CBSE Class 8 Mathematics Proportional Reasoning 1 Practice Paper Worksheet Set 03 with Answers

Practice Paper 26 of Class 8 Maths Ganita Prakash Chapter 7 Proportional Reasoning-1 carries 40 marks over 20 questions in Sections A–E, and every question has a clear step-by-step solution. Its focus is similar flags, the ratio 35 : 49, wheat for 240 people, filling time of a pipe, notebook prices, a flour–sugar mix, the cost of 8.4 m of cloth, sharing ₹6400 in the ratio 9 : 7, a pump's discharge rate, bolts made in a workshop, and case studies on issuing notebooks and cement bags for a concrete slab.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Reduce 96 : 144 to simplest form.
(b) Check whether 32 : 48 is proportional to it.
Answer:
(a) HCF of 96 and 144 is 48. So 96 : 144 = (96 ÷ 48) : (144 ÷ 48) = 2 : 3. Simplest form: 2 : 3.
(b) 32 : 48 = (32 ÷ 16) : (48 ÷ 16) = 2 : 3. Both ratios equal 2 : 3, so yes, 32 : 48 is proportional to 96 : 144.
Check by cross multiplication: 96 × 48 = 4608 and 144 × 32 = 4608, 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 (48) 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 10 kg rice costs ₹520, find cost per kg.
(b) Find cost of 18 kg rice.
Answer:
(a) Cost per kg = ₹520 ÷ 10 = ₹52.
(b) Cost of 18 kg = 18 × ₹52 = ₹936.
Check: 10 : 520 :: 18 : 936, since 10 × 936 = 520 × 18 = 9360.
In simple words: Find the price of 1 kg (₹52), then multiply by 18.

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 9 : 7.
(b) Divide ₹6400 in this ratio.
Answer:
(a) Total parts = 9 + 7 = 16.
(b) Value of 1 part = ₹6400 ÷ 16 = ₹400.
First share = 9 × ₹400 = ₹3600.
Second share = 7 × ₹400 = ₹2800.
Check: ₹3600 + ₹2800 = ₹6400.
In simple words: Split ₹6400 into 16 equal parts and give 9 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 train covers 210 km in 3 hours. Find speed.
(b) Distance covered in 5 hours.
Answer:
(a) Speed = distance ÷ time = 210 km ÷ 3 h = 70 km/h.
(b) At the same speed, distance in 5 hours = 70 × 5 = 350 km.
Check: 210 : 3 :: 350 : 5.
In simple words: The train goes 70 km every hour, so in 5 hours it goes 350 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 56 : 84
(b) Simplify 14 : 21
(c) State whether they are proportional
Answer:
(a) HCF of 56 and 84 is 28. 56 : 84 = (56 ÷ 28) : (84 ÷ 28) = 2 : 3. Answer: 2 : 3.
(b) HCF of 14 and 21 is 7. 14 : 21 = (14 ÷ 7) : (21 ÷ 7) = 2 : 3. Answer: 2 : 3.
(c) Both simplify to 2 : 3, so yes, they are proportional: 56 : 84 :: 14 : 21.
Cross-check: 56 × 21 = 1176 and 84 × 14 = 1176.
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 5 : 3
(b) Quantity of first part if total = 64 mL
(c) Quantity of second part
Answer:
(a) Total parts = 5 + 3 = 8.
(b) 1 part = 64 mL ÷ 8 = 8 mL. First part = 5 × 8 = 40 mL.
(c) Second part = 3 × 8 = 24 mL (or 64 − 40 = 24 mL).
Check: 40 : 24 = 5 : 3.
In simple words: Cut 64 mL into 8 equal parts of 8 mL; the first gets 5 parts and the second 3 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 pump discharges 480 L in 8 minutes. Rate per minute
(b) Time required to discharge 900 L
(c) Time required to discharge 1200 L
Answer:
(a) Rate = 480 L ÷ 8 min = 60 L per minute.
(b) Time for 900 L = 900 ÷ 60 = 15 minutes.
(c) Time for 1200 L = 1200 ÷ 60 = 20 minutes.
Check: 480 : 8 :: 900 : 15 :: 1200 : 20, since all equal 60 : 1.
In simple words: The pump lets out 60 litres each minute, so divide the litres by 60 to get the minutes.

Teacher's Note:
1. When the rate is fixed, time is directly proportional to the volume discharged.
2. To find time, divide the volume by the rate per minute.

Section D: Long Answer Question (5 Marks)

Question 18. A workshop manufactures 450 bolts in 9 hours.
(a) Bolts produced per hour
(b) Bolts produced in 2 hours
(c) Bolts produced in 15 hours
(d) Time required to produce 600 bolts
(e) Method used
Answer:
(a) Bolts produced per hour = 450 ÷ 9 = 50 bolts.
(b) In 2 hours = 50 × 2 = 100 bolts.
(c) In 15 hours = 50 × 15 = 750 bolts.
(d) Time = 600 ÷ 50 = 12 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, 9 : 450 :: 12 : 600.
In simple words: The workshop makes 50 bolts an hour; 2 hours give 100, 15 hours give 750, and 600 bolts take 12 hours.

Teacher's Note:
1. Find the hourly rate first; every other part follows from it.
2. Check (d): 12 hours × 50 bolts = 600 bolts.

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

Question 19. Case Study 1: School Library Books
A school library distributes notebooks equally among students. When 160 students are present, 320 notebooks are issued. During a special remedial programme, the number of students increases to 240. To maintain equal distribution without shortage, the librarian calculates the total number of notebooks required using proportional reasoning.
Answer the following questions:
(a) Notebooks per student
(b) Write the proportion
(c) Notebooks required for 240 students
(d) Additional notebooks required
Answer:
(a) Notebooks per student = 320 ÷ 160 = 2.
(b) Students : notebooks stays the same, so 160 : 320 :: 240 : x, where x notebooks are needed for 240 students.
(c) x = \( \frac{320 × 240}{160} \) = 480 notebooks (or 240 × 2 = 480).
(d) Additional notebooks = 480 − 320 = 160 notebooks.
In simple words: Each student gets 2 notebooks, so 240 students need 480, which is 160 more than before.

Teacher's Note:
1. Equal distribution means notebooks are directly proportional to the number of students.
2. 80 extra students × 2 notebooks = 160, a quick check of part (d).

Question 20. Case Study 2: Concrete Mixing
A builder uses 12 bags of cement to prepare concrete for a slab of area 30 m². For constructing a larger slab of area 50 m² with the same thickness and strength, the builder calculates the number of cement bags required using proportional reasoning to ensure structural safety.
Answer the following questions:
(a) Cement bags per m²
(b) Write the proportion
(c) Cement bags required for 50 m²
(d) Extra cement bags needed
Answer:
(a) Bags per m² = 12 ÷ 30 = \( \frac{2}{5} \) = 0.4 bag.
(b) Area : bags stays the same, so 30 : 12 :: 50 : x, where x bags are needed for 50 m².
(c) x = \( \frac{12 × 50}{30} \) = 20 bags (or 50 × 0.4 = 20).
(d) Extra bags = 20 − 12 = 8 bags.
In simple words: Each square metre needs 0.4 bag, so 50 m² needs 20 bags, which is 8 more than 12.

Teacher's Note:
1. With the same thickness and mix, cement is directly proportional to the slab area.
2. A rate per m² can be a decimal even though the final number of bags is whole.