Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 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 ā¹ 250 and sold for ā¹ 325.
(b) A refrigerate bought for ā¹ 12,000 and sold at ā¹ 13,500.
(c) A cupboard bought for ā¹ 2,500 and sold at ā¹ 3,000.
(d) A skirt bought for ā¹ 250 and sold at ā¹ 150.
Solution:
(a) Cost price (CP) = ā¹ 250
Selling price (SP) = ā¹ 325
āµ CP < SP so there is a profit.
ā“ Profit = SP – CP
= ā¹ 325 – ā¹ 250 = ā¹ 75
Now, Profit % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 75 }{ 250 }\) x 100 % = 30%
(b) Cost price (CP) = ā¹ 12,000
Selling price (SP) = ā¹ 13,500
āµ CP < SP, so there is a profit.
ā“ Profit = SP – CP = ā¹ 13,500 – ā¹ 12,000 = ā¹ 1,500
Now, Profit % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 1500 }{ 12000 }\) x 100%
= \(\frac { 150 }{ 12 }\)
= 12.5%
(c) Cost price (CP) = ā¹ 2500
Selling price (SP) = ā¹ 3000
āµ CP < SP, so there is a profit.
ā“ Profit = SP – CP = ā¹ 3000 – ā¹ 2500 = ā¹ 500
Now, Profit % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 500 }{ 2500 }\) x 100% = 20%
(d) Cost price (CP) = ā¹ 250
Selling price (SP) = ā¹ 150
āµ CP > SP,
so there is a loss.
ā“ Loss = CP – SP
= ā¹ 250 – ā¹ 150 = ā¹ 100
Now, Loss % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 100 }{ 250 }\) x 100% = 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
Total of the parts = 3 + 1 = 4
ā“ Percentage of the 1st part of the ratio
= \(\frac { 3 }{ 4 }\) x 100% = 75%
Percentage of the 2nd part of the ratio = \(\frac { 1 }{ 4 }\) x 100% = 25%
(b) 2 : 3 : 5
Total of the parts = 2 + 3 + 5 = 10
ā“ Percentage of 1st part of the ratio
= \(\frac { 2 }{ 10 }\) x 100% = (2 x 10)% = 20%
Percentage of 2nd part of the ratio
= \(\frac { 3 }{ 10 }\) x 100% = (3 x 10)% = 30%
Percentage of 3rd part of the ratio
= \(\frac { 5 }{ 10 }\) x 100% = (5 x 10)% = 50%
(c) 1 : 4
Total of parts = 1 + 4 = 5
ā“ Percentage of the 1st part of the ratio
= \(\frac { 1 }{ 5 }\) x 100% = (1 x 20)% = 20%
Percentage of the 2nd part of the ratio
= \(\frac { 4 }{ 5 }\) x 100% = (4 x 20)% = 80%
(d) 1 : 2 : 5
Total of the parts = 1 + 2 + 5 = 8
ā“ Percentage of the 1st part of the ratio
= \(\frac { 1 }{ 8 }\) x 100% = \(\frac { 25 }{ 2 }\) = 12.5%
Percentage of the 2nd part of the ratio
= \(\frac { 2 }{ 8 }\) x 100% = 25%
Percentage of the 3rd part of the ratio
= \(\frac { 5 }{ 8 }\) x 100% = \(\frac { 125 }{ 2 }\) % = 62.5%
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
Question 4.
Arun bought a car for ā¹ 3,50,000. The next year, the price went upto ā¹ 3,70,000. What was the percentage, of price increase?
Solution:
Initial cost = ā¹ 3,50,000
Increased cost = ā¹ 3,70,000
Increase in cost
= ā¹ 3,70,000 – ā¹ 3,50,000
= ā¹ 20,000
Question 5.
1 buy a T.V for ā¹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
CP of the T.V. = ā¹ 10,000
Profit percent = 20%
ā“ Profit = 20% of 10,000
= ā¹ \(\frac { 20 }{ 100 }\) x 10,000
= ā¹ 2,000
Now, SP = CP + Profit
Thus, I will get ā¹ 12,000 for the T.V.
= ā¹ 10,000 + ā¹ 2,000 = ā¹ 12,000
Thus, I will get ā¹ 12,000 for the T.V.
