### Comparing Quantities - Solutions 3

CBSE Class –VII Mathematics
NCERT 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 Rs. 250 and sold for Rs. 325.
(b) A refrigerator bought for Rs. 12,000 and sold for Rs. 13,500.
(c) A cupboard bought for Rs. 2,500 and sold for Rs. 3,000.
(d) A skirt bought for Rs. 250 and sold for Rs. 150.
Answer: (a) Cost price of gardening shears = Rs. 250
Selling price of gardening shears = Rs. 325
Since, S.P. > C.P., therefore here is profit.
$\therefore$ Profit = S.P. – C.P. = 325 – 250 = Rs. 75
Now Profit% = $\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}×100$ = $\frac{75}{250}×100$ = 30%
Therefore, Profit = Rs. 75 and Profit% = 30%
(b) Cost price of refrigerator = Rs. 12,000
Selling price of refrigerator = Rs. 13,500
Since, S.P. > C.P., therefore here is profit.
$\therefore$ Profit = S.P. – C.P. = 13500 – 12000 = Rs. 1,500
Now Profit% = $\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}×100$ = $\frac{1500}{12000}×100$ = 12.5%
Therefore, Profit = Rs. 1,500 and Profit% = 12.5%
(c) Cost price of cupboard = Rs. 2,500
Selling price of cupboard = Rs. 3,000
Since, S.P. > C.P., therefore here is profit.
$\therefore$ Profit = S.P. – C.P. = 3,000 – 2,500 = Rs. 500
Now Profit% = $\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}×100$ = $\frac{500}{2500}×100$ = 20%
Therefore, Profit = Rs. 500 and Profit% = 20%
(d) Cost price of skirt = Rs. 250
Selling price of skirt = Rs. 150
Since, C.P. > S.P., therefore here is loss.
$\therefore$ Loss = C.P. – S.P. = 250 – 150 = Rs. 100
Now Loss% = $\frac{\text{Loss}}{\text{C}\text{.P}\text{.}}×100$ = $\frac{100}{250}×100$ = 40%
Therefore, Profit = Rs. 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
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
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 population of a city decreased 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 Rs. 3,50,000. The next year, the price went up to Rs. 3,70,000. What was the percentage of price increase?
Answer: Increased in price of a car from Rs. 3,50,000 to Rs. 3,70,000.
Amount change = Rs. 3,70,000 – Rs. 3,50,000 = Rs. 20,000.
Therefore, Increased percentage = $\frac{\text{Amount of change}}{\text{Original amount}}×100$
$\frac{20000}{350000}×100$ = $5\frac{5}{7}\mathrm{%}$
Hence, the percentage of price increased is $5\frac{5}{7}\mathrm{%}.$
Question 5. I buy a T.V. for Rs. 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. = Rs. 10,000
Profit percent = 20%
Now, Profit = Profit% of C.P.
$\frac{20}{100}×10000$ = Rs. 2,000
Selling price = C.P. + Profit
= 10,000 + 2,000 = Rs. 12,000
Hence, he gets Rs. 12,000 on selling his T.V.
Question 6. Juhi sells a washing machine for Rs. 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Answer: Selling price of washing machine = Rs. 13,500
Loss percent = 20%
Let the cost price of washing machine be Rs.$x.$
Since, Loss = Loss% of C.P.
$⇒$ Loss = 20% of Rs.$x$ = $\frac{20}{100}×x=\frac{x}{5}$
Therefore, S.P. = C.P. – Loss
$⇒$ 13500 = $x-\frac{x}{5}$ $⇒$ 13500 = $\frac{4x}{5}$
$⇒$ $x=\frac{13500×5}{4}$ = Rs. 16,875
Hence, the cost price of washing machine is Rs. 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
$⇒$ $\frac{12}{100}×x=3$
$⇒$ $x=\frac{3×100}{12}$ = 25 g
Hence, the weight of chalk stick is 25 g.
Question 8. Amina buys a book for Rs. 275 and sells it at a loss of 15%. How much does she sell it for?
Answer: The cost price of a book = Rs. 275
Loss percent = 15%
Loss = Loss% of C.P. = 15% of Rs. 275 = $\frac{15}{100}×275$ = Rs. 41.25
Therefore, S.P. = C.P. – Loss = 275 – 41.25 = Rs. 233.75
Hence, Amina sells a book for Rs. 233.75.
Question 9. Find the amount to be paid at the end of 3 years in each case:
(a) Principal = Rs. 1,200 at 12% p.a.
(b) Principal = Rs. 7,500 at 5% p.a.
Answer: (a) Here, Principal (P) = Rs. 1,200, Rate (R) = 12% p.a., Time (T) = 3 years
Simple Interest =  = $\frac{1200×12×3}{100}$ = Rs. 432
Now, Amount = Principal + Simple Interest = 1200 + 432 = Rs. 1,632
(b) Here, Principal (P) = Rs. 7,500, Rate (R) = 5% p.a., Time (T) = 3 years
Simple Interest =  = $\frac{7500×5×3}{100}$ = Rs. 1,125
Now, Amount = Principal + Simple Interest = 7,500 + 1,125 = Rs. 8,625
Question 10. What rate gives Rs. 280 as interest on a sum of Rs. 56,000 in 2 years?
Answer: Here, Principal (P) = Rs. 56,000, Simple Interest (S.I.) = Rs. 280, Time (T) = 2 years
Simple Interest =
$⇒$ 280 = $\frac{56000×\text{R}×2}{100}$
$⇒$ R = $\frac{280×100}{56000×2}$
$⇒$ R = 0.25%
Hence, the rate of interest on sum is 0.25%.
Question 11. If Meena gives an interest of Rs. 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Answer: Simple Interest = Rs. 45, Rate (R) = 9% p.a., Time (T) = 1 year
Simple Interest =
$⇒$ 45 = $\frac{\text{P}×9×1}{100}$
$⇒$ P = $\frac{45×100}{9×1}$
$⇒$ P = Rs. 500
Hence, she borrowed Rs. 500.