GSEB Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

   

Gujarat Board 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

GSEB 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 ₹ 250 and sold for ₹ 325.
(b) A refrigerate bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:
(a) Cost price (CP) = ₹ 250
Selling price (SP) = ₹ 325
∵ CP < SP so there is a profit.
∴ Profit = SP – CP
= ₹ 325 – ₹ 250 = ₹ 75
Now, Profit % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 75 }{ 250 }\) x 100 % = 30%

(b) Cost price (CP) = ₹ 12,000
Selling price (SP) = ₹ 13,500
∵ CP < SP, so there is a profit.
∴ Profit = SP – CP = ₹ 13,500 – ₹ 12,000 = ₹ 1,500
Now, Profit % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 1500 }{ 12000 }\) x 100%
= \(\frac { 150 }{ 12 }\)
= 12.5%

(c) Cost price (CP) = ₹ 2500
Selling price (SP) = ₹ 3000
∵ CP < SP, so there is a profit.
∴ Profit = SP – CP = ₹ 3000 – ₹ 2500 = ₹ 500
Now, Profit % = \(\frac { Profit }{ CP }\) x 100%
= \(\frac { 500 }{ 2500 }\) x 100% = 20%

(d) Cost price (CP) = ₹ 250
Selling price (SP) = ₹ 150
∵ CP > SP,
so there is a loss.
∴ Loss = CP – SP
= ₹ 250 – ₹ 150 = ₹ 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
∴ 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
∴ 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
∴ 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
∴ 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%

GSEB Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

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
GSEB Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 1

Question 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:
Initial cost = ₹ 3,50,000
Increased cost = ₹ 3,70,000
Increase in cost
= ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000
GSEB Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 2

Question 5.
1 buy a T.V for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
CP of the T.V. = ₹ 10,000
Profit percent = 20%
∴ Profit = 20% of 10,000
= ₹ \(\frac { 20 }{ 100 }\) x 10,000
= ₹ 2,000
Now, SP = CP + Profit
Thus, I will get ₹ 12,000 for the T.V.
= ₹ 10,000 + ₹ 2,000 = ₹ 12,000
Thus, I will get ₹ 12,000 for the T.V.

Question 6.
Juki sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
Selling price (SP) = ₹ 13,500
Loss % = 20%
CP = ?
∴ CP – Loss = SP
∴ CP – (20% of CP) = ₹ 13,500
= ₹ 13,500
or CP – \(\frac { 20 }{ 100 }\)CP = ₹ 13,500
or \(\frac { 5CP – CP }{ 5 }\) = ₹ 13,500
or \(\frac { 4 }{ 5 }\)CP = ₹ 13,500
⇒ CP = ₹ 13,500 x \(\frac { 5 }{ 4 }\)
= ₹ 3,375 x 5
= ₹ 16,875
∴ Juhi bought the washing machine for ₹ 16,875.

GSEB Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

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) ∵ The ratio of calcium, carbon and oxygen in mixture is 10 : 3 : 12.
∴ Total of ratios = 10 + 3 + 12 = 25
∴ Percentage of carbon in the chalk mixture
= \(\frac { 3 }{ 25 }\)

(ii) Let the weight of the stick be x g.
∵ 12% of the chalk mixture is 3 g.
∴ 12% of x = 3
or \(\frac { 12 }{ 100 }\) × x = 3 ⇒ x = \(\frac { 3×100 }{ 12 }\) = 25
or Weight of the chalk stick = 25 g.

Question 8.
Amina buys a book for ₹ 215 and sells it at a loss of 15%. How much does she sell it for?
Solution:
Cost price of the book (CP) = ₹ 275
Loss % = 15%
∴ Loss = 15% of CP = ₹ \(\frac { 15 }{ 100 }\) x 275
= ₹ \(\frac { 15×11 }{ 4 }\)
= ₹ \(\frac { 165 }{ 4 }\)
= ₹ 41.25
Now SP = CP – Loss
= ₹ 275 – ₹ 41.25
= ₹ 233.75
Thus, Amina will sell 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.
(b) Principal = ₹ 7,500 at 5% p.a.
Solution:
(a) Here, Principal (P) = ₹ 1,200
Rate (R) = 12% p.a. and Time (T) = 3 years
∴ Interest = \(\frac { P×R×T }{ 100 }\) = ₹\(\frac { 1,200×12×3 }{ 100 }\)
₹ (12 x 12 x 3) = ₹ 432
Now, Amount = Principal + Interest
= ₹ 1,200 + ₹ 432 = ₹ 1,632

(b) Principal (P) = ₹ 7,500
Rate (R) = 5% p.a.
Time (T) = 3 years
∴ Interest = \(\frac { P×R×T }{ 100 }\) = ₹\(\frac { 7,500×5×3 }{ 100 }\)
= ₹ (75 x 5 x 3) = ₹ 1,125
Now, Amount = Principal + Interest
= ₹ 7,500 + ₹ 1,125 = ₹ 8,625

GSEB Solutions Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 year?
Solution:
Principal (P) = ₹ 56,000
Rate (R) = ?
Time (T) = 2 years
Interest = ₹ 280
∵ Interest = \(\frac { P×R×T }{ 100 }\)
∴ 280 = \(\frac { 56,000×R×2 }{ 100 }\)
or R = \(\frac { 280×100 }{ 2×56,000 }\) = \(\frac { 1 }{ 4 }\)% = 0.25%
Thus, 0.25% p.a. of interest rate will give the required interest.

Question 11.
If Meena gives an interest of ₹ 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 = ₹ 45
Time (T) = 1 year
Since Interest = \(\frac { P×R×T }{ 100 }\)
or ₹ 45 = \(\frac { 45×100 }{ 9 }\)
= ₹ 5 x 100
= ₹ 500

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