You can Download KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions and Answers to help you to revise the complete syllabus.

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

Try These (Page 119)

Question 1.

In a primary school, the parents were asked about the number of hours they spend per day 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 adjoining 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 more than 1\(\frac{1}{2}\) hours?

Solution:

(i) Let the total number of parents = x

Parents who helped \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours = 90

The percentage of these parents = 30%

∴ 30% of x = 90

⇒ \(\frac{30}{100} \times x\) = 90

⇒ x = \(\frac{90 \times 100}{30}\)

∴ x = 300

∴ Number of parents = 300

(ii) Percentage of parents who did not help = 50%

Number of parents who did not help = 50% of 300

= \(\frac{50}{100}\) × 300

= 150

(iii) Percentage of parents who helped more than 1\(\frac{1}{2}\) hours = 20%

Number of such parents = 20% of 300

= \(\frac{20}{100}\) × 300

= 60

Try These (Page 121)

Question 1.

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

(a) A dress marked at Rs. 120

(b) A pair of shoes marked at Rs. 750

(c) A bag marked at Rs. 250

Solution:

(a) Marked price of a dress = Rs. 120

Discount = 20%

Discount = 20% of 120

= \(\frac{20}{100} \times 120\)

= Rs. 24

S.P. = M.P – Discount

= 120 – 24

= Rs. 96

(b) Marked price of a pair of shoes = Rs. 750

Discount = 20% of 750

= \(\frac{20}{100} \times 750\)

= Rs. 150

∴ S.P. = M.P – Discount

= 750 – 150

= Rs. 600

(c) Marked price of a bag = Rs. 250

Discount = 20% of 250

= \(\frac{20}{100} \times 250\)

= Rs. 50

∴ S.P. = M.P – Discount

= 250 – 50

= Rs. 200

Question 2.

A table marked at Rs. 15,000 is available for Rs. 14,400. Find the discount given and the discount percent.

Solution:

Marked price of a table = Rs. 15,000

Sale price of a table = Rs. 14,400

Discount = M.P.- S.P

= 15,000 – 14,400

= Rs. 600

Discount % = \(\frac{\text { Discount }}{\text { M.P }} \times 100\)

= \(\frac{600}{15000} \times 100\)

= 4%

∴ Discount = Rs. 600 and Discount percent = 4

Question 3.

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

Solution:

Let Marked price = Rs. x

Sale price of a table = Rs. 5225

Discount = 5%

S.P. = M.P. – Discount

⇒ 5225 = x – 5% of x

⇒ 5225 = x – \(\frac{5}{100} \times x\)

⇒ 5225 = \(\frac{100 x-5 x}{100}\)

⇒ 5225 × 100 = 95x

⇒ x = \(\frac{5225 \times 100}{95}\)

⇒ x = 5500

∴ M.P. = Rs. 5500

Try These (Page 123)

Question 1.

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

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

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

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

Solution:

(a) Cost price of a cycle = Rs. 700

Overhead charges = Rs. 50

Total cost price = 700 + 50 = Rs. 750

Profit percent = 5%

Profit = 5% of total C.P.

= \(\frac{5}{100} \times 750\)

= Rs. 37.50

∴ S.P. = C.P. + Profit

= 750 + 37.50

= Rs. 787.50

(b) C.P. of lawn mower = Rs. 1150

Transportation charges = Rs. 50

Total C.P. = 1150 + 50 = Rs. 1200

Profit percent = 5%

Profit = 5% of total C.P.

= \(\frac{5}{100} \times 1200\)

= Rs. 60

∴ S.P. = C.P. + Profit

S.P. = 1200 + 60 = Rs. 1260

(c) C.P. of a fan = Rs. 560

Repair charges = Rs. 40

Total C.P. = 560 + 40 = Rs. 600

Profit % = 5%

∴ Profit = 5% of total C.P.

= \(\frac{5}{100} \times 600\)

= Rs. 30

∴ S.P. = C.P. + Profit

= 600 + 30

= Rs. 630

Try These (Page 123)

Question 1.

A shopkeeper bought two TV sets at Rs. 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:

C.P. of T.V. set = Rs. 10,000

Profit = 10%

∴ Profit = 10% of 10,000

= \(\frac{10}{100} \times 10,000\)

= Rs. 1000

S.P. = C.P. + Profit

= 10,000 + 1,000

= Rs. 11,000

Loss % on another T.V. set = 10%

Loss = 10% of C.P.

= \(\frac{10}{100} \times 10000\)

= Rs. 1000

∴ S.P. = C.P. – Loss

= 10,000 – 1,000

= Rs. 9,000

Total C.P. of both T.V. sets

= 10,000 + 10,000

= Rs. 20,000

Total S.P. of both T.V. sets = 11,000 + 9000

= Rs. 20,000

Hence C.P. = S.P.

