Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 1.

Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

(a) Gardening shears bought for ā¹ 250 and sold for ā¹ 325.

(b) A refrigerate bought for ā¹ 12,000 and sold at ā¹ 13,500.

(c) A cupboard bought for ā¹ 2,500 and sold at ā¹ 3,000.

(d) A skirt bought for ā¹ 250 and sold at ā¹ 150.

Solution:

(a) Cost price (CP) = ā¹ 250

Selling price (SP) = ā¹ 325

āµ CP < SP so there is a profit.

ā“ Profit = SP – CP

= ā¹ 325 – ā¹ 250 = ā¹ 75

Now, Profit % = \(\frac { Profit }{ CP }\) x 100%

= \(\frac { 75 }{ 250 }\) x 100 % = 30%

(b) Cost price (CP) = ā¹ 12,000

Selling price (SP) = ā¹ 13,500

āµ CP < SP, so there is a profit.

ā“ Profit = SP – CP = ā¹ 13,500 – ā¹ 12,000 = ā¹ 1,500

Now, Profit % = \(\frac { Profit }{ CP }\) x 100%

= \(\frac { 1500 }{ 12000 }\) x 100%

= \(\frac { 150 }{ 12 }\)

= 12.5%

(c) Cost price (CP) = ā¹ 2500

Selling price (SP) = ā¹ 3000

āµ CP < SP, so there is a profit.

ā“ Profit = SP – CP = ā¹ 3000 – ā¹ 2500 = ā¹ 500

Now, Profit % = \(\frac { Profit }{ CP }\) x 100%

= \(\frac { 500 }{ 2500 }\) x 100% = 20%

(d) Cost price (CP) = ā¹ 250

Selling price (SP) = ā¹ 150

āµ CP > SP,

so there is a loss.

ā“ Loss = CP – SP

= ā¹ 250 – ā¹ 150 = ā¹ 100

Now, Loss % = \(\frac { Profit }{ CP }\) x 100%

= \(\frac { 100 }{ 250 }\) x 100% = 40%

Question 2.

Convert each part of the ratio to percentage:

(a) 3 : 1

(b) 2 : 3 : 5

(c) 1 : 4

(d) 1 : 2 : 5

Solution:

(a) 3:1

Total of the parts = 3 + 1 = 4

ā“ Percentage of the 1st part of the ratio

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

Percentage of the 2nd part of the ratio = \(\frac { 1 }{ 4 }\) x 100% = 25%

(b) 2 : 3 : 5

Total of the parts = 2 + 3 + 5 = 10

ā“ Percentage of 1st part of the ratio

= \(\frac { 2 }{ 10 }\) x 100% = (2 x 10)% = 20%

Percentage of 2nd part of the ratio

= \(\frac { 3 }{ 10 }\) x 100% = (3 x 10)% = 30%

Percentage of 3rd part of the ratio

= \(\frac { 5 }{ 10 }\) x 100% = (5 x 10)% = 50%

(c) 1 : 4

Total of parts = 1 + 4 = 5

ā“ Percentage of the 1st part of the ratio

= \(\frac { 1 }{ 5 }\) x 100% = (1 x 20)% = 20%

Percentage of the 2nd part of the ratio

= \(\frac { 4 }{ 5 }\) x 100% = (4 x 20)% = 80%

(d) 1 : 2 : 5

Total of the parts = 1 + 2 + 5 = 8

ā“ Percentage of the 1st part of the ratio

= \(\frac { 1 }{ 8 }\) x 100% = \(\frac { 25 }{ 2 }\) = 12.5%

Percentage of the 2nd part of the ratio

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

Percentage of the 3rd part of the ratio

= \(\frac { 5 }{ 8 }\) x 100% = \(\frac { 125 }{ 2 }\) % = 62.5%

Question 3.

The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Solution:

Initial population = 25,000

Decreased population = 24,500

Decrease in population = 25,000 – 24,500 = 500

Question 4.

Arun bought a car for ā¹ 3,50,000. The next year, the price went upto ā¹ 3,70,000. What was the percentage, of price increase?

Solution:

Initial cost = ā¹ 3,50,000

Increased cost = ā¹ 3,70,000

Increase in cost

= ā¹ 3,70,000 – ā¹ 3,50,000

= ā¹ 20,000

Question 5.

1 buy a T.V for ā¹ 10,000 and sell it at a profit of 20%. How much money do I get for it?

