Number: approximation, error intervals and calculators – Year 9 maths lesson to download as PDF

Worked example 1: significant figures

Round 0.0047382 to 3 s.f. and 35 678 to 2 s.f.

0.0047382 to 3 s.f. is 0.00474. 35 678 to 2 s.f. is 36 000 (the zeros keep the place value).

Worked example 2: an estimate

Estimate \(\dfrac{19.8 \times 4.1}{0.52}\).

Round to 1 s.f.: \(\dfrac{20 \times 4}{0.5} = 160\) (dividing by 0.5 is the same as doubling). The exact answer is 156.1 to 4 s.f., so the estimate shows the calculator answer is sensible.

Worked example 3: error intervals

Write error intervals for (a) 7 cm to the nearest cm, (b) 250 g to the nearest 10 g, (c) 3.4 to 1 d.p.

(a) The rounding unit is 1, half is 0.5, so \(6.5 \le x < 7.5\). (b) Half of 10 is 5, so \(245 \le m < 255\). (c) The rounding unit is 0.1, half is 0.05, so \(3.35 \le x < 3.45\).

Worked example 4: area of a rectangle

A rectangle has sides 8 cm and 5 cm, each to the nearest cm. Find the greatest and least possible area.

Greatest: \(8.5 \times 5.5 = 46.75\ \text{cm}^2\). Least: \(7.5 \times 4.5 = 33.75\ \text{cm}^2\).

Worked example 5: fractions on a calculator

Work out \(\dfrac{4.7 + 3.2}{5.1 - 2.9}\) to 3 s.f.

A calculator follows BIDMAS but only does what you type, so use brackets or the fraction key: type (4.7 + 3.2) ÷ (5.1 − 2.9). The answer is 3.5909…, which is 3.59 to 3 s.f. Do not round in the middle of a calculation; keep the full display or use the answer memory and round only at the end.

Worked example 6: money and error intervals

A charity says it raised £4 300, correct to 2 significant figures. Write the error interval.

The second significant figure is the hundreds digit, so the accuracy is £100. Half of 100 is 50. The error interval is \(4250\le x<4350\). The charity could have raised as little as £4 250 and as much as £4 349.99.

Worked example 7: bounds of a speed

A car travels 150 km, to the nearest 10 km, in 2.0 hours, to 1 decimal place. Find the greatest possible average speed.

Distance: \(145\le d<155\). Time: \(1.95\le t<2.05\). For the greatest speed use the greatest distance and the least time: \(155\div1.95=79.5\) km/h (1 decimal place).

Worked example 8: choosing the accuracy of an answer

A rectangle is 6.3 cm by 4.8 cm, each to 1 decimal place. The area is 30.24 cm\(^2\) on a calculator. How accurately should it be given?

Lower bound: \(6.25\times4.75=29.6875\). Upper bound: \(6.35\times4.85=30.7975\). To 1 significant figure both are 30. So the area is 30 cm\(^2\) to 1 significant figure. Giving 30.24 would suggest an accuracy that the measurements cannot support.

Worked example 9: converting a recurring decimal

Show that \(0.\dot4\dot5=\dfrac5{11}\).

Let \(x=0.454545...\) Then \(100x=45.4545...\) Subtract: \(99x=45\), so \(x=\dfrac{45}{99}=\dfrac5{11}\). Because it can be written as a fraction, it is rational.

Worked example 10: an estimate catches a mistake

Ruby works out \(\dfrac{48.6+21.3}{6.8}\) and writes 51.7. Is she right?

Round each number to 1 significant figure: \(\dfrac{50+20}{7}=\dfrac{70}{7}=10\). Ruby's answer is about five times too big, so something has gone wrong. She probably typed \(48.6+21.3\div6.8\), which the calculator works out as \(48.6+3.13\ldots=51.7\) because division is done before addition. The correct calculation is \(69.9\div6.8=10.279\ldots\), which is \(10.3\) to 3 significant figures and agrees with the estimate.