NCERT Solutions for Maths Class 7 Chapter 7 Exercise 7.1 Comparing Quantities - FREE PDF Download
In NCERT Solution for Class 7th Exercise 7.1, Students will learn how to calculate and interpret ratios, convert them into percentages, and understand their practical applications. The problems in Exercise 7.1 are designed to build a strong foundation in comparing quantities, which is essential for solving more complex problems in later exercises and real-life scenarios. students will develop a solid understanding of calculating probabilities in complex situations, which is vital for their academic and competitive exam preparations. Students can access the revised Class 7 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.
- 3.1Exercise 7.1
Specially Class 7 Maths Chapter 7 Exercise 7.1 Question is important for students to understand it well. Students will learn how to calculate percentages, determine profit or loss in various transactions, and compute simple interest over different periods. Access Class 7 Maths Syllabus here.
Glance on NCERT Solutions Maths Chapter 7 Exercise 7.1 Class 7 | Vedantu
NCERT Solution for class 7th maths chapter 7 exercise 7.1 contains topics like the Meaning of Percentages, Converting Fractional Numbers to Percentages, Converting Decimals to Percentages, Converting Percentages to Fractions or Decimals, Fun with Estimation, and Use Of Percentages.
Meaning of Percentages: This section explains what percentages are and how they represent a part out of a whole (usually out of 100). Students will learn how the percentage symbol (%) is used.
Converting Fractional Numbers to Percentages: Here, Students will discover methods to convert fractions into percentages This might involve multiplying by 100 and adding a percent sign.
Converting Decimals to Percentages: Similar to fractions, this section teaches you how to convert decimal numbers into percentages. This might involve multiplying by 100 (or moving the decimal two places to the right) and adding a percent sign.
Converting Percentages to Fractions or Decimals: This section covers the reverse process - converting percentages back into fractions or decimals. You might learn to divide by 100 (or move the decimal two places to the left) and remove the percent sign.
Fun with Estimation: This section might introduce you to using percentages for estimation purposes. Students will learn to approximate values based on percentages.
Use of Percentages: This section explores real-life applications of percentages. You might see examples of how percentages are used in discounts, sales tax calculations, and other practical scenarios.
There are 10 questions in Exercise 7.1 Maths Class 7 Chapter 7 which experts at Vedantu fully solve.
Exercise 7.1
1. Convert the given fractional numbers to percent:
(a) $\frac{\text{1}}{\text{8}}$
Ans: A fractional number is given, $\frac{1}{8}$
$\frac{1}{8} =\frac{1}{8} \times 100 \%$
$=\frac{25}{2} \%=12.5 \%$
$\Rightarrow \frac{1}{8}=12.5 \%$
(b) $\frac{\text{5}}{\text{4}}$
Ans: A fractional number is given, $\frac{\text{5}}{\text{4}}$
$\frac{5}{4}=\frac{5}{4} \times 100 \%$
$=5 \times 25 \%$
$=125 \%$
$\Rightarrow \frac{5}{4}=125 \%$
(c) $\frac{\text{3}}{\text{40}}$
Ans: A fractional number is given, $\frac{\text{3}}{\text{40}}$
$\frac{3}{40}=\frac{3}{40} \times 100 \%$
$\quad=\frac{15}{2} \%$
$=7.5 \%$
$\Rightarrow \frac{3}{40}=7.5 \%$
(d) $\frac{\text{2}}{\text{7}}$
Ans: A fractional number is given, $\frac{\text{2}}{\text{7}}$
$\frac{2}{7}=\frac{2}{7} \times 100 \% $
$=\frac{200}{7} \%$
$=28 \frac{4}{7} \%$
$\Rightarrow \frac{2}{7}=28 \frac{4}{7} \%$
