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

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

Try These (Page No. 119)

Question 1.

In a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework.

There were 90 parents who helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours. The distribution of parents according to the time for which, they said they helped is given in the following figure: 20% helped for more than 1\(\frac{1}{2}\) hours per day; 30% helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours; 50% did not help at all.

Using this, answer the following:

(i) How many parents were surveyed?

(ii) How many said that they did not help?

(iii) How many said that they helped for man than 1\(\frac{1}{2}\) hours?

Solution:

(i) Since, 30% of total surveyed parents helped their children for “\(\frac{1}{2}\) hours to 1\(\frac{1}{2}\) hours”. And 90 parents helped their children for “\(\frac{1}{2}\) hours to 1\(\frac{1}{2}\) hours”.

30% [ surveyed parents] = 90

or \(\frac{30}{100}\) × [surveyed parents] = 90

or surveyed parents = 90 × 100/30

= 3 × 100

= 300

(ii) Since, 50% of surveyed parents did not help their children.

Number of parents who did not help = 50% of surveyed parents

= 50% of 300

= 50/100 × 300

= 150

(iii) Since, 20% of surveyed parents help their children for more than 1\(\frac{1}{2}\) hours, i.e. 20% of surveyed parents help for more than 1\(\frac{1}{2}\) hours.

Number of parents who helped for more than 1\(\frac{1}{2}\) hours = 20% of 300

= 20/100 × 300

= 20 × 3

= 60

Note: ‘of’ means multiplication

Try These (Page No. 121)

Question 2.

A shop gives a 20% discount. What would the sale price of each of these be?

(a) A dress marked at ₹ 120

(b) A pair of shoes marked at ₹ 750

(c) A bag marked at ₹ 250

Solution:

(a) Marked price of the dress = ₹ 120

Discount rate = 20%

Discount = 20% of ₹ 120

= ₹ \(\frac{20}{100}\) × 120

= ₹ 2 × 12

= ₹ 24

Sale price of the dress = [Marked Price] – [Discount]

= ₹ 120 – ₹ 24

= ₹ 96

(b) Marked price of the pair of shoes = ₹ 750

Discount rate = 20%

Discount = 20% of ₹ 750

= ₹ \(\frac{20}{100}\) × 750

= ₹ 2 × 75

= ₹ 150

Now, Sale price of the pair of shoes = [Marked Price] – [Discount]

= ₹ 750 – ₹ 150

= ₹ 600

(c) Marked price of the bag = ₹ 250

Discount rate = 20%

Discount = 20% of ₹ 250

= ₹ \(\frac{20}{100}\) × 250

= ₹ 2 × 25

= ₹ 50

Sale price of the bag = [Marked price] – [ Discount]

= ₹ 250 – ₹ 50

= ₹ 200

Question 3.

A table marked at ₹ 15,000 is available for ₹ 14,000. Find the discount given and the discount percent.

Solution:

The marked price of the table = ₹ 15000

Sale price of the table = ₹ 14400

Discount = [Marked Price] – [Sale price]

= ₹ 15000 – ₹ 14400

= ₹ 600

Discount percent = Discount/Marked Price × 100

Question 4.

An almirah is sold at ₹ 5,225 after allowing a discount of 5%. Find its marked price.

Solution:

Sale price of the almirah = ₹ 5255

Discount rate = 5%

Since, Discount = 5% of marked price

Marked Price – Discount = Sale price

or [Marked Price] – \(\frac{5}{100}\) × [Marked Price] = Sale price

or [Marked Price] × \(\frac{95}{100}\) = ₹ 5255

or [Marked Price] = ₹ 5255 × \(\frac{95}{100}\) = ₹ 5500

Try These (Page No. 123)

Question 5.

Find selling price (SP) if a profit of 5% is made on

(a) a cycle of ₹ 700 with ₹ 50 as overhead charges.

(b) a lawnmower bought at ₹ 1150 with ₹ 50 as transportation charges.

(c) a fan bought for ₹ 560 and expenses of ₹ 40 made on its repairs.

Solution:

(a) Total cost price = ₹ 700 + ₹ 50 (overhead expenses) = ₹ 750

profit = 5% of ₹ 750

= ₹ \(\frac{5}{100}\) × ₹ 780

= ₹ 5 × \(\frac{15}{100}\)

= ₹ \(\frac{75}{100}\)

= ₹ 37.50

Now, SP = CP + profit

= ₹ 750 + ₹ 37.50

= ₹ 787.50

(b) Total cost price = ₹ 1150 + ₹ 50 (Overhead expenses) = ₹ 1200

Profit = 5% of 1200

= \(\frac{5}{100}\) × ₹ 1200

= ₹ 5 × 12

= ₹ 60

Now, SP = CP + Profit

= ₹ 1200 + ₹ 60

= ₹ 1260

(c) Total cost price = ₹ 560 + ₹ 40 (Overhead expenses) = ₹ 600

Profit = 5% of ₹ 600 = \(\frac{5}{100}\) × ₹ 600

= ₹ 5 × 6

= ₹ 30

Now, SP = CP + profit

= ₹ 600 + ₹ 30

= ₹ 630

Try These (Page No. 123)

Question 6.

