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Direct and Inverse Proportions - Direct Proportion

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 in such a manner that the ratio of their corresponding values remains constant. For example, if the number of pens purchased increases, the total cost also increases in the same ratio.

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The constant ratio is denoted by k=xyk = \frac{x}{y}, where kk is called the constant of proportionality. This means that for any pair of values, xx is always a fixed multiple of yy.

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A practical way to identify direct proportion is through the cross-multiplication property: if x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2}, then x1y2=x2y1x_1 y_2 = x_2 y_1.

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Visually, if you plot the values of two directly proportional quantities on a graph with xx on the horizontal axis and yy on the vertical axis, the points will always lie on a straight line that passes through the origin (0,0)(0,0). The fact that it passes through (0,0)(0,0) shows that if xx is zero, yy must also be zero.

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In a direct proportion table, you can observe that as you move from one column to the next, the factor by which xx changes is the same factor by which yy changes. If xx is multiplied by 33, yy is also multiplied by 33.

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Direct proportion is commonly found in physics and daily life, such as the relationship between distance and time (at constant speed), or the relationship between the weight of an object and its cost.

📐Formulae

xy=k\frac{x}{y} = k (where kk is a constant)

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

x1y2=x2y1x_1 y_2 = x_2 y_1

💡Examples

Problem 1:

The cost of 77 kg of sugar is Rs280Rs 280. What is the cost of 1212 kg of sugar of the same quality?

Solution:

Let the weight of sugar be xx and the cost be yy. Here, x1=7x_1 = 7 kg and y1=Rs280y_1 = Rs 280. We need to find y2y_2 when x2=12x_2 = 12 kg. Since weight and cost are in direct proportion: x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2} 7280=12y2\frac{7}{280} = \frac{12}{y_2} By cross-multiplication: 7×y2=280×127 \times y_2 = 280 \times 12 y2=280×127y_2 = \frac{280 \times 12}{7} y2=40×12=480y_2 = 40 \times 12 = 480 The cost of 1212 kg of sugar is Rs480Rs 480.

Explanation:

We first identify that as the weight of sugar increases, the cost increases, confirming direct proportion. We then set up the ratio weightcost\frac{weight}{cost} and solve for the missing cost value using cross-multiplication.

Problem 2:

A car travels 432432 km on 4848 litres of petrol. How far would it travel on 2020 litres of petrol?

Solution:

Let the distance traveled be xx and the quantity of petrol be yy. Given: x1=432x_1 = 432 km, y1=48y_1 = 48 litres, and y2=20y_2 = 20 litres. Since distance and petrol consumption are in direct proportion: x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2} 43248=x220\frac{432}{48} = \frac{x_2}{20} Simplify the first ratio: 9=x2209 = \frac{x_2}{20} x2=9×20x_2 = 9 \times 20 x2=180x_2 = 180 The car will travel 180180 km on 2020 litres of petrol.

Explanation:

This problem uses the constant of proportionality approach. We find that the car travels 99 km per litre (the constant kk), then multiply this constant by the new amount of petrol (2020 litres) to find the total distance.