2. Convert the given decimal fractions to percents:
(a) $\text{0}\text{.65}$
Ans: A decimal fraction is given, needed to convert it into percentage from
$\Rightarrow 0.65=\frac{65}{100}$
$\Rightarrow 0.65=65 \%$
(b) $\text{2}\text{.1}$
Ans: A decimal fraction is given, needed to convert it into percentage from
$\Rightarrow 2.1=\frac{21}{10} \times 100 \%$
$\Rightarrow 2.1=210 \%$
(c) $\text{0}\text{.02}$
Ans: A decimal fraction is given, needed to convert it into percentage from
$\Rightarrow 0.02=\frac{2}{100}$
$\Rightarrow 0.02=2 \%$
(d) $\text{12}\text{.35}$
Ans: A decimal fraction is given, needed to convert it into percentage from
$\Rightarrow 12.35=\frac{1235}{100} \times 100 \%$
$\Rightarrow 12.35=1235 \%$
3. Estimate what part of the figure is coloured and hence find the percent which is coloured:
(i)
Ans: It is clear from the figure that the coloured part is $\text{= }\frac{\text{1}}{\text{4}}$
Therefore, the required percentage of coloured parts,
is $=\frac{1}{4} \times \frac{25}{25} $
$=\frac{25}{100} $
$=25 \%$
(ii)
Ans: It is clear from the figure that the coloured part is $\text{= }\frac{\text{3}}{\text{5}}$
Therefore, the required percentage of coloured parts,
is $\text{= }\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{20}}{\text{20}}$
$\text{= }\frac{\text{60}}{\text{100}}$
$\text{= 60 }\%$
(iii)
Ans: It is clear from the figure that the coloured part is $\text{= }\frac{\text{3}}{\text{8}}$
Therefore, the required percentage of coloured parts,
is $\text{= }\frac{\text{3}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{100}}{\text{100}}$
$\text{= }\frac{\text{300}}{\text{8}}\text{ }\%$
$\text{= 37}\text{.5 }\%$
4. Find:
(a) $\text{15 }\%$ of $\text{250}$
Ans: Needed to find the required percentage of the given number $\text{250}$ i.e.,
$\text{15 }\%\text{ }$ of $\text{250}$ $\text{= }\frac{\text{15}}{\text{100}}\text{ }\!\!\times\!\!\text{ 250}$
$\text{= 15 }\!\!\times\!\!\text{ 2}\text{.5}$
$\text{= 37}\text{.5}$
$\Rightarrow \text{ 15 }\%\text{ }$ of $\text{250}$$\text{= 37}\text{.5}$
(b). $\text{1 }\%$ of $\text{1}$ hour
Ans: Needed to find the required percentage i.e.,
$\text{1 }\%$ of $\text{1}$ hour $\text{= 1 }\%$ of $\text{60}$ minutes
$\text{= 1 }\%$ of $\text{60 }\!\!\times\!\!\text{ 60}$ seconds
$\text{= }\frac{\text{1}}{\text{100}}\text{ }\!\!\times\!\!\text{ 60 }\!\!\times\!\!\text{ 60}$ seconds
$\text{= 36}$ seconds
$\Rightarrow \text{1 }\%$ of $\text{1}$ hour $\text{= 36}$ seconds
(c). $\text{20 }\%$ of Rs$\text{2500}$
Ans: Needed to find the required percentage of the given currency Rs$\text{2500}$ i.e.,
$\text{20 }\%$ of Rs$\text{2500}$ $\text{= }\frac{\text{20}}{\text{100}}\text{ }\!\!\times\!\!\text{ 2500}$
$\text{= 20 }\!\!\times\!\!\text{ 25}$
$\text{=}$ Rs $\text{500}$
$\Rightarrow \text{20 }\%$ of Rs$\text{2500}$$\text{=}$Rs$\text{500}$
(d). $\text{75 }\%$ of $\text{1}$ kg
Ans: Needed to find the required percentage of the given quantity $\text{1}$ kg i.e.,
$\text{75 }\%$ of $\text{1}$ kg $\text{= }\frac{75}{\text{100}}\text{ }\!\!\times\!\!\text{ 1}$ kg
$\text{= 0}\text{.75}$kg
$\Rightarrow \text{75 }\%$ of $\text{1}$ kg $\text{= 0}\text{.75}$kg
5. Find the whole quantity if:
(a). $\text{5 }\%$ of it is $\text{600}$
Ans: Let the required whole quantity be $\text{x}$
Therefore, $\text{5 }\%$ of $\text{x}$ $\text{= 600}$
$\Rightarrow \text{ }\frac{\text{5}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 600}$
$\Rightarrow \text{ x = }\frac{\text{600 }\!\!\times\!\!\text{ 100}}{\text{5}}$
$\Rightarrow \text{ x = 12000}$
(b). $\text{12 }\%$ of it is Rs$\text{1080}$
