathematics – 6 109

Exercise 6.1

1. Identify whether the following statements are true or false.

a) If the cost price of an article is less than the selling price,
b) If CP and SP are equal, then there is profit
c) If the cost price of an article is less than selling price, then there is profit.
2. Find the profit or loss in each case, if

(a) CP = Rs. 1200, SP = Rs. 1000 (b) CP = Rs. 600, SP = Rs. 500

(c) CP = Rs. 800, SP = Rs. 995 (d) CP = Rs. 2500, SP = Rs. 3000

3. Find the selling price, if

(a) CP = Rs. 350, profit = Rs. 160 (b) CP = Rs. 2504, profit = Rs. 406
(c) CP = Rs. 3330, loss = Rs. 670 (d) CP = Rs. 6305, loss = Rs. 695

4. Find the cost price, if

(a) SP = Rs. 500, profit = Rs. 200 (b) SP = Rs. 375, gain = Rs. 75

(c) SP = Rs. 850, loss = Rs. 125 (d) SP = Rs. 650, loss = Rs. 50

5. (a) A shopkeeper buys an article for Rs. 5040 and sells it for Rs. 5160. Find his loss
or gain.

(b) Find the profit or loss when a chair bought for Rs. 325 is sold for Rs. 178.

6. (a) Cost price of an article is Rs. 550 and it is sold at a gain of Rs. 65. Find its selling price.
(b) Find the selling price of an article which is bought for Rs. 840 and sold at a loss of Rs. 45.

7. (a) A table is sold for Rs. 4500 at the gain of Rs. 750. Find its selling price.

(b) Selling price of an induction heater is Rs. 3500. If it is sold at a loss of Rs. 450,
find its cost price.

Answers

1. Consult your teacher.

1. (a) Loss = Rs.200 (b) Loss = Rs.100 (c) Profit = Rs.195 (d) Profit = Rs. 500

3. (a) Rs.510 (b) Rs. 2910 (c) Rs. 2660 (d) Rs. 5610. 4. (a) Rs.300 (b) Rs.300

(c) Rs. 975 (d) Rs. 700. 4. (a) Profit = Rs.120 (b) Loss = Rs. 147

6. (a) Rs.615 (b) Rs.795 7. (a) Rs.3750 (b) Rs.3950.

110 Oasis School Mathematics – 6

6.2 Profit and Loss Percent

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

\ There is the profit of 2%.

Loss percent means loss per hundred on the investment.

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

Here is 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.
Look at one more example

If an article costing Rs. 400 is sold at a loss of Rs. 20, find the loss percent,

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

If C.P. is 1, loss is Rs. 20 .
400
20
If C.P. is 100, loss is Rs. 400 × 100 = Rs. 5

\ Loss percent = 5%

We can use following formula to find profit or loss percent

Profit percent = PCr.oPfi.t× 100% Loss percent = Loss × 100%
C.P.

Worked Out Examples

Example: 1

If Cost price (C.P.) = Rs. 4500, Selling price (S.P.) = Rs. 5000, find profit
or loss percent.

Solution:

(a) Here, C.P. = Rs 4500

S.P. = Rs. 5000

Since S.P. > C.P. there is profit.

profit = S.P. – C.P.

= Rs. 5000 – Rs.4500 = Rs. 500

Oasis School Mathematics – 6 111

We have, Profit × 100% = 500 × 100%
Profit percent = C.P. 4500
1 1
= 9 × 100% = 11 9 %

Example: 2

An article is bought for Rs. 3000 and sold it for Rs. 2400. Find its (i) profit
(ii) profit percent.

Solution: C.P. = Rs. 3000

S.P. = Rs. 2400

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

∴ (i) Loss = C.P. – S.P.

= Rs. 3000 – Rs.2400 = Rs. 600.
Loss

We have, l oss percent = C.P. × 100%

= 600 × 100% = 1 × 100% = 20%
3000 5

Example: 3

A shopkeeper bought one dozen of pens for Rs. 240 and sold for Rs. 25 each.
Find his (i) profit (ii) profit percent.

Solution:

Here, Since, 1 dozen = 12

CP of 12 pens = Rs 240

SP of 1 pen = Rs 25

SP of 12 pens = Rs 25 × 12 = Rs 300.

Since SP > CP, there is profit.

We have, Profit = S.P. – C.P.

= 300 – 240 = Rs. 60

Again, profit percent = Profit × 100%
C.P.

= 60 × 100%
240

= 25%.

112 Oasis School Mathematics – 6

Exercise 6.2

1. Find the profit percent in each of the following cases.

(a) C.P. = Rs. 1200, Profit = Rs. 24. (b) C.P. = Rs. 450, Profit = Rs. 45.

(c) S.P. = Rs. 350, Profit = Rs. 50. (d) S.P. = Rs. 600, Profit = Rs. 100.

2. Find loss percent in each of the following cases.

(a) C.P. = Rs. 480, Loss = Rs. 120 (b) C.P. = Rs. 1200, Loss = Rs. 300

(c) S.P. = Rs. 600, Loss = Rs. 200 (d) S.P. = Rs. 900, Loss = Rs. 100

3. 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.

4. (a) An article is purchased at Rs. 350 and sold for Rs. 420. Find the profit percent.
(b) A chair is bought for Rs. 6000 and sold for Rs. 5000. Find the loss percent.

5. (a) A fruit seller bought 5 dozen of oranges for Rs. 600 and sold them for Rs. 9.50

each. Find profit or loss.

(b) A fruit seller bought 2 dozens of bananas at the rate of Rs. 120 per dozen and sold

each at Rs. 15.

