KCSE 2025 Maths P2 Q6 — Direct Variation

KCSE 2025 Form 4 Variation

Published

The Question

“Two quantities y and x are such that y varies directly as the square of (x + 1). Given that y = 200 when x = 4, determine the equation connecting the two quantities.”

1

Write the variation as an equation

Varies directly means the two quantities are proportional, so replace the proportionality with an equals sign and a constant k, called the constant of proportionality. Because y varies directly as the square of (x + 1), the variable part is (x + 1) squared.

y∝(x+1)2⇒y=k(x+1)2y \propto (x+1)^{2} \Rightarrow y = k(x+1)^{2}
2

Substitute the given values

Use the fact that y is 200 when x is 4 to find k. Put both values into the equation, then simplify inside the bracket first before squaring.

200=k(4+1)2200 = k(4+1)^{2}
200=k(5)2=25k200 = k(5)^{2} = 25k
3

Solve for the constant k

Divide both sides by 25 to isolate k. This gives the numerical value of the constant of proportionality.

k=20025=8k = \frac{200}{25} = 8
4

Write the connecting equation

Put the value k = 8 back into the general relationship. This gives the specific equation that connects y and x.

y=8(x+1)2y = 8(x+1)^{2}

Final Result

The constant of proportionality is k = 8, so the equation connecting the two quantities is y = 8(x + 1) squared.

Why this method works

Direct variation means one quantity is a fixed multiple of another, and that fixed multiple is the constant k. Introducing k turns the proportion statement into a solvable equation, and a single known pair of values (y = 200, x = 4) is enough to pin down k. Once k is known, the relationship holds for every pair of x and y, giving the connecting equation.

✓

Substitute x = 4 back into y = 8(x + 1)^2: 8 times (5)^2 = 8 times 25 = 200, which matches the given value of y.