Question 6.
Juki sells a washing machine for ā¹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
Selling price (SP) = ā¹ 13,500
Loss % = 20%
CP = ?
ā“ CP – Loss = SP
ā“ CP – (20% of CP) = ā¹ 13,500
= ā¹ 13,500
or CP – \(\frac { 20 }{ 100 }\)CP = ā¹ 13,500
or \(\frac { 5CP – CP }{ 5 }\) = ā¹ 13,500
or \(\frac { 4 }{ 5 }\)CP = ā¹ 13,500
ā CP = ā¹ 13,500 x \(\frac { 5 }{ 4 }\)
= ā¹ 3,375 x 5
= ā¹ 16,875
ā“ Juhi bought the washing machine for ā¹ 16,875.
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) āµ The ratio of calcium, carbon and oxygen in mixture is 10 : 3 : 12.
ā“ Total of ratios = 10 + 3 + 12 = 25
ā“ Percentage of carbon in the chalk mixture
= \(\frac { 3 }{ 25 }\)
(ii) Let the weight of the stick be x g.
āµ 12% of the chalk mixture is 3 g.
ā“ 12% of x = 3
or \(\frac { 12 }{ 100 }\) Ć x = 3 ā x = \(\frac { 3Ć100 }{ 12 }\) = 25
or Weight of the chalk stick = 25 g.
Question 8.
Amina buys a book for ā¹ 215 and sells it at a loss of 15%. How much does she sell it for?
Solution:
Cost price of the book (CP) = ā¹ 275
Loss % = 15%
ā“ Loss = 15% of CP = ā¹ \(\frac { 15 }{ 100 }\) x 275
= ā¹ \(\frac { 15Ć11 }{ 4 }\)
= ā¹ \(\frac { 165 }{ 4 }\)
= ā¹ 41.25
Now SP = CP – Loss
= ā¹ 275 – ā¹ 41.25
= ā¹ 233.75
Thus, Amina will sell the book for ā¹ 233.75.
Question 9.
Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ā¹ 1,200 at 12% p.a.
(b) Principal = ā¹ 7,500 at 5% p.a.
Solution:
(a) Here, Principal (P) = ā¹ 1,200
Rate (R) = 12% p.a. and Time (T) = 3 years
ā“ Interest = \(\frac { PĆRĆT }{ 100 }\) = ā¹\(\frac { 1,200Ć12Ć3 }{ 100 }\)
ā¹ (12 x 12 x 3) = ā¹ 432
Now, Amount = Principal + Interest
= ā¹ 1,200 + ā¹ 432 = ā¹ 1,632
(b) Principal (P) = ā¹ 7,500
Rate (R) = 5% p.a.
Time (T) = 3 years
ā“ Interest = \(\frac { PĆRĆT }{ 100 }\) = ā¹\(\frac { 7,500Ć5Ć3 }{ 100 }\)
= ā¹ (75 x 5 x 3) = ā¹ 1,125
Now, Amount = Principal + Interest
= ā¹ 7,500 + ā¹ 1,125 = ā¹ 8,625
Question 10.
What rate gives ā¹ 280 as interest on a sum of ā¹ 56,000 in 2 year?
Solution:
Principal (P) = ā¹ 56,000
Rate (R) = ?
Time (T) = 2 years
Interest = ā¹ 280
āµ Interest = \(\frac { PĆRĆT }{ 100 }\)
ā“ 280 = \(\frac { 56,000ĆRĆ2 }{ 100 }\)
or R = \(\frac { 280Ć100 }{ 2Ć56,000 }\) = \(\frac { 1 }{ 4 }\)% = 0.25%
Thus, 0.25% p.a. of interest rate will give the required interest.
Question 11.
If Meena gives an interest of ā¹ 45 for one year at 9% rate p.a. What is the sum she has borrowed.
Solution:
rincipal (P) = ?
Rate (R) = 9% p.a.
Interest = ā¹ 45
Time (T) = 1 year
Since Interest = \(\frac { PĆRĆT }{ 100 }\)
or ā¹ 45 = \(\frac { 45Ć100 }{ 9 }\)
= ā¹ 5 x 100
= ā¹ 500