(i) Find his total profit. (ii) Find his profit percent.

(c) A florist bought 240 roses for Rs.1800. If he sold all of them for Rs. 10 each, what

profit did he make? Also, find his profit percent.

(d) A shopkeeper bought 25 pens for Rs.1250 and sold them for Rs. 60 each. Find his

profit or loss percent.

Answers

1. (a) 2% (b) 10% (c) 16 2 % (d) 20% 2. (a) 25% (b) 25% (c) 25%
3
(d) 10% 3. (a) Loss = 10% (b) Loss = 20% (c) Profit = 20% (d) profit = 10%

4. (a) 20% (b) 16 2 % 5. (a) Loss = Rs. 30 (b) (i) Profit=Rs.120 (ii) Profit=Rs.50%
3
1
(c) Rs. 600, 33 3 % (d) Profit = 20%

Oasis School Mathematics – 6 113

Unit

7 Unitary Method

7.1 Introduction

Unit means one. The word unitary is derived from the word unit.

Consider the following example,

Cost of 1 orange = Rs. 7

Cost of 4 oranges = Rs. 7 + Rs. 7 + Rs. 7 + Rs. 7

= Rs. 4 × 7

= Rs. 28.

If the cost of 1 article = Rs. 5 Value of many items
Cost of 2 articles = Rs. 2 × 5 = number of items × unit value
Cost of 3 articles = Rs. 3 × 5
Cost of 4 articles = Rs. 4 × 5

Conversely, if the cost of 5 articles is Rs. 40, what is the cost of an article?

Cost of 5 articles = Rs. 40
Cost of 1 article = Rs. 40 ÷ 5 = Rs. 8

Value of one items = Value of given number of items
Number of items

Again, look at one more example:

If the cost of 8 pens is Rs. 56, what is the cost of 12 such pens?

Here, Cost of 8 pens = Rs. 56

Cost of 1 pen = Rs. 56
8

Cost of 12 pens = Rs. 56 × 12
8

= Rs. 84

Hence the method to find the cost of one article first and then the cost of any number
of articles of the same kind is called unitary method.

114 Oasis School Mathematics – 6

Study the given bill and discuss the questions given below in your class.
XYZ Stationery

Dhumbarahi, Kathmandu
Phone No. 01-433346

Cash Bill
Name : ....................................................................................................................

S.No. Item Rate Quantity Total

1. Pencil 15 6 90
2. Pen
3. Maths Book-6 50 10 500
4. Eraser 350 2 700

10 4 40

Total 1330/-

i. What is the cost of 1 pencil? ..................
Signature

ii. How the cost of 6 pencils is Rs. 90?

iii. What is the cost of 1 maths-6 Book?

iv. Discuss how the cost of 2 months-6 Book is Rs. 700?

v. If customer has bought 4 copies, with the rate Rs. 80, what would be the cost
of 4 copies?

Worked Out Examples

Example: 1

If 1 copy costs Rs. 40, how much will 3 copies cost?

Solution:

Here, cost of 1 copy is Rs. 40 More the number of copies,
∴ cost of 3 copies is Rs. 40 × 3 = Rs. 120 more the cost.
Hence, the cost of 3 copies = Rs. 120 So, I have to multiply.

Oasis School Mathematics – 6 115

Example: 2

If the cost of 15 pens is Rs. 150. What is the cost of 1 pen?

Solution: Less the number
of pens, less the
Here, the cost of 15 pens is Rs. 150 cost. So, I have to
divide.
∴ the cost of 1 pen is Rs. 150 = Rs. 10
15
Hence, the cost of 1 pen is Rs. 10

Exercise 7.1

1. Complete the given table:

Articles Unit cost Cost of 2 units Cost of 5 units
Rs. 2 × 6 = Rs. 12 Rs. 5 × 6 = Rs. 30
Pencil Rs. 6
Copy Rs. 15
Mango Rs. 18
Mobile Rs. 4, 500

2. From the information given below, find the total cost of articles.

(a) Cost of an article = Rs. 20, number of articles = 15
(b) Cost of an article = Rs.7, number of articles = 22

3. (a) If the cost of 1 pen is Rs. 30, find the cost of 18 pens.

(b) Sapana bought 1 kg of apple for Rs. 20, how much does 5 kg of apples cost?

(c) If a pen costs Rs. 28, how much would 18 pens cost?

4. Complete the given table:

Number of articles Total cost Unit cost
Rs. 120 ÷ 3 = Rs. 40
3 Rs. 120
5 Rs. 960
10 Rs. 840
12 Rs. 732

5. Find the unit cost of an article of the following information:
(a) number of objects = 15, total cost = Rs. 300
(b) number of objects = 22, total cost = Rs. 154

6. (a) If the cost of 10 pens is Rs. 200, what is the cost of 1 pen?
(b) If 7 books cost Rs. 490, find the cost of 1 book.
(c) If 12 quintals of sugar cost Rs. 1200, what will one quintal sugar cost?

116 Oasis School Mathematics – 6

Answers 2. (a) Rs.300 (b) Rs.154 3. (a) Rs.540 (b) Rs.100 (c) Rs.504
5. (a) Rs.20 (b) Rs.7 6. (a) Rs. 20 (b) Rs. 70 (c) Rs. 100.
1. Consult your teacher.
4. Consult your teacher

Direct Variation
Lets study the given example and draw out the conclusion
Cost of 1 pen is Rs. 45.
Cost of 2 pens is Rs. 2 × 45 = Rs. 90
Cost of 3 pens is Rs. 3 × 45 = Rs. 135
cost of 4 pens is Rs. 4 × 45 = Rs. 180

From the above example, it is clear that as the quantity increases, the price also
increases. Such relation is an example of direct variation.

