NCERT Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1

What is Direct Proportion?

Direct proportion is a mathematical relationship between two quantities where they increase or decrease together in a constant ratio. If we have two variables, say x (independent variable) and y (dependent variable), they are said to be in direct proportion if the ratio of their corresponding values remains constant. Mathematically, we express this as x/y = k, or x1/y1 = x2/y2 = k, where k is a positive constant called the constant of proportion. This means that if x doubles, y also doubles. If x is halved, y is also halved. A real-world example is buying apples: if 1 kg of apples costs ₹100, then 2 kg will cost ₹200. The ratio of weight to cost (1/100 or 2/200) remains constantly 0.01.

Steps to Verify Direct Proportion

1. Identify the Variables — Label the two given quantities as x and y. For example, let x be the number of parts of red pigment and y be the parts of base.
2. Calculate the Ratios — For each pair of given values (x1, y1), (x2, y2), and so on, calculate the individual ratios: x1/y1, x2/y2, etc.
3. Compare the Ratios — Check if all calculated ratios are equal to a single constant value, k.
4. Formulate the Conclusion — If all ratios are equal, then the quantities are in direct proportion. If even one ratio is different, they are not.

Worked Examples from Exercise 13.1

• Example 1: A mixture of paint is prepared by mixing 1 part of red pigment with 8 parts of base. Find the parts of base needed for 4 parts of red pigment. Let x be parts of red pigment and y be parts of base. Since they are in direct proportion, we use the formula x1/y1 = x2/y2. Given x1 = 1, y1 = 8, and x2 = 4. Substitute these values into the ratio equation: 1/8 = 4/y2 By cross-multiplying: y2 = 4 * 8 = 32 parts of base.
• Example 2: A railway station parking charges are: 4 hours = ₹60, 8 hours = ₹100, 12 hours = ₹140. Check if the parking charges are in direct proportion to the parking time. Let time be x and charges be y. We calculate the ratios: Ratio 1: x1/y1 = 4/60 = 1/15 Ratio 2: x2/y2 = 8/100 = 2/25 Ratio 3: x3/y3 = 12/140 = 3/35 Since 1/15 is not equal to 2/25 or 3/35, the ratios are not constant. Therefore, parking charges are NOT in direct proportion to the parking time.

Avoid This Common Proportionality Trap!

Students often make the mistake of assuming that if both variables increase, they must be in direct proportion. This is NOT always true! For variables to be in a direct proportion, they must increase at the exact same rate, meaning their ratio (x/y) must remain perfectly constant. Always calculate at least two or three ratios to verify consistency before writing your final exam response.

Practice Questions with Solutions

• Q: A machine in a soft drink factory fills 840 bottles in 6 hours. How many bottles will it fill in 5 hours? A: Step 1: Identify variables. Let the number of bottles filled be x and time taken in hours be y. Step 2: Since more time means more bottles filled, they are in direct proportion. Therefore, we can write: x1/y1 = x2/y2. Step 3: Substitute the given values: x1 = 840, y1 = 6, y2 = 5. We need to find x2. Step 4: Set up the equation: 840/6 = x2/5. Step 5: Solve for x2: 140 = x2/5 => x2 = 140 * 5 = 700. Final answer: The machine will fill 700 bottles in 5 hours.
• Q: If a photograph of a bacteria enlarged 50,000 times attains a length of 5 cm, what is the actual length of the bacteria? A: Step 1: Let the enlargement factor be x and the length be y (in cm). Step 2: They are in direct proportion because more enlargement leads to a proportional increase in observed length: x1/y1 = x2/y2. Step 3: Given x1 = 50,000, y1 = 5. We want to find the actual length (which corresponds to an enlargement factor of x2 = 1). Step 4: Substitute the values: 50,000/5 = 1/y2. Step 5: Simplify: 10,000 = 1/y2 => y2 = 1/10,000 = 0.0001 cm. Final answer: The actual length of the bacteria is 0.0001 cm (or 10^-4 cm).
• Q: In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship? A: Step 1: Let x be the dimensions of the model (in cm) and y be the dimensions of the actual ship (in m). Since this is a scale model, they are in direct proportion: x1/y1 = x2/y2. Step 2: Given x1 (model mast height) = 9 cm, y1 (actual mast height) = 12 m. For the length, actual ship length y2 = 28 m. We need to find the model length x2. Step 3: Set up the ratio: 9/12 = x2/28. Step 4: Simplify 9/12 to 3/4: 3/4 = x2/28. Step 5: Solve for x2: x2 = (3/4) 28 = 3 7 = 21 cm. Final answer: The length of the model ship is 21 cm.
• Q: Suppose 2 kg of sugar contains 9 10^6 crystals. How many sugar crystals are there in 5 kg of sugar? A: Step 1: Let weight of sugar be x (in kg) and number of crystals be y. Since weight and number of crystals are in direct proportion, we use x1/y1 = x2/y2. Step 2: Given x1 = 2, y1 = 9 10^6, and x2 = 5. We need to find y2. Step 3: Set up the equation: 2/(9 10^6) = 5/y2. Step 4: Cross-multiply: 2 y2 = 5 9 10^6 => 2 y2 = 45 10^6. Step 5: Solve for y2: y2 = (45 10^6)/2 = 22.5 10^6 = 2.25 10^7. Final answer: There are 2.25 10^7 crystals in 5 kg of sugar.

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