Well-designed AP 7th Class Maths Guide Chapter 7 Comparing Quantities Exercise 7.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Lines and Angles Class 7 Exercise 7.2 Solutions – 7th Class Maths 7.2 Exercise Solutions

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 ₹ 250 and sold for ₹ 325.

Solution:

CP = ₹250; SP = ₹325

SP > CP

Profit = SP – CP = ₹325 – ₹250 = ₹75

Profit % = \(\frac { P }{ CP }\) × 100 = \(\frac { 75 }{ 250 }\) × 100

= \(\frac { 75 }{ 25 }\) × 10 = 3 × 10 = 30 %

b) A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500.

Solution:

CP = ₹ 12,000 ; SP = ₹ 13,500

SP > CP

Profit = SP – CP

=₹ 13,500 – ₹ 12,000 = ₹ 1,500

Profit % = \(\frac { P }{ CP }\) × 100 = \(\frac { 1500 }{ 12000 }\) × 100

= \(\frac { 15 }{ 120 }\) × 100 = \(\frac{15 \times 10}{12}\) = \(\frac{5 \times 10}{4}\)

= \(\frac{5 \times 5}{2}\) = \(\frac { 25 }{ 2 }\) = 12\(\frac { 1 }{ 2 }\)% = 12.5%

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

Solution:

CP = ₹ 2,500; SP = ₹ 3,000

SP > CP

Profit = SP – CP

=₹ 3,000 – ₹ 2,500 = ₹ 500

Profit % = \(\frac { P }{ CP }\) × 100

= \(\frac { 500 }{ 2500 }\) 100 = \(\frac { 500 }{ 25 }\)

Profit % = 20 %

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

Solution:

CP = ₹250; SP = ₹ 150

SP < CP

Loss = CP – SP

= ₹250 – ₹ 150 = ₹ 100

loss % = \(\frac { l }{ CP }\) × 100 = \(\frac { 100 }{ 250 }\) × 100

= \(\frac { 100 }{ 25 }\) × 10 = 4 × 10 = 40%

Question 2.

Convert each part of the ratio to percentage :

a) 3 : 1

Solution:

Total Parts = 3 + 1 = 4

% of first part = \(\frac { 3 }{ 4 }\) × 100 = 3 × 25 = 75%

% of second part = \(\frac { 1 }{ 4 }\) × 100 = 25%

b) 2 : 3 : 5

Solution:

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

% of first part = \(\frac { 2 }{ 10 }\) × 100 = 2 × 10 = 20%

% of second part = \(\frac { 3 }{ 10 }\) × 100 = 30%

% of third part = \(\frac { 5 }{ 10 }\) × 100 = 50%

c) 1 : 4

Solution:

Total parts of the ratio = 1 + 4 = 5

% of first part = \(\frac { 1 }{ 5 }\) × 100 = 20%

% of second part = \(\frac { 4 }{ 5 }\) × 100 = 4 × 20 = 80%

d)1 : 2 : 5

Solution:

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

% of first part = \(\frac { 1 }{ 8 }\) × 100

\(\frac { 50 }{ 4 }\) = \(\frac { 25 }{ 2 }\) = 12\(\frac { 1 }{ 2 }\)% = 12.5%

% of second part = \(\frac { 2 }{ 8 }\) × 100

= \(\frac { 1 }{ 4 }\) × 100 = 25%

% of third part = \(\frac { 5 }{ 8 }\) × 100

= \(\frac{5 \times 50}{4}\) = \(\frac{5 \times 25}{2}\) = \(\frac { 125 }{ 2 }\) = 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

Changed population = ₹ 24,500

Difference = 25,000 – 24,500 = 500

Decrease % = \(\frac { 500 }{ 25,000 }\) × 100

= \(\frac { 5 }{ 250 }\) × 100

= \(\frac{5 \times 10}{25}\) = \(\frac { 50 }{ 25 }\) = 2%

Question 4.

