NCERT Solutions for Class 8 Maths Chapter 13: Direct and Inverse Proportions - Exercise 13.2

Exercise 13.2

Refer to page 9-17 for Exercise 13.2 in the PDF

1. Which of the following are in inverse proportion?

i. The number of workers on a job and the time to complete the job.

Ans: In this situation, observe that the more the workers, the time taken to complete the job will be less. Thus, this is a situation of an inverse proportion.

ii. The time taken for a journey and the distance travelled at a uniform speed.

Ans: If time taken for a journey increases, then the distance travelled at uniform speed remains same. So, it is not the Inverse proportion.

iii. Area of cultivated land and the crop harvested.

Ans: The given situation is not an example of inverse proportion because in more areas, more crops can be harvested.

iv. The time taken for a fixed journey and the speed of the vehicle.

Ans: This is an example of an inverse proportion because with more speed, we can complete a certain distance in less time.

v. The population of a country and the area of land per person.

Ans: If the population is increasing, then the area of the land per person will be decreasing accordingly. So, they are in inverse proportion.

2. In a Television game show, the prize money of Rs1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.

Ans: Assume the missing ratios as ${x_1}$,${x_2}$,${x_3}$,${x_4}$, and ${x_5}$.

Tabulate the data as follows.

From the table, we get $1 \times 100000 = 100000$ and

$2 \times 50000 = 100000$.

The number of winners and the prize money are inversely proportional to each other.

$1 \times 100000 = 4 \times {x_1}$

Divide both sides by 4 to solve the equation for ${x_1}$.

\[\dfrac{{1 \times 100000}}{4} = {x_1}\]

\[25000 = {x_1}\]

Now, find the value of ${x_2}$.

$1 \times 100000 = 5 \times {x_2}$

Divide both sides by 5 to solve the equation for ${x_2}$.

\[\dfrac{{1 \times 100000}}{5} = {x_2}\]

\[20000 = {x_2}\]

Now, find the value of ${x_3}$.

$1 \times 100000 = 8 \times {x_3}$

Divide both sides by 8 to solve the equation for ${x_3}$.

\[\dfrac{{1 \times 100000}}{8} = {x_3}\]

\[12500 = {x_3}\]

Now, find the value of ${x_4}$.

$1 \times 100000 = 10 \times {x_4}$

Divide both sides by 10 to solve the equation for ${x_4}$.

\[\dfrac{{1 \times 100000}}{{10}} = {x_4}\]

\[10000 = {x_4}\]

Now, find the value of ${x_5}$.

$1 \times 100000 = 20 \times {x_5}$

Divide both sides by 20 to solve the equation for ${x_5}$.

\[\dfrac{{1 \times 100000}}{{20}} = {x_5}\]

\[5000 = {x_5}\]

Thus, the complete table is as follows.

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.

Ans: From the given table, we obtain, 

$4 \times 90^\circ  = 360$ and

$6 \times 60^\circ $.

Thus, we can observe that the number of spokes and the angle between a pair of consecutive spokes are inversely proportional to each other.

$4 \times 90^\circ  = {x_1} \times 8$

Divide both sides by 8.

${x_1} = \dfrac{{4 \times 90^\circ }}{8}$

${x_1} = 45^\circ $

Similarly, find the value of ${x_2}$.

${x_2} = \dfrac{{4 \times 90^\circ }}{{10}}$

${x_2} = 36^\circ $

Now, find the value of ${x_3}$.

${x_3} = \dfrac{{4 \times 90^\circ }}{{12}}$

${x_3} = 30^\circ $

Thus, tabulate the table obtained.

i. Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?

Ans: Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.

ii. Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.

Ans: Let the angle between a pair of consecutive spokes on a wheel with 15 spokes be $x$.

$4 \times 90^\circ  = 15 \times x$

Divide both sides by 15.

$x = \dfrac{{4 \times 90^\circ }}{{15}}$

$x = 24^\circ $

Hence, the angle between the pair of consecutive spokes of a wheel, which has 15 spokes in it, is $24^\circ $.

iii. How many spokes would be needed, if the angle between a pair of consecutive spokes is $40^\circ $?

Ans: Let us assume that the number of spokes in a wheel, which has $40^\circ $ angles between a pair of spokes as $y$.

$4 \times 90^\circ  = y \times 40^\circ $

Divide both sides by $40^\circ $.

$y = \dfrac{{4 \times 90^\circ }}{{40^\circ }}$

$y = 9$

Hence, the number of spokes in such a wheel is 9.

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many each would get, if the number of the children is reduced by 4?

Ans: It is given that the number of children is reduced by 4 and the number of children is 24. So, the number of children will be, $\left( {24 - 4} \right)$.