Hence, a relation in which value of a variable increases or decreases in the same
proportion as the other variable increases or decreases is called a direct variation.

Look at the some example of direct variations:

(i) number of copies and their price.

(ii) time and distance covered by a vehicle.

Worked Out Examples

Example 1
If the cost of 60 apples is Rs. 540, what is the cost of 50 apples?
Solution:

Here, the cost of 60 apples is Rs. 540

∴ the cost of 1 apple = Rs. 540 [less article, less cost)
60

∴ the cost of 50 apples = Rs. 540 × 50 [more article, more costs]
60

Hence, the cost of 50 apples Rs. 450.

Exercise 7.2

1. Identify whether the given relations are direct variation or not
(i) number of books and their price.

Oasis School Mathematics – 6 117

(ii) number of workers and the work done by them.

(iii) number of pages and the time taken to read it.

(iv) number of people and time of given quantity of food lasts for.

2. Complete the given table:

Numbers of Total cost Unit cost Cost of 5 articles Cost of 8 articles
articles

4 Rs. 96 Rs. 96 ÷ 3 = Rs. Rs. 32 × 5 = Rs. Rs. 32 × 8 = Rs.
32 160 256

5 Rs. 240

8 Rs. 128

12 Rs. 600

3. (a) If the cost of 15 pens is Rs. 150, what is the cost of 30 pens?
(b) Manish purchased 5 pencils for Rs. 35. What will 25 pieces of pencils cost?
(c) 7 pieces of cloth cost Rs. 490. What will 25 pieces of cloth cost?
(d) 15 bags of cement cost Rs. 5,250, find the cost of 1 bag and also find the cost of
25 bags of cement.
(e) A man paid Rs. 180 for 5 dozen of bananas. Find the cost of 1 dozen and then find
the cost of 10 dozen of bananas.
(f) In 16 minutes Ram could write 6 pages. How many pages would he be able to
write in 24 minutes?
(g) If a man can walk to a distance of 18 km in 4 hours, how long can he walk in 6 hours?
(h) How many kilometres does Ram walk in 2 hours, if he can walk 8 km in 4 hours?

Answers

1. Consult your teacher. 2. Consult your teacher.

3. (a) Rs.300 (b) Rs.175 (c) Rs.1750 (d) Rs.350, Rs.8750
(h) 4km
(e) Rs.36, Rs.360 (f) 9 pages. (g) 27km

118 Oasis School Mathematics – 6

Objective Questions

Choose the correct alternatives:

1. Which of the following is not equal to 45%?
9

(i) 20 (ii) 4.5 (iii) 0.45
2. Cost price of an article is Rs 50 less than the selling price, then there is:

(i) Loss of Rs. 50 (ii) Profit of Rs. 50 (iii) neither profit not loss

3. Cost of 15 copies is Rs 300, then the cost of 1 copy is:

(i) Rs 20 (ii) Rs 4500 (iii) Rs. 45

4. Out of 150 students, 30 students failed in the examination, the percentage of passed
student is

(i) 20% (ii) 80% (iii) 75%

5. The cost of 5 kg sugar is Rs 400, then the cost of 2kg sugar is

(i) Rs. 80 (ii) 160 (iii) Rs 240

6. Which one of the following relation is not a direct variation?

(i) number of articles and their price.

(ii) number of workers and the time required to complete the given work.

(iii) time and distance covered

7. A man bought an article for Rs. 500 and sold it at the profit of Rs. 50, then the profit
percent is

(i) 50% (ii) 10% (iii) 20%

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

(i) Profit% = Profit × 100%
CP

(ii) Loss% = Loss × 100%
SP

(iii) Profit% = S.P.–C.P. × 100%
C.P.

Oasis School Mathematics – 6 119

Assessment Test Paper

Group "A" [3 × 1 = 3]

1. (a) Convert 20% into fraction.
(b) If the cost price (C.P.) = Rs. 550, selling price (S.P.) = Rs 750, find the profit.
(c) Write the formula to find the loss percent.

Group "B" [4 × 3 = 12]

2. (a) Out of 450 students in a school, 250 are girls. Find the percentage of boys.

(b) Cost price of an article is Rs. 600, it is sold at Rs. 750, find (i) profit (ii) profit percent

(c) If the cost of 10kg of rice is Rs. 980, what is the cost of 1 kg of rice?

(d) If the cost of 1 copy is Rs. 50, what is the cost of 12 copies?
Group "C" [4 × 2 = 8]

3. The cost of one dozen of pens is Rs. 2880.
(i) Find the cost of 1 pen. (ii) Find the cost of 8 pens.
4. Out of Rs. 30,000 a man spent 30% on food and 40% on cloths, find
(i) his expenditure on food
(ii) his expenditure on cloths.
(iii) his saving.

120 Oasis School Mathematics – 6

Mensuration

15Estimated Teaching Hours

Contents
• Distance
• Perimeter of Plane Figure
• Area of Plane Figure
• Volume of Cube and Cuboid

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following competencies:
• To convert the one unit of measurement of distance into another.
• To find the perimeter of plane figure by adding the length outer

boundary
• To find the perimeter rectangle and square using formula
• To compare the area of different shapes containing similar unit
• To approximate the area of irregular shapes
• To calculate the area of rectangle and square using formulae
• To calculate the area of plane shapes using the formula of area of

rectangle and square
• To calculate the volume of cube by counting the unit cube and

using formulae

Teaching Materials

• paper model of plane shapes, model of cube and cuboid, etc.

