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

 

Question 1. Which of the following are in inverse proportion?
1. The number of workers on a job and the time to complete the job.
2. The time taken for a journey and the distance travelled in a uniform speed.
3. Area of cultivated land and the crop harvested.
4. The time taken for a fixed journey and the speed of the vehicle.
5. The population of a country and the area of land per person.
Answer: If an increase in one quantity causes a corresponding decrease in the other, and vice versa, then the two quantities are said to vary inversely.
1. If the number of workers increases, the time required to complete the job will decrease. This is a clear case of inverse variation.
2. For a longer distance, more time will be needed. This is not an instance of inverse variation.
3. With a larger area of land, more crops can be harvested. This relationship does not represent inverse variation.
4. When the speed is higher, less time will be taken to cover a specific distance. This is an example of inverse variation.
5. For a greater population, less land area per person would be available. This illustrates a situation of inverse variation.
In simple words: Inverse proportion means that as one thing goes up, the other thing goes down. For these examples, we check if they follow that pattern. More workers mean less time (inverse). More distance means more time (not inverse). More land means more crops (not inverse). More speed means less time (inverse). More people mean less space per person (inverse).

Exam Tip: To identify inverse proportion, look for situations where one quantity's increase inherently leads to a decrease in another related quantity while the total or product remains constant.

 

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?

Answer: If the number of winners increases, the prize money for each individual decreases. Therefore, it is a situation of inverse proportion.
The total prize money is Rs. 1,00,000. Let the number of winners be \( n \) and the prize for each winner be \( x \). Then \( n \times x = 1,00,000 \).
We can find the missing values:
For 4 winners: \( 4 \times x_1 = 1 \times 1,00,000 \)
\( x_1 = \frac{1 \times 1,00,000}{4} = 25,000 \)
For 5 winners: \( 5 \times x_2 = 1 \times 1,00,000 \)
\( x_2 = \frac{1 \times 1,00,000}{5} = 20,000 \)
For 8 winners: \( 8 \times x_3 = 1 \times 1,00,000 \)
\( x_3 = \frac{1 \times 1,00,000}{8} = 12,500 \)
For 10 winners: \( 10 \times x_4 = 1 \times 1,00,000 \)
\( x_4 = \frac{1 \times 1,00,000}{10} = 10,000 \)
For 20 winners: \( 20 \times x_5 = 1 \times 1,00,000 \)
\( x_5 = \frac{1 \times 1,00,000}{20} = 5,000 \)
Thus, the completed table is shown below:

In simple words: When more people win, the prize money each person gets becomes smaller. This means it's an inverse proportion. We find each person's share by dividing the total prize (Rs. 1,00,000) by the number of winners.

Exam Tip: Remember that in inverse proportion, the product of the two quantities remains constant. Use this fact to solve for missing values in tables.

 

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?

1. Are the number of spokes and the angles formed between the pairs of consecutive spokes in in verse proportion?
2. Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes?
3. How many spokes would be needed, if the angle between a pair of consecutive spokes is 400?
Answer:
1. Clearly, as the number of spokes increases, the measure of the angle between consecutive spokes decreases. Thus, it is a situation of inverse variation.
We can find the missing angles using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
For 8 spokes: \( 4 \times 90 = 8 \times x_1 \)
\( x_1 = \frac{4 \times 90}{8} = 45^{\circ} \)
For 10 spokes: \( 4 \times 90 = 10 \times x_2 \)
\( x_2 = \frac{4 \times 90}{10} = 36^{\circ} \)
For 12 spokes: \( 4 \times 90 = 12 \times x_3 \)
\( x_3 = \frac{4 \times 90}{12} = 30^{\circ} \)
The completed table is as follows:

