7 th maths chapter 4 || 4.2

7th Standard Mathematics — Chapter 4: Direct and Inverse Proportion

Exercise 4.2 Solutions (Inverse Proportion)

1. Fill in the blanks

1. 16 taps can fill a petrol tank in 18 minutes. The time taken for 9 taps to fill the same tank will be 32 minutes.
   Working: x₁y₁ = x₂y₂ ⇒ 16 × 18 = 9 × y ⇒ y = (16 × 18) ÷ 9 = 32 minutes.
2. If 40 workers can do a project work in 8 days, then 80 workers can do it in 4 days.
   Working: x₁y₁ = x₂y₂ ⇒ 40 × 8 = x × 4 ⇒ x = 320 ÷ 4 = 80 workers.

2. Water Sump Filling Problem

6 pumps are required to fill a water sump in 1 hr 30 minutes. What will be the time taken to fill the sump if one pump is switched off?

Solution:

• Initial time = 1 hr 30 min = 90 minutes
• Initial number of pumps = 6
• Number of pumps remaining = 6 - 1 = 5 pumps

Since the number of pumps and time taken are in inverse proportion:

6 × 90 = 5 × Time

Time = 540 ÷ 5 = 108 minutes = 1 hour 48 minutes

Answer: The time taken will be 1 hour 48 minutes (108 minutes).

3. Duck Food Duration Problem

A farmer has enough food for 144 ducks for 28 days. If he sells 32 ducks, how long will the food last?

Solution:

• Initial number of ducks = 144
• Remaining ducks = 144 - 32 = 112 ducks

In inverse proportion: x₁y₁ = x₂y₂

144 × 28 = 112 × Days

Days = (144 × 28) ÷ 112 = 144 ÷ 4 = 36 days

Answer: The food will last for 36 days.

4. Machine Digging Problem

It takes 60 days for 10 machines to dig a hole. Assuming that all machines work at the same speed, how long will it take 30 machines to dig the same hole?

Solution:

• Machines × Days = Constant
• 10 × 60 = 30 × Days

Days = 600 ÷ 30 = 20 days

Answer: It will take 20 days.

5. Hostel Food Stock Problem

Forty students stay in a hostel. They had food stock for 30 days. If the students are doubled then for how many days the stock will last?

Solution:

• Initial students = 40
• Doubled students = 40 × 2 = 80 students

Since food stock and number of students are in inverse proportion:

40 × 30 = 80 × Days

Days = 1200 ÷ 80 = 15 days

Answer: The stock will last for 15 days.

6. Courier Parcel Weight Problem

Meena had enough money to send 8 parcels each weighing 500 grams through a courier service. What would be the weight of each parcel, if she has to send 40 parcels for the same money?

Solution:

• Number of parcels × Weight of each parcel = Total weight capability
• 8 × 500 g = 40 × Weight

Weight = 4000 ÷ 40 = 100 grams

Answer: The weight of each parcel would be 100 grams.

7. Garden Weeding Problem

It takes 120 minutes to weed a garden with 6 gardeners. If the same work is to be done in 30 minutes, how many more gardeners are needed?

Solution:

• Initial gardeners = 6
• Initial time = 120 minutes

Let total gardeners needed for 30 minutes be x:

6 × 120 = x × 30

x = 720 ÷ 30 = 24 gardeners

Additional Gardeners Needed:

More gardeners needed = 24 - 6 = 18 gardeners

Answer: 18 more gardeners are needed.

8. Bicycle Speed Increase Problem

Neelaveni goes by bi-cycle to her school every day. Her average speed is 12 km/hr and she reaches school in 20 minutes. What is the increase in speed, if she reaches the school in 15 minutes?

Solution:

• Initial speed = 12 km/hr
• Initial time = 20 minutes
• New time = 15 minutes

Speed and time are in inverse proportion:

12 × 20 = New Speed × 15

New Speed = 240 ÷ 15 = 16 km/hr

Increase in Speed:

Increase = 16 - 12 = 4 km/hr

Answer: The required increase in speed is 4 km/hr.

9. Toy Factory Machines Problem

A toy company requires 36 machines to produce car toys in 54 days. How many machines would be required to produce the same number of car toys in 81 days?

Solution:

• Machines × Days = Constant
• 36 × 54 = Machines × 81

Machines = (36 × 54) ÷ 81

Machines = (36 × 2) ÷ 3 = 12 × 2 = 24 machines

Answer: 24 machines would be required.