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Direct and Inverse Proportions - Concept and Problems of Direct Proportion

Grade 8ICSE

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

🔑Concepts

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Definition of Direct Proportion: Two quantities xx and yy are said to be in direct proportion if they increase or decrease together in such a way that the ratio of their corresponding values remains constant. Visually, if you imagine a scale, as one side goes up, the other side moves up at the exact same rate.

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The Constant of Proportionality: In a direct proportion, the ratio xy\frac{x}{y} is always a fixed value, denoted by the constant kk. This means x=kyx = ky. If you double the value of xx, the value of yy will also double to keep kk the same.

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Graphical Representation: When direct proportion is plotted on a coordinate plane with xx on the horizontal axis and yy on the vertical axis, the resulting graph is always a straight line that passes through the origin (0,0)(0,0). This straight line signifies that when one variable is zero, the other must also be zero.

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The Ratio Method: If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are two pairs of values in direct proportion, then the relationship is expressed as x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2}. This is the fundamental property used to find an unknown value when three other values are known.

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Identifying Direct Proportion: You can identify this relationship by checking if the 'more-more' or 'less-less' rule applies. For example, more articles purchased will result in more total cost, or less distance traveled will require less fuel. If the ratio remains identical across all data points, it is direct proportion.

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Tabular Consistency: In a data table showing direct proportion, dividing the value in the first row by the corresponding value in the second row for every column will always yield the same quotient. If any column results in a different quotient, the quantities are not in direct proportion.

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Cross-Multiplication Technique: To solve for an unknown in the proportion x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2}, we use cross-multiplication. Visually, this involves drawing an 'X' across the equals sign to multiply the numerator of one fraction by the denominator of the other, resulting in x1×y2=x2×y1x_1 \times y_2 = x_2 \times y_1.

📐Formulae

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

x=kyx = ky

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

x1⋅y2=x2⋅y1x_1 \cdot y_2 = x_2 \cdot y_1

💡Examples

Problem 1:

If the cost of 12 identical registers is ₹720, find the cost of 5 such registers.

Solution:

Step 1: Identify the quantities. Let the number of registers be xx and the cost be yy. Step 2: Set up the known values: x1=12,y1=720,x2=5x_1 = 12, y_1 = 720, x_2 = 5. Step 3: Let the unknown cost be y2y_2. Step 4: Use the direct proportion formula: x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2} Step 5: Substitute the values: 12720=5y2\frac{12}{720} = \frac{5}{y_2} Step 6: Cross-multiply: 12×y2=720×512 \times y_2 = 720 \times 5 Step 7: Solve for y2y_2: y2=720×512=60×5=300y_2 = \frac{720 \times 5}{12} = 60 \times 5 = 300. Final Answer: The cost of 5 registers is ₹300.

Explanation:

This is a direct proportion problem because as the number of registers decreases, the total cost also decreases. We equate the ratios of registers to cost to find the missing value.

Problem 2:

A car travels 150 km in 3 hours at a uniform speed. How far will it travel in 7 hours?

Solution:

Step 1: Let the distance be xx and time be yy. Step 2: Given x1=150x_1 = 150 km, y1=3y_1 = 3 hours, and y2=7y_2 = 7 hours. Step 3: Let the required distance be x2x_2. Step 4: Since speed is uniform, distance is directly proportional to time: x1y1=x2y2\frac{x_1}{y_1} = \frac{x_2}{y_2} Step 5: Substitute values: 1503=x27\frac{150}{3} = \frac{x_2}{7} Step 6: Simplify the first ratio: 50=x2750 = \frac{x_2}{7} Step 7: Solve for x2x_2: x2=50×7=350x_2 = 50 \times 7 = 350. Final Answer: The car will travel 350 km in 7 hours.

Explanation:

Since the speed is constant, the distance covered increases as the time increases. We used the constant ratio of distancetime\frac{distance}{time} to calculate the distance for the new time duration.