Solution:

CP of the T.V. = ā¹ 10,000

Profit percent = 20%

ā“ Profit = 20% of 10,000

= ā¹ \(\frac { 20 }{ 100 }\) x 10,000

= ā¹ 2,000

Now, SP = CP + Profit

Thus, I will get ā¹ 12,000 for the T.V.

= ā¹ 10,000 + ā¹ 2,000 = ā¹ 12,000

Thus, I will get ā¹ 12,000 for the T.V.

Question 6.

Juki sells a washing machine for ā¹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?

Solution:

Selling price (SP) = ā¹ 13,500

Loss % = 20%

CP = ?

ā“ CP – Loss = SP

ā“ CP – (20% of CP) = ā¹ 13,500

= ā¹ 13,500

or CP – \(\frac { 20 }{ 100 }\)CP = ā¹ 13,500

or \(\frac { 5CP – CP }{ 5 }\) = ā¹ 13,500

or \(\frac { 4 }{ 5 }\)CP = ā¹ 13,500

ā CP = ā¹ 13,500 x \(\frac { 5 }{ 4 }\)

= ā¹ 3,375 x 5

= ā¹ 16,875

ā“ Juhi bought the washing machine for ā¹ 16,875.

Question 7.

(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.

(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?

Solution:

(i) āµ The ratio of calcium, carbon and oxygen in mixture is 10 : 3 : 12.

ā“ Total of ratios = 10 + 3 + 12 = 25

ā“ Percentage of carbon in the chalk mixture

= \(\frac { 3 }{ 25 }\)

(ii) Let the weight of the stick be x g.

āµ 12% of the chalk mixture is 3 g.

ā“ 12% of x = 3

or \(\frac { 12 }{ 100 }\) Ć x = 3 ā x = \(\frac { 3Ć100 }{ 12 }\) = 25

or Weight of the chalk stick = 25 g.

Question 8.

Amina buys a book for ā¹ 215 and sells it at a loss of 15%. How much does she sell it for?

Solution:

Cost price of the book (CP) = ā¹ 275

Loss % = 15%

ā“ Loss = 15% of CP = ā¹ \(\frac { 15 }{ 100 }\) x 275

= ā¹ \(\frac { 15Ć11 }{ 4 }\)

= ā¹ \(\frac { 165 }{ 4 }\)

= ā¹ 41.25

Now SP = CP – Loss

= ā¹ 275 – ā¹ 41.25

= ā¹ 233.75

Thus, Amina will sell the book for ā¹ 233.75.

Question 9.

Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ā¹ 1,200 at 12% p.a.

(b) Principal = ā¹ 7,500 at 5% p.a.

Solution:

(a) Here, Principal (P) = ā¹ 1,200

Rate (R) = 12% p.a. and Time (T) = 3 years

ā“ Interest = \(\frac { PĆRĆT }{ 100 }\) = ā¹\(\frac { 1,200Ć12Ć3 }{ 100 }\)

ā¹ (12 x 12 x 3) = ā¹ 432

Now, Amount = Principal + Interest

= ā¹ 1,200 + ā¹ 432 = ā¹ 1,632

(b) Principal (P) = ā¹ 7,500

Rate (R) = 5% p.a.

Time (T) = 3 years

ā“ Interest = \(\frac { PĆRĆT }{ 100 }\) = ā¹\(\frac { 7,500Ć5Ć3 }{ 100 }\)

= ā¹ (75 x 5 x 3) = ā¹ 1,125

Now, Amount = Principal + Interest

= ā¹ 7,500 + ā¹ 1,125 = ā¹ 8,625

Question 10.

What rate gives ā¹ 280 as interest on a sum of ā¹ 56,000 in 2 year?

Solution:

Principal (P) = ā¹ 56,000

Rate (R) = ?

Time (T) = 2 years

Interest = ā¹ 280

āµ Interest = \(\frac { PĆRĆT }{ 100 }\)

ā“ 280 = \(\frac { 56,000ĆRĆ2 }{ 100 }\)

or R = \(\frac { 280Ć100 }{ 2Ć56,000 }\) = \(\frac { 1 }{ 4 }\)% = 0.25%

Thus, 0.25% p.a. of interest rate will give the required interest.

Question 11.

If Meena gives an interest of ā¹ 45 for one year at 9% rate p.a. What is the sum she has borrowed.

Solution:

rincipal (P) = ?

Rate (R) = 9% p.a.

Interest = ā¹ 45

Time (T) = 1 year

Since Interest = \(\frac { PĆRĆT }{ 100 }\)

or ā¹ 45 = \(\frac { 45Ć100 }{ 9 }\)

= ā¹ 5 x 100

= ā¹ 500