Direct And Inverse Proportions Class 8 NCERT

Understanding Direct and Inverse Proportion Relations

To understand proportions, we first analyze how two quantities $x$ and $y$ change relative to each other. In a Direct Proportion, if one variable increases, the other increases in such a way that their ratio remains constant. Mathematically, $x$ and $y$ are in direct proportion if $x / y = k$ (where $k$ is a non-zero constant). If $x_1, y_1$ and $x_2, y_2$ are corresponding initial and final values of these quantities, then we use the ratio equation:
$$\frac{x_1}{y_1} = \frac{x_2}{y_2}$$

In an Inverse Proportion, if one variable increases, the other decreases, and vice versa. For example, if more workers are hired to build a school, the days required to complete the project will decrease. Mathematically, $x$ and $y$ are in inverse proportion if their product remains constant: $x \cdot y = k$. If $x_1, y_1$ and $x_2, y_2$ are corresponding values, we use the product equation:
$$x_1 \cdot y_1 = x_2 \cdot y_2$$
Identifying the nature of variation is the critical first step to solving these questions.

Step-by-Step Method to Solve Proportion Problems

1. Create a Data Table — Identify the two quantities in the problem statement (e.g., $x$ and $y$). Draw a simple 2x2 grid table and fill in the known and unknown values ($x_1, y_1, x_2, y_2$).
2. Determine the Type of Proportion — Analyze the relationship. Ask yourself: 'If $x$ increases, will $y$ increase or decrease?' If $y$ increases, it is Direct. If $y$ decreases, it is Inverse.
3. Select and Set Up the Equation — For Direct Proportion, write $x_1 / y_1 = x_2 / y_2$. For Inverse Proportion, write $x_1 \cdot y_1 = x_2 \cdot y_2$.
4. Calculate and Solve — Substitute the given numerical values into your selected equation and solve for the unknown variable using basic cross-multiplication or division rules. Ensure units are consistent.

Real-Life Solved Examples

• Example 1 (Direct Proportion): A train travels 290 km using 10 litres of fuel. How far will it travel using 25 litres of fuel?
  Step 1: Set up the variables. Let distance be $x$ and fuel be $y$.
  Step 2: Since more fuel means more distance, they are in direct proportion. Formula: $x_1 / y_1 = x_2 / y_2$.
  Step 3: Substitute values: $290 / 10 = x_2 / 25$.
  Step 4: Solve for $x_2$: $x_2 = 29 \times 25 = 725$ km.
  Final Answer: The train travels 725 km.

• Example 2 (Inverse Proportion): If 36 workers can build a wall in 15 days, how many workers are needed to finish the same wall in 10 days?
  Step 1: Let the number of workers be $x$ and the number of days be $y$.
  Step 2: More workers will take fewer days, so they are in inverse proportion. Formula: $x_1 \cdot y_1 = x_2 \cdot y_2$.
  Step 3: Substitute values: $36 \cdot 15 = x_2 \cdot 10$.
  Step 4: Solve for $x_2$: $x_2 = (36 \cdot 15) / 10 = 540 / 10 = 54$ workers.
  Final Answer: 54 workers are required.

Common Mistakes & CBSE Exam Tips

1. Check Your Units: Always ensure that units match before plugging numbers into the formulas. If speed is in km/h, time must be in hours, not minutes!
2. Sanity Check: Before writing down your final answer, apply common sense. If a problem asks how long a faster train takes, your answer should be smaller than the initial time. If it is larger, you likely applied the direct proportion formula instead of inverse proportion.
3. Tabular Representation: Always draw the 2x2 table in your CBSE sheet. It helps prevent substitution errors and gets you step-wise marks!

Practice Questions with Solutions

• Q: A scale of a map is given as 1:20,000,000. If two cities are 5 cm apart on the map, what is the actual distance between them in kilometers?
  A: Step 1: Let the map distance be $x$ (in cm) and actual distance be $y$ (in cm). The proportion is direct.
  Step 2: Formula is $x_1/y_1 = x_2/y_2$.
  Here, $1 / 20,000,000 = 5 / y_2$.
  Step 3: Calculate $y_2 = 5 \times 20,000,000 = 100,000,000$ cm.
  Step 4: Convert cm to km. $100,000,000$ cm = $1,000,000$ m = $1000$ km.
  Final answer: The actual distance is 1000 km.

• Q: 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 total school hours remain the same?
  A: Step 1: Let the number of periods be $x$ and duration of each period be $y$.
  Step 2: Since total time is constant, increasing the number of periods will decrease the duration. This is an inverse proportion ($x_1 \cdot y_1 = x_2 \cdot y_2$).
  Step 3: Substitute values: $8 \times 45 = 9 \times y_2$.
  Step 4: Solve for $y_2 = (8 \times 45) / 9 = 8 \times 5 = 40$ minutes.
  Final answer: Each period would be 40 minutes.

• Q: If 12 sheets of paper weigh 40 grams, how many sheets of the same paper would weigh 2.5 kilograms?
  A: Step 1: Convert weight to the same unit. $2.5$ kg = $2500$ grams.
  Step 2: Let the number of sheets be $x$ and weight be $y$. Since sheets and weight are in direct proportion, $x_1 / y_1 = x_2 / y_2$.
  Step 3: Substitute values: $12 / 40 = x_2 / 2500$.
  Step 4: Solve for $x_2$: $x_2 = (12 \times 2500) / 40 = 30000 / 40 = 750$.
  Final answer: 750 sheets would weigh 2.5 kg.

• Q: A batch of sweets was packed in 24 boxes with 10 sweets in each box. If the same batch is packed using 8 sweets in each box, how many boxes would be filled?
  A: Step 1: Let the number of sweets in a box be $x$ and the number of boxes be $y$.
  Step 2: Since the total number of sweets is constant, fewer sweets per box means more boxes are needed. This is an inverse proportion ($x_1 \cdot y_1 = x_2 \cdot y_2$).
  Step 3: Substitute values: $10 \times 24 = 8 \times y_2$.
  Step 4: Solve for $y_2$: $y_2 = 240 / 8 = 30$ boxes.
  Final answer: 30 boxes would be filled.

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