Oasis School Mathematics – 6 121

Unit

8 Mensuration

8.1 Distance

As we know that length of an object is measured in centimeter and meter etc.
Distance between two objects can be measured in meter, km, etc.
Lets talk about other units of measurement of length.
Length of an object can be measured in 'Inch' and 'Feet' also.
• What is the relation of 'Inch' with cm?
• What is the relation of Inch and Feet?
Relation of Inch with feet.

From the figure, it is clear that

1 foot = 12 inches

2 feet = 2 × 12 inches = 24 inches

Hence, to convert feet into inches, multiply feet by 12.

Conversely, to convert inches into feet, divide inches by 12.

Lets be clear with an example,

Convert 5 feet into inches.

As we know that,

1 feet = 12 inches

5 feet = 5 × 12 inches

= 60 inches

Lets see another example,

Convert 48 inches into feet.

Here, 48 inches = 48 feet. = 4 feet.
12

122 Oasis School Mathematics – 6

Relation between Inch and Centimeters
Lets observe the given two ruler properly.

Lets compare the cm scale and inch scale of given ruler, we can see that
1 cm = 0.3937 inch

1 cm ≈ 0.4 inch.

1 cm is exactly equal to 0.3937 inch and it is approximately equal to 0.4 inch.

Again, lets compare the scale of 1 inch with cm,

then, 1 inch = 2.54 cm.

Lets be clear with given example:

Convert 5 inches into cm.

As 1 inch = 2.54 cm. • 1 cm = 0.3937 Inches

then, 5 inches = 5 × 2.54 cm • 1 Inch = 2.54 cm

= 12.7 cm. • To convert cm into inches
multiply it by 0.3937.
Again, lets convert cm into inches.
• To convert inches into cm,
Convert 25 cm into inches. multiply it by 2.54.

As, 1 cm = 0.3937 inches

25 cm = 25 × 0.3937 inches

= 9.8425 inches.

Relation of feet with cm

From the above ruler, it is clear that

1 feet = 30.48 cm.

2 feet = 2 × 30.48 cm.

= 60.96 cm

Oasis School Mathematics – 6 123

Hence, to convert feet into cm, multiply feet by 30.48 cm.
Lets be clear with an example,
Convert 6 ft. into cm.

As 1 feet = 30.48 cm • 1 feet = 30.48 cm

6 feet = 6 × 30.48 cm. • To convert feet into cm,
multiply it by 30.48.
= 182.88 cm
• To convert cm into feet, divide
Again, 30.48 cm = 1 ft. it by 30.48.

1 cm = 1 ft.
30.48

Then, 100 cm = 100 ft.
30.48

= 3.28 ft.

Relation between meter (m) and inches

1 m = 39.37 inches

Hence, to convert meter into inches, multiply meter by 39.37 inches.

Lets be clear with an example,

Convert 4 m into inches.

As we know that, • 1 m = 39.37 Inches

1m = 39.37 inches • To convert meter into inches,
4 m = 4 × 39.37 inches multiply it by 39.37.
Then,
= 157.48 inches • To convert inches into meter,
divide it by 39.37.
Again,

39.37 inches = 1 m

1 inch = 1 m.
39.37

Hence, to convert inch into meter, divide inches by 39.37.

Convert 90 inches into m.

Here, 39.37 inches = 1 m

1 inch = 1 m.
Then, 90 inches 39.37
1
= 39.37 × 90m

= 2.286 m.

124 Oasis School Mathematics – 6

Relation between meter (m) and feet (ft).

1 meter = 3.28 feet.

Hence, to convert meter into feet, multiply meter by 3.28.

Lets convert 4m into feet.

Lets convert 4m into feet.

Here, 4 m = 4 × 3.28 ft. • 1 meter = 3.28 feet

= 13.12 ft. • To convert meter into feet
Again, to convert ft. into meter (m), multiply it by 3.28.

3.28 feet = 1 meter • To convert feet into meter,
divide it by 3.28.
1 feet = 1 meter.
3.28

Hence, to convert feet in meter, divide feet by 3.28.

Convert 60 ft. into meter (m).

Here, 60 ft = 60 meter = 18.29 m.
3.28

Exercise 8.1

1. Fill in the blanks: (b) 1 feet = .......................cm
(a) 1 ft = ............... Inches (d) 1 cm = ................Inches
(c) 1 Inch = ................cm (f) 1 m = .................ft.
(e) 1 m = ....................Inches

2. Convert feet into Inches:
(a) 10 ft. (b) 6 ft. (c) 9 ft. (d) 8.5 ft. (e) 3.5 ft.

3. Convert inches into feet:
(a) 48 Inches ( b) 96 Inches (c) 40 Inches (d) 100 Inches

4. Convert Inches into cm:

(a) 8 Inches ( b) 12 Inches (c) 20 Inches (d) 25 Inches
(d) 35 cm
5. Convert cm into Inches: (c) 30 cm
(a) 20 cm (b) 15 cm

6. Convert feet into cm:

(a) 2 ft. ( b) 5 ft. (c) 4.5 ft. (d) 6.5 ft.

Oasis School Mathematics – 6 125

7. Convert cm into feet: (c) 100 cm (d) 120 cm
(a) 40 cm (b) 61 cm

8. Convert meter into inches:

(a) 2 m (b) 6 m (c) 7.5 m (d) 8.5 m

9. Convert inches into meter.
(a) 60 Inches (b) 84 Inches (c) 100 Inches (d) 120 Inches

10. Convert meter into feet.

(a) 4m (b) 7 m (c) 2.5 m (d) 8.5 cm

11. Convert feet into meter.

(a) 12 ft. (b) 15 ft. (c) 26 ft (d) 30 ft.

