Official Class 8 Mathematics Worksheets: Chapter 13 Direct and Inverse Proportions
Explore structured practice materials through the CBSE Class 8 Mathematics Direct and Inverse Proportions Practice Paper Worksheet Set 05 with Answers (Old NCERT Syllabus). Tailored for Class 8 learners, utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving accuracy before final school evaluations.
Solved Practice Worksheets for Mathematics
View or download the dedicated CBSE Class 8 Mathematics Direct and Inverse Proportions Practice Paper Worksheet Set 05 with Answers (Old NCERT Syllabus) resource below. Engaging with these practice papers under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum for Chapter 13 Direct and Inverse Proportions.
This Class 8 Maths Direct and Inverse Proportions practice paper carries 40 marks and has 17 questions in five sections (A to E), each with a step-by-step answer. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising this chapter. The paper tests deciding whether a situation is direct or inverse proportion, finding the constant of variation, completing proportion tables, and real-life problems on speed, pipes, food provisions, costs and shadows.
Section A: Multiple Choice Questions (1 Mark Each)
(a) \( \frac{250}{9} \) m
(b) \( \frac{500}{9} \) m
(c) 1000 m
(d) \( \frac{500}{3} \) m
Show Answer & Explanation
Answer: (d) \( \frac{500}{3} \) m
Explanation:
1. In 1 hour = 60 minutes the rickshaw covers 10 km = 10000 m.
2. Distance and time are in direct proportion (same speed), so in 1 minute it covers 10000 ÷ 60 m.
3. 10000 ÷ 60 = \( \frac{1000}{6} \) = \( \frac{500}{3} \) m.
4. So the answer is \( \frac{500}{3} \) m (about 166.7 m).
Teacher's Note:
1. Convert km to m and hours to minutes before dividing.
2. At a constant speed, distance is directly proportional to time.
(a) 200
(b) 84
(c) 99
(d) 70
Show Answer & Explanation
Answer: (b) 84
Explanation:
1. x ∝ y means \( \frac{x}{y} \) is constant, so \( \frac{x_1}{y_1} \) = \( \frac{x_2}{y_2} \).
2. \( \frac{5}{210} \) = \( \frac{2}{y_2} \), so 5y₂ = 2 × 210 = 420.
3. y₂ = 420 ÷ 5 = 84.
4. So y₂ = 84.
Teacher's Note:
1. For direct proportion, the ratio x : y stays the same.
2. Check: 5/210 = 1/42 and 2/84 = 1/42.
(a) 120 minutes
(b) 96 minutes
(c) 80 minutes
(d) 85 minutes
Show Answer & Explanation
Answer: (b) 96 minutes
Explanation:
1. 1 hour 20 minutes = 80 minutes.
2. Fewer pipes take more time, so the number of pipes and time are in inverse proportion.
3. 6 × 80 = 5 × t, so t = 480 ÷ 5 = 96 minutes.
4. So the tank fills in 96 minutes.
Teacher's Note:
1. In inverse proportion the product of the two quantities stays the same.
2. 96 minutes = 1 hour 36 minutes, which is longer than before, as expected.
(a) 30 days
(b) \( \frac{96}{5} \) days
(c) 120 days
(d) 40 days
Show Answer & Explanation
Answer: (a) 30 days
Explanation:
1. Persons left = 100 − 20 = 80.
2. Fewer persons make the food last longer, so this is inverse proportion.
3. 100 × 24 = 80 × d, so d = 2400 ÷ 80 = 30 days.
4. So the provision will last 30 days.
Teacher's Note:
1. Total food = 100 × 24 = 2400 person-days; divide by the new number of persons.
2. Option (b) comes from wrongly using direct proportion.
(a) 60 km/h
(b) 80 km/h
(c) 30 km/h
(d) 40 km/h
Show Answer & Explanation
Answer: (a) 60 km/h
Explanation:
1. Distance = 48 × 10 = 480 km.
2. Speed and time are in inverse proportion for a fixed distance.
3. New speed = 480 ÷ 8 = 60 km/h.
4. So the speed should be 60 km/h.
Teacher's Note:
1. For the same distance, less time needs more speed.
2. Check: 48 × 10 = 60 × 8 = 480.
(a) 100 litres
(b) 50 litres
(c) 25.16 litres
(d) 25 litres
Show Answer & Explanation
Answer: (a) 100 litres
Explanation:
1. Diesel used and distance covered are in direct proportion.
2. Distance per litre = 297 ÷ 54 = 5.5 km.
3. Diesel for 550 km = 550 ÷ 5.5 = 100 litres.
4. So the truck needs 100 litres.
Teacher's Note:
1. Find the distance for 1 litre (unitary method), then scale up.
2. Check: 54/297 = 100/550 since both equal 2/11.
Section B: Short Answer Questions (2 Marks Each)
Question 7. A batch of bottles was packed in 25 boxes, with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Answer:
More bottles in each box means fewer boxes, so the two are in inverse proportion.
