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Direct and Inverse Proportions - Concept and Problems of Inverse Proportion

Grade 8ICSE

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

🔑Concepts

Inverse Proportion refers to the relationship between two quantities xx and yy such that if one quantity increases, the other decreases in the same ratio, keeping their product constant. This is represented visually as a downward sloping curve on a graph (a rectangular hyperbola) that never touches the xx or yy axes.

The constant of variation, denoted by kk, is the fixed product of the two inversely related variables. For any pair of values (x,y)(x, y), the relationship is defined by the equation x×y=kx \times y = k. Visually, if you imagine a rectangle with sides xx and yy, the area of this rectangle remains constant regardless of how the individual sides change.

The inverse relationship can be written as x1yx \propto \frac{1}{y}, which is read as 'xx is inversely proportional to yy'. This means that xx is directly proportional to the reciprocal of yy.

In a tabular representation of inverse proportion, as you move from left to right, the values in the first row (xx) might increase while the values in the second row (yy) decrease, but multiplying the top value by the bottom value in any column will always yield the same result.

A common real-world application is the relationship between Speed and Time for a fixed distance. If you double your speed, the time taken is halved. On a speed-time graph, this is shown as a curve where the product of the coordinates at any point equals the total distance traveled.

Another classic example is Men and Days in work-related problems. If more workers are employed for a task, the number of days required to complete it decreases proportionally, assuming the rate of work remains constant.

When comparing two states of an inverse relationship, we use the ratio equation x1y1=x2y2x_1 y_1 = x_2 y_2. This can also be visualized as a balance: as the value of xx 'weight' increases, the 'distance' yy must decrease to keep the system in equilibrium.

📐Formulae

x1yx \propto \frac{1}{y}

x×y=kx \times y = k (where kk is a constant)

x1y1=x2y2x_1 y_1 = x_2 y_2

x1x2=y2y1\frac{x_1}{x_2} = \frac{y_2}{y_1}

Constant of Variation (k)=xy\text{Constant of Variation } (k) = x \cdot y

💡Examples

Problem 1:

If 35 men can finish a piece of work in 8 days, in how many days can 20 men finish the same work?

Solution:

Step 1: Identify the variables. Let the number of men be xx and the number of days be yy. Given x1=35x_1 = 35, y1=8y_1 = 8, and x2=20x_2 = 20. We need to find y2y_2. Step 2: Since more men will take fewer days, it is a case of inverse proportion. Use the formula x1y1=x2y2x_1 y_1 = x_2 y_2. Step 3: Substitute the values: 35×8=20×y235 \times 8 = 20 \times y_2. Step 4: Solve for y2y_2: 280=20y2    y2=28020=14280 = 20y_2 \implies y_2 = \frac{280}{20} = 14.

Explanation:

This problem uses the inverse relationship between labor and time. As the number of workers decreases from 35 to 20, the time required must increase. By maintaining the product of men and days constant (280280 man-days), we find the new duration.

Problem 2:

A car travels at a speed of 60 km/h60 \text{ km/h} and reaches its destination in 44 hours. If the speed is increased to 80 km/h80 \text{ km/h}, how much time will it take to cover the same distance?

Solution:

Step 1: Let speed be ss and time be tt. s1=60s_1 = 60, t1=4t_1 = 4, and s2=80s_2 = 80. Find t2t_2. Step 2: Speed and time are inversely proportional for a fixed distance (s1t1=s2t2s_1 t_1 = s_2 t_2). Step 3: 60×4=80×t260 \times 4 = 80 \times t_2. Step 4: 240=80×t2240 = 80 \times t_2. Step 5: t2=24080=3 hourst_2 = \frac{240}{80} = 3 \text{ hours}.

Explanation:

The distance (the product of speed and time) remains constant at 240 km240 \text{ km}. Since the speed increased, the time taken must decrease. Using the constant product rule for inverse proportion, we calculate the new time to be 3 hours.