Class 9 Mathematics Chapter 11: Direct and Inverse Proportions – Important Questions & Answers

Direct and Inverse Proportions holds significant weight in CBSE Class 9 Mathematics because it develops proportional reasoning—a skill essential for higher mathematics, physics, and real-world problem-solving. The CBSE board pattern allocates 8–10 marks to this chapter, distributed across multiple question formats. Examiners test students' ability to (1) identify proportional relationships from data or graphs, (2) set up and solve proportion equations, (3) apply inverse proportion to time-work and speed-distance-time problems, and (4) justify answers with mathematical reasoning. The 2024–25 syllabus emphasizes conceptual clarity over rote memorization: students must distinguish between y = kx (direct) and xy = k (inverse), and explain why a relationship is proportional or not. Time-work problems—where two workers together complete a job—frequently appear as 3-mark or 5-mark questions because they integrate multiple steps and demand logical setup. By mastering these question patterns, students strengthen algebraic thinking and develop confidence in handling unstructured word problems. This chapter also bridges Class 9 and Class 10 algebra, making thorough preparation crucial for sequential learning.

Multiple-choice questions in Class 9 Mathematics test quick recall, formula recognition, and conceptual discrimination. These 1-mark questions often appear in the first section of the question paper and set the tone for deeper problem-solving. **Q1. If x and y are directly proportional and y = 12 when x = 3, find the constant of proportionality (k). A) 4 B) 15 C) 36 D) 1 Answer: A) 4 Explanation: For direct proportion, y = kx. Substituting: 12 = k × 3 ⟹ k = 4. **Q2. If two quantities are in inverse proportion, which equation represents their relationship? A) y = kx B) y = k/x C) y + x = k D) y² = kx Answer: B) y = k/x Explanation: Inverse proportion means y ∝ 1/x, or xy = k (equivalent form). **Q3. A car travels 240 km in 4 hours. If distance and time are directly proportional, how far will it travel in 6 hours? A) 360 km B) 300 km C) 320 km D) 400 km Answer: A) 360 km Explanation: Speed = 240/4 = 60 km/h. Distance in 6 hours = 60 × 6 = 360 km. **Q4. If xy = 20 represents an inverse proportion, what is x when y = 5? A) 4 B) 25 C) 100 D) 15 Answer: A) 4 Explanation: xy = 20 ⟹ x × 5 = 20 ⟹ x = 4. **Q5. Which of the following is NOT an example of direct proportion? A) Distance travelled and time (at constant speed) B) Cost and quantity (fixed price per item) C) Speed and time taken (fixed distance) D) Earnings and hours worked (fixed wage) Answer: C) Speed and time taken (fixed distance) Explanation: Speed × Time = Distance (constant). As speed increases, time decreases—this is inverse proportion.

Two-mark questions test application and require justification or brief working. They often appear in the CBSE paper as follow-ups to 1-mark MCQs or standalone concept checks. **Q1. A photocopier machine prints 25 pages in 5 seconds. How long will it take to print 100 pages? Is this an example of direct or inverse proportion? Justify your answer.** Solution: Rate = 25/5 = 5 pages per second. Time for 100 pages = 100/5 = 20 seconds. This is direct proportion because as the number of pages increases, time increases proportionally (T = P/5, where T is time and P is pages). The constant ratio (pages per second) remains 5. **Q2. The speed of a vehicle is inversely proportional to the time taken to cover a fixed distance of 480 km. If speed is 60 km/h, find the time taken. Then, find the speed if time is 8 hours.** Solution: Given: Distance = 480 km (constant), Speed × Time = constant. S × T = 480 When S = 60 km/h: 60 × T = 480 ⟹ T = 8 hours. When T = 8 hours: S × 8 = 480 ⟹ S = 60 km/h. (Alternatively, when T = 8: S = 480/8 = 60 km/h.) **Q3. Two quantities x and y satisfy the relation xy = 72. Copy and complete the table: | x | 8 | 9 | 12 | 18 | | y | ? | 8 | ? | 4 | Identify the relationship and find missing values.** Solution: xy = 72 (inverse proportion). When x = 8: y = 72/8 = 9. When x = 12: y = 72/12 = 6. Completed table: y = 9, 8, 6, 4 for x = 8, 9, 12, 18 respectively. Relationship: x and y are inversely proportional with constant 72. **Q4. If 3 workers can complete a task in 12 days, how many days will 4 workers take to complete the same task? Assume all workers work at the same rate.** Solution: Total work = 3 × 12 = 36 worker-days. With 4 workers: Number of days = 36/4 = 9 days. This is inverse proportion: as the number of workers increases, days decrease (Workers × Days = constant). **Q5. Priya buys 5 notebooks for ₹75. Write an equation for the cost (C) as a function of notebooks (n). Is this direct or inverse proportion? How much will 12 notebooks cost?** Solution: Cost per notebook = 75/5 = ₹15. Equation: C = 15n (direct proportion). For 12 notebooks: C = 15 × 12 = ₹180. As the number of notebooks increases, cost increases proportionally (constant ratio = ₹15 per notebook).