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

Solution:

CP of car = ₹ 3,50,000

Price of car after I year = ₹ 3,70,000

Increased price = ₹ 3,70,000 – ₹ 3,50,000 = ₹ 20,000

% of increase = \(\frac { 20,000 }{ 3,50,000 }\) × 100

= \(\frac { 2 }{ 35 }\) × 100 = 5\(\frac { 5 }{ 7 }\)%

Question 5.

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

Solution:

CP of T.V. = ₹ 10,000; Profit = 20%

20% of 10,000 = \(\frac { 20 }{ 100 }\) × 10,000 = ₹ 2,000

Amount received after selling at 20%

profit = ₹ 10,000 + ₹ 2,000 = ₹ 12,000

Question 6.

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

Solution:

SP of washing machine = ₹ 13,500 Loss = 20%

Let the CP = ₹ x; CP – Loss = SP.

x – \(\frac { 20 }{ 100 }\) × x = 13,500

x – \(\frac { 1 }{ 5 }\)x = 13500 ⇒ \(\frac { 4x }{ 5 }\) = 13500

x = 13500 × \(\frac { 5 }{ 4 }\) ⇒ x = 16875

∴ Cost of washing machine = ₹ 16,875

Question 7.

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

Solution:

Given ratio = 10 : 3 : 12

Sum of all parts of ratio = 10 + 3 + 12 = 25

% of carbon in chalk = \(\frac { 3 }{ 25 }\) × 100 = 3 × 4 = 12%

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

Solution:

Carbon in the stick of chalk = 3g.

∴ 12% of carbon = 3g

Total weight of the chalk = x.

\(\frac { 12 }{ 100 }\) × x = 3 ⇒ x = \(\frac{3 \times 100}{12}\) ⇒ x = 25 g

Total weight of chalk stick = 25g.

Question 8.

Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?

Solution:

CP of the book = ₹ 275; Loss = 5%

15% of 275 = \(\frac { 15 }{ 100 }\) × 275

\(\frac { 15 }{ 4 }\) × 11 = ₹ 41.25

SP = CP – Loss

SP = 275 – 41.25

SP = ₹ 233.75

Question 9.

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

a) Principal = ₹ 1,200 at 12% p.a.

Solution:

P = ₹ 1,200; R = 12% p.a. ; T = 3 years

I = \(\frac { PTR }{ 100 }\) = \(\frac{1200 \times 3 \times 12}{100}\)

= 12 × 3 × 12 = 144 × 3 = ₹ 432

Arnount = P + 1

=1200 + 432 = ₹ 1,632

b) Principal = ₹7,500 at 5% p.a.

Solution:

P = ₹ 7,500 ; R = 5% p.a. ; T = 3 years

I = \(\frac { PTR }{ 100 }\) = \(\frac{7500 \times 3 \times 5}{100}\)

=75 × 15 = ₹ 1,125

Amount = P + I

= ₹ 7,500 + ₹ 1,125 = ₹ 8,625

Question 10.

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

Solution:

I = ₹ 280 ; P = ₹ 56,000 ; T = 3 years

I = \(\frac { PTR }{ 100 }\)

280 = \(\frac{56,000 \times 2 \times \mathrm{R}}{100}\)

R = \(\frac{28 \times 100}{5600 \times 2}\) = \(\frac{28}{56 \times 2}\)

R = \(\frac { 1 }{ 4 }\) ⇒ R = 0.25%

∴ Rate = 0.25%

Question 11.

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

Solution:

I = ₹45 ; T = 1 year; R = 9%

I = \(\frac { PTR }{ 100 }\)

45 = \(\frac{P \times 1 \times 9}{100}\)

45 × 100 = P × 9

P = \(\frac{45 \times 100}{9}\) = 5 × 100 ⇒ P = ₹ 500

∴ The sum Meena borrowed = ₹ 500