These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions

NCERT In-text Question Page No. 158

Question 1.

Find the percentage of children of different heights for the following data.

Answer:

Question 2.

A shop has the following number of shoe pairs of different sizes.

Size 2 : 20

Size 3 : 30

Size 4 : 28

Size 5 : 14

Size 6 : 8

Write this information in tabular form as done earlier and find the percentage of each shoe size available in the shop.

Answer:

NCERT In-text Question Page No. 159 & 160

Question 1.

A collection of 10 chips with different colours is given.

Answer:

Question 2.

Mala has a collection of bangles. She has 20 gold bangles and 10 silver bangles. What is the percentage of bangles of each type? Can you put it in the tabular form as done in the above example?

Answer:

We have

Note: We can use percentages for comparing various quantities.

NCERT In-text Question Page No. 160

Question 1.

Look at the examples below and in each of them, discuss which is better for comparison.

In the atmosphere, 1 g of air contains:

(i) .78g Nitrogen – 78 % Nitrogen

.21 g Oxygen or 21 % Oxygen

.01g Nitrogen – 1 % Other Gas

(ii) A shirt has:

\(\frac{2}{5}\) Cotton 60% Cotton

or

\(\frac{2}{5}\) Polyester 40% Polyester

Answer:

The second one (i.e. percentage) is better for comparison in both the cases.

NCERT In-text Question Page No. 161

Question 1.

Convert the following to per cents:

(a) \(\frac{12}{16}\)

(b) 3.5

(c) \(\frac{49}{50}\)

(d) \(\frac{2}{2}\)

(e) 0.05

Answer:

(a) \(\frac{12}{16}\) = (\(\frac{12}{16}\) x 100) % = ( \(\frac{3}{4}\) x 100) %

= (3 x 25)% = 75%

(b) 3.5 = (3.5 x 100)% = 350%

(c) \(\frac{49}{50}\) ( \(\frac{49}{50}\) x 100)%

= (\(\frac{49}{50}\) x 100)%

= (49 x 2)% = 98%

(d) \(\frac{2}{2}\) = ( \(\frac{2}{2}\) x 100)%

= (1 x 100)%

= 100%

(e) 0.05 = (0.05 x 100) % = 5 %

Question 2.

(i) Out of 32 students, 8 are absent.

What per cent of the students are absent?

(ii) There are 25 radios, 16 of them are out of order. What per cent of radios are out of order?

(iii) A shop has 500 parts, out of which 5 are defective. What per cent are defective?

(iv) There are 120 voters, 90 of them voted yes. What per cent voted yes?

Answer:

= (3 x 25)% = 75%

75% voters voted ‘yes’

NCERT In-text Question Page No. 162

Question 1.

Fill in the blanks

(i) 35% + ……………. % = 100%.

(ii) 64% + 20% + ……………. % = 100%

(iii) 45% = 100% – ……………. %.

(iv) 70% = ……………. % – 30%

Answer:

(i) 100% – 35% = 65%

35% + 65% = 100%

(ii) 64%+ 20% =84%

100% – 84% = 16%

64% + 20% + 16% = 100%

(iii) 100%-45% =55%

45% = 100% – 55%

(iv) 70% + 30% = 100%

70% = 100% – 30%

Question 16.

If 65% of students in a class have a bicycle, what per cent of the student do not have bicycles?

Answer:

Part of students having bicycles = 65%

Remaining part of students

= 100% – 65% = 35%

Thus, 35% of students do not have bicycles.

Question 17.

We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what per cent are mangoes?

Answer:

Quantity of apples = 50%

Quantity of oranges = 30%

Quantity of mangoes in the basket

= 100% – (50% + 30%)

= 100% -80% = 20%

Thus, the percent of mangoes in the basket = 20%

NCERT In-text Question Page No. 163

Question 1.

What per cent of these figures are shaded?

Answer:

(i) Fraction which is shaded

\(=\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)=\left(\frac{1+1+1}{4}\right)=\frac{3}{4}\)

Percentage of shaded part

= \(\frac { 3 }{ 4 }\) x 100% = 75%

(ii) Fraction of diagram which is shaded

\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{2+1+1}{8}=\frac{4}{8}=\frac{1}{2}\)

Percentage of diagram which is shaded = \(\frac { 1 }{ 2 }\) x 100% = 50%

NCERT In-text Question Page No. 164

Question 1.

