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.
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