Direct and Inverse Proportions Ex 13.2 for CBSE Class 8 Maths

Understanding Direct and Inverse Proportions

Proportion is a fundamental concept in mathematics that describes how two quantities are related. When two quantities are in direct proportion, it means that if one quantity increases, the other quantity also increases in the same ratio, and if one quantity decreases, the other quantity decreases proportionally. Imagine buying pens: if one pen costs ₹10, then two pens will cost ₹20, three pens will cost ₹30, and so on. The ratio of cost to the number of pens (10/1, 20/2, 30/3) remains constant, which is 10. Mathematically, if quantities 'x' and 'y' are directly proportional, we write x ∝ y, which means x = ky or x/y = k, where 'k' is the constant of proportionality.

On the other hand, inverse proportion describes a relationship where if one quantity increases, the other quantity decreases, and vice-versa, such that their product remains constant. Consider the speed of a car and the time taken to cover a fixed distance: if you drive faster (increase speed), you will take less time to reach your destination (decrease time). If quantities 'x' and 'y' are inversely proportional, we write x ∝ 1/y, which means x = k/y or xy = k, where 'k' is the constant of proportionality. Recognizing which type of proportion applies is the first crucial step in solving these problems!

How to Solve Problems with Direct Proportion

1. Step 1: Identify Direct Proportion — First, confirm that the two quantities increase or decrease together. For example, if 'more' of one means 'more' of the other, it's direct proportion. Let the two quantities be x and y. If x₁ corresponds to y₁ and x₂ corresponds to y₂, then x₁/y₁ = x₂/y₂.
2. Step 2: Set up the Proportion — Write down the known values and the unknown value. For direct proportion, the ratio of the quantities should be constant. So, x₁/y₁ = x₂/y₂ (where x₁ and y₁ are initial values, and x₂ and y₂ are final values, with one of x₂ or y₂ unknown).
3. Step 3: Solve for the Unknown — Use cross-multiplication to find the unknown value. Example: If 5 kg of sugar costs ₹200, what will 8 kg of sugar cost? (Let cost be 'y'). Initial: x₁ = 5 kg, y₁ = ₹200 Final: x₂ = 8 kg, y₂ = ? Setup: 5/200 = 8/y₂ Cross-multiply: 5 y₂ = 200 8 5y₂ = 1600 y₂ = 1600 / 5 y₂ = ₹320. So, 8 kg of sugar will cost ₹320.

How to Solve Problems with Inverse Proportion

1. Step 1: Identify Inverse Proportion — Confirm that if one quantity increases, the other decreases proportionally. If 'more' of one means 'less' of the other, it's inverse proportion. Let the two quantities be x and y. If x₁ corresponds to y₁ and x₂ corresponds to y₂, then x₁y₁ = x₂y₂.
2. Step 2: Set up the Product Relationship — For inverse proportion, the product of the quantities remains constant. So, x₁y₁ = x₂y₂ (where x₁ and y₁ are initial values, and x₂ and y₂ are final values, with one of x₂ or y₂ unknown).
3. Step 3: Solve for the Unknown — Divide to find the unknown value. Example: If 3 workers can paint a house in 10 days, how many days will 5 workers take? (Let days be 'y'). Initial: x₁ = 3 workers, y₁ = 10 days Final: x₂ = 5 workers, y₂ = ? Setup: 3 10 = 5 y₂ 30 = 5y₂ y₂ = 30 / 5 y₂ = 6 days. So, 5 workers will take 6 days to paint the house.

Key Differences and Practical Applications

• Direct Proportion: When one quantity increases, the other increases. When one decreases, the other decreases. The ratio x/y = k (constant). Examples: Distance travelled and petrol consumed; Number of items and total cost; Amount of work done and wages earned.
• Inverse Proportion: When one quantity increases, the other decreases. When one decreases, the other increases. The product xy = k (constant). Examples: Speed of a vehicle and time taken to cover a fixed distance; Number of workers and time taken to complete a job; Density of a substance and its volume for a fixed mass.
• Spotting the 'Constant': For direct proportion, look for a constant ratio (e.g., price per item). For inverse proportion, look for a constant product (e.g., total work units).

Exam Tip: Avoiding Common Mistakes

One of the most common mistakes students make is confusing direct and inverse proportion. Always start by asking yourself: 'If I increase one quantity, what happens to the other?' If it also increases, it's direct. If it decreases, it's inverse. Remember the fundamental relationships: for direct proportion, x/y = k, and for inverse proportion, xy = k. Clearly identify which quantities are x and y, and write down the initial and final states. Practice setting up the equations carefully. Also, ensure your units are consistent throughout the problem. Don't rush into calculations before correctly identifying the type of proportion!

Practice Questions with Solutions

• Q: A car travels 180 km in 3 hours. How far will it travel in 5 hours, assuming constant speed? A: Step 1: Identify the relationship. More time means more distance covered, so it's a direct proportion. Step 2: Set up the proportion. Let distance be d and time be t. d₁/t₁ = d₂/t₂. 180 km / 3 hours = d₂ / 5 hours Step 3: Solve for d₂. 60 = d₂ / 5 d₂ = 60 * 5 d₂ = 300 km Final answer: The car will travel 300 km in 5 hours.
• Q: 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes are used (assuming all pipes work at the same rate)? A: Step 1: Identify the relationship. Fewer pipes mean more time to fill the tank, so it's an inverse proportion. Convert 1 hour 20 minutes to minutes: 60 + 20 = 80 minutes. Step 2: Set up the inverse proportion. Let number of pipes be P and time be T. P₁T₁ = P₂T₂. 6 pipes 80 minutes = 5 pipes T₂ Step 3: Solve for T₂. 480 = 5 * T₂ T₂ = 480 / 5 T₂ = 96 minutes Final answer: It will take 96 minutes (or 1 hour 36 minutes) if only 5 pipes are used.
• Q: If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours? A: Step 1: Identify the relationship. To complete the work in less time, more workers will be needed, so it's an inverse proportion. Step 2: Set up the inverse proportion. Let number of workers be W and time be H. W₁H₁ = W₂H₂. 15 workers 48 hours = W₂ 30 hours Step 3: Solve for W₂. 720 = 30 * W₂ W₂ = 720 / 30 W₂ = 24 workers Final answer: 24 workers will be required to build the wall in 30 hours.
• Q: The cost of 12 pencils is ₹60. What would be the cost of 25 such pencils? A: Step 1: Identify the relationship. More pencils mean a higher cost, so it's a direct proportion. Step 2: Set up the proportion. Let number of pencils be N and cost be C. N₁/C₁ = N₂/C₂. 12 pencils / ₹60 = 25 pencils / C₂ Step 3: Solve for C₂. 12 C₂ = 60 25 12 * C₂ = 1500 C₂ = 1500 / 12 C₂ = ₹125 Final answer: The cost of 25 pencils would be ₹125.

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