2. Let the required angle measure be \( x^{\circ} \).
Using the inverse proportion relation \( n_1 x_1 = n_2 x_2 \):
\( 15 \times x^{\circ} = 4 \times 90^{\circ} \)
\( x^{\circ} = \frac{4 \times 90^{\circ}}{15} = 24^{\circ} \)
3. Let the required number of spokes be \( n \).
Using the inverse proportion relation \( n_1 x_1 = n_2 x_2 \):
\( n \times 40^{\circ} = 4 \times 90^{\circ} \)
\( n = \frac{4 \times 90^{\circ}}{40^{\circ}} = 9 \)
The number of spokes required is 9.
In simple words:
1. Yes, more spokes mean smaller angles between them, so it's inverse proportion. We use the rule (spokes × angle) to fill in the table.
2. For 15 spokes, we use the same rule: \( 15 \times x = 4 \times 90 \). Solving it shows the angle will be \( 24^{\circ} \).
3. If the angle is \( 40^{\circ} \), we need to find how many spokes it takes. We use \( n \times 40 = 4 \times 90 \), which means 9 spokes are required.

Exam Tip: Remember that the product of the number of spokes and the angle between them is always constant (equal to 360 degrees, which is a full circle). Use this constant to solve for unknown values.

 

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: The reduced number of children is \( 24 - 4 = 20 \).
When the number of children increases, the quantity of sweets each child gets decreases. Therefore, it is a situation of inverse variation.
Let \( x \) be the number of sweets each child would get if there were 20 children.
Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 24 \times 5 = 20 \times x \)
\( x = \frac{24 \times 5}{20} = \frac{120}{20} = 6 \)
Hence, each child will get 6 sweets.
In simple words: If you have 24 children, each gets 5 sweets. When 4 children leave, there are 20 children left. Since fewer children share the same total sweets, each child will get more. We calculate this by making sure (children × sweets per child) stays the same: \( 24 \times 5 = 20 \times x \), so \( x = 6 \) sweets.

Exam Tip: Always identify if it's a direct or inverse proportion first. For inverse proportion, the product of the two quantities remains constant. State the final answer clearly with units.

 

Question 5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last f there were 10 more animal in his cattle?
Answer: The number of animals added is 10.
So, the total number of animals now is \( 20 + 10 = 30 \).
If there are more animals, the food will last for a shorter number of days. This is a clear case of inverse variation.
Let \( x \) be the number of days the food will last for 30 animals.
Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 30 \times x = 20 \times 6 \)
\( x = \frac{20 \times 6}{30} = \frac{120}{30} = 4 \)
Therefore, the food will now last for 4 days.
In simple words: The farmer has food for 20 animals for 6 days. If 10 more animals come, there are 30 animals in total. More animals mean the food will run out faster. So, we find \( 30 \times x = 20 \times 6 \), which gives \( x = 4 \) days.

Exam Tip: In problems involving food supply, remember that the total amount of food is constant. So, the number of animals multiplied by the number of days the food lasts will remain the same for different animal counts.

 

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: If the number of persons increases, the time taken to complete the job will decrease. This is an instance of inverse variation.
Let \( x \) be the number of days 4 persons would take to complete the job.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 3 \times 4 = 4 \times x \)
\( x = \frac{3 \times 4}{4} = 3 \)
Therefore, the required number of days is 3.
In simple words: 3 people take 4 days to do a job. If you use 4 people (more workers), it will take less time. So, we set up \( 3 \times 4 = 4 \times x \) to find \( x \), which is 3 days.

Exam Tip: For "work and time" problems, assume the total work is constant. More workers usually mean less time for the same amount of work, indicating an inverse proportion.

 

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 required.
If more bottles are packed in one box, a fewer number of boxes will be needed to pack the same batch. This is an example of inverse variation.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 12 \times 25 = 20 \times x \)
\( x = \frac{12 \times 25}{20} = \frac{300}{20} = 15 \)
Thus, the required number of boxes is 15.
In simple words: We have a set number of bottles. If we put 12 bottles in each box, we need 25 boxes. If we put 20 bottles in each box (more bottles per box), we will need fewer boxes. We find this with \( 12 \times 25 = 20 \times x \), which means \( x = 15 \) boxes.