Find

(a) 50% of 164

(b) 75% of 12]

(c) 12 \(\frac { 1 }{ 2 }\) % of 64

Answer:

(a) 50% of 164 = \(\frac { 50 }{ 100 }\) x 164

= \(\frac { 1 }{ 2 }\) x 164 = 82

(b) 75% of 12 = \(\frac { 75 }{ 100 }\) x 12 = \(\frac { 3 }{ 4 }\) x 12

= 3 x 3 = 9

(c) 12\(\frac { 1 }{ 2 }\) % of 64 = \(\frac { 25 }{ 2 }\)% or 64

= \(\frac{2}{100} \times 64=\frac{25}{100} \times \frac{64}{2}\)

= \(\frac { 1 }{ 8 }\) x 64 = 8

Question 2.

8% children of a class of 25 like getting wet in the rain. How many children like getting wet in the rain?

Answer:

Number of children who like getting wet in rain

= 8% of 25 = \(\frac { 8 }{ 100 }\) x 25 = 2

Question 3.

9 is 25% of what number?

Answer:

Let the required be P

25% of P = 9

\(\frac { 25 }{ 100 }\) x P = 9

25 P = 9 x 100

P = \(\frac { 9 x 100 }{ 25 }\) = 9 x 4 = 36

The required number is 36.

Question 4.

75% of what number is 15?

Answer:

Let the required number be p

75% of P = 15

= \(\frac { 75 }{ 100 }\) x P = 15

\(\frac { 3 }{ 4 }\) x P = 15

\(\frac { 3P }{ 4 }\) = 15

3P = 15 x 4

P = \(\frac{15 \times 4}{3}\) = 5 x 4 = 20

The required number is 20

NCERT In-text Question Page No. 166

Question 1.

Divide 15 sweets between Manu and Sonu so that they get 20% and 80% of them respectively.

Answer:

Share of Manu = 20% of 15 sweets

= \(\frac { 20 }{ 100 }\) x 15 sweets

= 3 sweets

Share of Sonu = 80% of 15 sweets

= \(\frac { 80 }{ 100 }\) x 15

= \(\frac { 4 }{ 5 }\) x 15

= 4 x 3 = 12 sweets

Manu share = 3 sweets and Sonu share =12 sweets.

Question 1.

If the angles of a triangle are in the ratio 2:3:4, find the value of each angle.

Answer:

We know that sum of three angles of a triangle =180°

Let the three angles be 2x, 3x, and 4x

2x +3x + 4x = 180°

9x = 180°

x = \(\frac { 180° }{ 9 }\) = 20°

The three angles are 2(20° ); 3(20° ) and 4(20°)

i.e. 40°, 60° and 80°

NCERT In-text Question Page No. 167

Question 1.

Find percentage of increase or decrease

(a) Price of shirt decreased from ₹ 80 to ₹ 60

(b) Marks in a test increased from 20 to 30

Answer:

(a) Initial price of the shirt = ₹ 80

Decrease price of the shirt = ₹ 60

Decrease in the price

= ₹80 – ₹60 = ₹20

Per cent of decrease in price 20

= \(\frac { 20 }{ 80 }\) x 100%

= ₹ 25%

(b) Initial marks = 20

Increased marks = 30

Increase in marks = 30 – 20 = 10

Percent of increase in marks

= \(\frac { 10 }{ 20 }\) x 100% = 50%

Question 2.

My mother says in her childhood petrol was ₹ 1 a litre. It is ₹ 52 per litre today. By what per centage has the price gone up?

Answer:

Initial Price = ₹ 1

Increase price = ₹ 52

Increase in price = ₹52 – ₹1 = ₹51

Percentage of increase in petrol

= \(\frac { 51 }{ 1 }\) x 100%

= 5100%

NCERT In-text Question Page No. 169

Question 1.

A shopkeeper bought a chair for ₹ 375 and sold it for ₹ 400. Find the gain percentage.

Answer:

C.P of a chair = ₹ 375

S.P of a chair = ₹ 400

Profit = S.P – C.P

= ₹ 400 – ₹ 375 = ₹ 25

Profit percentage = \(\frac { Profit }{ C.P }\) x 100%

= \(\frac { 25 }{ 375 }\) x 100%

= \(\frac { 1 }{ 15 }\) x 100% = \(\frac { 20 }{ 3 }\)%

= 6\(\frac { 2 }{ 3 }\)%

Question 2.

Cost of an item is₹50 It was sold with a profit of 12%. Find the selling price.

Answer:

We have

CP of the item = ₹50

Profit % = 12%

Profit = 12% of ₹50

= ₹\(\frac { 12 }{ 100 }\) x 50 = ₹6

SP = CP + Profit

= ₹50 + ₹6 = ₹56

Question 3.