Ans: Let the required whole quantity be $\text{x}$
Therefore, $\text{12 }\%$ of $\text{x}$ $\text{= Rs 1080}$
$\Rightarrow \text{ }\frac{\text{12}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 1080}$
$\Rightarrow \text{ x = }\frac{\text{1080 }\!\!\times\!\!\text{ 100}}{\text{12}}$
$\Rightarrow \text{ x = Rs 9000}$
(c). $\text{40 }\%$ of it is $\text{500}$ km
Ans: Let the required whole quantity be $\text{x}$
Therefore, $\text{40 }\%$ of $\text{x}$ $\text{= 500}$km
$\Rightarrow \text{ }\frac{\text{40}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 500}$
$\Rightarrow \text{ x = }\frac{\text{500 }\!\!\times\!\!\text{ 100}}{\text{40}}$
$\Rightarrow \text{ x = 1250}$km
(d). $\text{70 }\%$ of it is $\text{14}$ minutes
Ans: Let the required whole quantity be $\text{x}$
Therefore, $\text{70 }\%$ of $\text{x}$ $\text{= 14}$minutes
$\Rightarrow \text{ }\frac{\text{70}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 14}$
$\Rightarrow \text{ x = }\frac{\text{14 }\!\!\times\!\!\text{ 100}}{\text{70}}$
$\Rightarrow \text{ x = 20}$ minutes
(e). $\text{8 }\%$ of it is $\text{40}$ liters
Ans: Let the required whole quantity be $\text{x}$
Therefore, $\text{8 }\%$ of $\text{x}$ $\text{= 40}$liters
$\Rightarrow \text{ }\frac{\text{8}}{\text{100}}\text{ }\!\!\times\!\!\text{ x = 40}$
$\Rightarrow \text{ x = }\frac{\text{40 }\!\!\times\!\!\text{ 100}}{\text{8}}$
$\Rightarrow \text{ x = 500}$ liters
6. Convert given percents to decimal fractions and also fractions to simplest form:
(a). $\text{25 }\%$
Ans: We have given a percent $\text{25 }\%$
Fraction form$\text{= }\frac{\text{25}}{\text{100}}$
Simplest fractional form $\text{= }\frac{\text{1}}{\text{4}}$
Decimal form $\text{= 0}\text{.25}$
(b). $\text{150 }\%$
Ans: We have given a percent $\text{150 }\%$
Fraction form $\text{= }\frac{\text{150}}{\text{100}}$
Simplest fractional form $\text{= }\frac{\text{3}}{\text{2}}$
Decimal form $\text{= 1}\text{.5}$
(c). $\text{20 }\%$
Ans: We have given a percent $\text{20 }\%$
Fraction form $\text{= }\frac{\text{20}}{\text{100}}$
Simplest fractional form $\text{= }\frac{\text{1}}{\text{5}}$
Decimal form $\text{= 0}\text{.2}$
(d). $\text{5 }\%$
Ans: We have given a percent $\text{5 }\%$
Fraction form $\text{= }\frac{\text{5}}{\text{100}}$
Simplest fractional form $\text{= }\frac{\text{1}}{\text{20}}$
Decimal form $\text{= 0}\text{.05}$
7. In a city, $\text{30 }\%$ are females, $\text{40 }\%$ are males and remaining are children. What percent are children?
Ans: Let the percentage of children be $\text{x }\%$
It is given that the percentage of females and males are $\text{30 }\%$ and $\text{40 }\%$ respectively.
And, the total percentage $\text{= 100 }\%\text{ =}$ Percentage of males and Percentage of females and Percentage of children
$\Rightarrow \text{ 100 }\%\text{ = 30 }\%\text{ + 40 }\%\text{ + x }\%$
$\Rightarrow \text{ 100 }\%\text{ = 70 }\%\text{ + x }\%$
$\Rightarrow \text{ x }\%\text{ = 100 }\%\text{ - 70 }\%\text{ }$
$\Rightarrow \text{ x }\%\text{ = 30 }\%$
Thus $\text{30 }\%\text{ }$is the population of children in the city.
8. Out of $\text{15,000}$ voters in a constituency, $\text{60 }\%$ voted. Find the percentage of voters who did not vote. Can you now find out how many did not vote?
Ans: The total number of voters $\text{= 15,000}$
The percentage of people who voted $\text{= 60 }\%$
So, the percentage of people who didn’t vote $\text{= 100 }\%\text{ - 60 }\%$
$\text{= 40 }\%$
And, the number of actual candidates, who didn’t vote $=40 \%$ of $\text{15000}$
$\text{= 6000}$
Thus, $\text{6,000}$ people out of $\text{15,000}$ did not vote.
