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# Important Questions for CBSE Class 7 Maths Chapter 7 - Comparing Quantities

Last updated date: 20th Sep 2024
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## CBSE Class 7 Maths Chapter - 7 Important Questions - Free PDF Download

If you're a Year 7 student studying with CBSE, in your NCERT Maths book, there's a chapter called "Comparing Quantities." It's like a special part of maths, and it's important to understand it because the things you learn here can help you in everyday life. This chapter is all about different ways to compare numbers. Knowing these ideas can help you solve more difficult maths problems later on. You can find important questions and answers for Chapter 7 on the Vedantu website. We explain everything in an easy way, so it's not hard to understand. If you need help with Class 7 Science, you can also sign up online for lessons on Vedantu.com.

## Study Important Questions for class 7 Mathematics Chapter 7 – Comparing Quantities

Very Short Answer Questions (1 Mark)

1. Find the ratio of $600\;{\text{g}}$ to $5\;{\text{kg}}$.

Ans: First we will convert masses into the same units.

$5\;{\text{kg}}\; = \;5\; \times \;1000\;{\text{g}}\; = \;5000\;{\text{g}}$

Now we will find ratio of $600\;{\text{g}}$ and $5000\;{\text{g}}$

$600\;{\text{g}}\;:\;5000\;{\text{g}}\;{\text{ = }}\;\dfrac{{600}}{{5000}}\; \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;\dfrac{6}{{50}}\; = \;\dfrac{3}{{25}} \\$

Required ratio is $3:25$

2. Write $\dfrac{2}{5}$ as percent.

Ans: We’ll multiply numerator and denominator by $20$ to make denominator $100$

$\dfrac{2}{5} \times \dfrac{{20}}{{20}} = \dfrac{{40}}{{100}} = 40\%$

3. Convert $0.25$ to percent.

Ans: To convert $0.25$ to percent, we have to multiply it by $100\%$`

$= 0.25 \times 100\% \\ = \dfrac{{25}}{{100}} \times 100\% \\ = 25\% \\$

4. Find $100\% - 55\% =$____

Ans: we have

$100\% = \dfrac{{100}}{{100}} \\ 55\% = \dfrac{{50}}{{100}} \\ 100\% - 55\% = \dfrac{{100}}{{100}} - \dfrac{{55}}{{100}} \\ = \dfrac{{100 - 55}}{{100}} \\ = \dfrac{{45}}{{100}} = 45\% \\$

5. Find $25\%$ of $150$.

Ans: By using formula, we have

$\dfrac{{25}}{{100}} \times 150 = 37.5$

6. In a class of $45$ students $40\%$ are girls. Find the no. of boys.

Ans: Given: Girls percentage in class= $40\%$

Total no. of students =$45$
No. of girls $= 45 \times \dfrac{{40}}{{100}} = 45 \times \dfrac{2}{5} = 18$

No. of boys = Total no. of students – No. of girls

$= 45 - 18 \\ = 27 \\$

7. Find the ratio of
a. $18\,{\text{m }}$ to $45\;{\text{cm}}$
Ans: Let’s convert both lengths into the same unit.
$1\;m$ = $100\;cm$

$\therefore 18\;m$ = $18\times 100\; cm$

$18\;m\:\text{to}\: 45\;cm$ = $\dfrac{1800}{45}$ = $\dfrac{40}{1}$ = $40:1$

b. $20$ days to $48$ hours
Ans: First convert both time into same units
\begin{align} & \text{1 day = 24 hours} \\ & \text{20 days = 20 }\!\!\times\!\!\text{ 24 = 480 hours }\!\!~\!\!\text{ } \\ & \text{20 days : 48 hours = 480 : 48} \\ & \text{ =}\frac{480}{48}\,=\ 10 \\ \end{align}

Required Ratio is $\text{10 : 1}$

8. In a city, $35\%$ of population are males and $25\%$ are females and the remaining are children. What is the percentage of children in the city?

