Class 8 Proportion Case Study Worksheet with Solutions

Class 8 Proportion Case Study Worksheet with Solutions

Class 8 Proportion Case Study Worksheet with Solutions

The Class 8 Proportion Case Study Worksheet with Solutions helps students understand proportion concepts through real-life examples. Moreover, each worksheet includes solved case studies that make learning easier. These problems encourage logical thinking and guide students to apply proportional reasoning effectively in different situations.

Why This Worksheet Is Useful

This worksheet strengthens conceptual understanding by offering clear, step-by-step solutions. Furthermore, the case study format improves analytical skills and boosts confidence. As a result, students learn how to compare quantities, identify relationships, and solve proportion problems accurately.

How to Practice Effectively

Students should read each case carefully, note the given data, and apply the correct proportion method. Additionally, solving these worksheets regularly helps improve exam performance and overall mathematical reasoning.

Case Study 2: Class 8 Proportion Case Study Worksheet with Solutions

A school is organizing an inter-school sports meet and must arrange transport and food for students and staff. The organising committee plans to prepare exactly **240 food packets** on the event day. For these 240 packets they use **30 kilograms of rice**, **18 kilograms of lentils**, and **12 litres of oil**. The standard recipe ratio to be maintained is rice : lentils : oil = **5 : 3 : 2**. For transport, the committee hires **8 buses**; each bus travels a route of **120 kilometres** to reach the venue.

The hiring agency charges an overall transportation cost of **Rs. 19200** for those 8 buses for the round trip. For food preparation, **6 volunteers** working together take **5 hours** to prepare all 240 packets. Due to late registrations the expected number of participants may increase; the committee needs quick calculations using **unitary method**, **direct proportion** and **inverse proportion** to scale ingredients, estimate extra cost if buses or distance increase, and compute change in preparation time if volunteers are added or removed. Use the given data to answer the questions that follow.

1. Are the quantities used for rice, lentils and oil for 240 packets in the correct standard ratio 5 : 3 : 2?
Solution:

The ratio of quantities used is $30 : 18 : 12$. Dividing each term by the greatest common divisor, 6, we get:

\(\frac{30}{6} : \frac{18}{6} : \frac{12}{6} = 5 : 3 : 2\)

Hence, the quantities are exactly in the required ratio.

Correct Answer: (a)

2. Using unitary method, how much rice will be needed if participants increase to 300 packets, assuming same rice per packet?
Solution:

Rice used for 240 packets = 30 kg. Rice per packet is:

\(\text{Rice per packet} = \frac{30}{240} = 0.125\text{ kg}\)

For 300 packets, required rice = $0.125 \times 300 = 37.5$ kg.

Correct Answer: (b)

3. If 6 volunteers take 5 hours to prepare 240 packets, how long will 10 volunteers take assuming work is shared equally (inverse proportion)?
Solution:

Time is inversely proportional to the number of volunteers. Let $t$ be the time for 10 volunteers. The product of (volunteers $\times$ time) is constant:

\(6 \times 5 = 10 \times t \quad\Rightarrow\quad t = \frac{30}{10} = 3\text{ hours}\)

Correct Answer: (a)

4. A bus covers 120 km at an average speed of 60 km per hour. If average speed drops by 20\%, what is the new travel time for the same 120 km?
Solution:

Original speed = 60 km/h. New speed after 20% drop:

\(\text{New Speed} = 60 – (0.20 \times 60) = 60 – 12 = 48\text{ km/h}\)

New travel time = Distance / New Speed.

\(\text{New Time} = \frac{120\text{ km}}{48\text{ km/h}} = 2.5\text{ hours}\)

Correct Answer: (b)

5. Transportation cost is proportional to the product of number of buses and route distance. If cost for 8 buses over 120 km is Rs. 19200, what will be the cost for 10 buses over 150 km?
Solution:

Cost $C$ is proportional to (Buses $\times$ Distance). $C = k \times (\text{Buses} \times \text{Distance})$.

From the first scenario: $19200 = k \times (8 \times 120) = k \times 960$. $\text{Constant } k = \frac{19200}{960} = 20$.

New scenario: Cost $C_{\text{new}} = 20 \times (10 \times 150)$.

\(C_{\text{new}} = 20 \times 1500 = 30000\)

So cost = Rs. 30000.

Correct Answer: (b)

Review your answers and solutions below:

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