NCERT Solutions for Class 7 Maths Chapter 8: Comparing Quantities - Exercise 8.1

Exercise 8.1

1. Find the ratio of:

1. Find the ratio of ₹ 5 to 50 paise.

Ans: ₹ 1 can be written as 100 paise. Therefore, ₹ 5 is equal to 500 paise.

Ratio of 500 paise to 50 paise is given as $R = \dfrac{{500}}{{50}}$.

Simplify $R = \dfrac{{500}}{{50}}$ using division.

$\Rightarrow R=\frac{500 \div 50}{50 \div 50}$

$\Rightarrow R = \frac{10}{1}$

Therefore, the ratio of ₹ 5 to 50pise is $10:1$.

2. Find the ratio of 15 kg to 210 g.

Ans: 1 kg can be written as 1000 g. Therefore, 15 kg is equal to 15000 g.

Ratio of 15000 g to 210 g is given as $R = \dfrac{{15000}}{{210}}$.

Simplify $R = \dfrac{{15000}}{{210}}$ using division.

$ \Rightarrow R = \dfrac{{15000 \div 30}}{{210 \div 30}}$

$\Rightarrow R = \dfrac{{500}}{7} $

Therefore, the ratio of 15 kg to 210 g is$500:7$.

3. Find the ratio of 9 m to 27 cm.

Ans: 1 m can be written as 100 cm. Therefore, 9 m is equal to 900 cm.

Ratio of 9 m to 27cm is given as $R = \dfrac{{900}}{{27}}$.

Simplify $R = \dfrac{{900}}{{27}}$ using division.

$ \Rightarrow R = \dfrac{{900 \div 9}}{{27 \div 9}}$

$  \Rightarrow R = \dfrac{{100}}{3} $

Therefore, the ratio of 9 m to 27 cm is $100:3$.

4. Find the ratio of 30days to 36hours.

Ans: 1 day can be written as 24 hours. Therefore, 30 days is equal to 720 hours.

Ratio of 720 hours to 36 hours is given as $R = \dfrac{{720}}{{36}}$.

Simplify $R = \dfrac{{720}}{{36}}$ using division.

$\Rightarrow R = \dfrac{{720 \div 36}}{{36 \div 36}} $ 

$\Rightarrow R = \dfrac{{20}}{1}$

Therefore, the ratio of 30 days to 36 hours is $20:1$.

2. In a computer lab there are 3 computers for every 6 students. How many computers will be needed for 24 students?

Ans: It is given that for 6 students there are 3 computers. Thus implies that for 1 student there is $\dfrac{3}{6}$ computer.

Thus the number of computers needed for 24 students is 24 multiplied by the computer needed for 1 student.

$\Rightarrow {\text{Computer for 24 students}} = \dfrac{3}{6} \times 24$

$\Rightarrow {\text{Computer for 24 students}} = 12$

Therefore, 12 computers are needed for 24 students.

3. Population of Rajasthan $ = $ 570 lakhs and population of U.P. $ = $ 1660 lakhs. Area of Rajasthan $ = $ 3 lakh $k{m^2}$ and area U.P. $ = $ 2 lakh $k{m^2}$.

1. How many people are there per $k{m^2}$ in both states?

Ans: People present per $k{m^2}$is given by $\dfrac{{{\text{Population}}}}{{{\text{Area}}}}$.

It is given that the population of Rajasthan is 570 lakhs and the area of Rajasthan is 3 lakh $k{m^2}$. 

Therefore,, People present per $k{m^2}$in Rajasthan $ = \dfrac{{570{\text{lakh}}}}{{3{\text{lakh per k}}{{\text{m}}^2}}}$.

Simplify $\dfrac{{570{\text{lakh}}}}{{3{\text{lakh per k}}{{\text{m}}^2}}}$ to get number of people per $k{m^2}$.

$\Rightarrow \dfrac{{570{\text{lakh}}}}{{3{\text{lakh per k}}{{\text{m}}^2}}}$

$\Rightarrow 190{\text{People per k}}{{\text{m}}^2}$

Number of people per ${\text{k}}{{\text{m}}^2}$ in Rajasthan is 190 lakh.

It is given that the population of U.P. is 1660 lakhs and area of U.P. is 2lakh $k{m^2}$. 

Therefore,, People present per $k{m^2}$in U.P.$ = \dfrac{{1660{\text{lakh}}}}{{{\text{2lakh per k}}{{\text{m}}^2}}}$.

Simplify $\dfrac{{1660{\text{lakh}}}}{{{\text{2lakh per k}}{{\text{m}}^2}}}$ to get number of people per $k{m^2}$.

$\Rightarrow \dfrac{{1660{\text{lakh}}}}{{{\text{2 lakh per k}}{{\text{m}}^2}}}$

$\Rightarrow 830{\text{People per k}}{{\text{m}}^2}$

Number of people per ${\text{k}}{{\text{m}}^2}$ in U.P. are 830 lakh.

2. Which state is less populated?

Ans: From part (a) of this question it is seen that number of people per ${\text{k}}{{\text{m}}^2}$ in U.P. is more than the number of people per ${\text{k}}{{\text{m}}^2}$ in Rajasthan.

Therefore, it can be concluded that Rajasthan is less populated.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.1

Opting for the NCERT solutions for Ex 8.1 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 8.1 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 8.1 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 8.1, 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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