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IB Mathematics: Applications and Interpretation SL · 12.2: Finance · 12.2.4
Rounding, bounds and percentage error Problem set A One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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Round each number as asked.
(a) 3.14159 to 2 decimal places
- The 3rd decimal place is 1, which is less than 5, so round down.
- 3.14
(b) 0.006083 to 2 significant figures
- The first two significant figures are 6 and 0 (the zeros before the 6 do not count). The next digit is 8, so round up.
- 0.0061
(c) 45 678 to 3 significant figures
- Keep 3 significant figures: 457. The next digit is 7, so round up.
- 45 700 (457 followed by two zeros)
(d) 19.96 to 1 decimal place
- The digit after the 1st decimal place is 6, so round up: 19.9 becomes 20.0.
- 20.0 (keep the zero)
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Find the percentage error, to 3 significant figures.
(a) Using \(\frac{22}{7}\) for \(\pi\)
- \(\text{percentage error} = \left|\dfrac{\text{approximate} - \text{exact}}{\text{exact}}\right| \times 100\%\)
- \(\left|\dfrac{3.142857 - 3.141593}{3.141593}\right| \times 100\% = 0.04025\ldots\%\)
- 0.0402%
(b) An estimate of 150 for a true value of 147.2
- \(\left|\dfrac{150 - 147.2}{147.2}\right| \times 100\%\)
- \(= 1.9022\ldots\%\)
- 1.90%
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Write an inequality for the possible values of each rounded number.
(a) 8.4, to 1 decimal place
- 1 d.p. means accurate to 0.1, so the error is at most half of that: 0.05.
- \(8.4 - 0.05 \le x < 8.4 + 0.05\)
- \(8.35 \le x < 8.45\)
(b) 250, to the nearest 10
- Nearest 10, so the error is at most 5.
- \(250 - 5 \le x < 250 + 5\)
- \(245 \le x < 255\)
(c) 0.37, to 2 decimal places
- 2 d.p. means accurate to 0.01, so the error is at most 0.005.
- \(0.37 - 0.005 \le x < 0.37 + 0.005\)
- \(0.365 \le x < 0.375\)
(d) 6000, to 2 significant figures
- 6000 to 2 s.f. is accurate to the nearest 100, so the error is at most 50.
- \(6000 - 50 \le x < 6000 + 50\)
- \(5950 \le x < 6050\)
In context
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A rectangle measures 9.3 cm by 4.7 cm, each to 1 decimal place. Find (a) the upper bound of its area, (b) the lower bound of its perimeter.
- \(9.3 \text{ to 1 d.p.: } 9.25 \le l < 9.35;\quad 4.7:\ 4.65 \le w < 4.75\)
- \(\text{(a) Upper bound of area} = 9.35 \times 4.75 = 44.4125\ \text{cm}^2\)
- \(\text{(b) Lower bound of perimeter} = 2(9.25 + 4.65) = 27.8\ \text{cm}\)
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A runner covers 200 m, to the nearest 10 m, in 24.6 s, to the nearest 0.1 s. Find the lower bound of the average speed.
- \(200 \text{ to the nearest 10: } 195 \le d < 205\)
- \(24.6 \text{ to the nearest 0.1: } 24.55 \le t < 24.65\)
- \(\text{Lowest speed: smallest distance, largest time}\)
- \(\dfrac{195}{24.65} = 7.9108\ldots \approx 7.91\ \text{m/s}\)
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A jogger estimates a route as 5 km. It is really 5.3 km. Find the percentage error in the estimate.
- \(\left|\dfrac{5 - 5.3}{5.3}\right| \times 100\%\)
- \(= 5.6604\ldots\%\)
- 5.66%
Stretch
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The side of a square is 12 cm, to the nearest centimetre. Find the maximum percentage error in the area if 12 cm is used.
- \(11.5 \le \text{side} < 12.5 \Rightarrow 11.5^2 \le \text{area} < 12.5^2\)
- \(132.25 \le \text{area} < 156.25\quad (\text{used } 12^2 = 144)\)
- \(\left|\dfrac{144 - 132.25}{132.25}\right| \times 100\% = 8.8847\ldots\%\)
- \(\left|\dfrac{144 - 156.25}{156.25}\right| \times 100\% = 7.8400\ldots\%\)
- Maximum percentage error: 8.88%
jmaths.xyz · AI SL · 12.2.4 Rounding and percentage error