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Proportional Reasoning-1 - Observing Similarity in Change

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Two quantities xx and yy are said to be in direct proportion if they increase (or decrease) together such that the ratio xy\frac{x}{y} remains constant. This constant is denoted by kk.

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In direct proportion, if y1y_1 is the value of yy corresponding to x1x_1, and y2y_2 is the value of yy corresponding to x2x_2, then x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2}.

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Two quantities xx and yy are said to be in inverse proportion if an increase in xx causes a proportional decrease in yy (and vice-versa) such that their product x×yx \times y remains constant.

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In inverse proportion, the relation is expressed as x1y1=x2y2=kx_1 y_1 = x_2 y_2 = k.

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Similarity in change refers to observing whether the change in one variable results in a consistent, predictable change in another variable, forming a linear or reciprocal relationship.

📐Formulae

xy=k (Constant for Direct Proportion)\frac{x}{y} = k \text{ (Constant for Direct Proportion)}

x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2}

x×y=k (Constant for Inverse Proportion)x \times y = k \text{ (Constant for Inverse Proportion)}

x1y1=x2y2x_1 y_1 = x_2 y_2

💡Examples

Problem 1:

If the cost of 8 kg8 \text{ kg} of sugar is ₹240₹ 240, find the cost of 15 kg15 \text{ kg} of sugar.

Solution:

Let the cost of 15 kg15 \text{ kg} sugar be ₹x₹ x. Since weight and cost are in direct proportion, we have: 8240=15x\frac{8}{240} = \frac{15}{x} Cross-multiplying: 8×x=15×2408 \times x = 15 \times 240 x=15×2408x = \frac{15 \times 240}{8} x=15×30x = 15 \times 30 x=450x = 450

Explanation:

As the quantity of sugar increases, the cost also increases. This is a case of direct proportion where x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2}.

Problem 2:

A car takes 2 hours2 \text{ hours} to reach a destination by travelling at the speed of 60 km/h60 \text{ km/h}. How long will it take when the car travels at the speed of 80 km/h80 \text{ km/h}?

Solution:

Let the time taken be xx hours. Speed and time are in inverse proportion. x1y1=x2y2x_1 y_1 = x_2 y_2 60×2=80×x60 \times 2 = 80 \times x 120=80x120 = 80x x=12080x = \frac{120}{80} x=1.5 hoursx = 1.5 \text{ hours}

Explanation:

If speed increases, the time taken to cover the same distance decreases. Therefore, speed and time are inversely proportional.

Problem 3:

Subtract the total cost of 5 items (₹1250₹ 1250) from a budget of ₹5000₹ 5000.

Solution:

5000−12503750\begin{array}{r} 5000 \\ - 1250 \\ \hline 3750 \end{array} The remaining budget is ₹3750₹ 3750.

Explanation:

Vertical subtraction to find the difference between the total budget and the cost of proportional items.