12. Solve the following problems.

(a) My height is 172 cm, convert my height into (i) inches, (ii) feet.

(b) The length of a table is 5 ft, what its length in (i) inches, (ii) cm (iii) m.

(c) Length and breadth of a book are 24 cm and 18 cm respectively. Find their length
and breadth in Inches.

Answers

1. Consult your teacher.

2. (a) 120 Inches (b) 72 Inches (c) 108 Inches (d) 102 Inches (e) 42 Inches

3. (a) 4 ft. (b) 8 ft. (c) 3.33 ft. (d) 8.33 ft.

4. (a) 20.32 cm (b) 30.48 cm (c) 50.8 cm (d) 63.5 cm

5. (a) 7.8 Inches (b) 5.9 Inches (c) 11.8 Inches (d) 13.77 Inches.

6. (a) 60.96 cm (b) 152.4 cm (c) 137.16 cm. (d) 198.12cm.

7. (a) 1.31 ft. (b) 2 ft. (c) 3.28 ft. (d) 3.94 ft.

8. (a) 78.74 Inches (b) 236.22 Inches (c) 295.27 Inches (d) 334.65 Inches

9. (a) 1.52 m (b) 2.13 m (c) 2.54 m (d) 3.05 m

10. (a) 13.12 ft. (b) 22.96 ft. (c) 21.32 ft. (d) 27.88 ft.

11. (a) 3.66 m (b) 4.57 m (c) 7.92 m (d) 17.38 m.

12. (a) 67.08 Inches, 5.64 ft. (b) 60 Inches, 152.4 cm, 1.524 m.

(c) 9.36 Inches, 7.02 Inches

126 Oasis School Mathematics – 6

8.2 Perimeter

Ramesh runs around the Shanka Park in his morning walk. How much distance does
he cover?

A school compound is to be enclosed by a wall. What is the length of the wall?

In both case we have to calculate the length of outer boundary. The length of outer
boundary is the perimeter.

In the given figure, what is the length of the outer boundary of ∆ ABC ?

AB + BC + AC

If AB = 4cm, BC = 6cm and AC = 5cm. A
Length of outer boundary
= AB + BC + AC fig (i) 4cm 5cm
= 4cm + 6cm + 5cm
= 15cm B 6cm C

Perimeter of rectangle A lD

In rectangle and parallelogram opposite sides are equal. So,

in the given figure, b b
C
AB = DC = b and AD = BC = l (suppose) B l
then the perimeter of rectangle or parallelogram Rectangle

= AB + BC + CD + DA

P = b + l + b + l = 2l + 2b = 2 (l + b)

Perimeter of a rectangle (p) = 2(l + b)

Perimeter of Square A B
D C
ABCD is a square. Its all sides are equal.
Let, AB = BC = CD = AD = l.
Now,
Perimeter of square ABCD

Oasis School Mathematics – 6 127

= AB + BC + CD + AD
or, P = 4l
Hence, perimeter of the square = 4l.

P = 4l.

Worked Out Examples

Example: 1

A rectangular garden is 15 m long and 8 m broad. Find its perimeter.

Solution: 15m

Here, Length of rectangular garden (l) = 15 m

Breadth of rectangular garden (b) = 8 m 8m

Perimeter (P) = ?

We have,

P = 2(l + b)

= 2(15 + 8)m

= 46m.

Example: 2

The perimeter of a rectangle is 60 m. Find the breadth of the field if its
length is 16 m.

Solution: A 16cm B
Here, Perimeter of a rectangle (P) = 60 m

Length of a rectangle (l) = 16 m

Breadth of a rectangle (b) = ? D C
Now, we know that,

Perimeter of rectangle (P) = 2(l + b)

or, 60 = 2(16 + b)

or, 60 = 16 + b
2

or, 30 = 16 + b

∴ b = 30 – 16 = 14 m

Hence, breadth of the parallelogram (b) = 14 m

Example: 3

If 80 m long wire is required to fence a square field one round, what is the
length of field?

Solution:

Here, ABCD is a square field whose length of each side is 'l'. Since length of wire

128 Oasis School Mathematics – 6

required to fence the square field is 80m, then AB
Perimeter of the field = 80 m DC
Length of each side (l) = ?
We have,
P = 4l

or, 80 = 4l
or,
∴ 80 = l
4

l = 20m

Hence, the length of each side = 20 m.

Exercise 8.2

1. Find the perimeter of each of the following rectangles:
(a) Write the formula to calculate the perimeter of a square.

(b) Write the formula to calculate the perimeter of a rectangle.

(c) If 'x' be the length of a side of a square, find its perimeter.

(d) If 'x' and 'y' be the length and breadth of a rectangle, then what is the perimeter?

2. Find the perimeter of the following parallelograms and rectangles:

(a) A B (b) P Q (c) E F (d) W X

2.5 cm
3 cm

10 ft.
2.5 cm

S 7.5 cm R H 3.5 cm G Z 15 ft. Y

D 5 cm C

3. Find the perimeter of each of the following rectangles:

(a) length = 5 cm, breadth = 4 cm

(b) length = 7.5 cm, breadth = 3.5 cm

(c) length = 8 ft., breadth = 6.5 ft.

