## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities (EX 8.3) Exercise 8.3

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## Access NCERT solutions for Class 7 Maths Chapter 8 – Comparing Quantities

Exercise 8.3

Refer for exercise 8.3 in the pdf

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.

## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3

Opting for the NCERT solutions for Ex 8.3 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.3 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 8 Exercise 8.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 8 Exercise 8.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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## FAQs on NCERT Solutions for Class 7 Maths Chapter 8 - Exercise

1. Do I need to practice all the questions provided in NCERT Solutions for Chapter 8 Comparing Quantities of Class 7 Maths?

Yes, practicing all the questions from the NCERT Solutions for Chapter 8 Comparing Quantities of Class 7 Maths is vital. This will give you a better understanding of the chapter. You will get insight into different varieties of questions, and the confidence to attempt them in the exam. Practice makes you perfect the methods you apply to solve the sums and that is the core to studying all chapters in Mathematics.

2. How many questions are there in NCERT Solutions for Chapter 8 Comparing Quantities of Class 7 Maths?

NCERT Solutions for Chapter 8 Comparing Quantities of Class 7 Maths has a total of 24 questions. They are divided between three exercises as follows: Exercise 8.1 - three Questions (two long answers), Exercise 8.2 - 10 Questions (four long answers), Exercise 8.3 - 11 Questions (seven long answers). Take a note to recheck all the steps required in the long answers with the help of NCERT Solutions so you do not miss marks designated to each step.

3. What does the term comparing quantities mean according to Chapter 8 of Class 7 Maths?

Comparing quantities as the name denotes is the method of comparing two or more quantities. However, both the quantities have to be in the same units. For example, you cannot compare the age of one person with the weight of another person.

The students can learn more about this Chapter in the NCERT Solutions for Chapter 8 of Class 7 Maths. The solutions are available in a chapter-wise format for all-encompassing learning of the topics covered in the course.

4. What is the formula for comparing quantities according to Chapter 8 of Class 7 Maths?

Chapter 8 Comparing Quantities of Class 7 Maths has a number of formulas. The students can find all the formulas on the NCERT Solutions for Chapter 8 of Class 7 Maths. These solutions explain the topics in a comprehensive manner for the students to be able to grasp the concepts. It is necessary that the students understand these formulas through the theory for better results instead of mugging them. Apprehension of the formulas helps you remember them.

5. What is PA in comparing quantities according to Chapter 8 of Class 7 Maths?

PA in Chapter 8 Comparing Quantities stands for Principal Amount. It is the initial amount borrowed that is added to the Simple Interest to determine the amount to be returned. For example, you borrow Rs. One thousand for a year at a 5% rate for the interest, the PA would be Rs. 1,000. Download NCERT Solutions for Chapter 8 of Class 7 Maths for better insight into all the concepts in the chapter.