Three-mark questions demand multi-step reasoning, application of formulas, and clear explanation. These are backbone questions in board exams and test depth of understanding. **Q1. A printing press can print 400 pages in 20 minutes. If the press works for 1 hour continuously, how many pages will it print? If the press prints 1600 pages, how long will it take?** Solution: Step 1: Find the rate (pages per minute). Rate = 400/20 = 20 pages per minute. Step 2: Pages printed in 1 hour (60 minutes). Pages = 20 × 60 = 1200 pages. Step 3: Time to print 1600 pages. Using Rate × Time = Pages: 20 × T = 1600 T = 1600/20 = 80 minutes = 1 hour 20 minutes. Answer: 1200 pages in 1 hour; 80 minutes (or 1 h 20 min) to print 1600 pages. **Q2. Ram and Shyam together can complete a job in 6 days. If Ram alone takes 10 days to complete the job, how many days will Shyam take working alone?** Solution: Step 1: Express work rates as fractions of the job per day. Ram's rate = 1/10 of the job per day. Combined rate = 1/6 of the job per day. Step 2: Find Shyam's rate. Shyam's rate = Combined rate − Ram's rate Shyam's rate = 1/6 − 1/10 Step 3: Find common denominator and subtract. 1/6 − 1/10 = 5/30 − 3/30 = 2/30 = 1/15 Step 4: If Shyam's rate is 1/15, he completes the job in 15 days. Answer: Shyam will take 15 days to complete the job alone. **Q3. The following table shows the relationship between the price of mangoes (per kg) and quantity bought (in kg) for a fixed budget of ₹300: | Price per kg (₹) | 10 | 15 | 20 | 30 | | Quantity (kg) | 30 | 20 | 15 | 10 | Verify that price and quantity are inversely proportional. Write the equation and find the price when quantity is 25 kg.** Solution: Step 1: Check if price × quantity = constant. 10 × 30 = 300 15 × 20 = 300 20 × 15 = 300 30 × 10 = 300 All products equal 300, confirming inverse proportion. Step 2: Write the equation. Price × Quantity = 300 P × Q = 300 or P = 300/Q Step 3: Find price when Q = 25 kg. P = 300/25 = ₹12 per kg. Answer: Price and quantity are inversely proportional (P × Q = 300). When quantity is 25 kg, price is ₹12 per kg. **Q4. A contractor estimates that 8 labourers working 6 hours daily can complete a construction project in 15 days. How many labourers working 8 hours daily are needed to complete the same project in 10 days?** Solution: Step 1: Calculate total work in labour-hours. Total work = Number of labourers × Hours per day × Number of days Total work = 8 × 6 × 15 = 720 labour-hours. Step 2: For the new scenario, express the equation. Let x = number of labourers needed. x × 8 × 10 = 720 Step 3: Solve for x. 80x = 720 x = 720/80 = 9 labourers. Answer: 9 labourers working 8 hours daily are needed to complete the project in 10 days.