Ans: The total percentage = $100$

$\%$ of males    =  $35\%$

$\%$ of females = $25\%$

Also, Total $\%$   =   $\%$ of (Females $+$ males $+$ children)

$\therefore$ $\%$ of children = Total $\%$ $-$ $\%$ of females $-$ $\%$ of males

$= (100\; - \;35\; - \;25)\;\% \\ = \;\left( {100 - \;60} \right)\;\% \\ = \;40\;\% \\$

9. Find:

a. $80\% \;{\text{of}}\;2\;{\text{kg's}}$

b. $25\% \;{\text{of}}\;3000$

Ans:

a. $80\% \;{\text{of}}\;2\;{\text{kg's}}$

We know, $2\;{\text{kg}}\; = \;2000\;{\text{g}}$

$80\% \;{\text{of}}\;2000\;{\text{g}}\;{\text{ = }}\;80\% \; \times \;2000\;{\text{g}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{{80}}{{100}} \times \;2000\;{\text{g}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;80\; \times \;20\;{\text{g}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;1600\;{\text{g}}\;{\text{ = }}\;{\text{1}}{\text{.6}}\;{\text{kg}} \\$

b. $25\% \;{\text{of}}\;3000$

$\dfrac{{25}}{{100}} \times 3000 \\ = 25 \times \;30 \\ = 750 \\$

10. Convert given percent to decimal fractions and also to fractions in simplest form

a. $10\%$

b. $40\%$

Ans: To convert percent into decimal fraction we will divide the number by $100$

a. $10\% \; = \;\dfrac{{10}}{{100}}\; = \;\dfrac{1}{{10}}\; = \;0.1$

b. $40\% \; = \;\dfrac{{40}}{{100}}\; = \;\dfrac{4}{{10}}\; = \;0.4$

11. Mithali buys a TV at $12,000$ and sells it at a profit of $20\%$. How much money does she get?
Ans: Mithali buys TV = Cost price of TV = $12,000$

Mithali sells TV at profit = $20\% \;{\text{of}}\;12000$

$= \dfrac{{20}}{{100}} \times 12,000\; \\ = \;20\, \times 120 \\ = {\text{Rs}}{\text{.}}\;2400 \\$

Now, Selling price = cost price $+$ profit

$= 12,000 + 2,400 \\ = {\text{ Rs}}{\text{. }}14,400 \\$

Therefore, Mithali get ${\text{Rs}}{\text{. }}14,400$

12. Find

a. $12\dfrac{1}{2}\% {\text{ of}}\;75$

b. $30\% {\text{ of }}150$

Ans:

a. $12\dfrac{1}{2}\% {\text{ of}}\;75$

We know, $12\dfrac{1}{2}\; = \;\dfrac{{25}}{2}$

$\therefore \dfrac{{25}}{2}\% {\text{ of}}\;75 = \dfrac{{25}}{{2 \times 100}} \times 75 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{1}{{2 \times 4}} \times 75 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{{75}}{8} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 9\dfrac{3}{8} \\$

b. $30\% {\text{ of }}150$

$= \dfrac{{30}}{{100}} \times 150\\ = \;3\; \times 15 \\ = \;45 \\$

13. Find the whole quantity if

a. $10\,\%$ of it is $800$

b. $15\%$ of it is $1050$

Ans:

a. $10\,\%$ of it is $800$

Let quantity is $x$

$\therefore \;10\;\% \;{\text{of}}\;x\; = \;800 \\ \Rightarrow \;10\;\% \; \times \;x\; = \;800 \\ \Rightarrow \;\;\dfrac{{10}}{{100}}\; \times \;x\; = \;800 \\ \Rightarrow \;\;\dfrac{1}{{10}}\; \times \;x\; = \;800 \\ \Rightarrow \;\;\;\;\;\;\;\;\;x\; = \;800\; \times \;10 \\ \;\;\;\;\;\therefore \;\;\;\;\;x = \;8000 \\$

