EXERCISE 8.3
1). Tell what is the profit or loss in the following transaction. Also find profit percent or loss percent in each case.
a) Gardening shears bought for ₹ 250 and sold for ₹ 325
Ans:
Cost price of gardening shears = ₹ 250
Selling price of gardening shears = ₹ 325
Since SP > CP, here there is profit
Profit = SP – CP
= 325 – 250 = ₹ 75
Profit % = Profit/CP × 100
= 75/250 ×100 = 30 % ∴ Profit =30 %
b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500
Ans: Cost price for refrigerator = ₹ 12,000
Selling price for refrigerator = ₹ 13,500
Since SP > CP, here there is profit
Profit = SP – CP
= 13,000 – 12,000 = ₹ 1500
Profit % = Profit/CP ×100
= 15000/12000 = 12.5 % ∴ Profit= 12.5 %
c) A cupboard bought for ₹ 2500 and sold at ₹ 3000.
Ans: Cost price of cupboard = ₹ 2,500
Selling price of cupboard = ₹ 3,000
Since, SP > CP, here there is profit
Profit = SP – CP
= 3,000 – 2,500 = 500
Profit % = Profit/CP ×100
= 500/2500 ×100
= 50000/2500 = 20% ∴ Profit = 20%
d) A skirt bought for ₹ 250 and sold at ₹ 150.
Ans: Cost price of skirt = ₹ 250
Selling price of skirt = ₹ 150
Since, CP > SP, here there is loss
Loss = CP – SP
= 250 – 150 = ₹ 100
Loss % = Loss/CP × 100
= 10000/250 = 40% ∴ Loss = 40%
2). Convert each part of the ratio to percentage.
a) 3 : 1
Ans: Ratio = 3 : 1
Total parts = 3 + = 4
1st part = 3/4 ×100 = 3 × 25% = 75%
2nd part = 1/4 ×100 = 1 × 25% = 25%
b) 2 : 3 : 5
Ans: Ratio = 2 : 3 : 5
Total parts = 2 + 3 + 5 = 10
1st part = 2/10 ×100%
= 2 ×10% =20%
2nd part = 3/10 ×100
= 3 × 10% = 30%
3rd part = 5/10 ×100
= 5 ×10% = 50%
c) 1:4
Ans: Ratio = 1 : 4
Total parts = 1 + 4 = 5
1st part = 1/5 ×100 = 1 × 20% = 20%
2nd part = 4/5 ×100 = 4 × 20% = 80%
d) 1 : 2 : 5
Ans: Ratio = 1 : 2 : 5
1st part = 1/8 ×100 = (100%)/8 = 12.5%
2nd part = 2/8 ×100 = (100%)/4 = 25%
3rd part = 5/8 ×100 = (500%)/8 = 62.5%
3). The particular of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Ans: Original Amount = initial population = 25,000
Amount Change = decrease in population = 25,000 – 24,500 = 500
Percentage of decrease = (amount change)/(original amount ) ×100
= 500/25,000 × 100 = 50,000/25,000 = 2 % The percentage decrease was 2 %
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?
Ans: Original Amount = ₹ 3, 70, 000 – ₹ 3, 50, 000 = ₹ 20, 000
Percentage increase = (amount change)/(original amount ) ×100
= 20,000/3,50,000 × 100
= 20,000/3,50,000 = (200 )/35 = 40/7 % The percentage of price increase was 40/7 %
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?
Ans: Cost price of the T.V = ₹ 10,000
Percentage of profit = 20%
Profit % = Profit/CP ×100
Profit = ( Profit % × CP)/100
= (20 × 10,000 )/( 100) = ₹ 2000
Selling price = Cost price + profit
= 10,000 + 2000
= ₹ 12,000
∴ I get a profit of ₹ 12,000 for it.
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6). Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Ans: The selling price of washing machine = ₹ 13,500
Percentage of loss = 20%
Using the formula to find the cost price
CP = 100/(100 − loss% ) × SP
= 100/( 100 – 20 ) × 13,500
= 100/80 × 13,500 = 1350000/80 = ₹ 16875 The price she bought it for is ₹ 16875
7).i) Chalk contain calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage carbon in the chalk.
Ans: Ratio : 10 : 3 : 12
Total parts = 10 + 3 + 12 = 25
Part of carbon = 3/25
So, percentage of carbon = 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?
Ans: Let us assume the weight of the chalk stick to weight of carbon = 3 g
So, 12% of x = 3 g
12/100 × x = 3
x = ( 300 × 100)/12 = 300/12 = 25
∴ The weight of the chalk stick is 25 g.
8). Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Ans: Cost price of the book = ₹ 275
Loss % = 15% and SP = ?
CP = 100/(100 – loss% ) × SP
SP = CP × (100 – loss% )/100
= 275 × (100 – 15 @ )/100
= 275 × 85/100 = 23375/100 = ₹ 233.75 She sells the book for ₹ 233.75.
9). Find the amount to be paid at the end of 3 years in each case:
a) Principal= ₹ 1,200 at 12% P.a.
Ans: I = (P × T × R)/100
= (1200 × 3 × 12 )/100 = 36 ×12 = ₹ 430
b) Principal = ₹ 7500 at 5% p.a.
Ans: I = (P × T × R)/100
= (7500 × 3 × 5 )/100 = 75 ×15 = ₹ 1,125
10). What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Ans: I = (P ×T × R)/100
R = (I × 100)/(P × T) = (280 × 100)/(56,000 × 2 ) = 14/56 = 1/4 = 0.25
11). If Meena gives an interest of ₹ 45for one year at 9% rate p.a. what is the sum she has borrowed?
Ans: I = (P ×R× T)/100
P = (I × 100)/(R × T) = (45 × 100)/(9 × 1 ) = ₹ 500
Hence Meena borrowed ₹ 500.