4. Find the perimeter of the following squares:
(a) A B (b) P Q (c) W X

D 5 cm C S 2.5 cm R ZY
5 cm

Oasis School Mathematics – 6 129

5. Find the perimeter of each of the following squares:
(a) length = 3 cm (b) length = 6 cm (c) length = 7 cm (d) length = 8 m

6.

(a) Perimeter of a square is 24 cm, find the length of its side.

(b) If the perimeter of a square feild is 32m, find the length of each side.

(c) Perimeter of a rectangle is 28 cm. If its length is 10 cm, find it breadth.

(d) Perimeter of a rectangular field is 48 m, if the breadth of the field is 8m, find its
length.

7.
(a) A rectangular field is 20m long and 15 m wide. Find the length of fencing material
to fence the field.

(b) A square field of length of each side 30m is to be fenced, find the length of fencing
material to fence the field.

8. Solve the following problems.

(a) If 100m long fencing material is required to fence a square field. Find the length
of each side.

(b) If 48 m long wire is required to fence a square field one round, what is the length
of the field?

Answers

1. Consult your teacher.

2. (a) 16cm (b) 20cm (c) 12cm (d) 50 ft.

3. (a) 18cm (b) 22cm (c) 29 ft. 4. (a) 20cm (b) 10cm (c) 30 ft.
(b) 8m (c) 4cm (d) 16m
5. (a) 12cm (b) 24cm (c) 30cm 6. (a) 6cm (b) 12m

7. (a) 70m (b) 120 m 8. (a) 25m

Project Work
I. Measure the length and breadth of basketball court, your classroom, your

kitchen, your bedroom and find their respective perimeter in feet.

Length (l) Breadth (b) P = 2(l+b)

Basketball Court
Your Classroom
Your Kitchen
Your Bedroom

130 Oasis School Mathematics – 6

8.3 Area

Area of regular and irregular shapes

If we paste a picture on the wall, it covers some space of the
wall; space covered by the surface of an object is its area.

Let's compare the area
of given two objects.

Which figure has more It is difficult to identify
area ? which figure has more area.

Again,

Figure (i) Figure (ii)

Which figure has more Both figures have similar
area ? shape so, we can compare them

second figure has more area.

In the first case, two objects have different shapes so it is difficult to compare their area.
In the second case, two objects have same shape, so it is easier to compare their area.

Let's compare the area of two given figures: Figure (i)
Which figure has more area ?
Both figures contain similar shape of triangles.
How many similar triangles does figure (i) have ?
4 triangles.

How many similar triangles does figure (ii) have ? Figure (ii)
4 triangles.

Oasis School Mathematics – 6 131

Their areas are equal. i.e. we can compare their area.

I understand ! we can compare the area of
two objects of different shapes containing

similar units.

13 13 5
24 24
Figure (i) Figure (ii)

Here, figure (ii) has more area than figure (i) because figure (i) contains 4 boxes and
figure (ii) contains 5 boxes of same size.

Exercise 8.3

1. Identify whether the area of the given pair of objects can easily be compared or not ?
(a) (b)

(c) (d)

2. Compare the following objects of different shapes containing similar units.
(a) (b)

(i) and (ii) (i) and (ii)

(c) (d)

(i) and (ii) (i) and (ii)

132 Oasis School Mathematics – 6

3. Identify how many objects having shape are necessary to cover the given space.
(a) (b) (c)

Answers 2. Consult your teacher 3. (a) 16 (b) 19 (c) 32

1. Consult your teacher

8.4 The Unit of Area 1cm

Given square is 1cm in length and 1cm in breadth. 1cm
Then its area is 1 cm² or 1 square cm.
Hence, cm² or square cm is the unit of area.

The area of a figure formed by the squares of side 1 unit each

The area of figure formed by the squares of side
1unit each can be obtained by counting the number
of squares.

The given figure is composed by many squares

having area 1 sq. unit each.

REMEMBER area = 1 sq. unit

area = 1 sq. unit
2

There are 11 squares which are shaded.
∴ Area of shaded portion = 11 square units

Lets observe the given figure. This figure contains full square boxes as well as half
square boxes.

Lets count the total square boxes in the figure.
Number of square boxes = 9
Number of half square boxes = 3
Area of 8 squares boxes = 8 square units

Oasis School Mathematics – 6 133

Area of 6 half square boxes = 1
332×sq2uasrqeuuarneitus.nit.
=

Hence, the area of shaded part = 9 square units + 3 square units
2
21
= 2 square units.

Activity:

Find the area of given figure using Geoboard.

The first figure is the half part of the rectangle having area 2 sq. units.
1
Hence, area of shaded part = 2 × 2 sq. units = 1 sq. units.

How to find the area of an irregular object ?

In the given figure,
the number of completely shaded square boxes = 3
∴ Their area = 3 square units
Number of square boxes, whose more than half part is
shaded = 5
∴ Their approximate area = 5 square units
The number of square boxes whose less than half part is
shaded = 8

134 Oasis School Mathematics – 6

∴ Their approximate area = 8 × 0 = 0
∴ Approximate area of shaded part = (3+5+0) square units = 8 square units

REMEMBER Approximate area of more than half of unit square = 1 square unit

Approximate area of shaded part = 1 square unit

Approximate area of less than half of unit square = 0 square unit

Approximate area of shaded part = 0 square unit



Worked Out Examples

Example: 1

Find the area of given figure by counting the unit squares.
Solution:
Here, The number of square box = 11
Since the area of 1 square box is 1 square unit area of
given figure = 11 square units.

Example: 2

Find the area of the shaded part of the given figure.

