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Number - Direct and Inverse Variation

Grade 8IB

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

🔑Concepts

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Direct Variation: Two variables xx and yy are in direct variation if their ratio remains constant. As one variable increases, the other increases at a constant rate. This is written as y∝xy \propto x.

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The Constant of Proportionality (kk): In any variation, kk is the non-zero constant that relates the two variables. For direct variation, y=kxy = kx, and for inverse variation, y=kxy = \frac{k}{x}.

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Inverse Variation: Two variables xx and yy are in inverse variation if their product remains constant. As one variable increases, the other decreases. This is written as y∝1xy \propto \frac{1}{x}.

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Graphical Representation: The graph of a direct variation is always a straight line passing through the origin (0,0)(0, 0). The graph of an inverse variation is a curve called a hyperbola that approaches but never touches the xx and yy axes.

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Word Problems: Direct variation often involves units and costs (e.g., more items cost more), while inverse variation often involves speed and time or sharing work (e.g., more workers take less time).

📐Formulae

y=kxy = kx

k=yxk = \frac{y}{x}

y=kxy = \frac{k}{x}

k=x⋅yk = x \cdot y

y1x1=y2x2 (for direct variation)\frac{y_1}{x_1} = \frac{y_2}{x_2} \text{ (for direct variation)}

x1y1=x2y2 (for inverse variation)x_1 y_1 = x_2 y_2 \text{ (for inverse variation)}

💡Examples

Problem 1:

If yy varies directly as xx, and y=20y = 20 when x=5x = 5, find the value of yy when x=12x = 12.

Solution:

  1. Find the constant kk: k=yx=205=4k = \frac{y}{x} = \frac{20}{5} = 4 2. Write the equation: y=4xy = 4x 3. Substitute x=12x = 12: y=4×12=48y = 4 \times 12 = 48

Explanation:

Since the variation is direct, we use the ratio of yy to xx to find the constant multiplier kk and then apply it to the new xx value.

Problem 2:

If aa is inversely proportional to bb, and a=10a = 10 when b=4b = 4, find aa when b=8b = 8.

Solution:

  1. Find the constant kk: k=a×b=10×4=40k = a \times b = 10 \times 4 = 40 2. Write the equation: a=40ba = \frac{40}{b} 3. Substitute b=8b = 8: a=408=5a = \frac{40}{8} = 5

Explanation:

In inverse variation, the product of the two variables remains the same. Here, 10×4=4010 \times 4 = 40, so when bb increases to 88, aa must decrease to 55 so that their product remains 4040.

Problem 3:

A car travels at a constant speed. If it covers 150150 km in 33 hours, how far will it travel in 77 hours?

Solution:

This is direct variation because distance (dd) is directly proportional to time (tt).

  1. Find speed (kk): k=dt=1503=50 km/hk = \frac{d}{t} = \frac{150}{3} = 50 \text{ km/h} 2. Calculate distance for 77 hours: d=50×7=350 kmd = 50 \times 7 = 350 \text{ km}

Explanation:

The distance increases as time increases, keeping the speed constant. We calculate the speed first and then multiply by the new time.