Practise 19 exam-style questions on Estimation and approximation for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
19 questions
1–5 marks each
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19 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
2 marks
Work out an estimate for \[ \dfrac{612 \times 4.87}{0.193} \]
Round every number to 1 significant figure, then work out the easier calculation. Dividing by 0.2 is the same as multiplying by 5, so \(3000 \div 0.2 = 15000\). A common mistake is to think dividing by 0.2 makes the answer smaller.
Mark scheme
1 mark
M1 for rounding at least two of the numbers correctly to 1 significant figure, e.g. 600, 5 or 0.2
allow other sensible rounding, e.g. 610; correct answer with no working scores full marks
1 mark
A1 for 15 000
accept 14 000 to 16 000 from sensible rounding (e.g. \(610 \times 5 \div 0.2 = 15250\)); do not accept the exact value 15 442.69... without an estimate shown
Question 2
2 marks
Callum pays £18.95 every week for 52 weeks. He uses a spreadsheet to work out the total he pays. The spreadsheet shows a total of £9854
Use estimation to check whether the spreadsheet total is correct. You must show your working.
Model answer
£18.95 ≈ £20 and 52 ≈ 50 \(50 \times 20 = 1000\) The total should be about £1000, but the spreadsheet shows £9854, which is about 10 times too big. The spreadsheet total is not correct (the correct total is £985.40).
Explanation
Round to easy numbers and multiply: about £20 a week for about 50 weeks is about £1000. Answers from technology can be wrong because of a typing error (here 189.5 instead of 18.95), so an estimate is a quick check of the size of the answer.
Mark scheme
1 mark
M1 for a correct estimate, e.g. \(50 \times 20 = 1000\) or \(52 \times 20 = 1040\)
oe
1 mark
C1 for a statement that the total should be about £1000, so £9854 is not correct
oe, e.g. "it is about 10 times too big"
Question 3
3 marks
Work out an estimate for \[ \dfrac{(6.12 \times 10^{-3})^2 \times 1.97 \times 10^{8}}{\sqrt{0.0398}} \] Give your answer in standard form.
Round each number to 1 significant figure, choosing 0.04 under the square root because it is a square number (\(0.2^2 = 0.04\)). Square both parts of \(6 \times 10^{-3}\): \(6^2 = 36\) and \((10^{-3})^2 = 10^{-6}\). Dividing by 0.2 is the same as multiplying by 5.
Mark scheme
1 mark
M1 for rounding to \(6 \times 10^{-3}\), \(2 \times 10^{8}\) and 0.04
oe; correct answer with no working scores full marks
1 mark
M1 for \((6 \times 10^{-3})^2 = 3.6 \times 10^{-5}\) and \(\sqrt{0.04} = 0.2\), or for a numerator of 7200
oe
1 mark
A1 for \(3.6 \times 10^{4}\)
cao in standard form (36 000 scores M1 M1 A0); the exact value \(3.698... \times 10^{4}\) alone scores 0
Question 4
4 marks
By rounding each number to a suitable value, show that \[ \dfrac{\sqrt{48.7} \times 3.12^2}{0.523} < 150 \] You must explain how your rounding shows that the result is less than 150.
Model answer
\(\sqrt{48.7} < \sqrt{49} = 7\) \(3.12^2 < 3.2^2 = 10.24\) 0.523 > 0.5, so dividing by 0.523 gives a smaller result than dividing by 0.5 So \(\dfrac{\sqrt{48.7} \times 3.12^2}{0.523} < \dfrac{7 \times 10.24}{0.5} = 143.36\) 143.36 < 150, so the expression is less than 150.
Explanation
To prove a value is less than 150, every rounding must make the result bigger: round the numerator up and the denominator down. Then the true value is less than your estimate, and the estimate is less than 150. Rounding 48.7 to 50 or 0.523 to 0.52 also works if you keep track of the directions.
Mark scheme
1 mark
B1 for replacing \(\sqrt{48.7}\) by a larger value, e.g. \(\sqrt{49} = 7\)
must be a value at least as large
1 mark
B1 for replacing \(3.12^2\) by a larger value, e.g. \(3.2^2 = 10.24\) or 10
must be a value at least as large
1 mark
B1 for replacing 0.523 by a smaller value, e.g. 0.5, with the reason that dividing by a smaller number gives a larger result
oe
1 mark
C1 for a complete argument: the expression is less than their overestimate, e.g. \(\dfrac{7 \times 10.24}{0.5} = 143.36\) (or \(\dfrac{7 \times 10}{0.5} = 140\)), which is less than 150
working out the exact value (129.88...) alone scores 0
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