Estimating
Number
- Foundation
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Estimate $48.7 \times 3.1$ by rounding each number to $1$ significant figure.
2 marks
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Estimate $612 \div 19$ by rounding each number to $1$ significant figure.
2 marks
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By rounding each number to $1$ significant figure, estimate $\dfrac{4.87 \times 21.3}{0.53}$.
3 marks
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Estimate the cost of $38$ notebooks at £$4.15$ each by rounding each number to $1$ significant figure.
3 marks
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Show that an estimate of $\dfrac{312 \times 4.8}{0.21}$, using $1$ significant figure, is $7500$.
4 marks
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A car travels $186$ miles in $3.9$ hours. By rounding each number to $1$ significant figure, estimate the average speed in miles per hour.
4 marks
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Use approximations to $1$ significant figure to estimate $\dfrac{7.8^{2} \times 0.46}{0.021}$.
5 marks
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A rectangular field is $47$ m by $28$ m. By rounding each length to $1$ significant figure, estimate the area. Then work out the exact area and the difference between the estimate and the exact area. Explain the direction of the error.
6 marks
Mark scheme
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Answer. $150$
- $50 \times 3$
- $150$
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Answer. $30$
- $600 \div 20$
- $30$
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Answer. $200$
- $\dfrac{5 \times 20}{0.5}$
- $100 \div 0.5 = 200$
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Answer. £$160$
- $40 \times 4$
- £$160$
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Answer. Estimate $= 7500$
- $312$ rounds to $300$, $4.8$ rounds to $5$, $0.21$ rounds to $0.2$.
- $\dfrac{300 \times 5}{0.2} = \dfrac{1500}{0.2}$
- $1500 \div 0.2 = 7500$
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Answer. $50$ mph
- $200 \div 4$
- $50$ mph
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Answer. $1600$
- $7.8$ rounds to $8$, $0.46$ rounds to $0.5$, $0.021$ rounds to $0.02$.
- $\dfrac{8^{2} \times 0.5}{0.02} = \dfrac{32}{0.02}$
- $1600$
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Answer. Estimate $1500$ m$^{2}$, exact $1316$ m$^{2}$, difference $184$ m$^{2}$. The estimate is larger because both lengths were rounded up.
- Estimate: $50 \times 30 = 1500$ m$^{2}$.
- Exact: $47 \times 28 = 1316$ m$^{2}$.
- Difference: $1500 - 1316 = 184$ m$^{2}$.
- Both $47$ and $28$ were rounded up, so the estimate is larger than the true area.