Inverse Square Variation: Why a 50% Increase Isn't a 50% Decrease

The question

The brightness, $I$ lux, of the light from a desk lamp on a sheet of paper varies inversely as the square of the distance, $d$ cm, between the lamp and the paper. (In IGCSE wording: $I$ is inversely proportional to $d^2$ .)

When $d = 50$ , $I = 360$ .

(a) Express $I$ in terms of $d$ . [2 marks]

(b) Find the value of $d$ when $I = 250$ . [2 marks]

(c) The lamp is moved so that the distance $d$ increases by 50%. Find the percentage change in the brightness $I$ . [3 marks]

Cambridge IGCSE note: direct and inverse proportion is Extended content in 0580.

The common wrong answer

Parts (a) and (b) usually go well. Part (c) is where most students lose marks. The two answers I see most often are:

” $d$ goes up by 50%, so $I$ goes down by 50%.”

I_{\text{new}} = \frac{I}{1.5} = 0.667I \quad\Rightarrow\quad \text{a decrease of } 33.3\%

The first treats “inverse” as “the opposite percentage”. The second divides by the multiplier 1.5 but forgets the square. Both ignore the actual relationship from part (a). Cambridge examiner reports describe exactly this: when asked how one quantity changes after the other is multiplied, many candidates guessed “it halves” or “it doubles”, or forgot the square or square root, instead of using the equation.

The correct working

(a) Write the variation as an equation with a constant $k$ :

I = \frac{k}{d^2}

Substitute $d = 50$ and $I = 360$ :

360 = \frac{k}{50^2} \quad\Rightarrow\quad k = 360 \times 2500 = 900\,000

So $I = \dfrac{900\,000}{d^2}$ .

(b) Substitute $I = 250$ :

250 = \frac{900\,000}{d^2} \quad\Rightarrow\quad d^2 = \frac{900\,000}{250} = 3600

d = \sqrt{3600} = 60

So $d = 60$ (a distance can’t be negative, so we take the positive root).

(c) An increase of 50% means the new distance is 1.5 times the old one. Replace $d$ with $1.5d$ in the equation:

I_{\text{new}} = \frac{k}{(1.5d)^2} = \frac{k}{2.25d^2} = \frac{1}{2.25} \times \frac{k}{d^2}

I_{\text{new}} = \frac{4}{9} \times I_{\text{old}} \approx 0.444 \times I_{\text{old}}

The brightness keeps $\frac{4}{9}$ of its old value, so it loses $\frac{5}{9}$ of it:

\text{percentage change} = \left(\frac{4}{9} - 1\right) \times 100\% = -55.6\%

So the brightness decreases by 55.6% (3 s.f.).

Check with real numbers from part (b): at $d = 60$ , $I = 250$ . Increase $d$ by 50% to 90:

I = \frac{900\,000}{90^2} = \frac{900\,000}{8100} = 111.1\ldots

\frac{250 - 111.1\ldots}{250} \times 100\% = 55.6\%

Same answer, so the multiplier method works.

Why it works

Percentages describe multipliers, and variation equations tell you what happens to multipliers. “Increase by 50%” means “multiply by 1.5”. Because $I$ varies inversely as $d^2$ , multiplying $d$ by 1.5 multiplies $d^2$ by $1.5^2 = 2.25$ , and so divides $I$ by 2.25.

Percentages don’t flip or cancel neatly. A 50% rise followed by a 50% fall doesn’t take you back to where you started, so there is no reason a 50% rise in $d$ should give a 50% fall in $I$ . The only safe route is through the multiplier.

How to spot it next time

Turn every percentage change into a multiplier first: up 50% is $\times 1.5$ , down 20% is $\times 0.8$ .
Put the multiplier into the variation equation, including the power: here the change in $I$ is $\frac{1}{1.5^2}$ , not $\frac{1}{1.5}$ .
If you’re unsure, test it with actual numbers (like $d = 60$ and $d = 90$ above). It takes a minute and catches the mistake every time.