CBSE Class 8 Mathematics Proportional Reasoning 1 Practice Paper Worksheet Set 02 with Answers

Here is Practice Paper 24 for Class 8 Maths Ganita Prakash Chapter 7 Proportional Reasoning-1: 40 marks, 20 questions over Sections A to E, with step-by-step answers. The questions cover similar hoardings, ratios equivalent to 54 : 81, pulses for 350 people, pump time, iron–carbon and sand-weight problems, sharing ₹10000 in the ratio 13 : 7, components made in 7.5 hours, a dairy's milk supply, and case studies on flood relief drinking water and gravel for a road extension.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Reduce 168 : 252 to simplest form.
(b) Check whether 14 : 21 is proportional to it.
Answer:
(a) HCF of 168 and 252 is 84. So 168 : 252 = (168 ÷ 84) : (252 ÷ 84) = 2 : 3. Simplest form: 2 : 3.
(b) 14 : 21 = (14 ÷ 7) : (21 ÷ 7) = 2 : 3. Both ratios equal 2 : 3, so yes, 14 : 21 is proportional to 168 : 252.
Check by cross multiplication: 168 × 21 = 3528 and 252 × 14 = 3528, which are equal.
In simple words: Divide by the HCF to simplify; both ratios become 2 : 3, so they are proportional.

Teacher's Note:
1. Finding the HCF (84) in one step is faster than dividing again and again by small numbers.
2. Proportional means the two ratios are equal; cross products are equal too.

Question 12. (a) If 18 kg onions cost ₹540, find cost per kg.
(b) Find cost of 26 kg onions.
Answer:
(a) Cost per kg = ₹540 ÷ 18 = ₹30.
(b) Cost of 26 kg = 26 × ₹30 = ₹780.
Check: 18 : 540 :: 26 : 780, since 18 × 780 = 540 × 26 = 14040.
In simple words: Find the price of 1 kg (₹30), then multiply by 26.

Teacher's Note:
1. This is the unitary method: first find the value of one unit, then of many units.
2. Cost and quantity are in direct proportion: double the kg, double the cost.

Question 13. (a) Find total parts in the ratio 13 : 7.
(b) Divide ₹10000 in this ratio.
Answer:
(a) Total parts = 13 + 7 = 20.
(b) Value of 1 part = ₹10000 ÷ 20 = ₹500.
First share = 13 × ₹500 = ₹6500.
Second share = 7 × ₹500 = ₹3500.
Check: ₹6500 + ₹3500 = ₹10000.
In simple words: Split ₹10000 into 20 equal parts and give 13 parts to the first and 7 parts to the second.

Teacher's Note:
1. To divide a quantity in a given ratio, first add the terms of the ratio.
2. Always check that the shares add up to the whole amount.

Question 14. (a) A bus travels 280 km in 4 hours. Find speed.
(b) Distance travelled in 9 hours.
Answer:
(a) Speed = distance ÷ time = 280 km ÷ 4 h = 70 km/h.
(b) At the same speed, distance in 9 hours = 70 × 9 = 630 km.
Check: 280 : 4 :: 630 : 9.
In simple words: The bus goes 70 km every hour, so in 9 hours it goes 630 km.

Teacher's Note:
1. At a constant speed, distance is directly proportional to time.
2. Write the unit km/h with speed and km with distance.

Section C: Short Answer Questions (3 Marks Each)

Question 15. (a) Simplify 90 : 135
(b) Simplify 20 : 30
(c) Check whether the two ratios are proportional
Answer:
(a) HCF of 90 and 135 is 45. 90 : 135 = (90 ÷ 45) : (135 ÷ 45) = 2 : 3. Answer: 2 : 3.
(b) HCF of 20 and 30 is 10. 20 : 30 = (20 ÷ 10) : (30 ÷ 10) = 2 : 3. Answer: 2 : 3.
(c) Both simplify to 2 : 3, so yes, they are proportional: 90 : 135 :: 20 : 30.
Cross-check: 90 × 30 = 2700 and 135 × 20 = 2700.
In simple words: Both ratios shrink to 2 : 3, so they are equal (proportional).

Teacher's Note:
1. Simplify each ratio fully before comparing.
2. Equal simplest forms, or equal cross products, both prove proportionality.