Exam Tip: When dealing with items packed into containers, remember that the total number of items remains constant. The number of items per container multiplied by the number of containers will always be the same.

 

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 \( x \) be the number of machines required.
To produce the same number of articles in fewer days, more machines would be needed. This is a case of inverse variation.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 42 \times 63 = x \times 54 \)
\( x = \frac{42 \times 63}{54} = \frac{2646}{54} = 49 \)
Thus, the required number of machines is 49.
In simple words: A factory uses 42 machines for 63 days. If they want to finish the same work faster, in 54 days, they will need more machines. We find this using \( 42 \times 63 = x \times 54 \), which gives \( x = 49 \) machines.

Exam Tip: When work needs to be completed within a specific timeframe, the number of resources (like machines) and the time taken are inversely proportional. Always ensure the "work" or "articles produced" remains constant in such problems.

 

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: If the speed is higher, the number of hours needed to travel a fixed distance will be less. This represents a case of inverse variation.
Let \( x \) be the time taken when the car travels at 80 km/h.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 60 \times 2 = 80 \times x \)
\( x = \frac{60 \times 2}{80} = \frac{120}{80} = \frac{3}{2} \) hours
This can also be written as \( 1\frac{1}{2} \) hours.
Thus, the required number of hours is \( 1\frac{1}{2} \).
In simple words: A car goes 60 km/h and takes 2 hours. If it goes faster, at 80 km/h, it will take less time. We find this by \( 60 \times 2 = 80 \times x \), which means \( x = 1.5 \) hours.

Exam Tip: Remember the relationship: Distance = Speed × Time. For a fixed distance, speed and time are inversely proportional. Always convert fractions to clear time units if possible, like 1.5 hours or 1 hour 30 minutes.

 

Question 10. Two persons could fit new window in a house in 3 days.
1. One of the persons fell ill before the work started. How long would the job take now?
2. How many persons would be needed to fit the windows in one day?
Answer:
1. Let \( x \) be the time taken by the remaining persons to complete the job.
If one person fell ill, then \( 2 - 1 = 1 \) person remains. When the number of persons is fewer, more days will be required to complete the job. This is a case of inverse proportion.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 2 \times 3 = 1 \times x \)
\( x = \frac{2 \times 3}{1} = 6 \)
Thus, 1 person will complete the job in 6 days.
2. We need to find how many persons are required to fit the windows in one day. Let \( x \) be the number of persons.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 2 \times 3 = x \times 1 \)
\( x = \frac{2 \times 3}{1} = 6 \)
Therefore, 6 persons will be required to complete the job in 1 day.
In simple words:
1. Two people can install windows in 3 days. If one person gets sick, only one person is left. With fewer workers, it will take longer. Using the inverse rule, \( 2 \times 3 = 1 \times x \), so \( x = 6 \) days.
2. If we want the job done in just 1 day (less time), we will need more people. Using the inverse rule, \( 2 \times 3 = x \times 1 \), so \( x = 6 \) people are needed.

Exam Tip: In "men-work-day" problems, assume total work remains constant. If time decreases, the number of workers must increase proportionately, following inverse variation principles.

 

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 houres to be the same?
Answer: Let \( x \) be the duration per period in minutes for 9 periods.
For a fixed total duration of school hours, if there are more periods, the duration of each period will be shorter. This is an instance of inverse proportion.

Using the inverse proportion rule \( n_1 x_1 = n_2 x_2 \):
\( 8 \times 45 = 9 \times x \)
\( x = \frac{8 \times 45}{9} = \frac{360}{9} = 40 \)
Thus, the required duration per period is 40 minutes.
In simple words: A school has 8 periods, each 45 minutes long. If they decide to have 9 periods in the same total school time, each period will need to be shorter. We find this using \( 8 \times 45 = 9 \times x \), which means \( x = 40 \) minutes for each period.

Exam Tip: When the total time (total school hours) is constant, the number of periods and the duration of each period are inversely proportional. Multiply the number of periods by their duration to find the total time, and use this constant for calculations.