Class 8 Inverse Proportion Case Study Questions

Class 8 Inverse Proportion Case Study Questions

Class 8 Inverse Proportion Case Study Questions

Class 8 Inverse Proportion Case Study Questions help students understand how one quantity increases when another decreases. Moreover, these case studies include real-life situations that make learning meaningful. Each problem guides students to compare values and apply inverse proportion effectively.

Why These Case Studies Are Important

These worksheets improve logical reasoning and boost concept clarity. Furthermore, the step-by-step answers help learners understand how inverse relationships work in different situations. As a result, students gain confidence in solving numerical problems.

How to Practice These Questions

Students should observe how values change and identify the correct inverse relation. Additionally, solving these questions regularly enhances accuracy and prepares students well for exams.

Case Study 5: Class 8 Inverse Proportion Case Study Questions

A district education office organises a one-day learning camp for 150 students. The kitchen team prepares standardized food packets. For 150 packets they used **12.5 kilograms of rice**, **7.5 kilograms of lentils** and **5 litres of oil**. The standard recipe ratio to be maintained is **rice : lentils : oil = 5 : 3 : 2**.

Transportation to the camp uses hired buses; **3 buses** each travelled **80 km** (round trip) and the total transport charge paid was **Rs. 7200**. **Five volunteers** worked together in the kitchen and took **4 hours** to prepare the 150 food packets.

During planning the team must verify ratios, compute using the unitary method, and apply direct and inverse proportion to adjust ingredients, time, and cost for potential changes.

1. Do the quantities used for 150 packets (12.5 kg rice, 7.5 kg lentils, 5 L oil) match the standard ratio $5:3:2$?
Solution:

To simplify the ratio $12.5 : 7.5 : 5$, find the largest common factor that turns them into integers. Dividing by $2.5$ (since $5/2.5=2$):

\(\frac{12.5}{2.5} : \frac{7.5}{2.5} : \frac{5}{2.5} = 5 : 3 : 2\)

The quantities match the standard recipe exactly.

Correct Answer: (a)

2. Using the unitary method, how much rice is required if the team needs to prepare 220 packets (same rice per packet as in the case)?
Solution:

This is a direct proportion problem. Rice for 150 packets = $12.5$ kg. Rice per packet is:

\(\text{Rice per packet} = \frac{12.5}{150} = \frac{1}{12}\text{ kg}\)

For 220 packets, required rice:

\(\frac{1}{12} \times 220 = \frac{220}{12} = \frac{55}{3} = 18.\overline{3}\text{ kg}\)

Correct Answer: (b)

3. Rice costs Rs. 60 per kilogram. Using the value from question 2, what is the cost of rice needed for 220 packets?
Solution:

Rice required (from Q2) $= \frac{55}{3}$ kg. Cost per kilogram = Rs. 60.

\(\text{Total Cost} = \frac{55}{3} \times 60 = 55 \times 20 = 1100\text{ Rs.}\)

Correct Answer: (b)

4. If 5 volunteers take 4 hours to prepare 150 packets, how many volunteers are required to finish the same work in 2 hours (assume perfect inverse proportion)?
Solution:

This is an inverse proportion problem: (Volunteers $\times$ Time) is constant. Let $v$ be the required volunteers:

\(5 \times 4 = v \times 2 \quad\Rightarrow\quad v = \frac{20}{2} = 10\)

Thus, 10 volunteers are needed.

Correct Answer: (a)

5. Transport cost is directly proportional to (number of buses $\times$ distance). If 3 buses for 80 km cost Rs. 7200, what will be the transport cost for 5 buses for 120 km?
Solution:

The cost is proportional to the combined factor (Buses $\times$ Distance).

Initial factor = $3 \times 80 = 240$. Cost = Rs. 7200.

New factor = $5 \times 120 = 600$.

The scaling factor for the cost is:

\(\text{Scale Factor} = \frac{600}{240} = 2.5\)

New cost $=$ Initial Cost $\times$ Scale Factor:

\(7200 \times 2.5 = 18000\text{ Rs.}\)

Correct Answer: (b)

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