Math In Real World Context

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Percentage increase, then a percentage cut

Question: In 2018 a car cost $45000\usd{}45\,000; in 2025 the same car costs $85000\usd{}85\,000. (a) Find the percentage increase, to 1 d.p. (b) The dealer then cuts the 2025 price by 5%5\% — find the new price.

Step 1. (a) Percentage change always divides by the original value:

Increase=8500045000=$40000\text{Increase} = 85\,000 - 45\,000 = \usd{}40\,000

Step 2.

% increase=4000045000×100%=88.9%  (1 d.p.)\%\text{ increase} = \dfrac{40\,000}{45\,000} \times 100\% = 88.9\% \; (1 \text{ d.p.})

Step 3. (b) A 5%5\% cut leaves 95%95\%:

New price=95100×85000=$80750\text{New price} = \dfrac{95}{100} \times 85\,000 = \usd{}80\,750

Step 4. Note the asymmetry: the cut is 5%5\% of the 2025 price ($4250\usd{}4\,250), which is much more than 5%5\% of the 2018 price — every percentage lives on its own base.

The identical arithmetic runs profit and loss: profit% =sellingcostcost×100%= \dfrac{\text{selling} - \text{cost}}{\text{cost}} \times 100\% — cost price on the bottom.

⚠ Watch out: Dividing by the new value (4000085000=47.1%\frac{40000}{85000} = 47.1\%) is the standard error — the original is always the base. And a 88.9%88.9\% rise followed by a 5%5\% cut is not an 83.9%83.9\% net rise; multipliers (×1.889×0.95\times 1.889 \times 0.95), not additions.

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