NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities Ex 7.2

Exercise 7.2

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.

Ans: It is given that the gardening shears are bought for ₹250 and sold for ₹325. This means that the cost price is ₹250 and the selling price is ₹325.

It is observed that the selling price is greater than the cost price, that is, ${\text{S}}{\text{.P}} > {\text{C}}{\text{.P}}$

Thus, a profit is occurring, which is calculated as cost price subtracted from the selling price.

${\text{Profit}} = {\text{S}}{\text{.P}} - {\text{C}}{\text{.P}}$

${\text{Profit}} = 325 - 250$

${\text{Profit}} = 75$

Thus, the profit is found to be ₹75.

Now, the profit percent is given by the formula,

${\text{Profit }}\%  = \dfrac{{{\text{Profit}}}}{{{\text{C}}{\text{.P}}}} \times 100$

Substitute 75 for profit and 250 for C.P in the above formula.

${\text{Profit }}\%  = \dfrac{{{\text{75}}}}{{250}} \times 100$

${\text{Profit }}\%  = 30\% $

Therefore, the profit is found to be ₹75 and profit $\% $ is $30\% $.

b. A refrigerator bought ₹12,000 and sold at ₹13,500.

Ans: It is given that a refrigerator is bought for ₹12,000 and sold for ₹13,500. This means that the cost price is ₹12,000 and the selling price is ₹13,500.

It is observed that the selling price is greater than the cost price, that is, ${\text{S}}{\text{.P}} > {\text{C}}{\text{.P}}$.

Thus, a profit is occurring, which is calculated as cost price subtracted from the selling price.

${\text{Profit}} = {\text{S}}{\text{.P}} - {\text{C}}{\text{.P}}$

${\text{Profit}} = 13500 - 12000$

${\text{Profit}} = 1500$

Thus, the profit is found to be ₹1,500.

Now, the profit percent is given by the formula,

${\text{Profit }}\%  = \dfrac{{{\text{Profit}}}}{{{\text{C}}{\text{.P}}}} \times 100$

Substitute 1500 for profit and 12000 for C.P in the above formula.

${\text{Profit }}\%  = \dfrac{{{\text{1500}}}}{{12000}} \times 100$

${\text{Profit }}\%  = 12.5\% $

Therefore, the profit is found to be ₹1500 and profit $\% $ is $12.5\% $.

c. A cupboard bought for ₹2,500 and sold at ₹3,000.

Ans: It is given that a cupboard is bought for ₹2,500 and sold for ₹3,000. This means that the cost price is ₹2,500 and the selling price is ₹3,000.

It is observed that the selling price is greater than the cost price, that is, ${\text{S}}{\text{.P}} > {\text{C}}{\text{.P}}$.

Thus, a profit is occurring, which is calculated as cost price subtracted from the selling price.

${\text{Profit}} = {\text{S}}{\text{.P}} - {\text{C}}{\text{.P}}$

${\text{Profit}} = 3000 - 2500$

${\text{Profit}} = 500$

Thus, the profit is found to be ₹500.

Now, the profit percent is given by the formula,

${\text{Profit }}\%  = \dfrac{{{\text{Profit}}}}{{{\text{C}}{\text{.P}}}} \times 100$

Substitute 500 for profit and 2500 for C.P in the above formula.

${\text{Profit }}\%  = \dfrac{{{\text{500}}}}{{2500}} \times 100$

${\text{Profit }}\%  = 20\% $

Therefore, the profit is found to be ₹500 and profit$\% $ is $20\% $.

d. A skirt bought for ₹250 and sold at ₹150.

Ans:It is given that a skirt is bought for ₹250 and sold for ₹150. This means that the cost price is ₹250 and the selling price is ₹150.

It is observed that the selling price is lesser than the cost price, that is, ${\text{S}}{\text{.P}} < {\text{C}}{\text{.P}}$.

Thus, a loss is being occurred, which is calculated as the selling price subtracted from the cost price.

${\text{Loss}} = {\text{C}}{\text{.P}} - {\text{S}}{\text{.P}}$

${\text{Loss}} = 250 - 150$

${\text{Loss}} = 100$

Thus, the loss is found to be ₹100.

Now, the profit percent is given by the formula,

${\text{Loss}}\%  = \dfrac{{{\text{Loss}}}}{{{\text{C}}{\text{.P}}}} \times 100$

Substitute 100 for loss and 250 for C.P in the above formula.

