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

Haryana State Board HBSE 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Exercise 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:
Two quantities are said to vary inversely proportional if the increase (or decrease) in one quantity causes the decrease (or increase) in the other quantity.
Hence,
(i) is inverse proportional.
(ii) is direct proportional.
(iii) is inverse proportional.
(iv) is inverse pooportional.
(v) is inverse proportional.

Question 2.
In a Television game show, the prize money of Rs. 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 ?

Solution:
If xy = K, then x and y are said to vary inversely.
xy = x₁y₁ = x₂y₂ = x₃y₃ = ….
= 1 × 1,00,000
= 2 × 50,000 = 1,00,000
or \(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{1}}{y_{2}}\)
⇒ \(\frac{1}{4}\) = \(\frac{x}{1,00,000}\)
⇒ 4x = 1,00,000
∴ x = \(\frac{1,00,000}{4}\) = 25000
Similarly, y = 20000, z = 12500
a = 10000, b = 6000.

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° ?
Solution:
In inversely proportional,

(i) Since, number of spokes are increases and angle between a pair of consecutive spokes are decreases. Hence it is inversely proportional.
(ii) \(\frac{4}{15}\) = \(\frac{a}{90}\) ⇒ a = \(\frac{4 \times 90}{15}\) = 24°
⇒ Angle = 24°
(iii) \(\frac{4}{x}\) = \(\frac{40}{90}\) ⇒ 40 × x = 4 × 90
x = \(\frac{4 \times 90}{40}\) = 9
No. of spokes = 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 ?
Solution:
Let each get sweets = x

Now, \(\frac{5}{x}\) = \(\frac{20}{24}\)
⇒ 20x = 5 × 24
⇒ x = \(\frac{5 \times 24}{20}\) = 6
Hence required each get sweets = 6.

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 ?
Solution:
Let the No. of days = x

Now, \(\frac{20}{30}\) = \(\frac{x}{6}\)
⇒ 30 × x = 20 × 6
⇒ x = \(\frac{20 \times 6}{30}\) = 4
Hence required days = 4.

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 ?
Solution:
Let the days = x

Now, \(\frac{3}{4}\) = \(\frac{x}{4}\)
⇒ 4 × x = 3 × 4
⇒ x = \(\frac{3 \times 4}{4}\) = 3 days
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 ?

Solution:
Let the no. of boxes = x

Now, \(\frac{25}{x}\) = \(\frac{20}{12}\)
⇒ 20 × x = 25 × 12
⇒ x = \(\frac{25 \times 12}{20}\) = 15 boxes
Hence, required boxes = 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 ?
Solution:
Let the number of machines = x

Now, \(\frac{42}{x}\) = \(\frac{54}{63}\)
⇒ 54 × x = 42 × 63
⇒ x = \(\frac{42 \times 63}{54}\) = 49
Hence, the required 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 ?
Solution:
Let the time = x

Now, \(\frac{2}{x}\) = \(\frac{80}{60}\)
⇒ 80 × x = 2 × 60
⇒ 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 window in one day ?
Solution:
(i) Let the no. of days = x

Now, \(\frac{2}{1}\) = \(\frac{x}{3}\)
⇒ 1 × x = 2 × 3
⇒ x = 6 days
Hence required days = 6.

(ii) Let the no. of person = y

Now, \(\frac{2}{x}\) = \(\frac{1}{3}\)
⇒ 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 ?
Solution:
Let the no. of duration = x

Now, \(\frac{45}{x}\) = \(\frac{9}{8}\)
⇒ 9 × x = 8 × 45
⇒ x = \(\frac{8 \times 45}{9}\) = 40
Hence, requied time duration = 40 minutes.