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This Class 8 Maths Direct and Inverse Proportions practice paper (Paper 02) carries 40 marks and has 17 questions in five sections (A to E), all with step-by-step answers. It follows the earlier (old) NCERT Class 8 Maths syllabus, which makes it useful for state-board students and for revising this topic. The paper tests identifying direct and inverse variation, completing x–y tables using the constant of variation, and word problems on work-force, speed, springs, shelves and food supplies in a fort.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 25 and 35
(b) 35 and 25
(c) 35 and 49
(d) 15 and 21
Show Answer & Explanation
Answer: (b) 35 and 25
Explanation:
1. x and y vary directly, so \( \frac{x}{y} \) = \( \frac{10}{14} \) = \( \frac{5}{7} \) for every pair.
2. (a) \( \frac{25}{35} \) = \( \frac{5}{7} \), (c) \( \frac{35}{49} \) = \( \frac{5}{7} \), (d) \( \frac{15}{21} \) = \( \frac{5}{7} \).
3. (b) \( \frac{35}{25} \) = \( \frac{7}{5} \), which is not \( \frac{5}{7} \).
4. So the pair that is not possible is 35 and 25.
Teacher's Note:
1. In direct variation x must stay smaller than y here, since 10 < 14.
2. Option (b) swaps the numbers of a correct pair; watch the order of x and y.
(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 = speed × time = 48 × 10 = 480 km.
2. For a fixed distance, speed and time are in inverse proportion.
3. Required speed = 480 ÷ 8 = 60 km/h.
4. So the speed should be 60 km/h.
Teacher's Note:
1. Less time for the same distance needs a higher speed.
2. Check: 48 × 10 = 60 × 8.
(a) speed and distance covered
(b) distance covered and taxi fare
(c) distance travelled and time taken
(d) speed and time taken
Show Answer & Explanation
Answer: (d) speed and time taken
Explanation:
1. For a fixed distance, time = distance ÷ speed.
2. If the speed increases, the time taken decreases in the same ratio, and speed × time stays constant.
3. The other pairs increase together, so they vary directly.
4. So speed and time taken vary inversely.
Teacher's Note:
1. Inverse variation: one goes up, the other goes down, and their product is constant.
2. Distance–fare and distance–time (at fixed speed) are direct variations.
(a) \( \frac{20}{3} \) km/h
(b) 16 km/h
(c) 20 km/h
(d) 15 km/h
Show Answer & Explanation
Answer: (c) 20 km/h
Explanation:
1. 20 minutes = \( \frac{1}{3} \) hour, so the distance = 12 × \( \frac{1}{3} \) = 4 km.
2. 12 minutes = \( \frac{12}{60} \) = \( \frac{1}{5} \) hour.
3. Required speed = 4 ÷ \( \frac{1}{5} \) = 20 km/h.
4. So her speed should be 20 km/h.
Teacher's Note:
1. Shortcut with inverse proportion: 12 × 20 = s × 12, so s = 20.
2. Change minutes into hours when the speed is in km/h.
(a) 30 days
(b) \( \frac{96}{5} \) days
(c) 60 days
(d) 40 days
Show Answer & Explanation
Answer: (c) 60 days
Explanation:
1. Fewer men take more days, so men and days are in inverse proportion.
2. 36 × 25 = 15 × x.
3. x = 900 ÷ 15 = 60.
4. So 15 men will take 60 days.
Teacher's Note:
1. The total work is 36 × 25 = 900 man-days.
2. Using direct proportion by mistake would give 10.4 days, which is less than 25 and clearly wrong.
(a) directly with each other.
(b) inversely with each other.
(c) neither directly nor inversely with each other.
(d) sometimes directly and sometimes inversely with each other.
Show Answer & Explanation
Answer: (a) directly with each other.
Explanation:
1. If the land is equally fertile everywhere, each square metre gives the same yield.
2. Doubling the area doubles the yield, so yield ÷ area stays constant.
3. So they vary directly with each other.
Teacher's Note:
1. The word 'uniformly' is what makes the variation exact.
2. Direct variation: both quantities change in the same ratio.
Section B: Short Answer Questions (2 Marks Each)
Question 7. If the cost of 93 m of a certain kind of plastic sheet is Rs 1395, then what would it cost to buy 105 m of such plastic sheet?
Answer:
Let the cost of 105 m be Rs x.
A longer sheet costs more in the same ratio, so this is direct proportion.
\( \frac{93}{1395} \) = \( \frac{105}{x} \), so x = \( \frac{105 × 1395}{93} \) = 105 × 15 = 1575.
So 105 m of plastic sheet costs Rs 1575.
In simple words: Each metre costs Rs 15, so 105 m costs Rs 1575.