Solution :

In the given figure,the number of completely shaded

square box = 8

Their area = 8 square units

Number of half shaded square box = 8
1

Their area = 8 × 2 square units = 4 square units
∴ Area of the shaded part = 8 square units + 4 square units

= 12 square units.

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Example 3

Find the approximate area of the shaded part.
Solution :
The number of completely shaded square = 1
∴ Its area = 1 square units
The number of square whose more than half part
is shaded = 6 square units
Their approximate area = 6 square units
The number of squares whose less than half part is shaded = 4
Their approximate area = 0 square units
∴ Approximate area of shaded part = (1+6+0) square units = 7 square units

Exercise 8.4

1. Find the area of the following figures by counting the squares.
(a) (b) (c)

2. Find the area of given figure by counting the boxes.

(a) (b)

(c) (d)

(f) (g)
(e)

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3. Find the approximate area of given irregular shapes.

(a)
b)
(c) (d)

(e) (f)

(g)

Answers (b) 11 square units (c) 1512 square units 2. (a) 7 square units

1. (a) 19 square units (c) 9 square units (d) 6 square units (e) 12 1 square units
(b) 10 square units 2
(f) 11 square units
(c) 12 square units (g) 26 1 square units 3. (a) 20 square units (b) 18 square units
2

(d) 23 sq. units (e) 13 sq. units (f) 28 sq. units (g) 12

8.5 Area of Rectangle and Square A 5 units B

Area of rectangle 1 234 5

In the given figure, rectangle ABCD is formed by the 6 7 8 9 10 3 units
combination of 15 squares having each area 1 square
unit. 11 12 13 14 15
DC
Therefore, area of rectangle ABCD = 15 square units.

Again, the number of squares along the length of
ABCD = 5

The number of squares along the breadth = 3

Now,

Area of a rectangle ABCD = 15 square units.
= 5 × 3 square units.
= length × breadth

∴ Area of the rectangle = length × breadth = l × b

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Area of a square

A square is also a rectangle having the length equal to the breadth.

Since, length = breadth

Area of a square = length × breadth
= length × length
= (length)2

∴ Area of a square = l2

Worked Out Examples

Example: 1

Find the area of given rectangle. 6 cm

Solution:

Here, 9 cm
Length of the rectangle (l) = 9cm

Breadth of the rectangle (b) = 6cm

We have, area of a rectangle = l × b = 9cm × 6cm = 54cm2

Example: 2

Find the area of given square. 3 cm

Solution:

Here, 3 cm

Length of the square (l) = 3cm

We have,

Area of a square = l 2

= (3cm)2 = 9cm2

Example: 3

Find the area of shaded part in the given figure. 5cm
5cm 10 cm
Solution:
12cm
Here,

Area of the outer rectangle =l×b

= 12 × 10 cm2

= 120 cm2

Again, area of square inside it = l2

= 52

= 25cm2

Now, area of shaded part = 120cm2 – 25cm2 = 95cm2

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Example: 4 A 5cm B

Find the area of given figure. 3cm
Solution :
Produce the side BC to F of GE. 7cm 6cm D
For the rectangle ABFG, C
length (l) = 7cm 4cm
breadth (b) = 5cm
We have, F 11cm E

Area of a rectangle = l × b A 5cm B D
3cm
= 5cm × 7cm C 6cm
7cm
4cm
= 35cm2
Again, for the rectangle CDEF, G 11Fcm E

Length (l) = 6cm

Breadth (b) = 4cm

We have,

Area of a rectangle = l × b
= 6cm × 4cm = 24cm2

∴ Area of the given figure = 35cm2+24cm2 = 59cm2

Example: 5

Find the area of rectangular field whose perimeter is 60m and the length is 20m.
Solution :
Given,
Perimeter (P) = 60m
Length (l) = 20m
Area (A) = ?
We have,
P = 2(l + b)
or, 60 = 2(20 + b)
or, 30 = 20 + b
or, b = (30 – 20)
∴ b = 10m
We have,
A = l × b

= 20m × 10m = 200m²

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Exercise 8.5

1. Find the area of the following figures. (By using formula)

(a) (b) (c) (d) 5.6 ft.

5 cm 3 cm

4 ft.

3.6 cm
5.6 ft.
5.5 cm
4 ft.
2. Find the area of the rectangles whose sides are:

(a) 10 cm and 8 cm (b) 9 cm and 6 cm (c) 4.5 cm and 1.5 cm

(d) 10 m and 1.6 m (e) 7.5 m and 70 cm

3. Find the area of the given square whose each side is

(a) 5cm (b) 7cm (c) 3.5cm (d) 7.5cm

4. Find the length of the following rectangles.

(a) Area = 120cm² and breadth = 12cm (b) Area = 36m² and breadth = 4m.

5. Find the breadth of the following rectangles.

(a) Area = 120cm² and length = 12 cm (b) Area = 27m² and length = 9m.

6. Find the length of the following squares.

(a) Area = 25 cm² (b) Area = 81 ft² (c) Area = 100 cm² (d) Area = 121 m²

7. Find the area of the shaded portion in the following figures.

(a) (b) 6 cm (c) 20 cm

3 cm 2 cm 2 cm
5 cm 12 cm
2 cm 5 cm
2 cm
6 cm

12 cm 2 cm

(d) 10 cm (e) 12 cm (f) 16 cm

8 cm

8 cm 10cm

6cm

10cm
6 cm

3cm

8. Find the area of the following figures.

(a) 2 cm 2 cm (b) 6 cm

4cm 2 cm
2 cm
2 cm

4 cm
3cm

1 cm
2 cm
2 cm
6 cm 8 cm

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(c) 1 cm (d) 10 cm

2 cm 6 cm
2 cm
2 cm 6 cm
2 cm

2 cm
3 cm 5 cm

4 cm
12 cm
5 cm

9. Solve the following: 10 cm

(a) Area of a rectangular field is 160 m2 and the breadth is 10 m, find its length and
its perimeter.