So, whole quantity = $8000$

b. $15\%$ of it is $1050$

Let the quantity is $x$

$\therefore \;15\;\% \;{\text{of}}\;x\; = \;1050 \\ \Rightarrow \;15\;\% \; \times \;x\; = \;1050 \\ \Rightarrow \;\;\dfrac{{15}}{{100}}\; \times \;x\; = \;1050 \\ \Rightarrow \;\;\;\;\dfrac{3}{{20}}\; \times \;x\; = \;1050 \\ \Rightarrow \;\;\;\;\;\;\;\;\;\;x\; = \;1050\; \times \;\dfrac{{20}}{3}\\ \;\;\;\;\;\therefore \;\;\;\;\;\;x = \;350\; \times \;20 \\ \;\;\;\;\;\;\; \Rightarrow \;\;x\; = \;7000 \\$

14. Manvita saves ${\mathbf{Rs}}.{\text{ }}{\mathbf{9000}}$ from her salary every month. If it is ${\mathbf{5}}\%$ of her salary. Find her salary.

Ans: Let Manvita’s salary = $x\;{\text{Rs}}{\text{.}}$

${\mathbf{5}}\%$ of Manvita’s Salary  = ${\mathbf{Rs}}.{\text{ }}{\mathbf{9000}}$

$\therefore 5\% {\text{ of x}} = 9000 \\ \Rightarrow \dfrac{5}{{100}} \times x = 9000 \\ \;\;\;\;\;\; \Rightarrow x = \dfrac{{9000 \times 100}}{5}\\ \;\;\;\;\; \Rightarrow \;x\; = \;1800\; \times 100 \\ \;\;\;\;\;\therefore \;\;x = 1,80,000\\$

Therefore, Manvita’s salary is $1,80,000$

15. convert the following fractions to percent.

a. $\dfrac{3}{4}$

b. $\dfrac{2}{5}$

Ans:

a. We’ll multiply the numerator and denominator by $25$to make the denominator $100$.

$\dfrac{3}{4} \times \dfrac{{25}}{{25}}\; = \;\dfrac{{75}}{{100}}\; = \;75\;\%$

b. We’ll multiply the numerator and denominator by $20$to make the denominator $100$.

$\dfrac{2}{5} \times \dfrac{{20}}{{20}}\; = \;\dfrac{{40}}{{100}}\; = \;40\;\%$

16. Convert the following into percent

a. $\dfrac{{34}}{{50}}$

b. $\dfrac{1}{4}$

c. $0.03$

Ans: First, we will make denominator $100$, and then we will convert that number into percent

a. $\dfrac{{34}}{{50}} \times \dfrac{2}{2} = \dfrac{{68}}{{100}} = 68\%$

b. $\dfrac{1}{4} \times \dfrac{{25}}{{25}} = \dfrac{{25}}{{100}} = 25\%$

c. $0.03\;\; = \;\dfrac{3}{{100}} = 3\%$

17. Out of $24,000$voters in a constituency $48\%$ voted. Find the no. of voters who did not vote.

Ans: Total voters = $24,000$

Vote percentage = $48\%$

No. of Voters voted = $48\,\% \;{\text{of}}\;12,000$

$= \dfrac{{48}}{{100}} \times 24,000 \\ = \;48 \times 240 \\ = 11,520 \\$

No. of voters did not vote = Total voters – No. of voters voted

$= 24,000 - 11,520 \\ = 12,480 \\$

Therefore, $12,480$ people did not vote.

18. If ${\text{Rs}}{\text{. }}500$has to be divided among Mishala, Manvita and Meera in the ratio $1{\text{ }}:{\text{ }}3{\text{ }}:{\text{ }}6$. Then how much money will each get and what will be the percentages.