Teacher's Note:
1. 1395 ÷ 93 = 15 makes the working short.
2. Keep lengths together and costs together when writing the proportion.
Question 8. A work-force of 420 men with a contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months?
Answer:
More men finish the work in less time, so men and months are in inverse proportion.
Let the number of men needed for 7 months be y. Then 420 × 9 = 7 × y.
y = 3780 ÷ 7 = 540.
Extra men = 540 − 420 = 120.
So the contractor must employ 120 extra men.
In simple words: The job needs 3780 man-months; to finish in 7 months he needs 540 men, i.e. 120 more.
Teacher's Note:
1. The question asks for extra men, so subtract the men already working.
2. Many students stop at 540; read the last line of the question again.
Question 9. Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes?
Answer:
Let the distance covered in 315 minutes be x m.
At the same speed, more time means more distance: direct proportion.
\( \frac{125}{100} \) = \( \frac{315}{x} \), so x = \( \frac{100 × 315}{125} \) = 252.
So she would cover 252 metres.
In simple words: She walks 0.8 m every minute, so in 315 minutes she covers 252 m.
Teacher's Note:
1. Unitary method: 100 ÷ 125 = 0.8 m per minute.
2. Check: 252 ÷ 315 = 0.8 as well.
Section C: Short Answer Questions (3 Marks Each)
Answer:
Let the extension for 700 gm be x cm.
| Weight (in gm) | 250 | 700 |
| Extension (in cm) | 3.5 | x |
Extension varies directly with weight, so \( \frac{250}{3.5} \) = \( \frac{700}{x} \).
250x = 3.5 × 700 = 2450, so x = 2450 ÷ 250 = 9.8.
So the extension is 9.8 cm.
In simple words: 700 gm is 2.8 times 250 gm, so the spring stretches 2.8 × 3.5 = 9.8 cm.
Teacher's Note:
1. Cross-multiply carefully with decimals.
2. Check: 3.5 ÷ 250 = 0.014 and 9.8 ÷ 700 = 0.014.
Answer:
Let x boxes fit on 20.4 m of shelf.
| Number of boxes | 68 | x |
| Shelf-length (in m) | 13.6 | 20.4 |
A longer shelf holds more boxes in the same ratio: direct proportion.
\( \frac{68}{13.6} \) = \( \frac{x}{20.4} \), so x = \( \frac{68 × 20.4}{13.6} \) = \( \frac{1387.2}{13.6} \) = 102.
So 102 boxes would occupy a shelf length of 20.4 m.
In simple words: Each metre of shelf holds 5 boxes, so 20.4 m holds 102 boxes.
Teacher's Note:
1. 68 ÷ 13.6 = 5 boxes per metre makes it quick.
2. 20.4 is 1.5 times 13.6, so the boxes are 1.5 × 68 = 102.
Question 12. Which of the following quantities vary inversely as each other?
(i) The number x of men hired to construct a wall and the time y taken to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.
Answer:
(i) More men finish the wall in less time; xy stays constant. So this is inverse variation.
(ii) A longer journey costs more, so the two increase together. This is not inverse variation.
(iii) A longer journey uses more petrol, so the two increase together. This is not inverse variation.
So only (i) varies inversely.
In simple words: Only men and time work against each other; the other two pairs grow together.
Teacher's Note:
1. Ask: when x increases, does y decrease? If yes, test whether xy is constant.
2. Bus fares and petrol use grow with distance, so they are (roughly) direct variations.
Question 13. 1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 days. How many soldiers left the fort?
Answer:
After 4 days, the food left is enough for 1200 soldiers for 28 − 4 = 24 days.
Let x soldiers leave. Then 1200 − x soldiers remain and the food lasts 32 days.
Soldiers and days are in inverse proportion: 1200 × 24 = (1200 − x) × 32.
1200 − x = \( \frac{1200 × 24}{32} \) = 900.
x = 1200 − 900 = 300.
So 300 soldiers left the fort.
In simple words: The leftover food was 28800 soldier-days; lasting 32 days means 900 soldiers stayed, so 300 left.
Teacher's Note:
1. Use the food remaining after 4 days, i.e. 24 days for 1200 soldiers.
2. Check: 900 × 32 = 28800 = 1200 × 24.
Section D: Long Answer Questions (4 Marks Each)
| x | 5 | … | 10 | 35 | 25 | … |
| y | 8 | 12 | … | … | … | 32 |
Answer:
x varies directly as y, so \( \frac{x}{y} \) = k. From x = 5, y = 8: k = \( \frac{5}{8} \).
For y = 12: x = \( \frac{5}{8} \) × 12 = \( \frac{60}{8} \) = 7.5.
For x = 10: y = 10 × \( \frac{8}{5} \) = 16.
