GCSE Maths Practice: percentages

Question 4 of 10

A phone is on sale with 20% off the normal price. What percentage of the sale price is the original price?

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\( \begin{array}{l} \textbf{A phone is discounted by 20%.} \\ \textbf{What % of the sale price is the original?} \end{array} \)

Choose one option:

Use multipliers: new = 0.8 × original, so original = new ÷ 0.8 = 1.25 × new = 125% of new.

Reverse comparison after a discount

After a 20% reduction, the sale price is 80% of the original. To compare the original to the new, divide by the remaining percentage: \(\dfrac{100\%}{80\%}=1.25=125\%\). This is a classic Higher Tier reverse-percentage comparison: you are not finding the discount, but the original as a percentage of the reduced value.

Method 1: Multiplier algebra

Write \(N=0.8O\). Rearranging, \(\dfrac{O}{N}=\dfrac{1}{0.8}=1.25\Rightarrow O=1.25N\). Converting \(1.25\) to a percentage gives \(125\%\).

Method 2: Proportion bar

Think of a bar for \(O\) split into 100 equal parts. A 20% reduction leaves 80 parts (the sale price \(N\)). How many ‘sale-price units’ fit into \(O\)? \(\dfrac{100}{80}=1.25\) sale-price units; i.e. \(125\%\).

Common traps

  • Answering “80%” (that’s new as a % of original, not the question asked).
  • Adding 20 to 100 then comparing incorrectly (the task is a ratio of \(O\) to \(N\)).
  • Using \(O=N\times1.2\): that would be correct for reversing a 20% increase, not a decrease.

Check with numbers (different from the question)

  • If an item costs £200, 20% off gives £160. Original as % of new: \(200/160=1.25=125\%\).
  • If an item costs £50, 20% off gives £40. Then \(50/40=1.25=125\%\).

Extend your thinking

If the reduction is \(r\%\), the new price is \((1-r/100)O\). Then \(\dfrac{O}{N}=\dfrac{1}{1-r/100}=\dfrac{100}{100-r}\). So the original as a percentage of the new is \(\dfrac{100}{100-r}\times100\%\). For example, \(r=25\%\Rightarrow\) original is \(\dfrac{100}{75}\times100\%=\dfrac{400}{3}\%\approx133.\overline{3}\%\).