Ans:  Total Amount =${\text{Rs}}{\text{. }}500$

Given, ${\text{Rs}}{\text{. }}500$has to be divided among Mishala, Manvita and Meera in the ratio $1{\text{ }}:{\text{ }}3{\text{ }}:{\text{ }}6$

Let common ratio be $x$

Mishala’s share = $x$

Manvita’s share = $3x$

Meera’s share = $6x$

According to question

$x + 3x + 6x\; = \;500 \\ \;\;\;\; \Rightarrow \;10\;x\; = \;500 \\ \;\;\;\;\;\;\;\;\;\;\therefore x = \dfrac{{500}}{{10}}\; = \;50 \\$

Mishala’s share = $50\;{\text{Rs}}.$

$\% = \dfrac{1}{{10}} \times 100 = 10\%$

Manvita’s share = $3x\; = \;3 \times \;50\; = \;150\;{\text{Rs}}.$

$\% = \dfrac{3}{{10}} \times 100 = 30\%$

Meera’s share = $6x\, = \;6\; \times \;\;50\; = \;300\;{\text{Rs}}.$

$\% = \dfrac{6}{{10}} \times 100 = 60\%$

19.Find the amount to be paid at the end of $4$ years at each case

a. Principal = ${\text{Rs}}{\text{. }}1500{\text{ at }}8\%$ PA

b. Principal = ${\text{Rs}}{\text{. }}7500{\text{ at }}5\%$ PA

Ans:

a. Given,

${\text{P}} = 1500,\;{\text{R}} = 8\% \;\,{\text{PA}}\;,\;{\text{T}} = 4{\text{ years }}$

We know, ${\text{Amount}}\;{\text{ = }}\;{\text{principal}}\;{\text{ + }}\;{\text{interest}}$

And, $\text{Interest}\;$ = $\dfrac{\text{Principal}\; \times \text{Rate}\;\times\;\text{Time}}{{100}}$

${\text{I}} = \dfrac{{1500 \times 8 \times 4}}{{100}} = 15 \times 8 \times 4\; = 480\;{\text{Rs}}.$

Amount = ${\text{P}}\;{\text{ + }}\;{\text{I}}\;{\text{ = }}\;{\text{1500}}\;{\text{ + }}\;{\text{480}}\;{\text{ = }}\;{\text{1980}}\;{\text{Rs}}{\text{.}}$

b. ${\text{P}} = 7500,\;{\text{R}} = 5\% \;{\text{PA}},\;\;{\text{T}} = 4{\text{ years }}$

Now, interest ${\text{I = }}\dfrac{{{\text{PRT}}}}{{{\text{100}}}} = \dfrac{7500 \times 5 \times 4}{100} = \;75 \times 5 \times 4\; = \;1500\;{\text{Rs}}.$

Therefore, Amount = ${\text{P}}\;{\text{ + }}\;{\text{I}}\; = \,{\text{7500}}\;{\text{ + }}\;{\text{1500}}\;{\text{ = }}\;{\text{9000}}\;{\text{Rs}}{\text{.}}$

20. What rate gives an interest of ${\text{Rs}}{\text{. }}540$on a sum of ${\text{Rs}}{\text{. }}18000$in $3{\text{ years}}$?

Ans:  Given, Interest, ${\text{I}}\;{\text{ = }}\;{\text{Rs}}{\text{. }}540$

Principal,${\text{P}}\;{\text{ = }}\;{\text{Rs}}{\text{. }}18000$

Time, ${\text{T}}\; = \;3{\text{ years}}$

We know, $\text{Interest}\;$ = $\dfrac{\text{Principal}\; \times \text{Rate}\;\times\;\text{Time}}{{100}}$

$\therefore {\text{Rate}}\;{\text{ = }}\;\dfrac{\text{Interest}\;\times\;\text{100}}{\text{Principal}\,\times\;\text{Time}}$

$= \dfrac{{540 \times 100}}{{18000 \times 3}} = \dfrac{{180}}{{180}}\;\; = \;1\%$

## CBSE Important Questions Class 7 Maths Chapter 7

### Solved Examples

1. There are 45 students in a class, and 40% are girls. What is the number of boys?

Solution:

Total number of students in a class: 45

Percentage of girls

So, the number of girls in the class are:

40/100 x 45

= 2 x 9

= 18

So, the total number of boys is 45 - 18 = 27

2. What is 2/5  as a percent?

Solution:

⅖ x 20/20 = 40/100

= 40%

## Why you should Use Vedantu’s Important Questions for Class 7 Math?

Benefits of Important Questions from Vedantu for Class 7 Chapter 7 - Comparing Quantities”:

• Focus on key topics for efficient studying.

