RBSE Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

Question 1.
Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed, of the vehicle.
(v) The population of a country and the area of land per person.
Answer:
(i) If we increase the number of workers on a job the work will be finished in less time and vice versa. So, when one quantity increases the other decreases. Hence, it is in inverse proportion.
(ii) For longer distance, more time would be required.
∴ It is not a case of inverse proportion.
(iii) If cultivate land is more, then crop harvested is more. So, it is inverse in direct proportion.
(iv) More the speed, less would be the time to cover a fixed distance. So, it is a case of inverse proportion.
(v) If the population is more then area of land per person would be less. So, it is a case of inverse proportion.
So, (i), (ii), (v) are in inverse proportions.

Question 2.
In a Television game show, the prize money of ₹ 1,00,000 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?

Answer:
Here, more the number of winners, less is the prize for each winner. So it is a case of inverse proportion.
∴ 4 × x = 1 × 100000
or x = \(\frac{1,00,000}{4}\) = 25,000
Thus, for 4, it is 25,000.
5 × y = 1 × 100000
or y = \(\frac{1,00,000}{5}\) = 2
8 × z = 1 × 100000
or z = \(\frac{1,00,000}{8}\) = 12,500
Thus, for 8, it is 12,500.
10 × r = 1 ×100000
or r = \(\frac{1,00,000}{10}\) = 10,000
Thus, for 10, it is 10,000.
20 × p = 1 × 100000
or p = \(\frac{1,00,000}{20}\) = 5 000
Thus, for 20, it is 5,000.

Question 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.

(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Answer:
Here, more is the number of spokes, less is the measure of angle between a pair of consecutive spokes.
Here, we observe
4 × 90°= 6 × 60° = 360°
So, it is a case of inverse proportion.
∴ 8 × x = 4 ×x 90°
or x = \(\frac{4 \times 90^{\circ}}{8}\) = 45°
Thus, for 8, it is 45°.
10 × y = 4 × 90°
or y = \(\frac{4 \times 90^{\circ}}{10}\) = 36°
Thus, for 10, it is 36°.
12 × z = 4 × 90°
or z = \(\frac{4 \times 90^{\circ}}{12}\) = 30°
Thus, for 12, it is 30°.

(i) Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.

(ii) Let x° be the angle between a pair of consecutive spokes on a wheel with 15 spokes.
∴ 15 × x = 4 × 90°
or x = \(\frac{4 \times 90^{\circ}}{15}\) = 24°
Thus, the required angle is 24°

(iii) Let x spokes are needed on a wheel if the angle between a pair of consecutive spokes is 40°.
∴ x × 40° = 4 × 90°
or x = \(\frac{4 \times 90^{\circ}}{40}\) = 9°
Thus, the required number of spokes is 9.

Question 4.
If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Answer:
Initially the no. of children = 24
If we reduce the no. of children by 4 then 24 - 4 = 20 children.

Since more the no. of children less no. of sweets each would get i.e. this is the case of inverse proportion.
Hence, x₁y₁ = x₂y₂
⇒ 5 × 24 = x × 20
⇒ x = \(\frac{5 \times 24}{20}\) = 6
∴ Each Children will get 6 sweets.

Question 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?
Answer:
If we add 10 animals then the total number of animals = 10 + 20 = 30

More the number of animals, less number of days the food will last.
∴ It is a case of inverse variation.
Thus, we have
30 × x = 20 × 6
or x = \(\frac{20 \times 6}{30}\) = 4
Therefore, the food will now last for 4 days.

Question 6.
A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Answer:
Let the no. of days to complete the job = x
∵ More the no. of persons, less time would take to finish the same work.
i.e. the given two quantities are in inverse proportion.

Hence, required days = 3.

Question 7.
A batch of bottles were 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?

Answer:
Let x be the number of boxes to be needed when 20 bottles are packed in each box. Then,

Clearly, more is the number of bottles, less will be the number of boxes needed for packing.
So, it is a case of inverse proportion.
∴ 25 × 12 = x × 20
⇒ x = \(\frac{25 \times 12}{20}\) = 15
Thus, the boxes needed for packing is 15.

Question 8.
A factory requires 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?
Answer:
Let the number of machines required be JC. We know that more the no. of machines, less no. of days will take.

x₁y₁ = x₂y₂
⇒ 42 × 63 = x × 54
or x = \(\frac{42 \times 63}{54}\) = 7 × 7 = 49
Thus, the required number of machines = 49.

Question 9.
A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Answer:
Let the time taken to complete the journey = x

Since distance is fixed.
∴ Distance = Speed × Time
Speed × Time = Fixed
More speed, less time to cover the same distance
\(\frac{x_{1}}{x_{2}}=\frac{y_{2}}{y_{1}}\) (Inverse Proportion)
Now, \(\frac{2}{x}=\frac{80}{60}\)
or 2 × 60 = 80 × x
or x = \(\frac{2 \times 60}{80}\) = 1.5 hours
Hence, required time =1.5 hours.

Question 10.
Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Answer:
(i) Let the no of days = x
or 2 × 3 = 1 × x
or x = 6
Hence, required no. of days = 6

(ii) Let the no. of person = x

\(\frac{x_{1}}{x_{2}}=\frac{y_{2}}{y_{1}}\)
Now, \(\frac{2}{x} = \frac{1}{3}\) (Inverse Proportion)
or x × 1 = 2 × 3 = 6 persons
Hence, required persons = 6

Question 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?
Answer:
Let the no. of duration of each period = x minutes. We know that the no. of school hours is fixed. So, more periods in a day means less duration of each period.
Hence, it is a case of inverse proportion.

Clearly, more the periods, less will be the duration of the period.
So, it is a case of inverse proportion.
∴ 8 × 45 = 9 × x
⇒ x = \(\frac{8 \times 45}{9}\) = 40