CBSE Class 7 Maths Important Questions 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 

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

Short Answer Questions (2 Marks)

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\;\% $

Long Answer Questions (3 Marks)

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

Importance of Comparing Quantities

Understanding how to compare quantities is super important! In the Class 7 CBSE NCERT Maths book, there's a chapter all about it. This chapter talks about different kinds of maths, like ratio, percentage, decimals, profit and loss, and interests. There are many ideas in this chapter, so it's good to learn them well. If you ever come across tricky problems, Vedantu's website has solutions for them. They're like answers to the uncommon problems you might find in this chapter.

You can also find examples of important questions for Class 7 Maths, especially in the chapter about comparing quantities. These examples come with solved papers, which means the answers are already there. It's a good idea to look at these examples, get some guidance, and then try solving similar problems by yourself. This way, you can learn even better!

Main Subtypes of Comparing Quantities and its Relevance

Let's talk about why these maths concepts are super useful in real life!

• Ratio: When you practise problems involving ratio, you're basically getting really good at estimating things. This skill helps not only in planning your day-to-day activities but also in understanding other cool subjects like Chemistry and Physics.

• Decimals: In the important questions of Chapter 7 in Class 7 Maths, you'll find lots of sums about decimals. Pay attention because these problems could show up in your final exam, and they're worth a big part of your marks. Beyond exams, knowing decimals helps you handle money transactions correctly and make sure you're getting the right amount of stuff for the money you spend.

• Percentage: Learning about percentages is another smart way to compare quantities. If you nail the percentage problems, comparing numbers becomes a breeze. It's like having a special knowledge about how percentages, fractions, and decimals are all connected. Once you've got these down, you might encounter problems that involve converting between them, and these could earn you lots of marks. 

• Problems of Interest: There are generally two types of interests and comparing quantities Class 7 important questions have a high chance of being based on one of these concepts. A clear concept of interest can enhance your knowledge about money matters. You can gradually learn various banking procedures and manage your own pocket money.

Apart from interests, you can also do sums on the concepts of gain or loss. This also has relevance in terms of reality as you can brush-up your business skills right from a very young age. Moreover, by doing the sums on comparing qualities, you can develop an interest in solving mathematical problems that can help you face more complex problems of different subclasses of Mathematics in the higher standards.

Remember, there's no better way to get good at maths than by practising it a lot. The more you practise, the better you become! Keep at it, and you'll see yourself getting better and better at solving those tricky maths problems. Practice makes perfect!

5 Important Formulas of Class 7 Chapter 7 Comparing Quantities You Shouldn’t Miss!

Understanding these key formulas in Chapter 7, "Comparing Quantities," is essential for solving problems accurately. Here are five formulas that you should focus on:

1. Ratio Formula

A ratio compares two quantities of the same kind by division.

$\text{Ratio of } a \text{ to } b = \dfrac{a}{b} \text{ or } a:b$

2. Percentage Formula

Percentage represents a number as a fraction of 100.

$\text{Percentage} = \dfrac{\text{Part}}{\text{Whole}} \times 100$

3. Simple Interest Formula

Simple interest is calculated based on the initial principal amount, rate, and time.

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

where $P$ is the principal amount, $R$ is the rate, and $T$ is the time.

4. Amount Formula (in Simple Interest)

The total amount after adding simple interest to the principal.

$\text{Amount (A)} = P + SI$

5. Profit and Loss Percentage Formula

These formulas help in calculating profit or loss percentages based on cost and selling prices.

$\text{Profit Percentage} = \dfrac{\text{Profit}}{\text{Cost Price}} \times 100$

$\text{Loss Percentage} = \dfrac{\text{Loss}}{\text{Cost Price}} \times 100$

Benefits of Important Questions for Class 7 Maths Chapter 7 Comparing Quantities

• Important Questions for Class 7 Maths Chapter 7 Comparing Quantities help break down complex ideas like ratios, percentages, and simple interests, making them easier to understand and apply.

• Regular practice with these practice questions improves students' ability to handle calculations, which is essential for solving problems accurately and quickly.

• These questions give students insight into the types of questions commonly asked in exams, building familiarity with the pattern and difficulty level.

• Practising a variety of questions enhances students' approach to different types of problems, boosting their problem-solving skills.

• These important questions cover the main topics, making it easier for students to revise the chapter effectively before exams.

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 analysing 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.

Important Study Materials for Class 7 Maths Chapter 7

CBSE Class 7 Maths Important Questions for All Chapters

Class 7 Maths Important Questions and Answers cover key topics, aiding in thorough preparation and making revision simpler.

Important Study Materials for Class 7 Maths