Rounding, bounds and percentage error Problem set B Worked solutions

Round each number as asked.

1. (a) 2.71828 to 2 decimal places

   • The third decimal place is 8, so round up.
   • 2.72

2. (b) 0.004718 to 2 significant figures

   • The first two significant figures are 4 and 7 (the zeros before the 4 do not count). The next digit is 1, so round down.
   • 0.0047

3. (c) 72 436 to 3 significant figures

   • The first 3 significant figures are 724 and the next digit is 3, so round down.
   • 72 400

4. (d) 49.97 to 1 decimal place

   • The second decimal place is 7, so round up: 49.9 becomes 50.0.
   • 50.0 (keep the zero to show 1 decimal place)

Find the percentage error, to 3 significant figures.

1. (a) Using 3.14 for \(\pi\)

   • \(\text{percentage error} = \left|\dfrac{\text{estimate} - \text{true}}{\text{true}}\right| \times 100\)
   • \(\left|\dfrac{3.14 - \pi}{\pi}\right| \times 100 = 0.0507\ldots\)
   • 0.0507%

2. (b) An estimate of 85 for a true value of 81.6

   • \(\left|\dfrac{85 - 81.6}{81.6}\right| \times 100 = \dfrac{3.4}{81.6} \times 100 = 4.1666\ldots\)
   • 4.17%

Write an inequality for the possible values of each rounded number.

1. (a) 6.7, to 1 decimal place

   • The rounding unit is 0.1, so go 0.05 either side of 6.7.
   • \(6.65 \le x < 6.75\)

2. (b) 340, to the nearest 10

   • The rounding unit is 10, so go 5 either side of 340.
   • \(335 \le x < 345\)

3. (c) 0.58, to 2 decimal places

   • The rounding unit is 0.01, so go 0.005 either side of 0.58.
   • \(0.575 \le x < 0.585\)

4. (d) 8000, to 2 significant figures

   • 2 significant figures of 8000 means the nearest 100, so go 50 either side.
   • \(7950 \le x < 8050\)

A rectangular garden measures 12.6 m by 7.4 m, each to 1 decimal place. Find (a) the upper bound of its area, (b) the lower bound of its perimeter.

• Each length is within 0.05 m: \(12.55 \le L < 12.65,\ 7.35 \le W < 7.45\)
• (a) Largest area uses both upper bounds: \(12.65 \times 7.45 = 94.2425\) m²
• (b) Smallest perimeter uses both lower bounds: \(2(12.55 + 7.35) = 39.8\) m

A cyclist covers 600 m, to the nearest 10 m, in 58.7 s, to the nearest 0.1 s. Find the lower bound of the average speed.

• Distance: \(595 \le d < 605\). Time: \(58.65 \le t < 58.75\).
• Smallest speed: smallest distance, largest time.
• \(\dfrac{595}{58.75} = 10.127\ldots\), so 10.1 m/s

A builder estimates a wall as 8 m long. It is really 8.6 m. Find the percentage error in the estimate.

• \(\left|\dfrac{8 - 8.6}{8.6}\right| \times 100\)
• \(= \dfrac{0.6}{8.6} \times 100 = 6.976\ldots\)
• 6.98%

The side of a square is 15 cm, to the nearest centimetre. Find the maximum percentage error in the area if 15 cm is used.

• The side is between 14.5 and 15.5, so the area is between \(14.5^2 = 210.25\) and \(15.5^2 = 240.25\).
• Used area: \(15^2 = 225\).
• \(\dfrac{225 - 210.25}{210.25} \times 100 = 7.02\%\) and \(\dfrac{240.25 - 225}{240.25} \times 100 = 6.35\%\)
• The maximum percentage error is 7.02%.