Exercise 7.2

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 :

Here, CP = ₹ 250 , SP = ₹ 325 , Since SP > CP

∴ Profit = SP – CP

= ₹ – ₹ 250 = ₹ 756566

= Profit % = Profitx100CP = 75100 × 100 = 304050%

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

Solution :

Here, CP = ₹ 12,000 ; SP = ₹ 13,500 ; Since SP > CP

∴ Profit = SP – CP

= ₹ – ₹ 12,000 = ₹ 1,5002,0001,800

= Profit % = ProfitCP × = 1500×10012000

= 252% = 121210141516%

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

Solution :

Here, CP = ₹ 2500 ; SP = ₹ ; Since SP > CP

∴ Profit = SP – CP

= ₹ – ₹ 2500 = ₹ 50010001500

= Profit % = ProfitCp × 100 = 5002500 × 100 = 203040%

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

Solution :

Here, CP = ₹ ; SP = ₹ 150 ; Here CP > SP

∴ Loss = CP –

= ₹ 250 – ₹ 150 = ₹ 100150300

= Loss % = Loss×100CP = 100×100250 = 405060%

2. Convert each part of the ratio to percentage.

Solution :

(a) 3 : 1

Sum of the ratio parts = 3 + 1 = 4

Percentage of first part = 34 × 100 = 756555%

Percentage of second part = 14 × 100 = 253515%

(b) 2 : 3 : 5

Percentage of first part = 210 × 100 = 205060%

Percentage of second part = 310 × 100 = 308090%

Percentage of third part = 510 × 100 = 50100150%

(c) 1 : 4

Percentage of first part = 15 × 100 = 201030%

Percentage of second part = 45 × 100 = 8090100%

(d) 1 : 2 : 5

Percentage of first part = 18 × 100 = 121210151514%

Percentage of second part = 28 × 100 = 253545%

Percentage of third part = 58 × 100 = 621255151018%

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

Solution :

Instruction

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 :

Instruction

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 :

Here, CP = ₹ 10,000 , Profit = % , SP = ?

SP = CP (1+Profit100) = (1+20100)

= 10,000 × 65 = ₹ 12,00010,000

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 = % , CP = ?

⇒ SP = CP(1 - Loss100)

⇒ = CP (1 - 20100)

⇒ 13500 = CP (1 - )

⇒ 13500 = CP × 5

CP = 13500 × 4

CP = 3375 × 5 = ₹ 168751600016580

7. (i) Chalk contains calcium, carbon and oxygen in the ratio 1012. Find the percentage of carbon in chalk.

Solution :

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

∴ Percentage of carbon in chalk

= 325 × 100% = %

Hence, the Percentage of carbon in chalk = 121015%

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

Solution :

Weight of carbon = g

∴ Weight of chalk = 33 × 25 g = g

Hence, the weight of chalk = 253520g

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 = ₹ , Loss = %

SP = CP(1-Loss100) = 275(1 - 100)

= 275 × 85100 = ₹ 233.75250234.6

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

Solution :

(a) Principal = ₹ 1,200 at 12% p.a.

Given: Principal = ₹ , Rate of interest = 12% p.a , T = 3 years

= 1200×12×3100

Amount = Principal + Interest

= ₹ 1200 + ₹ = ₹ 163220001806

Hence, the required amount = ₹ 1632

(b) Principal = ₹ 7,500 at 5% p.a.

Given: Principal = ₹ , Rate = 5% p.a , Time = 3 years

= 7500×5×3100

Amount = Principal + Interest

= ₹ 7500 + = ₹ 862585008636

Hence, the required amount = ₹ 8625.

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

Solution :

Given: Principal = ₹ , Interest = ₹ 280 , Time = years , Rate = ?

Rate = 100×IP×T = 100×28056000×2 = 141615

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 = ₹ , Time = year , Rate = 9% p.a.

Principal = 100×IR×T = 100×459×1 = ₹500550450