• Prepares students for exams and reduces anxiety.

• Reinforces understanding of fundamental concepts.

• Teaches effective time management.

• Enables self-assessment and progress tracking.

• Strategic approach for higher scores.

• Covers a wide range of topics for comprehensive understanding.

• Supports exam preparation and boosts confidence.

## Conclusion

Engaging with Important Questions for CBSE Class 7 Maths Chapter 7 - Comparing Quantities serves as a pivotal step toward mastering the fundamental concepts of quantitative comparison. These questions offer students a comprehensive review of various scenarios, honing their skills in analyzing and contrasting quantities. By tackling these exercises, students develop a solid foundation in mathematical comparisons, fostering their ability to apply these skills in practical scenarios. This practice ensures a thorough understanding of the chapter, empowering students to confidently tackle quantitative challenges and excel in their mathematical journey.

## FAQs on Important Questions for CBSE Class 7 Maths Chapter 7 - Comparing Quantities

1. How to Quickly Grasp the Sums Based on the Concept of Interest?

There are mainly two subtypes of interest and sums are based on various operations on them. These two variants are simple interest and compound interest. Although compound interest is more complicated than simple interest and it is taught in higher standards. You need to learn all the formulas related to simple interest and solve the sums in different procedures. The main formula of simple interest is SI= (PxRxT)/100. Here, P is the principal amount, R denotes the urate of interest and T denotes time. Now, you have to derive either of these three while doing a sum on interest belonging to the exercises of Chapter 7.

2. What is the Right Way to Learn the Sums on Decimals Easily?

The right way to learn the sums on decimal is simple as you need to study the basic concepts of decimal units at first. You can differentiate between integers and decimal numbers and learn how to denote decimal numbers verbally. In the next step, you will come to know about adding two decimal numbers, subtracting them, or multiplying and dividing. Furthermore, decimals’ advanced problems relate to conversations where you have to learn to change the decimal numbers to fractions and work with both decimals and percentages. As per the sums on percentage are concerned, you can learn the basic mathematical works with them like the decimals.

3. What do you mean by comparing quantities?

Two values must have the same units to be compared, and two ratios can be compared by turning them into like fractions. We say the two provided ratios are comparable if the two fractions are equal. Comparing Quantities refers to the quantitative relationship between two quantities that indicates their respective sizes. It is merely a tool for comparing amounts. It has multiple benefits and uses in our day to day life, and hence it is very crucial to learn about the concepts related to it.

4. What is the use of comparing quantities Class 7?

To compare is to perform an analysis between the differences in numbers, amounts, or values in order to validate if one quantity is larger than, lower than, or equal to another. We can define or determine how much a number is bigger or less by comparing. We may also compare decimals and fractions in this way. If you want to know how much something weighs, you may do that by measuring it against another standard or reference unit.

5. What is the use of percentage Class 7?

To make computations easier, we utilise percentages. Working with parts of 100 is considerably easier than working with thirds, twelfths, and so on, especially since many fractions lack a precise (non-recurring) decimal counterpart. When the number of individuals in the groups are not the same, it is critical to compare using percentages. Comparing counts and percentages are similar when the number of persons in the categories is the same. Hence, using percentages can come in handy in multiple operations and can help address different confusions and problems.

6. What does it mean by interest in maths?

The cost of borrowing money is interest, which is paid to the lender as a charge for the loan. Simple interest is a fixed proportion of the principal amount borrowed or lent that is paid or received over a period of time. The amount of money paid for the use of someone else's money is known as interest. It is a very crucial agent in transactions carried out in today’s market. Hence it is important to have a good knowledge of its concepts, processes and uses. To know more, solve the important questions by visiting the page Important questions for Class 7 Maths and download a free PDF of the same.

7. Where can I find the Solutions of Class 7 Maths Chapter 7?

The solutions are easily available on the Vedantu site.

• Click on the link- NCERT Solutions for Class 7 maths chapter 7.

• The webpage with Vedantu’s solutions for Class 7 Maths Chapter 7 will open.