Bradley's Maths

Inverse Proportion - Non-Calculator

GCSE all examination boards and IGCSE Cambridge (0580)

Exam Question

$$y\propto\frac{1}{\sqrt{x}}$$

When $x=9$, $y=7$

Find $y$ when $x=25$

Solution

When dealing with inverse proportion, as $x$ increases, $y$ decreases. This provides a useful check: since $x$ increases from 9 to 25, our final value of $y$ must be less than 7.

$$y\propto\frac{1}{\sqrt{x}}\rightarrow y=\frac{k}{\sqrt{x}}$$

for some unknown value of $k$. Our first task is to find the value of $k$ and we use the initial values of $x$ and $y$ to do that

\begin{align*} y &= \frac{k}{\sqrt{x}} \\ 7 &= \frac{k}{\sqrt{9}} \\ 3 \times 7 &= k \\ k &= 21 \end{align*}

Now that we have the value of our 'Constant of Proportionality', $k$, it is now a simple matter to substitute this and the new value of $x$ into our equation to find the new value of $y$

When $x=25$:

\begin{align*} y&=\frac{21}{\sqrt{25}}\\ y&=\frac{21}{5} \end{align*}

Final answer:

$$y=4.2$$

Check: As $x$ increased, $y$ decreased — so our answer is sensible.

The Head Teacher's Eye: Is the answer sensible

Always check: Is the answer sensible? With inverse proportion as $x$ increases so $y$ decreases and vice-versa. A check takes just seconds and can be the difference between 3 marks and 0 marks - beware!

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Topics covered in this question

  • Inverse proportion
  • Finding the constant of proportionality
  • Manipulating algebraic expressions

Video Explanation