${\text{Loss}}\%  = \dfrac{{{\text{100}}}}{{250}} \times 100$

${\text{Loss}}\%  = 40\% $

Therefore, the loss is found to be ₹100 and loss$\% $ is $40\% $.

2. Convert each part of the ratio to percentage:

a. The ratio is $3:1$.

Ans: The total part is found to be $3 + 1 = 4$.

Thus, the parts in fractional form can be written as $\dfrac{3}{4}$ and $\dfrac{1}{4}$.

Therefore, the percentage of each part is obtained by multiplying the fractional part by 100.

${\text{Percentages of parts}} = \dfrac{3}{4} \times 100:\dfrac{1}{4} \times 100$

On evaluating further,

${\text{Percentages of parts}} = \dfrac{{300}}{4}:\dfrac{{100}}{4}$

${\text{Percentages of parts}} = 75:25$

Therefore, the part of the ratio to percentage is $75\% :25\% $.

b. The ratio is $2:3:5$.

Ans: The total part is found to be $2 + 3 + 5 = 10$.

Thus, the parts in fractional form can be written as $\dfrac{2}{{10}}$, $\dfrac{3}{{10}}$ and $\dfrac{5}{{10}}$.

Therefore, the percentage of each part is obtained by multiplying the fractional part by 100.

${\text{Percentages of parts}} = \dfrac{2}{{10}} \times 100:\dfrac{3}{{10}} \times 100:\dfrac{5}{{10}} \times 100$

On evaluating further,

${\text{Percentages of parts}} = \dfrac{{200}}{{10}}:\dfrac{{300}}{{10}}:\dfrac{{500}}{{10}}$

${\text{Percentages of parts}} = 20:30:50$

Therefore, the part of the ratio to percentage is $20\% :30\% :50\% $.

c. The ratio is $1:4$.

Ans: The total part is found to be $1 + 4 = 5$.

Thus, the parts in fractional form can be written as $\dfrac{1}{5}$ and $\dfrac{4}{5}$.

Therefore, the percentage of each part is obtained by multiplying the fractional part by 100.

${\text{Percentages of parts}} = \dfrac{1}{5} \times 100:\dfrac{4}{5} \times 100$

On evaluating further,

${\text{Percentages of parts}} = \dfrac{{100}}{5}:\dfrac{{400}}{5}$

${\text{Percentages of parts}} = 20:80$

Therefore, the part of the ratio to percentage is $20\% :80\% $.

d. The ratio is $1:2:5$.

Ans: The total part is found to be $1 + 2 + 5 = 8$.

Thus, the parts in fractional form can be written as $\dfrac{1}{8}$, $\dfrac{2}{8}$ and $\dfrac{5}{8}$.

Therefore, the percentage of each part is obtained by multiplying the fractional part by 100.

${\text{Percentages of parts}} = \dfrac{1}{8} \times 100:\dfrac{2}{8} \times 100:\dfrac{5}{8} \times 100$

On evaluating further,

${\text{Percentages of parts}} = \dfrac{{100}}{8}:\dfrac{{200}}{8}:\dfrac{{500}}{8}$

${\text{Percentages of parts}} = 12.5:25:62.5$

Therefore, the part of the ratio to percentage is $12.5\% :25\% :62.5\% $.

3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

Ans: It is known from the given information that the population of a city has decreased from 25,000 to 24,500. 

Thus, the population decreased by $\left( {25,000 - 24,500} \right) = 500$.

The formula to find the decreased percentage is given as follows,

${\text{Decreased Percentage}} = \dfrac{{{\text{Population decreased}}}}{{{\text{Original population}}}} \times 100$

Substitute 500 for population decrease and 25000 for original population in the above formula.

${\text{Decreased Percentage}} = \dfrac{{{\text{500}}}}{{{\text{25000}}}} \times 100$

${\text{Decreased Percentage}} = \dfrac{{{\text{500}}}}{{{\text{25000}}}} \times 100$

${\text{Decreased Percentage}} = 2\% $

Therefore, the decreased percentage is found to be $2\% $.

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?

Ans: It is given that Arun bought a car for ₹3,50,000 and the next year the prices went up to ₹3,70,000.

The price of a car increases from ₹3,50,000 to ₹3,70,000. Thus, the amount by which the price is increased is,

${\text{Amount increase}} = 370000 - 350000$

${\text{Amount increase}} = 20000$

Therefore, the amount increased to ₹20,000.