An article was sold for ₹250 with a profit of 5%. What was its cost price?

Answer:

Let the cost price be ₹ x.

Profit % = 5%

∴ Profit = 5% of x = ₹ \(\frac { 5 }{ 100 }\) x x =₹ \(\frac { x }{ 20 }\)

∴ SP = CP + Profit

∴ ₹ 250 = ₹ x + \(\frac { x }{ 20 }\) [∵ ₹ = ₹ 250]

or ₹ 250 = ₹ \(\frac { 21 }{ 20 }\)x

or x = \(\frac{250 \times 20}{21}=\frac{5000}{21}=238 \frac{2}{21}\)

Thus cost price of the article = ₹ 238 \(\frac { 2 }{ 21 }\)

Question 4.

An item was sold for ₹ 540 at a loss of 5% what was its cost price?

Answer:

Let the cost rice be ₹ x

Loss% = 5%

Loss = 5% of ₹ x

= \(\frac { 5 }{ 100 }\) x x = \(\frac { x }{ 20 }\)

Selling price = Cost price – Loss

540 = x – \(\frac { x }{ 20 }\)

540 = \(\frac{20 x-x}{20}=\frac{19 x}{20}\)

19x = 540 x 20

x = \(\frac{540 \times 20}{19}\)

= \(\frac{10800}{19}=568 \frac{8}{19}\)

or ₹ 568.40 (approx.)

Cost price of the article

= ₹ 568\(\frac{8}{19}\) or ₹ 568.42

NCERT In-text Question Page No. 170

Question 1.

₹ 10,000 is invested at 5% interest rate p.a. Find the interest at the end of one year.

Answer:

Principal (P) = ₹ 10,000

Rate (R) = 5% p.a.

Time (T) = 1 year

Question 2.

₹ 3,500 is given at 7% p.a. rate of interest. Find the interest which will be received at the end of two years.

Answer:

Principal (P) = ₹ 3,500; Rate (R) = 7% p.a. Time (T) = 2 years

∴ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= ₹ \(\frac{3,500 \times 7 \times 2}{100}\)

= ₹ 35 x 7 x 2 = ₹ 490.

Question 3.

₹ 6,050 is borrowed at 6.5% rate of interest p.a. Find the interest and the amount to be paid at the end of 3 years.

Answer:

Principal (P) = ₹ 60,50

Rate (R) = 6.5% p.a.

Time (T) = 3% years

∴ Simple interest

Since, Amount = Principal + Interest

= ₹ 6,050 + ₹ 1,17.75 = ₹ 7,229.75

Thus, the amount to be paid at the end of 3 years = 7,229.75.

Question 4.

₹ 7000 is borrowed at 3.5% rate of interest p.a. borrowed for 2 years. Find the amount to be paid at the end of the second year.

Answer:

Principal (P) = ₹ 7,000

Rate = 3.5% p.a.

Time (T) = 2 years

∴ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= ₹ \(\begin{aligned}

&7,000 \times 3.5 \times 2\\

&0

\end{aligned}\) = ₹ 70 x \(\frac{35}{10}\) x 2

= ₹ 7 x 35 x 2 = ₹ 490

Since, Amount = Principal + Interest

Amount to be paid at th end of 2nd year = ₹ 7,000 + ₹ 490 = ₹ 7,490.

NCERT In-text Question Page No. 171

Question 1.

You have ₹ 2,400 in you account and the interest rate is 5%. After how many years would you earn ₹ 240 as interest.

Answer:

Here, principal (P) = 2,400

Rate (R) = 5% p.a.

Time (T) = ?

Simple interest = ₹ 2,40

∵ Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

∴ 240 = \(\frac{2400 \times 5 \times \mathrm{T}}{100}\)

∴ T = \(\frac{240 \times 100}{2400 \times 5}\) = 2 years

Thus, an interest of ? 240 will be obtained after 2 years.

Question 2.

On a certain sum, the interest paid after 3 years is ₹ 450 at 5% rate of interest per annum. Find the sum.

Answer:

Let the sum be ‘P’

450 = \(\frac{3 \mathrm{P}}{20}\)

Interest (I) = ₹ 450

3P = 450 x 20

Rate (R) = 5% p.a

P = \(\)

Time (T) = 3 years = 150 x 20

I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

= ₹ 3000

450 = \(\frac{\mathrm{P} \times 5 \times 3}{100}\)

The required sum is ₹ 3000.