Numbers - Limits of Accuracy
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calculator
Simple
Total: 3 marks
A rectangle measures \(8.5\text{ cm}\) by \(10.7\text{ cm}\), both correct to \(1\) decimal place.
Calculate the upper bound of the perimeter of the rectangle.
\[
\underline{\hspace{4cm}}\text{ cm}
\]
Answer
\[
P=2(l+w)
\]
\[
=2(10.75+8.55)
\]
\[
=2(19.3)
\]
\[
=38.6
\]
\[
\boxed{38.6\text{ cm}}
\]
calculator
Simple
Total: 1 mark
A cuboid measures \(10\text{ cm}\) by \(4\text{ cm}\) by \(6\text{ cm}\).
Each side is measured correct to the nearest centimetre.
Complete the inequality for the volume, \(V\), of this cuboid.
\[
\cdots\text{ cm}^3 \le V < \cdots\text{ cm}^3
\]
Answer
\[
9.5 \times 3.5 \times 5.5 = 182.875
\]
\[
10.5 \times 4.5 \times 6.5 = 307.125
\]
\[
\boxed{182.875\text{ cm}^3 \le V < 307.125\text{ cm}^3}
\]
calculator
Simple
Total: 1 mark
A car travels \(14\text{ km}\), correct to the nearest kilometre.
This takes \(12\) minutes, correct to the nearest minute.
Calculate the lower bound of the speed of the car.
Give your answer in kilometres per minute.
\[
\underline{\hspace{4cm}}\text{ km/min}
\]
Answer
\[
\text{Lower distance}=13.5
\]
\[
\text{Upper time}=12.5
\]
\[
\text{Lower speed}
=\frac{13.5}{12.5}
\]
\[
=1.08
\]
\[
\boxed{1.08\text{ km/min}}
\]
calculator
Simple
Total: 2 marks
The length, \(l\) m, of a rope is \(18.7\text{ m}\), correct to the nearest \(10\) centimetres.
Complete this statement about the value of \(l\).
\[
\cdots \le l < \cdots
\]
Answer
\[
10\text{ cm}=0.1\text{ m}
\]
\[
\frac{0.1}{2}=0.05
\]
\[
18.7-0.05=18.65
\]
\[
18.7+0.05=18.75
\]
\[
\boxed{18.65 \le l < 18.75}
\]
calculator
Medium
Total: 3 marks
\[
P=2w+2h
\]
\(w=11\) and \(h=9.5\), both correct to \(2\) significant figures.
Find the lower bound and the upper bound for \(P\).
Lower bound = __________
Upper bound = __________
Answer
\[
10.5 \le w < 11.5
\]
\[
9.45 \le h < 9.55
\]
\[
\text{Lower bound}
=2(10.5)+2(9.45)
\]
\[
=39.9
\]
\[
\text{Upper bound}
=2(11.5)+2(9.55)
\]
\[
=42.1
\]
\[
\boxed{\text{Lower bound }=39.9}
\]
\[
\boxed{\text{Upper bound }=42.1}
\]
calculator
Medium
Total: 1 mark
Virat has \(200\) metres of wire, correct to the nearest metre.
He cuts the wire into \(n\) pieces of length \(3\) metres, correct to the nearest \(20\) centimetres.
Calculate the largest possible value of \(n\).
\[
n=\underline{\hspace{3cm}}
\]
Answer
\[
\text{Upper bound of wire}=200.5
\]
\[
\text{Lower bound of piece length}=2.9
\]
\[
n=\frac{200.5}{2.9}
\]
\[
=69.1379\ldots
\]
\[
\boxed{n=69}
\]
non-calculator
Simple
Total: 1 mark
Ella's height is \(175\text{ cm}\), correct to the nearest \(5\text{ cm}\).
Write down the upper bound of Ella's height.
\[
\underline{\hspace{3cm}}\text{ cm}
\]
Answer
Short Method:
$$
\text{Upper Bound} = \text{Given Value} + \frac{\text{Rounding Unit}}{2}
$$
$$
= 175 + \frac{5}{2}
$$
$$
= 175 + 2.5
$$
$$
= 177.5
$$
Answer:
$$
\boxed{177.5\text{ cm}}
$$
non-calculator
Simple
Total: 2 marks
The sides of a square are \(15.1\text{ cm}\), correct to \(1\) decimal place.
Find the upper bound of the area of the square.
\[
\underline{\hspace{4cm}}\text{ cm}^2
\]
Answer
Method:
\[
\text{Upper bound of side} = 15.1 + 0.05 = 15.15\text{ cm}
\]
\[
\text{Upper bound of area}
= (15.15)^2
\]
\[
= 229.5225
\]
Answer:
\[
\boxed{229.5225\text{ cm}^2}
\]
non-calculator
Simple
Total: 3 marks
The sides of an isosceles triangle are measured correct to the nearest millimetre.
One side has a length of \(8.2\text{ cm}\) and another has a length of \(9.4\text{ cm}\).
Find the largest possible value of the perimeter of this triangle.
\[
\underline{\hspace{4cm}}\text{ cm}
\]
Answer
\[
8.2 \rightarrow 8.25
\]
\[
9.4 \rightarrow 9.45
\]
\[
\text{Isosceles sides }=9.45,\;9.45,\;8.25
\]
\[
P=9.45+9.45+8.25
\]
\[
=27.15
\]
\[
\boxed{27.15\text{ cm}}
\]
non-calculator
Simple
Total: 2 marks
Violet and Wilfred recorded their times to run \(200\text{ m}\), correct to the nearest second.
Violet took \(36\) seconds and Wilfred took \(39\) seconds.
Work out the upper bound of the difference between their times.
\[
\underline{\hspace{4cm}}\text{ s}
\]
Answer
\[
\text{Violet: }35.5 \le t_V < 36.5
\]
\[
\text{Wilfred: }38.5 \le t_W < 39.5
\]
\[
\text{Upper bound of difference}
=39.5-35.5
\]
\[
=4
\]
\[
\boxed{4\text{ s}}
\]