Five-mark questions are comprehensive, combining multiple concepts, and require detailed justification. These are highest-weight questions and demand structured reasoning. **Q1. A farmer has enough hay to feed 20 cows for 30 days. If the farmer purchases 10 additional cows, how many days will the hay last? Explain your reasoning and identify the type of proportion involved. If the farmer wants the hay to last 50 days, how many cows should he have? Show all working.** Solution: Step 1: Calculate total hay in 'cow-days' (a measure of hay quantity). Total hay = Number of cows × Number of days Total hay = 20 × 30 = 600 cow-days. Step 2: Find days hay will last with 30 cows. With 30 cows: 30 × Days = 600 Days = 600/30 = 20 days. Identification: Number of cows and duration are inversely proportional because the total hay (constant) is the product of cows and days. As cows increase, days decrease. Step 3: Find number of cows if hay lasts 50 days. Number of cows × 50 = 600 Number of cows = 600/50 = 12 cows. Answer: With 30 cows, hay lasts 20 days. This is inverse proportion (Cows × Days = 600). For hay to last 50 days, the farmer should have 12 cows. **Q2. A printing company charges a fee based on the number of pages printed and the quality of paper. The cost is directly proportional to the number of pages. For 500 pages on standard paper, the cost is ₹1,500. (a) Write the equation relating cost to pages. (b) Calculate the cost for 1,200 pages. (c) If the budget is ₹3,000, how many pages can be printed? (d) Create a table for costs at 100, 200, 300, 400, 500 pages and verify the direct proportion constant.** Solution: Step 1(a): Find the constant of proportionality (k). Cost = k × Pages 1,500 = k × 500 k = 1,500/500 = ₹3 per page. Equation: C = 3P (where C is cost and P is pages). Step 2(b): Calculate cost for 1,200 pages. C = 3 × 1,200 = ₹3,600. Step 3(c): Find pages for ₹3,000 budget. 3,000 = 3 × P P = 3,000/3 = 1,000 pages. Step 4(d): Create and verify table. | Pages (P) | 100 | 200 | 300 | 400 | 500 | | Cost (₹) | 300 | 600 | 900 | 1,200 | 1,500 | Verification of constant k: 300/100 = 3, 600/200 = 3, 900/300 = 3, 1,200/400 = 3, 1,500/500 = 3. All ratios equal 3, confirming direct proportion with k = 3. Answer: (a) C = 3P. (b) Cost for 1,200 pages is ₹3,600. (c) 1,000 pages can be printed with ₹3,000. (d) Table shows consistent constant k = 3 (cost per page). **Q3. Two pipes A and B fill a tank. Pipe A alone fills the tank in 6 hours, and pipe B alone fills it in 9 hours. (a) If both pipes work together, how long will they take to fill the tank? (b) If both pipes are open for 2 hours and then pipe B is closed, how much time will pipe A take to complete filling? (c) If a drain pipe C empties the full tank in 12 hours, and all three are open together, how long will it take to fill the tank?** Solution: Step 1(a): Find combined filling rate. Rate of A = 1/6 per hour. Rate of B = 1/9 per hour. Combined rate = 1/6 + 1/9. Finding common denominator (LCM of 6 and 9 is 18): 1/6 + 1/9 = 3/18 + 2/18 = 5/18 per hour. Time to fill = 1 ÷ (5/18) = 18/5 = 3.6 hours = 3 hours 36 minutes. Step 2(b): Tank filled in 2 hours by both pipes. Work done = (5/18) × 2 = 10/18 = 5/9 of tank. Remaining work = 1 − 5/9 = 4/9 of tank. Time for pipe A to fill remaining part: (1/6) × Time = 4/9 Time = (4/9) ÷ (1/6) = (4/9) × 6 = 24/9 = 8/3 = 2⅔ hours = 2 hours 40 minutes. Step 3(c): All three pipes together. Rate of C (draining) = −1/12 per hour. Net rate = 1/6 + 1/9 − 1/12. Finding common denominator (LCM of 6, 9, 12 is 36): 1/6 = 6/36, 1/9 = 4/36, 1/12 = 3/36. Net rate = 6/36 + 4/36 − 3/36 = 7/36 per hour. Time to fill = 1 ÷ (7/36) = 36/7 ≈ 5.14 hours = 5 hours 8.6 minutes. Answer: (a) Both pipes together: 3.6 hours (or 3 h 36 min). (b) Pipe A needs 2 h 40 min to complete after both ran for 2 hours. (c) All three open: 36/7 hours ≈ 5 hours 9 minutes.

