RBSE Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325. 
(b) A refrigerator bought ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 25,00 and sold at ₹ 30,00.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Answer:
(a) Cost price of gardening shears = ₹ 250
Selling price of gardening shears = ₹ 325
Since, SP > CP, therefore here is profit.
₹ Profit = SP - CP = ₹ 325 - ₹ 250 = ₹ 75 Profit
Now, Profit% =\( \frac{\text { Profit }}{\text { CP }}\) × 100
= \(\frac{75}{250}\) × 100 = 30%
Therefore, Profit = 7 75 and Profit % = 30%

(b) Cost price of refrigerator = ₹ 12,000
Selling price of refrigerator = ₹ 13,500
Since, SP > CP, therefore here is profit.
∴ Profit = SP - CP = ₹ 13,50012,000 = ₹ 15,00
Now Profit % = \(\frac{\text { Profit }}{\text { CP }}\) × 100
= \(\frac{1500}{12000}\) × 100 = 12.5%
Therefore, Profit = ₹ 15,00 and Profit % = 12.5%.

(c) Cost price of cupboard = ₹ 2,500 
Selling price of cupboard = ₹ 3,000 
Since, SP > CP, therefore it is a profit.
∴ Profit = SP - CP
= ₹ 3,000 - ₹ 2,500 = ₹ 500
Now Profit% = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100
= \(\frac{500}{2,500}\) × 100 = 20%
Therefore, Profit = ₹ 500 and Profit % = 20%

(d) Cost price of skirt = ₹ 250
Selling price of skirt = ₹ 150 
Since, CP > SP, therefore here is loss.
∴ Lose = CP - SP
= ₹ 250 - ₹ 150 = ₹ 100
Now Loss % = \(\frac{\mathrm{Loss}}{\mathrm{CP}}\) × 100 
= \(\frac{100}{250}\) × 100% = 40%
Therefore, Profit = ₹ 100 and Profit % = 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
Answer:
(a) 3 :1
Total part = 3 + 1 = 4
Therefore, fractional part = \(\frac{3}{4}: \frac{1}{4}\)
⇒ Percentage of parts = \(\frac{3}{4}\) × 100 : \(\frac{1}{4}\) × 100
⇒ Percentage of parts = 75% : 25%

(b) 2:3:5
Total part = 2 + 3 + 5 = 10
Therefore, Fractional part = \(\frac{2}{10}: \frac{3}{10}: \frac{5}{10}\)
⇒ Percentage of parts = \(\frac{2}{10}\) × 100: \(\frac{3}{10}\) × 100: \(\frac{5}{10}\) × 100
⇒ Percentage of parts = 20%: 30%: 50%

(c) 1 : 4 .
Total part = 1 + 4 = 5
Therefore, Fractional part = \(\frac{1}{5}: \frac{4}{5}\)
⇒ Percentage of parts = \(\frac{1}{5}\) × 100: \(\frac{4}{5}\) × 100
⇒ Percentage of parts = 20% : 80%

(d) 1 : 2 : 5
Total part = 1 + 2 + 5 = 8 12 5
Therefore, Fractional part = \(\frac{1}{8}: \frac{2}{8}: \frac{5}{8}\)
⇒ Percentage of parts = \(\frac{1}{8}\) × 100: \(\frac{2}{8}\) × 100: \(\frac{5}{8}\) × 100
⇒ Percentage of parts = 12.5% : 25% : 62.5% 

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Answer:
The decreased population of a city from 25,000 to 24,500.
Population decreased = 25,000 - 24,500 = 500
Decreased Percentage = \(\frac{\text { Population decreased }}{\text { Original population }}\) × 100
= \(\frac{500}{25000}\) × 100 = 2%
Hence, the percentage decreased is 2%.

