CBSE Class 8 Mathematics Proportional Reasoning 1 Practice Paper Worksheet Set 01 with Answers

This is Class 8 Maths Ganita Prakash Chapter 7 Proportional Reasoning-1 Practice Paper 22: a 40-mark test of 20 questions in Sections A to E, with a step-by-step answer under every question. It checks whether two display boards are similar, finding a proportional ratio, the unitary method (rice for 450 people, a water tanker, geometry boxes, gravel by volume), a nickel–zinc mixture, dividing ₹15000 in the ratio 21 : 9, speed, machine and warehouse rates, and two case studies on Anganwadi food supply and painting a school.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Reduce 294 : 441 to simplest form.
(b) Check whether 14 : 21 is proportional to it.
Answer:
(a) HCF of 294 and 441 is 147. So 294 : 441 = (294 ÷ 147) : (441 ÷ 147) = 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 294 : 441.
Check by cross multiplication: 294 × 21 = 6174 and 441 × 14 = 6174, 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 (147) 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 25 kg wheat costs ₹875, find cost per kg.
(b) Find cost of 42 kg wheat.
Answer:
(a) Cost per kg = ₹875 ÷ 25 = ₹35.
(b) Cost of 42 kg = 42 × ₹35 = ₹1470.
Check: 25 : 875 :: 42 : 1470, since 25 × 1470 = 875 × 42 = 36750.
In simple words: Find the price of 1 kg (₹35), then multiply by 42.

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 21 : 9.
(b) Divide ₹15000 in this ratio.
Answer:
(a) Total parts = 21 + 9 = 30.
(b) Value of 1 part = ₹15000 ÷ 30 = ₹500.
First share = 21 × ₹500 = ₹10500.
Second share = 9 × ₹500 = ₹4500.
Check: ₹10500 + ₹4500 = ₹15000.
In simple words: Split ₹15000 into 30 equal parts and give 21 parts to the first and 9 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 truck travels 360 km in 6 hours. Find speed.
(b) Distance travelled in 11 hours.
Answer:
(a) Speed = distance ÷ time = 360 km ÷ 6 h = 60 km/h.
(b) At the same speed, distance in 11 hours = 60 × 11 = 660 km.
Check: 360 : 6 :: 660 : 11.
In simple words: The truck goes 60 km every hour, so in 11 hours it goes 660 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 156 : 234
(b) Simplify 20 : 30
(c) Check whether the two ratios are proportional
Answer:
(a) HCF of 156 and 234 is 78. 156 : 234 = (156 ÷ 78) : (234 ÷ 78) = 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: 156 : 234 :: 20 : 30.
Cross-check: 156 × 30 = 4680 and 234 × 20 = 4680.
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 15 : 5
(b) Quantity of first part if total = 200 mL
(c) Quantity of second part
Answer:
(a) Total parts = 15 + 5 = 20.
(b) 1 part = 200 mL ÷ 20 = 10 mL. First part = 15 × 10 = 150 mL.
(c) Second part = 5 × 10 = 50 mL (or 200 − 150 = 50 mL).
Check: 150 : 50 = 15 : 5 (both simplify to 3 : 1).
In simple words: Cut 200 mL into 20 equal parts of 10 mL; the first gets 15 parts and the second 5 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 1680 units in 21 hours. Rate per hour
(b) Units produced in 10 hours
(c) Time required to produce 2400 units
Answer:
(a) Rate = 1680 ÷ 21 = 80 units per hour.
(b) In 10 hours: 80 × 10 = 800 units.
(c) Time = 2400 ÷ 80 = 30 hours.
Check: 1680 : 21 :: 2400 : 30, since 1680 × 30 = 21 × 2400 = 50400.
In simple words: The machine makes 80 units each hour; multiply by hours to get units, divide units by 80 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 warehouse packs 2100 cartons in 30 hours.
(a) Cartons packed per hour
(b) Cartons packed in 6 hours
(c) Cartons packed in 45 hours
(d) Time required to pack 3150 cartons
(e) Method used
Answer:
(a) Cartons packed per hour = 2100 ÷ 30 = 70 cartons.
(b) In 6 hours = 70 × 6 = 420 cartons.
(c) In 45 hours = 70 × 45 = 3150 cartons.
(d) Time = 3150 ÷ 70 = 45 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, 30 : 2100 :: 45 : 3150.
In simple words: The warehouse packs 70 cartons every hour; multiply by the hours to get cartons, or divide cartons by 70 to get hours.

Teacher's Note:
1. Parts (c) and (d) are the same fact read both ways: 45 hours ↔ 3150 cartons.
2. Once the rate per hour is known, every other part is a single multiplication or division.

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

Question 19. Case Study 1: Anganwadi Nutrition Supply
An Anganwadi centre prepares food daily. When 280 children are present, 42 kg of food grains are used. During a vaccination drive, the number of children increases to 360. To ensure every child receives the same quantity of food, the worker calculates the required food grains using proportional reasoning.
Answer the following questions:
(a) Food grains used per child
(b) Write the proportion
(c) Food grains required for 360 children
(d) Extra food grains required
Answer:
(a) Per child = 42 kg ÷ 280 = \( \frac{3}{20} \) kg = 0.15 kg (150 g).
(b) Children : food grains stays the same, so 280 : 42 :: 360 : x, where x kg is needed for 360 children.
(c) x = \( \frac{42 × 360}{280} \) = 54 kg (or 360 × 0.15 = 54 kg).
(d) Extra grains = 54 − 42 = 12 kg.
Check: 80 more children × 0.15 kg = 12 kg.
In simple words: Each child gets 0.15 kg, so 360 children need 54 kg, which is 12 kg more than before.

Teacher's Note:
1. Equal share per child means the food grains are in direct proportion to the number of children.
2. The extra amount can be found directly: (360 − 280) × 0.15 = 12 kg.

Question 20. Case Study 2: Painting a School Building
A team of painters uses 18 litres of paint to paint 120 m² of wall area. For repainting a larger school building having 200 m² wall area with the same thickness of paint, the supervisor calculates the paint required using proportional reasoning to avoid wastage or shortage.
Answer the following questions:
(a) Paint used per m²
(b) Write the proportion
(c) Paint required for 200 m²
(d) Extra paint required
Answer:
(a) Paint per m² = 18 L ÷ 120 = \( \frac{3}{20} \) L = 0.15 L (150 mL).
(b) Area : paint stays the same, so 120 : 18 :: 200 : x, where x litres is needed for 200 m².
(c) x = \( \frac{18 × 200}{120} \) = 30 litres (or 200 × 0.15 = 30 L).
(d) Extra paint = 30 − 18 = 12 litres.
In simple words: Each square metre takes 0.15 L, so 200 m² takes 30 L, which is 12 L more than 18 L.

Teacher's Note:
1. With the same thickness, paint used is directly proportional to the area painted.
2. Keep units: litres for paint, m² for area.