## Chapter 8 Comparing Quantities Exercise 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 ₹ 250 and sold for ₹ 325.Answer:**

Cost Price = ₹ 250

Selling Price = ₹ 325

Selling Price > Cost Price

Therefore, there is a profit.

Profit = SP - CP

= 325 - 250 = ₹ 75

Profit Percent

= Profit/CP x 100 = 75/250 x 100 = 7500/250 = 30%

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

Answer:

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

Answer:

Cost Price = ₹ 12000

Selling Price = ₹ 13500

Selling Price > Cost Price

Therefore, there is a profit.

Profit = SP - CP

= 13500 - 12000 = ₹ 1500

Profit Percent

= Profit/CP x 100 = 1500/12000 x 100 = 150000/12000 = 12.5%

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

Answer:

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

Answer:

Cost Price = ₹ 2500

Selling Price = ₹ 3000

Selling Price > Cost Price

Therefore, there is a profit.

Profit = SP - CP

= 3000 - 2500 = ₹ 500

Profit Percent

= Profit/CP x 100 = 500/2500 x 100 = 50000/2500 = 20%

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

Answer:

Cost Price = ₹ 250

Answer:

Selling Price = ₹ 150

Selling Price < Cost Price

Therefore, there is a loss.

Loss = CP - SP

= 250 - 150 = ₹ 100

Profit Percent

= Loss/CP x 100 = 100/250 x 100 = 10000/250 = 40%

Question 2: Convert each part of the ratio to percentage:

(a) 3 : 1

Answer:

Question 2: Convert each part of the ratio to percentage:

(a) 3 : 1

Answer:

= 3/(3+1) = 3/4 = 3/4 x 100 = 300/4 = 75%

= 1/(3+1) = 1/4 = 1/4 x 100 = 100/4 = 25%

**(b) 2: 3: 5**

Answer:

Answer:

= 2/(2+3+5) = 2/10 = 2/10 x 100 = 200/10 = 20%

= 3/(2+3+5) = 3/10 = 3/10 x 100 = 300/10 = 30%

= 5/(2+3+5) = 5/10 = 5/10 x 100 = 500/10 = 50%

(c) 1:4

Answer:

(c) 1:4

Answer:

= 1/(1+4) = 1/5 = 1/5 x 100 = 100/5 = 20%

= 4/(1+4) = 4/5 = 4/5 x 100 = 400/5 = 80%

(d) 1: 2: 5

Answer:

(d) 1: 2: 5

Answer:

= 1/(1+2+5) = 1/8 = 1/8 x 100 = 100/8 = 12.5%

= 2/(1+2+5) = 2/8 = 2/8 x 100 = 200/8 = 25%

= 5/(1+2+5) = 5/8 = 5/8 x 100 = 500/8 = 62.5%

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

Answer:

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

Answer:

Original population = 25000

Population after decrease = 24500

Decrease = 25000 - 24500 = 500

Decrease percentage = decrease/(original value) x 100

= 500/25000 x 100 = 50000/25000 = 2%

Therefore, the percentage decreased is 2%.

**Question 4: Arun bought a car for ₹ 350000. The next year, the price went upto ₹ 370000. What was the percentage of price increase?**

Answer:

Answer:

Original price of car = ₹ 350000

Price of car after increase in price = ₹ 370000

Increase in price = 370000 - 350000 = ₹ 20000

Increase percentage = increse/(original value) x 100

= 20000/350000 x 100 = 2000000/350000 = 40/7 = 5 5/7%

**Question 5: I buy a T.V. for ₹ 10000 and sell it at a profit of 20%. How much money do I get for it?**

Answer:

Answer:

Cost Price = ₹ 10000

Profit Percent = 20%

Selling Price = ?

Profit 20% of 10000

= 20/100 x 10000 = ₹ 2000

Therefore, selling price = cost price + profit

= 10000 + 2000 = ₹ 12000

Therefore, he get ₹ 12000 money for TV.

**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?**

Answer:

Answer:

Selling Price = ₹ 13500

Loss percent = 20%

Cost Price = ?

Cost Price = 100/(100- Loss percent) x SP

= 100/(100-20) x 13500

= 100/80 x 13500 = ₹ 16875

Therefore, Juhi brought it for ₹ 16875.

**Question 7: (i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.**

Answer:

Answer:

Calcium : Carbon : Oxygen

= 10 : 3 : 12

Total parts = 10 + 3 + 12 = 25

Fraction of carbon = 3/25

Percentage of carbon = 3/25 x 100 = 300/25 = 12%

Therefore, the percentage of carbon in chalk is 12%.

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

Answer:

Let the weight of chalk = x grams

Answer:

Weight of carbon = 3 grams

12% of x = 3

= 12/100 x x = 3

x = (3 x 100)/12 = 300/12 = 25 grams

Therefore, the weight of the chalk is 25 grams.

**Question 8: Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?**

Answer:

Answer:

Cost Price = ₹ 275

Loss percent = 15%

Selling Price = ?

Selling Price = (100- Loss percent)/100 x Cost Price

= (100-15)/100 x 275 = 23375/100 = 935/4 = 233 3/4 = ₹ 233.75

Therefore, she sold the book for ₹ 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.

Answer:

(a) Principal = ₹ 1,200 at 12% p.a.

Answer:

Principal (P) = ₹ 1200

Rate of Interest (R) = 12%

Time (T) = 3 years

Interest = (Principal x Time x Rate)/100

= (1200 x 3 x 12)/100 = ₹432

Amount = Principal + Interest

= 1200 + 432

= ₹ 1632

(b) Principal = ₹ 7,500 at 5% p.a.

Answer:

(b) Principal = ₹ 7,500 at 5% p.a.

Answer:

Principal (P) = ₹ 7500

Rate of Interest (R) = 5%

Time (T) = 3 years

Interest = (Principal x Time x Rate)/100

= (7500 x 3 x 5)/100 = ₹1125

Amount = Principal + Interest

= 7500 + 1125

= ₹ 8625

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

Answer:

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

Answer:

Rate = ?

Interest = ₹ 280

Time = 2 years

Rate = (100 x Interest)/(Principal x Time)

= (100 x 280)/(56000 x 2) = 28000/112000 = 1/4%

Question 11: If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?

Answer:

Question 11: If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?

Answer:

Interest = ₹ 45

Rate = 9%

Time = 1 year

Principal = ?

Simple Interest = (Principal x Rate x Time)/100

= 45 = (p x 9 x 1)/100

= p = (45 x 100)/9 = 4500/9

= ₹ 500

Therefore, she borrowed ₹ 500.

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