Now, the increased percentage can be given by the formula,

${\text{Increased percentage}} = \dfrac{{{\text{Amount increased}}}}{{{\text{Original amount }}}} \times 100$

Substitute 20000 for Amount increased and 350000 for the original amount in the above formula.

${\text{Increased percentage}} = \dfrac{{{\text{20000}}}}{{{\text{350000}}}} \times 100$

${\text{Increased percentage}} = 5\dfrac{5}{7}\% $

Hence, the percentage of price increase is found to be $5\dfrac{5}{7}\% $.

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?

Ans: The T.V. is bought for ₹10,000 and is sold at a profit of $20\% $. This means that the cost price of the T.V. is ₹10,000 and the profit percent is $20\% $.

Thus, profit will be the profit $\% $ of C.P which is given as,

${\text{Profit}} = {\text{Profit }}\% {\text{ of C}}{\text{.P}}$

${\text{Profit}} = \dfrac{{20}}{{100}}{\text{of 10,000}}$

${\text{Profit}} = \dfrac{{20}}{{100}} \times {\text{ 10,000}}$

${\text{Profit}} = 2,000$

Now, the selling price is the sum of cost price and the profit.

${\text{Selling price}} = {\text{Cost price}} + {\text{Profit}}$

${\text{Selling price}} = 10,000 + 2,000$

${\text{Selling price}} = 12,000$

Therefore, he gets ₹12,000 for selling the T.V he bought for ₹10,000.

6. Juhi sells a washing machine for ₹13,500. She loses $20\% $ in the bargain. What was the price at which she bought it?

Ans: The washing machine is sold by Juhi for ₹13,500 and she loses $20\% $ in the bargain. This means that the selling price of the washing machine is ₹13,500 and the loss percent is $20\% $.

Assume the cost price of the washing machine to be $x$.

Thus, loss will be the loss $\% $ of C.P which is given as,

${\text{Loss}} = {\text{Loss }}\% {\text{ of C}}{\text{.P}}$

${\text{Loss}} = 20\% {\text{ of }}x$

${\text{Loss}} = \dfrac{{20}}{{100}} \times x$

${\text{Loss}} = \dfrac{{20}}{{100}} \times x$

${\text{Loss}} = \dfrac{x}{5}$

It is known that, ${\text{S}}{\text{.P}} = {\text{C}}{\text{.P}} - {\text{Loss}}$.

Thus,

$13500 = x - \dfrac{x}{5}$

$13500 = \dfrac{{4x}}{5}$

$x = \dfrac{{13500 \times 5}}{4}$

$x = 16,875$

Therefore, the cost price of the washing machine is found to be ₹16,875.

7. (i) Chalk contains Calcium, Carbon and Oxygen in the ratio 10 : 3 : 12. Find the percentage of Carbon in chalk.

Ans: The ratio of Calcium, Carbon and Oxygen in the chalk is $10:3:12$. 

Thus, the total part will be the addition of the ratios, that is, $\left( {10 + 3 + 12} \right) = 15$.

The part of the carbon is found to be 3 from the given ratio.

The fractional part can be written as $\dfrac{3}{{25}}$.

Thus, the percentage of Carbon part in the chalk is given by,

${\text{Percentage of Carbon part in the chalk}} = \dfrac{3}{{25}} \times 100$

${\text{Percentage of Carbon part in the chalk}} = 12\% $

Therefore, the percentage of Carbon in the chalk is found to be $12\% $.

(ii) If in a stick of chalk, Carbon is 3 g, what is the weight of the chalk stick?

Ans: It is given that the quantity of Carbon in the chalk stick is 3g. Assume the weight of chalk is $x$ g.

The percentage of Carbon part in the chalk is found to be $12\% $. 

Thus,

$12\% {\text{ of }}x = 3$

$\dfrac{{12}}{{100}} \times x = 3$

$x = \dfrac{{3 \times 100}}{{12}}$

$x = 25$

Therefore, the weight of the chalk is obtained to be 25g.

8. Amina buys a book for ₹275 and sells it at a loss of $15\% $. How much does she sell it for?

Ans: The book bought by Amine costs ₹275 and it is sold at a loss of $15\% $. This means that the cost price of the book is ₹275 and the loss percent is $15\% $.

Thus, loss will be the loss $\% $ of C.P which is given as,

${\text{Loss}} = {\text{Loss }}\% {\text{ of C}}{\text{.P}}$

${\text{Loss}} = 15\% {\text{ of 275}}$

${\text{Loss}} = \dfrac{{15}}{{100}} \times 275$

${\text{Loss}} = 41.25$

Thus, there was a loss of ₹41.25.

