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Direct and Inverse Proportions - Inverse Proportion

Grade 8CBSE

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

🔑Concepts

Definition of Inverse Proportion: 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. Visually, imagine a balance scale where adding weight to one side requires moving the weight closer to the center to maintain equilibrium.

The Constant Product Rule: For any two quantities in inverse proportion, the product of their corresponding values is always equal to a non-zero constant kk. This is written as xy=kxy = k. In a table of values, multiplying the top value by the bottom value in any column will always yield the same result.

Graphical Representation: When values of an inverse proportion are plotted on a graph where the horizontal axis represents xx and the vertical axis represents yy, the points form a smooth, downward-sloping curve called a hyperbola. This curve approaches the axes but never touches them, illustrating that as one variable approaches infinity, the other approaches zero.

The Ratio Relationship: If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are two pairs of values in an inverse proportion, the relationship can be expressed as x1y1=x2y2x_1 y_1 = x_2 y_2. This can also be written as a ratio: x1x2=y2y1\frac{x_1}{x_2} = \frac{y_2}{y_1}, showing that the ratio of xx values is the inverse of the ratio of yy values.

Real-world Application - Speed and Time: For a fixed distance, speed and time are inversely proportional. If you double your speed, the time taken for the journey is halved. Visually, a faster-moving object covers the same distance line in a shorter duration.

Real-world Application - Work and Men: The number of workers and the time taken to complete a task are inversely proportional. More workers mean less time is required. If you visualize a wall being built, more hands on the bricks make the height grow faster, reducing the total days on the calendar.

📐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}

💡Examples

Problem 1:

If 3636 men can finish a piece of work in 2525 days, how many days will 1515 men take to do the same work?

Solution:

Let the number of men be xx and the number of days be yy. Since fewer men will take more days, this is a case of inverse proportion. We are given: x1=36x_1 = 36, y1=25y_1 = 25, x2=15x_2 = 15, and we need to find y2y_2. Using the formula x1y1=x2y2x_1 y_1 = x_2 y_2: 36×25=15×y236 \times 25 = 15 \times y_2 900=15y2900 = 15y_2 y2=90015y_2 = \frac{900}{15} y2=60y_2 = 60

Explanation:

We identified the relationship as inverse because decreasing the number of workers increases the time required. We then equated the products of the two sets of values to solve for the unknown number of days.

Problem 2:

A car travels at a speed of 6060 km/h to reach a destination in 22 hours. How much time will it take when the car travels at a speed of 8080 km/h?

Solution:

Let speed be xx and time be yy. Since distance is constant, speed and time are inversely proportional. Given: x1=60x_1 = 60 km/h, y1=2y_1 = 2 hours, x2=80x_2 = 80 km/h. We need to find y2y_2. Using the inverse proportion relation: x1y1=x2y2x_1 y_1 = x_2 y_2 60×2=80×y260 \times 2 = 80 \times y_2 120=80y2120 = 80y_2 y2=12080=32=1.5y_2 = \frac{120}{80} = \frac{3}{2} = 1.5 hours.

Explanation:

As the speed increases from 6060 km/h to 8080 km/h, the time taken must decrease. By keeping the product of speed and time constant (which represents the total distance), we calculated the new time as 1.51.5 hours or 11 hour 3030 minutes.