iGCSE Mathematics (0580) :E2.8 Express direct and inverse proportion in algebraic terms and use this form of expression to find unknown quantities.iGCSE Style Questions Paper 2

Question

$y$ is inversely proportional to the square root of $(x – 2)$.
When $x = 4.25$, $y = 12$.

Find $x$ when $y = 3$.

▶️ Answer/Explanation

Solution

Ans: $x = 38$

1. Set up inverse proportion: $y = \frac{k}{\sqrt{x-2}}$

2. Find k using given values: $12 = \frac{k}{\sqrt{4.25-2}} ⇒ k = 12×1.5 = 18$

3. For y=3: $3 = \frac{18}{\sqrt{x-2}}$

4. Solve: $\sqrt{x-2} = 6 ⇒ x-2 = 36 ⇒ x = 38$

Question

$y$ is directly proportional to the square root of $(x-3)$.
When $x = 28$, $y = 20$.

Find $y$ when $x = 39$.

▶️ Answer/Explanation

Solution

First find constant of proportionality: $y = k\sqrt{x-3}$ → $20 = k\sqrt{25}$ → $k = 4$.

Now find y when x=39: $y = 4\sqrt{39-3} = 4×6 = 24$.

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