Thus, the number of children left are 20.

Now, tabulate the data as follows.

If the number of students is less, then each student will get a greater number of sweets. Thus, the given problem follows an inverse proportion.

$24 \times 5 = 20 \times x$

Divide both sides by 20.

$\dfrac{{24 \times 5}}{{20}} = x$

$6 = x$

Hence, each student will get 6 sweets.

5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?

Ans: Let the number of days that the food will last if there were 10 more animals in the cattle be $x$.

Tabulate the data as follows.

In the given scenario we can see that, more the number of animals, less will be the number of days for which the food will last. Thus, the number of days the food will last and the number of animals is inversely proportional to each other.

$20 \times 6 = 30 \times x$

Divide both sides by 30.

$\dfrac{{20 \times 6}}{{30}} = x$

$\dfrac{{120}}{{30}} = x$

$4 = x$

Thus, the food will last for 4 days.

6. A contractor estimates that 3 persons could rewire Jaswinder’s house in 4 days. If he uses 4 persons instead of three, how long should they take to complete the join?

Ans: Let us assume that the number of days required by 4 persons to complete the job be $x$.Tabulate the data as follows.

We can see that if the number of persons is more, then it will take less time to complete the job. Therefore, the number of days and the persons required are inversely proportional to each other.

$4 \times 3 = x \times 4$

Divide both sides by 4.

$\dfrac{{4 \times 3}}{4} = x$

$\dfrac{{12}}{4} = x$

$3 = x$

Hence, the number of days required to complete the job is 3.

7. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Ans: Assume the number of boxes filles, by using 20 bottles in each box be $x$.

Tabulate the data as follows.

Observe that more the number of bottles, lesser the number of boxes required. Thus, the number of bottles and boxes are inversely proportional to each other.

$12 \times 25 = 20 \times x$

Divide both sides by 20.

$\dfrac{{12 \times 25}}{{20}} = x$

$15 = x$

Hence, the number of boxes required to pack these bottles is 15.

8. A factory required 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Ans: Let the number of machines required be $x$.

Tabulate the data as follows.

Observe that more the number of machines, less will be the number of days that it will take to produce the given number of articles.

Thus, the given problem follows an inverse proportion.

$42 \times 63 = 54 \times x$

Divide both sides by 54.

$\dfrac{{42 \times 63}}{{54}} = x$

$49 = x$

Hence, the number of machines required to produce the given number of articles in 54 days is 49.

9. A car takes 2 hours to reach a destination by travelling at the speed of 60km/hr. How long will it take when the car travels at the speed of 80km/hr?

Ans: Let the time taken by the car to reach the destination by travelling at the speed of 60 km/hr be $x$ hours.

Tabulate the data as follows.

We can see that if the speed of the car is more then, less time will be taken. Thus, the speed of the car and time taken by the car are inversely proportional to each other.

$60 \times 2 = 80 \times x$

Divide both sides by 80.

$\dfrac{{60 \times 2}}{{80}} = x$

$\dfrac{{120}}{{80}} = x$

$\dfrac{3}{2} = x$

Thus, the time required by the car to reach the given destination is $\dfrac{3}{2}$ hours.

10. Two persons could fit new windows in house in 3 days.

i. One of the people fell ill before the work started. How long would the job take now?

Ans: Let the number of days required by 1 man to fit all the windows be $x$.

Tabulate the data as follows.

We can see that a greater number of days will be required if the number of persons are less. Thus, this represents an inverse proportion. Therefore, 

$2 \times 3 = 1 \times x$

Solve the equation.

$6 = x$

Hence, the number of days taken by one man to fit all the given windows is 6.

ii. How many persons would be needed to fit the windows in one day?

Ans: Let the number of persons required to fit all the windows in one day be $y$.

Tabulate the data as follows.

We can see that a lesser number of days will be required if the number of persons are more. Thus, this represents an inverse proportion. Therefore, 

$2 \times 3 = 1 \times y$

Solve the equation.

$6 = y$

Hence, the number of persons required to fit all the windows in one day is 6.

11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?

Ans: Let the duration of each period be $x$ minutes.

Tabulate the data as follows.

Observe that if a greater number of periods will be held in a day, then the duration of each period will be lesser.

Thus, this is the case of inverse proportion.

$45 \times 8 = x \times 9$

Divide both sides by 9 and solve for $x$.

$x = \dfrac{{45 \times 8}}{9}$

$x = 5 \times 8$

$x = 40$

Hence, the duration of each period will be 40 minutes.

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 8 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 13.2 Class 8 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 8 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 8 Maths Chapter 13 Exercise 13.2 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.

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