Question 16. (a) Total parts in the ratio 11 : 4
(b) Quantity of first part if total = 150 mL
(c) Quantity of second part
Answer:
(a) Total parts = 11 + 4 = 15.
(b) 1 part = 150 mL ÷ 15 = 10 mL. First part = 11 × 10 = 110 mL.
(c) Second part = 4 × 10 = 40 mL (or 150 − 110 = 40 mL).
Check: 110 : 40 = 11 : 4.
In simple words: Cut 150 mL into 15 equal parts of 10 mL; the first gets 11 parts and the second 4 parts.

Teacher's Note:
1. The fraction of the total for the first part is (first term) ÷ (sum of terms).
2. The second part can be found by subtracting the first part from the total.

Question 17. (a) A machine produces 720 components in 12 hours. Rate per hour
(b) Components produced in 7.5 hours
(c) Time required to produce 900 components
Answer:
(a) Rate = 720 ÷ 12 = 60 components per hour.
(b) In 7.5 hours: 60 × 7.5 = 450 components.
(c) Time = 900 ÷ 60 = 15 hours.
Check: 720 : 12 :: 900 : 15, since 720 × 15 = 12 × 900 = 10800.
In simple words: The machine makes 60 components each hour; multiply by hours to get components, divide components by 60 to get hours.

Teacher's Note:
1. Production and time are in direct proportion when the rate stays the same.
2. To find time, divide the work by the rate; to find work, multiply the rate by time.

Section D: Long Answer Question (5 Marks)

Question 18. A dairy supplies 960 litres of milk in 8 hours.
(a) Milk supplied per hour
(b) Milk supplied in 3 hours
(c) Milk supplied in 15 hours
(d) Time required to supply 1800 litres
(e) Method used
Answer:
(a) Milk supplied per hour = 960 ÷ 8 = 120 litres.
(b) In 3 hours = 120 × 3 = 360 litres.
(c) In 15 hours = 120 × 15 = 1800 litres.
(d) Time = 1800 ÷ 120 = 15 hours.
(e) Unitary method (proportional reasoning): the quantity is directly proportional to the time, so we found the amount for 1 hour and then scaled it. For example, 8 : 960 :: 15 : 1800.
In simple words: The dairy sends out 120 litres every hour, so 3 hours give 360 L, 15 hours give 1800 L, and 1800 L take 15 hours.

Teacher's Note:
1. Milk supplied and time are in direct proportion at a steady rate.
2. Write the unit (litres or hours) with every answer.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: Flood Relief Water Supply
During a flood relief operation, 500 people receive drinking water using 750 litres per day. As more families arrive, the number of people increases to 680. To ensure uninterrupted water supply and equal distribution, the relief coordinator calculates the total quantity of water required using proportional reasoning.
Answer the following questions:
(a) Water supplied per person
(b) Write the proportion
(c) Water required for 680 people
(d) Extra water required
Answer:
(a) Water per person = 750 L ÷ 500 = \( \frac{3}{2} \) L = 1.5 litres per day.
(b) People : water stays the same, so 500 : 750 :: 680 : x, where x litres is needed for 680 people.
(c) x = \( \frac{750 × 680}{500} \) = 1020 litres (or 680 × 1.5 = 1020 L).
(d) Extra water = 1020 − 750 = 270 litres.
Check: 180 more people × 1.5 L = 270 L.
In simple words: Each person gets 1.5 L a day, so 680 people need 1020 L, which is 270 L more.

Teacher's Note:
1. Equal sharing means water is directly proportional to the number of people.
2. Extra water can be found directly from the extra people: 180 × 1.5 = 270 L.

Question 20. Case Study 2: Road Construction
A contractor uses 180 tonnes of gravel to construct 300 metres of road. For extending the road to 520 metres with the same width and thickness, the contractor estimates the gravel required using proportional reasoning to avoid shortage and delay.
Answer the following questions:
(a) Gravel used per metre
(b) Write the proportion
(c) Gravel required for 520 metres
(d) Extra gravel required
Answer:
(a) Gravel per metre = 180 t ÷ 300 = \( \frac{3}{5} \) t = 0.6 tonne.
(b) Length : gravel stays the same, so 300 : 180 :: 520 : x, where x tonnes is needed for 520 m.
(c) x = \( \frac{180 × 520}{300} \) = 312 tonnes (or 520 × 0.6 = 312 t).
(d) Extra gravel = 312 − 180 = 132 tonnes.
In simple words: Each metre of road needs 0.6 tonne, so 520 m need 312 tonnes, which is 132 tonnes more.

Teacher's Note:
1. Same width and thickness means gravel is directly proportional to the length of road.
2. Extra gravel = extra length × 0.6 = 220 × 0.6 = 132 t, a quick check.