Now, 

${\text{S}}{\text{.P = C}}{\text{.P}} - {\text{Loss}}$

Substitute 275 for C.P and $41.25$ for loss in the above formula.

${\text{S}}{\text{.P}} = 275 - 41.25$

${\text{S}}{\text{.P}} = 233.75$

Therefore, the price at which Amina sells her book is ₹$233.75$.

9. Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹1,200 at $12\% $ p.a. 

Ans: The principal $\left( P \right) = $₹1,200, the rate $\left( R \right) = 12\% {\text{ p}}{\text{.a}}$ and the time $\left( T \right) = 3{\text{ years}}$.

The formula to evaluate the Simple Interest is given by,

${\text{Simple Interest}} = \dfrac{{P \times R \times T}}{{100}}$

${\text{Simple Interest}} = \dfrac{{1200 \times 12 \times 3}}{{100}}$

${\text{Simple Interest}} = 432$

Thus, the simple interest is found to be ₹432.

Now, the amount is the addition of Principal and Simple Interest.

${\text{Amount}} = {\text{Principal}} + {\text{Simple Interest}}$

${\text{Amount}} = {\text{1200}} + {\text{432}}$

${\text{Amount}} = Rs.1,632$

Therefore, the amount is calculated to be ₹1,632.

(b) Principal = ₹7,500 at $5\% $ p.a.

Ans: It is given that the principal $\left( P \right) = $₹7,500, the rate $\left( R \right) = 5\% {\text{ p}}{\text{.a}}$ and the time $\left( T \right) = 3{\text{ years}}$.

The formula to evaluate the Simple Interest is given by,

${\text{Simple Interest}} = \dfrac{{P \times R \times T}}{{100}}$

${\text{Simple Interest}} = \dfrac{{7500 \times 5 \times 3}}{{100}}$

${\text{Simple Interest}} = Rs.1,125$

Thus, the simple interest is found to be ₹1,125.

Now, the amount is the addition of Principal and Simple Interest.

${\text{Amount}} = {\text{Principal}} + {\text{Simple Interest}}$

${\text{Amount}} = 7500 + 1125$

${\text{Amount}} = Rs.8,625$

Therefore, the amount is calculated to be ₹ 8,625.

10. What rate gives ₹280 as interest on a sum of ₹56,000 in 2 years?

Ans: It is given that the principal $\left( P \right) = $₹56,000, the simple interest $\left( {{\text{S}}{\text{.I}}} \right) = $₹280, the time $\left( T \right) = 2{\text{ years}}$.

The simple interest is formulated as follows,

${\text{Simple Interest}} = \dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}$

Substitute 280 for the Simple interest, 56000 for $P$ and 2 for $T$

${\text{280}} = \dfrac{{56000 \times {\text{R}} \times 2}}{{100}}$

${\text{R}} = \dfrac{{280 \times 100}}{{56000 \times 2}}$

${\text{R}} = 0.25\% $

Therefore, the rate of interest that gives ₹280 as interest on a sum of ₹56,000 in 2 years is $0.25\% $.

11. If Meena gives an interest of ₹45 for one year at $9\% $ rate p.a. What is the sum she has borrowed?

Ans: It is given that Meena gives an interest of ₹45 for a year and the rate of interest is $9\% $ p.a.

The simple interest is formulated as follows,

${\text{Simple Interest}} = \dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}$

$45 = \dfrac{{{\text{P}} \times {\text{9}} \times {\text{1}}}}{{100}}$

${\text{P}} = \dfrac{{45 \times 100}}{{9 \times 1}}$

${\text{P}} = 500$

Therefore, the amount that is borrowed by her is found to be ₹500.

Conclusion

NCERT Solutions for Class 7 Maths Chapter 7 Exercise 7.2 – Comparing Quantities by Vedantu offers a thorough understanding of essential financial concepts. This exercise focuses on practical applications such as calculating profit, loss, and simple interest, which are important for grasping the fundamentals of financial mathematics.

Students should pay special attention to understanding the formulas used for these calculations and practice solving a variety of problems in class 7 math 7.2 to reinforce their learning. Key areas to focus on include accurately determining percentages for profit and loss, as well as computing simple interest over multiple years. By mastering these concepts, students will confidently handle real-life financial calculations and be ready for advanced mathematical topics.

Class 7 Maths Chapter 7: Exercises Breakdown

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