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Non-calculator 2 marks
Estimate the value of \(73.3 \times 8.3\).
Round each number to 1 significant figure.
Show answer
\(560\)Round each number to 1 significant figure:
\(73.3 \approx 70\) and \(8.3 \approx 8\)
\(70 \times 8 = 560\)
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Non-calculator 3 marks
Three warehouses hold these numbers of boxes.
\(2,347\), \(3,180\), \(5,946\)
(a)Work out an estimate of the total, by rounding each number to the nearest 1000.
(b)Is your estimate bigger or smaller than the exact total?
Give a reason for your answer.
Show answer
(a) \(11000\)
(b) Smaller — the exact total is 11,473, so the estimate is out by 473.\(2,347\) rounds to \(2,000\), \(3,180\) to \(3,000\) and \(5,946\) to \(6,000\).
\(2,000 + 3,000 + 6,000 = 11,000\)
The exact total is \(2,347 + 3,180 + 5,946 = 11,473\).
The estimate of \(11,000\) is smaller than that, by \(473\).
- (a) M1each rounded to the nearest 1000, \(2,000\), \(3,000\) and \(6,000\)
- (a) A1estimate \(11,000\)
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Non-calculator 3 marks
Tomas tries to work out \(\frac{ 87.87 + 69.63 }{ 14.07 }\).
He writes down the answer \(92.82\).
By rounding each number to the nearest 10, show that his answer is incorrect.
Show answer
Estimate \(\frac{ 90 + 70 }{10} = 16\); Tomas wrote \(92.82\), which is far from \(16\), so the answer is incorrect.Round each number to the nearest 10:
\(87.87 \approx 90\), \(69.63 \approx 70\) and \(14.07 \approx 10\)
Add inside the bracket first, then divide:
\(\frac{ 90 + 70 }{10} = \frac{ 160 }{10} = 16\)
The estimate is \(16\), but Tomas wrote \(92.82\), which is far too big.
So Tomas's answer is incorrect.
- M1rounding all three numbers: \(90\), \(70\) and \(10\)
- M1sum first then divide: \(160 \div 10 = 16\)
- A1compares the estimate \(16\) with Tomas's \(92.82\) and concludes the answer is incorrect
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Non-calculator 3 marks
Estimate the value of \(\frac{ 76.8 \times 64.8 }{ 5.1 }\).
Round each number to 1 significant figure.
Show answer
\(960\)Round each number to 1 significant figure:
\(76.8 \approx 80\), \(64.8 \approx 60\) and \(5.1 \approx 5\)
\(\frac{ 80 \times 60 }{ 5 } = \frac{ 4800 }{ 5 } = 960\)
- M1rounds to 1 significant figure: \(80\), \(60\), \(5\)
- M1product \(4800\) divided by \(5\)
- A1estimate \(960\)
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Non-calculator 3 marks
Work out an estimate for the value of \(\sqrt{ 28.3 \times 29.5 }\).
Round each number to 1 significant figure.
Show answer
\(30\)Round each number to 1 significant figure:
\(28.3 \approx 30\) and \(29.5 \approx 30\)
\(\sqrt{ 30 \times 30 } = \sqrt{ 900 } = 30\)
- M1rounds to 1 significant figure: \(30\) and \(30\)
- M1square root of \(900\)
- A1estimate \(30\)
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Non-calculator 3 marks
A school buys 54 calculators costing £6.20 each.
(a)Work out an estimate for the total cost. Round each number to 1 significant figure.
(b)Is your estimate an over-estimate or an under-estimate of the true cost? Give a reason for your answer.
Show answer
(a) \(\text{£}300\)
(b) An under-estimateRound each number to 1 significant figure:
\(54 \approx 50\) and \(£6.20 \approx £6\)
\(50 \times £6 = £300\)
54 was rounded down to 50, and £6.20 was rounded down to £6.
The true cost is \(£334.80\), so £300 is an under-estimate.
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Non-calculator 3 marks
A stadium holds 3106 people.
Find an estimate of 41% of 3106.
You must show your working.
Show answer
\(1200\ \text{people}\)Round both numbers to 1 significant figure:
41% becomes 40%, and 3106 becomes 3000.
10% of 3000 is \(300\), so 40% is 4 lots of that:
\(4 \times 300 = 1200\)
- M1rounds to 1 significant figure: 40% and 3000
- M110% of 3000 is \(300\), then times \(4\)
- A1estimate \(1200\)
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Non-calculator 2 marks
A roll of ribbon is 71.19 m long.
It is to be cut into pieces that are each 0.69 m long.
Calculate an estimate of how many pieces can be cut from the roll.
You must show your working.
Show answer
\(100\)Round both lengths to 1 significant figure:
\(71.19\) becomes \(70\), and \(0.69\) becomes \(0.7\).
\(70 \div 0.7 = 100\)
So about 100 pieces.
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Non-calculator 3 marks
Nadia uses a calculator to work out \(63.6 \times 24\).
She writes down the answer \(152640\).
(a)Calculate an estimate of \(63.6 \times 24\) by rounding each number to 1 significant figure.
(b)Explain how your estimate shows that Nadia's answer is wrong.
Show answer
(a) \(1200\)
(b) The estimate is about 1200, but Nadia wrote \(152640\), which is far bigger than that.\(63.6\) rounds to \(60\) and \(24\) rounds to \(20\).
\(60 \times 20 = 1200\)
The answer should be somewhere near \(1200\).
\(152640\) is far bigger than \(1200\), so the decimal point in Nadia's answer is in the wrong place.
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Non-calculator 2 marks
An office buys 45 printers for a total of £1953.00.
Calculate an estimate of the cost of one printer.
You must show your working.
Show answer
\(\text{£}40\)Round both numbers to 1 significant figure:
£1953.00 becomes £2000, and 45 becomes 50.
\(2000 \div 50 = 40\)
So each one costs about £40.