Class 7 Comparing Quantities Ex 7.2

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.

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

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

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

Solution:

(a) Here, CP = ₹ 250

SP = ₹ 325

Since SP > CP

∴ Profit = SP – CP

= ₹ 325 – ₹ 250 = ₹ 75

Profit %=\({\text{Profit} \times 100 \over \text{CP}}={75 \times 250 \over 100}=30\%\)

Hence, the required profit = ₹ 75

(b) Here, CP = ₹ 12,000

SP = ₹ 13,500

Since SP > CP

∴ Profit = SP – CP = ₹ 13,500 – ₹ 12,000 = ₹ 1,500

Profit %=\({\text{Profit} \times 100 \over \text{CP}}={1500 \times 100 \over 12000}={25 \over 2}\%=12{1 \over 2}\%\)

profit % = 12.5%

(c) Here, CP = ₹ 2500

SP = ₹ 3000

Since SP > CP

∴ Profit = SP – CP

= ₹ 3000 – ₹ 2500 = ₹ 500

Profit %=\({\text{Profit} \times 100 \over \text{CP}}={500 \times 2500 \over 100}=20\%\)

Hence, the required profit = ₹ 500 and profit% = 20%

(d) Here, CP =₹ 250

SP = ₹150

Here CP > SP

∴ Loss = CP – SP

= ₹ 250 – ₹ 150 = ₹ 100

Loss %=\({\text{Loss} \times 100 \over \text{CP}}={100 \times 100 \over 250}=40\%\)

Hence, the required loss = ₹ 100 and loss% = 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

Sum of the ratio parts = 3 + 1 = 4

Percentage of first part=\({3 \over 4} \times 100 =75\%\)

Percentage of Second part=\({1 \over 4} \times 100 =25\%\)

(b) 2 : 3 : 5

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

Percentage of first part=\({2 \over 10} \times 100 =20\%\)

Percentage of Second part=\({3 \over 10} \times 100 =30\%\)

Percentage of Third part=\({5 \over 10} \times 100 =50\%\)

(c) 1 : 4

Sum of the ratio parts =1 + 4 = 5

Percentage of first part=\({1 \over 5} \times 100 =20\%\)

Percentage of Second part=\({4 \over 5} \times 100 =80\%\)

(d) 1 : 2 : 5

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

Percentage of first part=\({1 \over 8} \times 100 =12{1 \over 2}\%\)

Percentage of Second part=\({2 \over 8} \times 100 =25\%\)

Percentage of Third part=\({5 \over 8} \times 100 =62{1 \over 2}\%\)

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

Percentage of decrease in population =\({500 \times 100 \over 25000}= 2\%\)

Hence the Percentage of decrease in population = 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:

Original price of the car = ₹ 3,50,000

Price increased next year = ₹ 3,70,000

Increase in price = ₹ 3,70,000 – ₹ 3,50,000

= ₹ 20,000

∴ Percentage of the increase in the price

=\[{20000 \times 100 \over 350000}={40 \over 7}\%=5{5\over 7}\%\]

Hence, the Percentage of increase in price = \(5{5\over 7}\)% 

Question 5.

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

Solution:

Here, CP = ₹ 10,000

Profit = 20%

SP = ?

SP=CP(1+\(\text{Profit}\over 100\))=10000(1+\({20 \over 100}\))

=\(10000 \times {6 \over 5}\)=₹12000

Hence, the required money got by me = ₹ 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 the washing machine =₹ 13,500

Loss = 20%

CP = ?

SP=CP(1-\(\text{Loss} \over 100\))

13500=CP(1-\(20 \over 100\))

13500=CP(1-\(1 \over 5\))

13500=CP(\(5-1 \over 5\))

13500=CP(\(4 \over 5\))

CP=\(13500 \times {5 \over 4}\)=₹16875

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) Sum of the ratio parts = 10 + 3 + 12 = 25

∴ Percentage of carbon in chalk =\({3 \over 25 }\times 100\%=12\%\)

Hence, the Percentage of carbon in chalk = 12%

weight of the chalk stick=\({100\over 12} \times 3\)=25 g

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 book = ₹ 275

Loss = 15%

SP=CP\((1-{\text{Loss} \over 100})=275(1-{15 \over 100})\)

=\(275 \times {85 \over 100}\)=₹233.75

Hence, the required selling price = ₹ 233.75

Question 9.

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

(a) Principal = ₹ 1200 at 12% p.a.

(b) Principal = ₹ 7500 at 5% p.a.

Solution:

(a) Given: Principal = ₹ 1200

Rate of interest = 12% p.a., T = 3 years

Interest=\({P \times R \times T \over 100}={1200 \times 12 \times 3 \over 100}=432\)

Amount = Principal + Interest

= ₹ 1200 + ₹ 432 =₹ 1632

Hence, the required amount = ₹ 1632

(b) Given: Principal = ₹ 7500

Rate = 5% p.a.

Time = 3 years

Interest=\({P \times R \times T \over 100}={7500 \times 5 \times 3 \over 100}\)

= ₹1125

Amount = Principal + Interest

= ₹ 7500 + 1125 = ₹ 8625

Hence, the required amount = ₹ 8625.

Question 10.

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

Solution:

Given: Principal = ₹ 56,000

Interest = ₹ 280

Time = 2 years

Rate = ?

Rate=\({100 \times I \over P \times T}={100 \times 280 \over 56000 \times 2}={1 \over 4}\%=0.25\%\)

Hence, the required 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:

Given: Interest = ₹ 45

Time = 1 year

Rate = 9% p.a.

Principal=\({100 \times I \over R \times T}={100 \times 45 \over 9 \times 1}\)

=₹ 500

Hence, the required sum = ₹ 500.