For x = 35: y = 35 × \( \frac{8}{5} \) = 56.
For x = 25: y = 25 × \( \frac{8}{5} \) = 40.
For y = 32: x = \( \frac{5}{8} \) × 32 = 20.
Completed table:
| x | 5 | 7.5 | 10 | 35 | 25 | 20 |
| y | 8 | 12 | 16 | 56 | 40 | 32 |
In simple words: Every x is five-eighths of its y, so multiply or divide by 5/8 to fill each gap.
Teacher's Note:
1. To find y from x, multiply by \( \frac{8}{5} \); to find x from y, multiply by \( \frac{5}{8} \).
2. Check any column: 7.5 ÷ 12 = 0.625 = 5 ÷ 8.
| x | 12 | 16 | … | 8 | … |
| y | … | 6 | 4 | … | 0.25 |
Answer:
x varies inversely as y, so xy = k. From x = 16, y = 6: k = 96.
For x = 12: 12y = 96, so y = 8.
For y = 4: 4x = 96, so x = 24.
For x = 8: 8y = 96, so y = 12.
For y = 0.25: 0.25x = 96, so x = 96 ÷ 0.25 = 384.
Completed table:
| x | 12 | 16 | 24 | 8 | 384 |
| y | 8 | 6 | 4 | 12 | 0.25 |
In simple words: In every column x × y must be 96, so each gap is 96 divided by the number given.
Teacher's Note:
1. Dividing by 0.25 is the same as multiplying by 4.
2. Check every column: the product should be 96.
Question 16. If x and y vary inversely as each other and
(i) x = 3 when y = 8, find y when x = 4
(ii) x = 5 when y = 15, find x when y = 12
(iii) x = 30, find y when constant of variation = 900.
(iv) y = 35, find x when constant of variation = 7.
Answer:
Since x and y vary inversely, xy = k.
(i) k = 3 × 8 = 24. When x = 4: 4y = 24, so y = 6.
(ii) k = 5 × 15 = 75. When y = 12: 12x = 75, so x = \( \frac{75}{12} \) = \( \frac{25}{4} \) (= 6.25).
(iii) 30y = 900, so y = 30.
(iv) 35x = 7, so x = \( \frac{7}{35} \) = \( \frac{1}{5} \).
In simple words: Multiply the known pair to get the constant, then divide the constant by the new value.
Teacher's Note:
1. The constant of variation in inverse variation is the product xy.
2. Always simplify fractions such as 75/12 = 25/4.
Section E: Case Study Based Question (4 Marks)
| x (metres) | 2 | 4 | 5 | 10 | 13 |
| y (Rs) | y₂ | y₃ | 210 | y₄ | y₅ |
(a) Find the value of y₂. (1)
(b) Find the value of y₃. (1)
(c) Find the value of y₄. (1)
(d) Find the value of y₅. (1)
Answer:
Length and cost are in direct proportion, so \( \frac{x}{y} \) = \( \frac{5}{210} \) in every column. Cost of 1 m = 210 ÷ 5 = Rs 42.
(a) y₂ = \( \frac{2 × 210}{5} \) = Rs 84.
(b) y₃ = \( \frac{4 × 210}{5} \) = Rs 168.
(c) y₄ = \( \frac{10 × 210}{5} \) = Rs 420.
(d) y₅ = \( \frac{13 × 210}{5} \) = Rs 546.
In simple words: Cloth costs Rs 42 a metre, so multiply 42 by the number of metres.
Teacher's Note:
1. 10 m is twice 5 m, so its cost is simply twice Rs 210.
2. Check (d): 546 ÷ 13 = 42.
Free study material for Mathematics
CBSE Class 8 Mathematics Assignments for Chapter 13 Direct and Inverse Proportions
Revision Assignment: Chapter 13 Direct and Inverse Proportions (CBSE)
Access structured practice assignments for Chapter 13 Direct and Inverse Proportions designed in alignment with the latest CBSE curriculum for Class 8 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.
Key Advantages of Solving Chapter 13 Direct and Inverse Proportions Assignments
- Exam Alignment: Questions strictly follow contemporary CBSE sample papers and grading schemes.
- Comprehensive Coverage: Features objective drills, case studies, and structured descriptive problems for Chapter 13 Direct and Inverse Proportions.
- Pacing & Precision: Regular problem-solving builds critical calculation speed and test-taking accuracy.
How to Approach Mathematics Chapter 13 Direct and Inverse Proportions Assignments
- Concept Foundation: Review the NCERT book for Class 8 Mathematics thoroughly before diving into assignment tasks.
- Self-Evaluation: Solve exercises independently before inspecting professional answer guides.
- Progress Monitoring: Note down complex formulas or concepts, clearing them up using available online practice aids.
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