12 × 25 = 20 × x.
x = 300 ÷ 20 = 15.
So 15 boxes would be filled.
In simple words: There are 300 bottles in all; packing 20 per box needs 300 ÷ 20 = 15 boxes.
Teacher's Note:
1. The total number of bottles (300) does not change.
2. In inverse proportion, multiply across: x₁y₁ = x₂y₂.
Question 8. A school has 8 periods a day, each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Answer:
The school time is fixed, so more periods means shorter periods: inverse proportion.
8 × 45 = 9 × x.
x = 360 ÷ 9 = 40.
So each period would be 40 minutes long.
In simple words: The school day has 360 minutes of periods; split into 9 periods, each is 40 minutes.
Teacher's Note:
1. Total time = 8 × 45 = 360 minutes stays the same.
2. Always ask: if one quantity goes up, does the other go up or down?
Question 9. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Answer:
Let the number of bottles filled in 5 hours be x.
More time means more bottles, so this is direct proportion.
\( \frac{6}{840} \) = \( \frac{5}{x} \), so 6x = 5 × 840 = 4200.
x = 4200 ÷ 6 = 700.
So the machine fills 700 bottles in five hours.
In simple words: The machine fills 140 bottles an hour, so 5 hours gives 700 bottles.
Teacher's Note:
1. Unitary method: 840 ÷ 6 = 140 bottles per hour.
2. In direct proportion, the ratio of the two quantities stays the same.
Section C: Short Answer Questions (3 Marks Each)
Question 10. If y varies inversely as x, and if y = 4 when x = 72, find the following:
(a) the constant of variation
(b) the value of y when x = 9.
Answer:
(a) Inverse variation means xy = k, where k is the constant of variation.
k = 72 × 4 = 288. So the constant of variation is 288.
(b) When x = 9: y = \( \frac{k}{x} \) = \( \frac{288}{9} \) = 32.
So y = 32 when x = 9.
In simple words: The product xy is always 288, so when x is 9, y must be 32.
Teacher's Note:
1. In inverse variation, the product (not the ratio) is constant.
2. x became 8 times smaller (72 → 9), so y became 8 times bigger (4 → 32).
Answer:
Let the number of men needed be x.
| Number of men | 11 | x |
| Length of trench (in m) | \( 6\frac{3}{4} \) | 27 |
More men can dig a longer trench in one day, so this is direct proportion.
\( \frac{11}{6\frac{3}{4}} = \frac{x}{27} \), i.e. \( \frac{11}{27/4} = \frac{x}{27} \).
x = 11 × 27 × \( \frac{4}{27} \) = 11 × 4 = 44.
So 44 men should be employed.
In simple words: 27 m is exactly 4 times \( 6\frac{3}{4} \) m, so 4 times as many men are needed: 44.
Teacher's Note:
1. Change the mixed number to an improper fraction: \( 6\frac{3}{4} \) = \( \frac{27}{4} \).
2. Spotting that 27 ÷ \( \frac{27}{4} \) = 4 makes the sum very quick.
Question 12. It is given that s varies directly as q.
(a) Write an equation that specifies the variation and find the constant of variation if s is 4 when q is 6.
(b) Find s when q is 15.
(c) Find q when s is 12.
Answer:
(a) s varies directly as q, so s = kq, where k is the constant of variation.
4 = k × 6, so k = \( \frac{4}{6} \) = \( \frac{2}{3} \). The equation is s = \( \frac{2}{3} \)q.
(b) s = \( \frac{2}{3} \) × 15 = 10.
(c) 12 = \( \frac{2}{3} \)q, so q = 12 × \( \frac{3}{2} \) = 18.
In simple words: s is always two-thirds of q, so q = 15 gives s = 10, and s = 12 needs q = 18.
Teacher's Note:
1. Find k first; every other part uses the same k.
2. Check (c): \( \frac{2}{3} \) × 18 = 12.
Question 13. There are 100 students in a hostel. Food provision for them is for 20 days. How long will these provisions last, if 25 more students join the group?
Answer:
Number of students now = 100 + 25 = 125. Let the food last y days.
More students finish the food sooner, so this is inverse proportion.
100 × 20 = 125 × y.
y = 2000 ÷ 125 = 16.
So the provisions will last 16 days.
In simple words: The food is enough for 2000 student-days; shared by 125 students it lasts 16 days.
Teacher's Note:
1. Remember to add the new students to the old number.
2. Check: 125 × 16 = 2000 = 100 × 20.
Section D: Long Answer Questions (4 Marks Each)
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Answer:
Let the distance in 20 minutes be x km and the time for 250 km be y minutes.
| Distance travelled (in km) | 75 | x | 250 |
| Time taken (in minutes) | 60 | 20 | y |
At a uniform speed, distance is directly proportional to time.
(i) \( \frac{75}{60} \) = \( \frac{x}{20} \), so x = \( \frac{75}{60} \) × 20 = 25.
The train travels 25 km in 20 minutes.
