CHAPTER 11 - DIRECT AND INVERSE PROPORTIONS

Exercise 11.2

1. Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.

Inverse Proportion: As the number of workers increases, the time to complete the job decreases. This is a classic example of inverse proportion.

(ii) The time taken for a journey and the distance travelled in a uniform speed.

Direct Proportion: As the time increases, the distance travelled also increases, given a uniform speed. This is direct proportion.

(iii) Area of cultivated land and the crop harvested.

Direct Proportion: As the area of cultivated land increases, the amount of crop harvested also increases. This is direct proportion.

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

Inverse Proportion: As the speed of the vehicle increases, the time taken for a fixed journey decreases. This is inverse proportion.

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

Inverse Proportion: As the population of a country increases, the area of land available per person decreases. This is inverse proportion.

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?

The total prize money is fixed at ₹ 1,00,000. When the number of winners increases, the prize money per winner decreases. Therefore, it is an inverse proportion.

To complete the table, divide the total prize money by the number of winners:

Number of winners: 1, 2, 4, 5, 8, 10, 20

Prize for each winner (in ₹):
1 winner: \( \frac{1,00,000}{1} = 1,00,000 \)
2 winners: \( \frac{1,00,000}{2} = 50,000 \)
4 winners: \( \frac{1,00,000}{4} = 25,000 \)
5 winners: \( \frac{1,00,000}{5} = 20,000 \)
8 winners: \( \frac{1,00,000}{8} = 12,500 \)
10 winners: \( \frac{1,00,000}{10} = 10,000 \)
20 winners: \( \frac{1,00,000}{20} = 5,000 \)

Final Answer: The prize money for each winner is inversely proportional to the number of winners. The completed table is:

Number of winners: 1, 2, 4, 5, 8, 10, 20
Prize for each winner (in ₹): 1,00,000, 50,000, 25,000, 20,000, 12,500, 10,000, 5,000
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.

The total angle of a wheel is \(360^\circ\). The angle between consecutive spokes is inversely proportional to the number of spokes.

Complete the table:
Angle = \( \frac{360^\circ}{\text{Number of spokes}} \)
Number of spokes: 4, 6, 8, 10, 12
Angle between a pair of consecutive spokes:
4 spokes: \( \frac{360^\circ}{4} = 90^\circ \)
6 spokes: \( \frac{360^\circ}{6} = 60^\circ \)
8 spokes: \( \frac{360^\circ}{8} = 45^\circ \)
10 spokes: \( \frac{360^\circ}{10} = 36^\circ \)
12 spokes: \( \frac{360^\circ}{12} = 30^\circ \)
(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?

Yes, the number of spokes and the angles formed are in inverse proportion because as the number of spokes increases, the angle between them decreases.

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

Angle = \( \frac{360^\circ}{15} = 24^\circ \)

Final answer: \(24^\circ\)
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is \(40^\circ\)?

Number of spokes = \( \frac{360^\circ}{\text{Angle}} = \frac{360^\circ}{40^\circ} = 9 \)

Final answer: 9 spokes
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?

This is an inverse proportion problem. Let \(x_1 = 24\) children, \(y_1 = 5\) sweets. The new number of children is \(x_2 = 24 - 4 = 20\). We need to find the new number of sweets per child, \(y_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 24 \times 5 = 20 \times y_2 \]

\[ 120 = 20 y_2 \]

\[ y_2 = \frac{120}{20} = 6 \]

Final answer: Each child would 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?

This is an inverse proportion problem. Let \(x_1 = 20\) animals, \(y_1 = 6\) days. The new number of animals is \(x_2 = 20 + 10 = 30\). We need to find the new duration, \(y_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 20 \times 6 = 30 \times y_2 \]

\[ 120 = 30 y_2 \]

\[ y_2 = \frac{120}{30} = 4 \]

Final answer: The food would last for 4 days.
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?

This is an inverse proportion problem. Let \(x_1 = 3\) persons, \(y_1 = 4\) days. The new number of persons is \(x_2 = 4\). We need to find the new time, \(y_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 3 \times 4 = 4 \times y_2 \]

\[ 12 = 4 y_2 \]

\[ y_2 = \frac{12}{4} = 3 \]

Final answer: It would take them 3 days.
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?

This is an inverse proportion problem. First, find the total number of bottles: \(25 \text{ boxes} \times 12 \text{ bottles/box} = 300 \text{ bottles}\). Now, divide the total number of bottles by the new number of bottles per box to find the number of boxes needed.

Number of boxes = \( \frac{\text{Total bottles}}{\text{Bottles per box}} = \frac{300}{20} = 15 \)

Final answer: 15 boxes would be filled.
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?

This is an inverse proportion problem. Let \(x_1 = 42\) machines, \(y_1 = 63\) days. The new time is \(y_2 = 54\) days. We need to find the new number of machines, \(x_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 42 \times 63 = x_2 \times 54 \]

\[ 2646 = 54 x_2 \]

\[ x_2 = \frac{2646}{54} = 49 \]

Final answer: 49 machines would be required.
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?

This is an inverse proportion problem. Let \(x_1 = 60\) km/h, \(y_1 = 2\) hours. The new speed is \(x_2 = 80\) km/h. We need to find the new time, \(y_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 60 \times 2 = 80 \times y_2 \]

\[ 120 = 80 y_2 \]

\[ y_2 = \frac{120}{80} = 1.5 \]

Final answer: It will take 1.5 hours (or 1 hour and 30 minutes).
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?

This is an inverse proportion problem. Originally, there were \(x_1 = 2\) persons and it took \(y_1 = 3\) days. Now, the number of persons is \(x_2 = 2 - 1 = 1\). We need to find the new time, \(y_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 2 \times 3 = 1 \times y_2 \]

\[ 6 = y_2 \]

Final answer: The job would take 6 days.
(ii) How many persons would be needed to fit the windows in one day?

This is also an inverse proportion problem. We know \(x_1 = 2\) persons and \(y_1 = 3\) days. The new time is \(y_2 = 1\) day. We need to find the new number of persons, \(x_2\).

Using the inverse proportion formula: \(x_1 y_1 = x_2 y_2\)

\[ 2 \times 3 = x_2 \times 1 \]

\[ 6 = x_2 \]

Final answer: 6 persons would be needed.
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?

This is an inverse proportion problem. First, find the total time of the school day: \(8 \text{ periods} \times 45 \text{ minutes/period} = 360 \text{ minutes}\). Now, divide the total time by the new number of periods to find the duration of each period.

Duration of each period = \( \frac{\text{Total school time}}{\text{Number of periods}} = \frac{360}{9} = 40 \)

Each period would be 40 minutes long.

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