**Case Study: School Canteen Supply Chain** A school canteen supplies lunch to students. The cost per meal (C) is inversely proportional to the number of meals prepared daily (M) due to bulk purchasing and fixed overhead costs. On a day when 200 meals are prepared, the cost per meal is ₹50. **(a) Write an equation relating cost per meal to number of meals. Identify the constant.** Solution: Since C is inversely proportional to M: C = k/M Substituting C = 50 and M = 200: 50 = k/200 k = 50 × 200 = 10,000. Equation: C = 10,000/M. **(b) Calculate the cost per meal if 250 meals are prepared on a particular day.** Solution: C = 10,000/250 = ₹40 per meal. **(c) The canteen budget is fixed at ₹8,000 per day. If the canteen operates on days when at least 160 meals are prepared, will it stay within budget on a day when 160 meals are prepared?** Solution: Cost per meal when M = 160: C = 10,000/160 = ₹62.50 per meal. Total cost = 62.50 × 160 = ₹10,000. Budget = ₹8,000. Total cost (₹10,000) exceeds budget (₹8,000), so NO, the canteen will NOT stay within budget. **(d) What is the minimum number of meals the canteen must prepare to keep the total cost within ₹8,000? (HOTS: Requires understanding of constraints and inverse relationships)** Solution: Let total cost = C per meal × number of meals = (10,000/M) × M = 10,000. Wait—this simplifies to constant 10,000 (total cost is always the product k = 10,000, not variable). Actually, re-read: 'cost per meal' is inversely proportional. Total daily cost = (cost per meal) × (number of meals). Total cost = (10,000/M) × M = 10,000 (constant, always ₹10,000). This means the canteen spends ₹10,000 daily regardless of portion size—the inverse relationship balances out. To keep within ₹8,000 budget, the school's initial assumption is flawed. However, if overhead costs are fixed and reduce per unit: Let fixed cost = F, variable cost per meal = V. Realistic model: C = (F + V×M)/M = F/M + V. As M increases, C decreases (approaching V). For this textbook problem: The canteen cannot stay within ₹8,000 if the true cost structure yields k = 10,000, as total = 10,000 always. The school must either increase budget to ₹10,000 or renegotiate supplier terms (change k). Answer (d): Based on the inverse model C = 10,000/M, total daily cost is always ₹10,000 (constant). The canteen cannot meet an ₹8,000 budget. The school must increase budget or reduce the constant through bulk supplier discounts.

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**Direct Proportion** Definition: y is directly proportional to x if y = kx, where k is a non-zero constant. Symbol: y ∝ x Example: Distance = Speed × Time (at constant speed, distance ∝ time) **Inverse Proportion** Definition: y is inversely proportional to x if xy = k or y = k/x. Symbol: y ∝ 1/x Example: Speed × Time = Distance (fixed); as speed increases, time decreases inversely. **Time-Work Problems** Concept: Work = Rate × Time. If A does 1/a of work in 1 day, rate = 1/a. Combined rate (A and B together) = 1/a + 1/b. Time to complete = 1 ÷ (combined rate). **Key Properties** 1. In direct proportion, ratio of quantities remains constant: y/x = k (always). 2. In inverse proportion, product of quantities remains constant: xy = k (always). 3. Graphically, direct proportion is a straight line through origin; inverse is a hyperbola. 4. For multiple workers/machines, work is additive: Total rate = Rate₁ + Rate₂ + …