Comparing Quantities - Solutions 3

CBSE Class –VII Mathematics
NCERT Solutions
Chapter 8 Comparing Quantities (Ex. 8.3)

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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% = ProfitC.P.×100$\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}\times 100$ = 75250×100$\frac{75}{250}\times 100$ = 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% = ProfitC.P.×100$\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}\times 100$ = 150012000×100$\frac{1500}{12000}\times 100$ = 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% = ProfitC.P.×100$\frac{\text{Profit}}{\text{C}\text{.P}\text{.}}\times 100$ = 5002500×100$\frac{500}{2500}\times 100$ = 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% = LossC.P.×100$\frac{\text{Loss}}{\text{C}\text{.P}\text{.}}\times 100$ = 100250×100$\frac{100}{250}\times 100$ = 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

Answer: (a) 3 : 1
Total part = 3 + 1 = 4

Therefore, Fractional part = 34:14$\frac{3}{4}:\frac{1}{4}$

⇒$\Rightarrow$ Percentage of parts = 34×100:14×100$\frac{3}{4}\times 100:\frac{1}{4}\times 100$

⇒$\Rightarrow$ Percentage of parts = 75% : 25%

(b) 2 : 3 : 5
Total part = 2 + 3 + 5 = 10

Therefore, Fractional part = 210:310:510$\frac{2}{10}:\frac{3}{10}:\frac{5}{10}$

⇒$\Rightarrow$ Percentage of parts = 210×100:310×100:510×100$\frac{2}{10}\times 100:\frac{3}{10}\times 100:\frac{5}{10}\times 100$

⇒$\Rightarrow$ Percentage of parts = 20% : 30% : 50%

(c) 1 : 4
Total part = 1 + 4 = 5

Therefore, Fractional part = 15:45$\frac{1}{5}:\frac{4}{5}$

⇒$\Rightarrow$ Percentage of parts = 15×100:45×100$\frac{1}{5}\times 100:\frac{4}{5}\times 100$

⇒$\Rightarrow$ Percentage of parts = 20% : 80%

(d) 1 : 2 : 5
Total part = 1 + 2 + 5 = 8

Therefore, Fractional part = 18:28:58$\frac{1}{8}:\frac{2}{8}:\frac{5}{8}$

⇒$\Rightarrow$ Percentage of parts = 18×100:28×100:58×100$\frac{1}{8}\times 100:\frac{2}{8}\times 100:\frac{5}{8}\times 100$

⇒$\Rightarrow$ 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 = Population decreasedOriginal population×100$\frac{\text{Population decreased}}{\text{Original population}}\times 100$ = 50025000×100$\frac{500}{25000}\times 100$ = 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 = Amount of changeOriginal amount×100$\frac{\text{Amount of change}}{\text{Original amount}}\times 100$

= 20000350000×100$\frac{20000}{350000}\times 100$ = 557%$5\frac{5}{7}\mathrm{\%}$

Hence, the percentage of price increased is 557%.$5\frac{5}{7}\mathrm{\%}.$

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.

= 20100×10000$\frac{20}{100}\times 10000$ = 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.$x.$

Since, Loss = Loss% of C.P.

⇒$\Rightarrow$ Loss = 20% of Rs.x$x$ = 20100×x=x5$\frac{20}{100}\times x=\frac{x}{5}$

Therefore, S.P. = C.P. – Loss

⇒$\Rightarrow$ 13500 = x−x5$x-\frac{x}{5}$ ⇒$\Rightarrow$ 13500 = 4x5$\frac{4x}{5}$

⇒$\Rightarrow$ x=13500×54$x=\frac{13500\times 5}{4}$ = 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 = 325$\frac{3}{25}$

Percentage of Carbon part in chalk = 325×100$\frac{3}{25}\times 100$ = 12%

(ii) Quantity of Carbon in chalk stick = 3 g

Let the weight of chalk be x$x$ g.

Then, 12% of x$x$ = 3

⇒$\Rightarrow$ 12100×x=3$\frac{12}{100}\times x=3$

⇒$\Rightarrow$ x=3×10012$x=\frac{3\times 100}{12}$ = 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 = 15100×275$\frac{15}{100}\times 275$ = 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 = P × R × T100$\frac{\text{P }\times \text{ R }\times \text{ T}}{100}$ = 1200×12×3100$\frac{1200\times 12\times 3}{100}$ = 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 = P × R × T100$\frac{\text{P }\times \text{ R }\times \text{ T}}{100}$ = 7500×5×3100$\frac{7500\times 5\times 3}{100}$ = 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 = P × R × T100$\frac{\text{P }\times \text{ R }\times \text{ T}}{100}$

⇒$\Rightarrow$ 280 = 56000×R×2100$\frac{56000\times \text{R}\times 2}{100}$

⇒$\Rightarrow$ R = 280×10056000×2$\frac{280\times 100}{56000\times 2}$

⇒$\Rightarrow$ 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 = P × R × T100$\frac{\text{P }\times \text{ R }\times \text{ T}}{100}$

⇒$\Rightarrow$ 45 = P×9×1100$\frac{\text{P}\times 9\times 1}{100}$

⇒$\Rightarrow$ P = 45×1009×1$\frac{45\times 100}{9\times 1}$

⇒$\Rightarrow$ P = Rs. 500

Hence, she borrowed Rs. 500.