9. Meeta saves Rs $\text{400}$ from her salary. If this is $\text{10 }\%$ of her salary. What is her salary?
Ans: Let $\text{x}$ be the salary of Meeta.
Since $\text{10 }\%$ of her salary $\text{= Rs 400}$
$\Rightarrow \text{ 10 }\%\text{ of x = 400}$
$\Rightarrow \text{ 10 }\%\text{ }\!\!\times\!\!\text{ x = 400}$
$\Rightarrow \text{ }\frac{\text{10}}{\text{100}}\text{x = 400}$
$\Rightarrow \text{ x = Rs 4,000}$
Therefore, the salary of Meeta is $\text{Rs 4,000}$.
10. A local cricket team played $\text{20}$ matches in one season. It won $\text{25 }\%$ of them. How many matches did they win?
Ans: The local cricket team played $\text{20}$ matches.
They won $\text{25 }\%$ of matches out of $\text{20}$
Therefore, the number of matches the cricket team won $\text{= 25 }\%\text{ of 20}$
$\text{= }\frac{\text{25}}{\text{100}}\text{ }\!\!\times\!\!\text{ 20}$
$\text{= 5}$ matches
So, the local cricket team won $\text{5}$ matches out of $\text{20}$.
Conclusion
In Class 7 Maths Chapter 7 Exercise 7.1 provides a solid understanding of comparing quantities using ratios and percentages. Maths Class 7 Chapter 7 Exercise 7.1 helps students grasp the basics of these concepts and apply them in various contexts. By working through the problems, students build confidence in their ability to compare different quantities accurately, setting a strong foundation for more advanced topics.
Class 7 Maths Chapter 7: Exercises Breakdown
CBSE Class 7 Maths Chapter 7 Other Study Materials
Chapter-Specific NCERT Solutions for Class 7 Maths
Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
Important Related Links for NCERT Class 7 Maths
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FAQs on NCERT Solutions for Class 7 Maths Chapter 7 Comparing Quantities Ex 7.1
1. Do I need to practice all the questions provided in class 7 math exercise 7.1?
Yes, practising all the questions from the NCERT Solutions for Chapter 7 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 class 7 maths chapter 7.1?
NCERT Solutions for Chapter 7 Comparing Quantities of Class 7 Maths has a total of 24 questions. They are divided into three exercises as follows: Exercise 7.1 - 10 Questions (four long answers), Exercise 7.1 - 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 for each step.
3. What does the term comparing quantities mean according to class 7 math exercise 7.1?
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 7 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 class 7 maths chapter 7.1?
Chapter 7 Comparing Quantities of Class 7 Maths has several formulas. The students can find all the formulas on the NCERT Solutions for Chapter 7 of Class 7 Maths. These solutions explain the topics in a comprehensive manner for the students to be able to grasp the concepts. The students must understand these formulas through the theory for better results instead of mugging them. Apprehension of the formulas helps you remember them.
5. What is a ratio in class 7 math 7.1 Exercise?
A ratio is a way to compare two quantities by dividing one quantity by the other. It shows how many times one quantity is contained in another. Ratios are usually expressed in the form a or a/b.
6. How do you convert a ratio to a percentage?
According to class 7 math 7.1 Exercise, To convert a ratio to a percentage, you divide the first quantity by the second quantity, multiply the result by 100, and add the percentage symbol (%). For example, to convert the ratio 3:4 to a percentage, you calculate (3/4) * 100 = 75%.
7. Why is it important to learn about comparing quantities?
Learning to compare quantities is important because it helps in understanding relationships between different values. It is useful in everyday situations like comparing prices, calculating discounts, understanding statistics, and making informed decisions.
8. What types of questions are included in Exercise 7.1?
Exercise 7.1 includes questions on calculating ratios, converting ratios to percentages, and interpreting these comparisons in various contexts. These questions help students build a foundational understanding of comparing quantities.
9. How can I improve my skills in Class 7th Exercise 7.1 comparing quantities?
To improve your skills in comparing quantities, practice solving a variety of problems, understand the basic concepts of ratios and percentages and apply these concepts in real-life situations. Reviewing work examples and seeking help from teachers or peers can also be beneficial.
10. Can ratios be simplified answer according to class 7th maths chapter 7 exercise 7.1?
Yes, ratios can be simplified by dividing both the numerator and the denominator by their greatest common divisor (GCD). For example, the ratio 6:9 can be simplified to 2:3 by dividing both numbers by 3.