A shopkeeper bought two TV sets at ₹ 10,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.

Solution:

(a) Marked price of the dress = ₹ 120

Discount rate = 20%

Discount = 20% of ₹ 120

= ₹ \(\frac{20}{100}\) × 120

= ₹ 2 × 12

= ₹ 24

Sale price of the dress = [Marked Price] – [Discount]

= ₹ 120 – ₹ 24

= ₹ 96

(b) Marked price of the pair of shoes = ₹ 750

Discount rate = 20%

Discount = 20% of ₹ 750

= ₹ \(\frac{20}{100}\) × 750

= ₹ 2 × 75

= ₹ 150

Now, Sale price of the pair of shoes = [Marked Price] – [Discount]

= ₹ 750 – ₹ 150

= ₹ 600

(c) Marked price of the bag = ₹ 250

Discount rate = 20%

Discount = 20% of ₹ 250

= ₹ \(\frac{20}{100}\) × 250

= ₹ 2 × 25

= ₹ 50

Sale price of the bag = [Marked price] – [Discount]

= ₹ 250 – ₹ 50

= ₹ 200

Try These (Page No. 126)

Question 7.

Find the interest and amount to be paid on ₹ 15000 at 5% per annum after 2 years.

Solution:

Here P = ₹ 15000, R% = 5%, T = 2 years

Simple Interest = \(\frac{\text { PRT }}{100}\)

= \(\frac{15000 \times 5 \times 2}{100}\)

= ₹ 1500

Amount = ₹ 15000 + ₹ 1500 = ₹ 16,500

Try These (Page No. 129)

Question 8.

Find Cl on a sum of ₹ 8000 for 2 years at 5% per annum compounded annually.

Solution:

We have

P = ₹ 8000, R = 55 p.a., T = 2 years

∴ \(A=P\left[1+\frac{R}{100}\right]^{n}\)

= ₹ 8000 \(\left[1+\frac{5}{100}\right]^{2}\)

= ₹ \(8000\left[\frac{21}{20}\right]^{2}\)

= ₹ \(8000 \times \frac{21}{20} \times \frac{21}{20}\)

= ₹ (20 × 21 × 21)

= ₹ 8820

Now, compound interest = A – P

= ₹ 8820 – ₹ 8000

= ₹ 820

Try These (Page No. 130)

Question 9.

Find the time period and rate for each.

1. A sum taken for 1\(\frac{1}{2}\) years at 8% per annum is compounded half-yearly.

2. A sum taken for 2 years at 4% per annum compounded half-yearly.

Solution:

1. We have interest rate 8% per annum 1\(\frac{1}{2}\) year.

It is compounded half-yearly.

Time period (n) = 2(1\(\frac{1}{2}\)) = 3 half yearly

Rate (R) = \(\frac{1}{2}\) (8%) = 4% per half yearly.

2. We have an interest rate of 4% per annum for the year

Time period (n) = 2(2) = 4 half-yearly

Rate (R) = \(\frac{1}{2}\) (4%) = 2% per half yearly.

Try These (Page No. 131)

Question 10.

Find the amount to be paid

1. At the end of 2 years on ₹ 2,400 at 5% per annum compounded annually.

2. At the end of 1 year on ₹ 1,800 at 8% per annum compounded quarterly.

Solution:

1. We have:

P = ₹ 2400, R = 5% p.a., T = 2 years

∴ Interest is compounded annually i.e, n = 2

2. Here, interest compounded quarterly.

R = 8%, p.a. = \(\frac{8}{4}\), i.e., 2% per quarter

T = 1 year = 4 × 1, i.e., 4 quarters on

n = 4

Try These (Page No. 133)

Question 11.

A machinery worth ₹ 10,500 depreciated by 5%. Find its value after one year?

Solution:

Here, P = 10, 500, R = 5% p.a.

T = 1 year, n = 1

∴ A = \(\mathrm{P}\left[1+\frac{5}{100}\right]^{1}\)

[∴ Depreciation is there, r = -5%]

= ₹ \(10500\left[1+\frac{5}{100}\right]^{1}\)

= ₹ 10500 × \(\frac{19}{20}\)

= ₹ 525 × 19

= ₹ 9975

Thus, machinery value after 1 year = ₹ 9975

Note: For depreciation, we use the formula as A = \(\mathrm{P}\left[1-\frac{\mathrm{R}}{100}\right]^{\mathrm{n}}\)

Question 12.

Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.

Solution:

Present population, P = 12 lakh

Rate of increase, R = 4% p.a.

Time, T = 2 years

n = 2

Population after 2 years = \(\mathrm{P}\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{n}}\)

= 12 lakh \(\left[1+\frac{4}{100}\right]^{2}\)

= 12 lakh \(\left[\frac{26}{25}\right]^{2}\)

= 12,00, 000 × \(\frac{26}{25} \times \frac{26}{25}\)

= 1920 × 26 × 26

= 12,97,920

Thus, the population of the town will be 12,97,920 after 2 years.