(b) Area of a rectangular field is 250 m2 and the length is 25 m, find its breadth and
the perimeter of the field.

10. Solve the following:

(a) Perimeter of a rectangular sheet of paper is 100 cm and the length is 30 cm, find
its breadth and area.

(b) If the perimeter of a rectangle is 80 cm, its breadth is 16 cm, find its length and its
area.

11. Solve the following:
(a) If the area of a square field is 256 m2, find its perimeter.

(b) The area of a square sheet of paper is 144cm2, find its perimeter.

12. Solve the following:

(a) If the perimeter of a square field is 52m, find its area.

(b) The perimeter of a square sheet of paper is 60 cm, find its area.
13. The map of a home designed by an engineer is given below:

(a) Find the area of each room.
(b) Find the total area occupied by a house.

Bed room II Sitting room 15 ft.

Library Hall Bath room 6 ft.
Bed room
Kitchen 8 ft.
6 ft. 6 ft. 12 ft.
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Answers

1. (a) 15cm² (b) 16 ft² (c) 19.8cm² (d) 31.36 ft²

2. (a) 80cm² (b) 54cm² (c) 6.75cm² (d) 16m² (e) 5.25m²

3. (a) 25cm² (b) 49cm² (c) 12.25cm² (d) 56.25cm²

4. (a) 10cm, (b) 9cm 5. (a) 10cm (b) 3cm

6. (a) 5m (b) 9cm (c) 10cm (d) 11m.

7. (a) 51cm² (b) 32cm² (c) 230cm² (d) 56cm² (e) 72cm² (f) 64cm²

8. (a) 20cm² (b) 22cm² (c) 12cm² (d) 96cm²

9. (a) 52 cm (b) 70 cm 10. (a) 20 cm, 600 cm2 (b) 24 cm, 384 cm2

11. (a) 64 cm (b) 48 cm 12. (a) 169 cm2 (b) 225 cm2

13. (a) Kitchen= 96 square ft., Bath room = 72 square ft.

Bedroom I = 84 square ft., Library Hall = 84 square ft.

Bedroom II = 180 square ft. sitting room = 180 square ft. (b) 696 square ft.

8.6 Volume

Look at the given examples, how much space is occupied by the given box?
How much space is occupied by given cupboard?

Space occupied by the objects in the above examples are their respective volume.
∴ Volume is the space occupied by the given object.

Unit of volume

Standard unit of volume is m³ or cubic metre. The smaller unit of volume is cm³ or
cubic cm.

Volume of a unit cube 1 unit 1 unit 1 unit

The given figure is a unit cube. Its length, breadth and height 1 unit
each is equal to 1 unit.
∴ Volume of this cube is 1 cubic unit. 1 unit

Volume of a cuboid

It is a unit cube. 1 unit

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Its volume is 1 cubic units.
This figure contains 12 unit cubes.
So, its volume is 12 cubic units.

This figure contains 2 layers of 12 units. 3 unit
So, its volume is 2 × 12 cubic units = 24 cubic units.
1 unit
4 unit

Now,

Volume of above cuboid = 24 cubic units

= 4 × 3 × 2 cubic units

=l×b×h 2 unit 3 unit

∴ Volume of a cuboid = l × b × h 4 unit

Volume of a cube REMEMBER

In a cube, Length, breadth and height
length = breadth = height of a cube are equal.
l=b=h

We have, volume of the cube = l × b × h
= l × l × l = l3

∴ Volume of the cube = l3

Note: 1 cu.cm = 1 cm3

Worked Out Examples

Example : 1

Find the volume of following cuboid by counting the
unit cubes of volume 1 cubic cm.

Solution:

In the given figure,
Number of unit cubes = 8
∴ Volume of the given solid = 8 cubic cm.

Example: 2 9 cm

Find the volume of given cuboid.
Solution:

Here, Length of the cuboid (l) = 10 cm 10cm 8 cm
breadth of the cuboid (b) = 8 cm

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height of the cuboid (h) = 9 cm

We have,

Volume (v) =l×b×h

= 10cm × 8cm × 9cm = 720cm3

Example: 3

Find the volume of a room whose length, breadth and height are 5 m, 4 m
and 3 m respectively.
Solution:

Here, Length of the room (l) = 5 m

Breadth of the room (b) = 4 m

Height of the room (h) = 3 m

We know that,

Volume of the room = l × b × h = 5m × 4m × 3m = 60 m³

∴ The volume of the room is 60 m³.

Example: 4

If the volume of cube is 125 cm³, find the length of each side.

Solution: Here, Volume of the cube = 125cm³

or, l³ = 125cm³

l³ = (5 × 5 × 5)cm³

∴ l = 5cm.

Example: 5

Capacity of a rectangular tank is 60000 cm3. If its length and breadth are 2
m and 2.5m, find its height.

Solution:
Here, Length of the rectangular tank (l) = 2m = 2 × 100 cm = 200 cm

Breadth of the rectangular tank (b) = 2.5 m = 2.5 × 100 cm = 250 cm

Volume of the rectangular tank (V) = 60000 cm3

We have,

V = l × b × h = 6000 = 200 × 250 × h

or, h= 60000
200×250

∴ h = 1.2 cm

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