AI SL · 12.2
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Rounding and percentage error
12.2.4
A rectangular vegetable plot measures 12.4 m by 7.8 m, each measured to the nearest 0.1 m. (a) Write down the lower and upper bounds of the length. (b) Find the upper bound of the area. (c) Leila estimates the area by rounding each length to the nearest metre. Taking 12.4 m by 7.8 m as the true dimensions, find the percentage error in her estimate.
Have a go first. Then open the working one line at a time.
12.35 m and 12.45 m
Half of 0.1 m either side of 12.4 m.
12.45 × 7.85 = 97.7325
Upper bound of the length times upper bound of the width: 97.7 m².
Estimate 12 × 8 = 96 m². True 12.4 × 7.8 = 96.72 m²
Round each length to the nearest metre for her estimate.
|96 − 96.72| ÷ 96.72 × 100 = 0.744%
The difference, over the true value, times 100.
Write 0.040862 correct to (a) 2 decimal places and (b) 3 significant figures.
Have a go first. Then open the working one line at a time.
\(0.04\)
Two digits after the point. The next digit is 0, so it stays.
\(0.0409\)
The first significant figure is the 4. Three of them are 4, 0, 8, and the next digit, 6, rounds the 8 up.
\(x = 6.3\) to one decimal place. Write down an inequality for the possible values of \(x\).
Have a go first. Then open the working one line at a time.
6.2 and 6.4 are the next values to one decimal place.
Halfway to each is where the rounding changes.
\(6.25 \le x < 6.35\)
6.25 rounds up to 6.3, so it is included. 6.35 rounds up to 6.4, so it is not.
A sprinter runs 120 m, measured to the nearest 10 m, in 14 s, measured to the nearest second. Find the upper bound of the average speed.
Have a go first. Then open the working one line at a time.
\(115 \le d < 125\) and \(13.5 \le t < 14.5\)
Half of 10 m and half of 1 s either side.
\(\frac{125}{13.5} = 9.259\ldots\)
Speed is distance ÷ time. For the biggest answer, take the biggest distance and the smallest time.
\(9.26\) m/s
To 3 significant figures.
Find the percentage error in using 3.16 as an estimate of \(\sqrt{10}\).
Have a go first. Then open the working one line at a time.
\(\left|\frac{3.16 - \sqrt{10}}{\sqrt{10}}\right| \times 100\)
The true value is \(\sqrt{10}\), from the calculator.
\(0.0720\%\)
Three significant figures: 7, 2, 0.
The side of a square is 8 cm to the nearest centimetre. Find the maximum percentage error in the area if 8 cm is used.
Have a go first. Then open the working one line at a time.
\(7.5 \le s < 8.5\)
Half a centimetre either side of 8.
\(56.25 \le A < 72.25\)
Square each bound. The area used is \(8^2 = 64\).
\(\left|\frac{64 - 56.25}{56.25}\right| \times 100 = 13.8\%\) and \(\left|\frac{64 - 72.25}{72.25}\right| \times 100 = 11.4\%\)
The true area could be at either end.
\(13.8\%\)
The larger of the two.
A rectangular gate is 2.4 m wide and 1.1 m high, each measured to the nearest 0.1 m. A brace is fixed along its diagonal. (a) Find the upper bound of the length of the diagonal, to 3 significant figures. (b) A piece of timber is exactly 2.7 m long. Is it certain to be long enough for the brace?
Have a go first. Then open the working one line at a time.
2.45 m and 1.15 m
Half of 0.1 m more than each measurement. The longest diagonal comes from the longest sides.
\(\sqrt{2.45^2 + 1.15^2} = 2.7065\ldots\)
The diagonal is less than 2.71 m, to 3 significant figures.
2.7065… > 2.7
No. The diagonal could be longer than 2.7 m, so the timber is not certain to be long enough.
A cylindrical candle has radius 3.2 cm and height 11.5 cm, each measured to 1 decimal place. Find the largest possible volume of the candle, to 3 significant figures. Use \(V = \pi r^2 h\).
Have a go first. Then open the working one line at a time.
\(r < 3.25\) and \(h < 11.55\)
Half of 0.1 cm more than each measurement.
\(\pi \times 3.25^2 \times 11.55 = 383.26\ldots\)
Multiplying: biggest with biggest. Put the bounds into the formula, not the measured values.
\(383\) cm³
The measured values give 370 cm³, so rounding could hide about 13 cm³.
A kitchen scale gives readings with a percentage error of 4%. (a) A bag of flour has a true mass of 2.5 kg. Find the least and the greatest reading the scale could give. (b) The scale reads 5.2 kg for a bag of rice, and this reading is too high. Find the true mass of the rice.
Have a go first. Then open the working one line at a time.
4% of 2.5 = 0.1 kg
Percentage error is measured against the true value.
2.5 − 0.1 = 2.4 kg and 2.5 + 0.1 = 2.6 kg
The reading is between 2.4 kg and 2.6 kg.
reading = 1.04 × true mass
Too high by 4% of the true mass, which is not known yet.
5.2 ÷ 1.04 = 5 kg
Check: 4% of 5 is 0.2, and 5 + 0.2 = 5.2.
A school buys 38 calculators at £11.80 each. The bursar estimates the cost by rounding 38 to the nearest ten and £11.80 to the nearest pound. (a) Find the estimate. (b) State whether it is an overestimate or an underestimate. (c) Find the percentage error in the estimate. (d) Explain why an estimate is appropriate here.
Have a go first. Then open the working one line at a time.
40 × 12 = 480
38 rounds to 40 and £11.80 rounds to £12. The estimate is £480.
38 × 11.80 = 448.40
The true cost is £448.40. The estimate is more than this, so it is an overestimate. Both numbers were rounded up.
\(\left|\frac{480 - 448.40}{448.40}\right| \times 100 = 7.047\ldots\)
7.05%, to 3 significant figures.
The bursar only needs a rough figure to plan a budget. A small error does no harm, and an overestimate means enough money is set aside. An estimate is not appropriate where a small error could cause harm, such as the dose of a medicine.