∴ No profit or loss.

Try These (Pages 124 – 125)

Question 1.

Find the buying price of each of the following when 5% S.T. is added on the purchase of

(a) A towel at Rs. 50

(b) Two bars of soap at Rs. 35 each

(c) 5 kg of flour at Rs. 15 per kg.

Solution:

(a) Price of a towel = Rs. 50

S.T. = 5%

Amount of S.T. = 5% of price of towel

= \(\frac{5}{100} \times 50\)

= Rs. 2.50

∴ Buying price = Price + S.T.

= 50 + 2.50

= Rs. 52.50

(b) Price of two bars of soap = 35 × 2 = Rs. 70

S.T. = 5%

∴ Amount of S.T. = 5% of price of soap

= \(\frac{5}{100} \times 70\)

= Rs. 3.50

∴ Total price = Price + S.T.

= 70 + 3.50

= Rs. 73.50

(c) Price of 1 kg of flour = Rs. 15

Price of 5 kg of flour = 15 × 5 = Rs. 75

S.T. = 5%

∴ Amount of S.T. = 5% of price of flour

= \(\frac{5}{100} \times 75\)

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

∴ Total price of flour = Price + S.T.

= 75 + 3.75

= Rs. 78.75

Question 2.

If 8% VAT is included in the price, find the original price of

(a) A TV bought for Rs. 13,500

(b) A shampoo bottle bought for Rs. 180.

Solution:

(a) Let marked price = Rs. 100

V.A.T. = 8%

Amount of V.A.T. = 8% of 100

= \(\frac{8}{100} \times 100\)

= Rs. 8

Total price = 100 + 8 = Rs.108

∴ Price of T.V. included V.A.T. = 13,500

Original price = \(\frac{100}{108} \times 13500\) = Rs. 12,500

(b) Let marked price = Rs. 100

V.A.T. = 8%

Amount of V.A.T. = 8% of 100

= \(\frac{8}{100} \times 100\)

= Rs. 8

Total price = 100 + 8 = Rs. 108

∴ Price of Shampoo = Rs. 180

Original price = \(\frac{100}{108} \times 180\) = Rs. 166.67

Try These (Page 126)

Question 1.

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

Solution:

P = Rs. 15,000

R = 5% (p.a)

T = 2 years

S.I. = \(\frac{\mathbf{P} \times \mathbf{R} \times \mathbf{T}}{100}\)

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

= Rs. 1500

Amount = S.I. + P

= 15000 + 1500

= Rs. 16,500

Try These (Page 129)

Question 2.

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

Solution:

P = Rs. 8000

Time = 2 years

Rate = 5%

A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)

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

= \(8000\left(1+\frac{1}{20}\right)^{2}\)

= \(8000 \times\left(\frac{21}{20}\right)^{2}\)

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

= 20 × 21 × 21

= 20 × 441

= Rs. 8820

C.I. = A – P

= 8820 – 8000

= Rs. 820

Try These (Page 130)

Question 1.

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. Time = \(\frac{3}{2}\) years

= \(\frac{3}{2}\) × 2

= 3 (half years)

Rate = 8 (p.a) = \(\frac{8}{2}\) = 4 (half yearly)

2. Time = 2 years

= 2 × 2

= 4 (half years)

Rate = 4% (p.a)

= \(\frac{4}{2}\)

= 2% (half yearly)

Try These (Page 131)

Question 1.

Find the amount to be paid

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

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

Solution:

1. P = Rs. 2400

R = 5% (p.a)

T = 2

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(2400\left(1+\frac{5}{100}\right)^{2}\)

= \(2400 \times \frac{21}{20} \times \frac{21}{20}\)

= 6 × 21 × 21

= 6 × 441

= Rs. 2646

2. P = Rs. 1800

R = 8% (p.a), or (\(\frac{8}{4}\) = 2 % quarterly)

Time = 1 year = 4 quarters

A = \(P\left(1+\frac{R}{100}\right)^{T}\)

= \(1800\left(1+\frac{2}{100}\right)^{4}\)

= \(1800 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}\)

= \(\frac{121773618}{62500}\)

= Rs. 1948.38

Try These (Page 133)

Question 1.

Machinery worth Rs. 10,500 depreciated by 5%. Find its value after one year.

Solution:

P = Rs. 10,500

R = 5% (p.a)

Time = 1 year

Machineis depreciated

A = \(P\left(1-\frac{R}{100}\right)^{T}\)

= \(10,500\left(1-\frac{5}{100}\right)^{1}\)

= \(10,500 \times \frac{95}{100}\)

= 105 × 95

= Rs. 9975

Question 2.

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 = 12 lakh

Time = 2 years

Rate of increase in population = 4% (p.a.)

Population after 2 years = 12,00,000 \(\left(1+\frac{4}{100}\right)^{2}\)

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

= 1297920