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went up to ₹ 3,70,000. What was the percentage of price increase?
Answer:
Increased in price of a car from ₹ 3,50,000 to ₹ 3,70,000 .
Amount change = ₹ 3,70,000 - ₹ 3,50,000
= ₹ 20,000
Therefore, Increased percentage = \(\frac{\text { Amount of change }}{\text { Original amount }}\) × 100
= \(\frac{20000}{350000}\) × 100 = 5\(\frac{5}{7}\)%
Hence, the percentage of price increased is 5\(\frac{5}{7}\)%

Question 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?
Answer:
The cost price of T.V. = ₹ 10,000 Profit percent = 20%
Now, Profit = Profit % of CP
= \(\frac{20}{100}\) × 10,000 = ₹ 2,000
Selling price = CP + Profit
= ₹ 10,000 + ₹ 2,000 = ₹ 12,000
Hence, he gets ₹ 12,000 on selling his T.V.

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:
Selling price of washing machine = ₹ 13,500 
Loss percent = 20%
Let the cost price of washing machine be ₹ x. 
Since, Loss = Loss % of CP
⇒ Loss = 20% of ₹ x = \(\frac{20}{100}\) × x = \(\frac{x}{5}\)
Therefore, SP = CP - Loss
13500 = x - \(\frac{x}{5}=\frac{4 x}{5}\)
x = \(\frac{13500 \times 5}{4}\) = ₹ 16,875
Hence, the cost price of washing machine is ₹ 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?
Answer: 
(i) Given ratio = 10 : 3: 12
Total part = 10 + 3 + 12 = 25
Part of Carbon = \(\frac{3}{25}\)
Percentage of Carbon part in chalk = \(\frac{3}{25}\) × 100 = 12%

(ii) Quantity of Carbon in chalk stick = 3 g
Let the weight of chalk be x g.
Then, 12% of x = 3 = 3
⇒ \(\frac{12}{100}\) × x = 3
⇒ x = \(\frac{3 \times 100}{12}\) = 25g
Hence, the weight of chalk stick is 25 g.

Question 8. 
Amina huys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Answer:
The cost of a book = ₹ 275 
Loss percent = 15% 
Loss = Loss % of CP
= 15% of ₹ 275
= \(\frac{15}{100}\) × 275 = ₹ 41.25
Therefore, SP = CP - Loss
= ₹ 275 - ₹ 41.25 = ₹ 233.75
Hence, Amina sells a 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% pa
(b) Principal = ₹ 7,500 at 5% p.a.
Answer:
(a) Here, Principal (P) = ₹ 1,200,
Rate (R) = 12% p.a., Time (T) = 3 years
Simple Interest = \(\frac{P \times R \times T}{100}=\frac{1200 \times 12 \times 3}{100}\)
= ₹ 432
Now, Amount = Principal + Simple Interest
= ₹ 12,00 + ₹ 432 = ₹ 1,632

(b) Here, Principal (P) = ₹ 7,500, Rate (R) = 5%p.a., Time(T) = 3years
Simple Interest = \(\frac{P \times R \times T}{100}\)
= \(\frac{7500 \times 5 \times 3}{100}\) = ₹ 1,125
Now, Amount = Principal + Simple Interest = ₹ 7,500 + ₹ 1,125 = ₹ 8,625

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 yearS?
Answer:
Here, Principal (P) = ₹ 56,000
Simple Interest (SI) = ₹ 280
Time (T) = 2years
Simple Interest = \(\frac{P \times R \times T}{100}\)
⇒ 280 = \(\frac{56000 \times R \times 2}{100}\)
⇒ R = \(\frac{280 \times 100}{56000 \times 2}\)
⇒ R = 0.25%
Hence, the rate of interest on sum is 0.25%.

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:
Here, Simple Interest = ₹ 45
Rate (R) = 9%
Time (T) = 1 year
⇒ 45 = \(\frac{P \times 9 \times 1}{100}\)
⇒ P = \(\frac{45 \times 100}{9}\) = ₹ 500
Hence, she borrowed ₹ 500