ExpertMedium3 marksStructured
AQA GCSE · Question 10 · Number
384 000 electric cars were sold this year.
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This is 20% more than last year.
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How many were sold last year?
384 000 electric cars were sold this year.
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This is 20% more than last year.
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How many were sold last year?
How to approach this question
1. Identify that this is a reverse percentage problem. You are given the final amount after a percentage increase and need to find the original amount.
2. Determine the multiplier for a 20% increase. A 20% increase means you have 100% + 20% = 120% of the original amount.
3. Convert this percentage to a decimal multiplier (120 / 100 = 1.2).
4. Set up an equation: Original Amount × 1.2 = 384 000.
5. Solve for the Original Amount by dividing 384 000 by the multiplier.
Full Answer
Let the number of cars sold last year be L.
This year's sales are 20% more than last year's. This means this year's sales are 100% + 20% = 120% of last year's sales.
As a decimal multiplier, 120% is 1.20.
So, we can set up the equation:
L × 1.20 = 384 000
To find L, we need to divide this year's sales by the multiplier:
L = 384 000 / 1.20
L = 320 000
**Answer: 320 000**
This problem requires us to find the original value before a percentage increase. This is known as a reverse percentage calculation.
The number of cars sold this year (384,000) is 20% *more* than last year.
This means that 384,000 represents 100% (last year's amount) + 20% (the increase).
So, 384,000 = 120% of last year's sales.
To find 100%, we can first find 1% and then multiply by 100, or use a decimal multiplier.
Using the multiplier method is faster:
120% as a decimal is 1.2.
Let L be the number of cars sold last year.
L × 1.2 = 384,000
L = 384,000 ÷ 1.2
L = 320,000
So, 320,000 cars were sold last year.
Common mistakes
✗ The most common mistake is to calculate 20% of 384,000 and subtract it. This would be finding 80% of the current year's sales, not reversing a 20% increase from the previous year. (384,000 × 0.8 = 307,200, which is incorrect).
✗ Dividing by 0.20 instead of 1.20.
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