(ii) \( \frac{75}{60} \) = \( \frac{250}{y} \), so y = \( \frac{250 × 60}{75} \) = 200 minutes.
So the time needed is 200 minutes = 3 hours 20 minutes.
In simple words: The train goes 75 km every 60 minutes, so 25 km in 20 minutes, and 250 km takes 200 minutes.
Teacher's Note:
1. Write the speed in matching units (km per minute or per hour) before comparing.
2. Check (ii): 250 ÷ 75 = 3⅓ hours = 3 hours 20 minutes.
| (a) | (b) | (c) | (d) | (e) | |
| V (in cm³) | ? | 48 | 60 | ? | 100 |
| P (in atmosphere) | 2 | ? | \( \frac{3}{2} \) | 1 | ? |
Answer:
V varies inversely as P, so PV = k (a constant).
From column (c): k = \( \frac{3}{2} \) × 60 = 90.
(a) P = 2: 2 × V = 90, so V = 45 cm³.
(b) V = 48: P × 48 = 90, so P = \( \frac{90}{48} \) = \( \frac{15}{8} \) atmosphere.
(d) P = 1: 1 × V = 90, so V = 90 cm³.
(e) V = 100: P × 100 = 90, so P = \( \frac{90}{100} \) = \( \frac{9}{10} \) = 0.9 atmosphere.
Completed table:
| (a) | (b) | (c) | (d) | (e) | |
| V (in cm³) | 45 | 48 | 60 | 90 | 100 |
| P (in atmosphere) | 2 | \( \frac{15}{8} \) | \( \frac{3}{2} \) | 1 | \( \frac{9}{10} \) |
In simple words: Pressure × volume is always 90, so each missing value is 90 divided by the given one.
Teacher's Note:
1. Use the only column where both values are known to find the constant.
2. As V goes up across the table, P goes down: a quick check for inverse proportion.
Answer:
Let the length of cloth be x metres and its cost be Rs y.
A longer piece costs more in the same ratio, so this is direct proportion: \( \frac{x}{y} \) is constant.
Cost of 1 metre = 210 ÷ 5 = Rs 42.
For 2 m: \( \frac{5}{210} \) = \( \frac{2}{y_2} \), so y₂ = \( \frac{2 × 210}{5} \) = 84.
For 4 m: y₃ = \( \frac{4 × 210}{5} \) = 168.
For 10 m: y₄ = \( \frac{10 × 210}{5} \) = 420.
For 13 m: y₅ = \( \frac{13 × 210}{5} \) = 546.
| Length of cloth x (in m) | 2 | 4 | 5 | 10 | 13 |
| Cost y (in Rs) | 84 | 168 | 210 | 420 | 546 |
In simple words: Each metre costs Rs 42, so multiply 42 by the number of metres.
Teacher's Note:
1. The unitary method (Rs 42 per metre) is the fastest way to fill such a table.
2. Check: 546 ÷ 13 = 42.
Section E: Case Study Based Question (4 Marks)
Question 17. One day the Maths teacher was explaining Direct and Inverse Proportion with the help of many examples. He gave the example of the relation between the length of a shadow and the height of the object casting it. A water tank casts a shadow 21 m long. A tree of height 9.6 m casts a shadow 8 m long at the same time.
Answer the following questions based on the above information:
(a) Find the height of the water tank in the above situation. (1)
(b) A water tank casts a shadow 27 m long. A tree of height 10.5 m casts a shadow 9 m long at the same time. The lengths of the shadows are directly proportional to their heights. Find the height of the tank. (1)
(c) A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. (1)
(d) A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time, a tower casts a shadow of length 40 m on the ground. Determine the height of the tower. (1)
Answer:
At the same time of day, the length of a shadow is directly proportional to the height of the object.
(a) \( \frac{8}{9.6} \) = \( \frac{21}{x} \), so x = \( \frac{21 × 9.6}{8} \) = \( \frac{201.6}{8} \) = 25.2 m.
(b) \( \frac{9}{10.5} \) = \( \frac{27}{x} \), so x = \( \frac{27 × 10.5}{9} \) = \( \frac{283.5}{9} \) = 31.5 m.
(c) \( \frac{6}{4} \) = \( \frac{x}{28} \), so x = \( \frac{28 × 6}{4} \) = 42 m.
(d) \( \frac{12}{8} \) = \( \frac{x}{40} \), so x = \( \frac{12 × 40}{8} \) = 60 m.
In simple words: Height ÷ shadow is the same for everything at the same moment, so each tall object's height is its shadow times that ratio.
Teacher's Note:
1. Keep heights together and shadows together in the proportion.
2. In (a), the shadow is 21 ÷ 8 times longer, so the tank is 2.625 × 9.6 = 25.2 m tall.
Free study material for Mathematics
Chapter 13 Direct and Inverse Proportions Printable Worksheets and Exercises for Class 8 Mathematics
Mastering Chapter 13 Direct and Inverse Proportions with Printable Worksheets
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