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## 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.

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.

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.

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.
Cost Price = ₹ 250
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

= 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

= 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

= 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

= 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.

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?

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?

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?

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.

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?
Let the weight of chalk = x grams
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?

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.

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.

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?

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?

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.