A school sports field has area \(A=l\times w\), where the length is \(l=2.4\times10^2\) m and the width \(w\) is unknown. The reported area is \(A=3.36\time...
Question 1 Report
A school sports field has area \(A=l\times w\), where the length is \(l=2.4\times10^2\) m and the width \(w\) is unknown. The reported area is \(A=3.36\times10^4\) m\(^2\).
- Work out \(w\), giving your answer in standard form. (2)
- A mower covers \(8\times10^2\) m\(^2\) per hour. Work out the time needed to mow the whole field, giving your answer in standard form. (1)
- The mowing budget for the week is \(4\times10^1\) hours. State, with a reason, whether this is enough time. (1)
- Mowing costs \(\pounds1.5\times10^1\) per hour. Work out the total cost of mowing the field, giving your answer in standard form. (2)
Answer Details
Rearranging the area formula \(A=l\times w\) finds an unknown side length, and once the field's dimensions and area are known, mowing time and cost follow from further standard-form division and multiplication.
(a) Rearranging for \(w\) by dividing the area by the length:
\[w=3.36\times10^{4}\div2.4\times10^{2}=33600\div240=140=1.4\times10^{2} \text{ m}\] [2 marks](b) Dividing the total area by the mower's rate:
\[3.36\times10^{4}\div8\times10^{2}=33600\div800=42=4.2\times10^{1} \text{ hours}\] [1 mark](c) Comparing the 42 hours needed with the budget of \(4\times10^{1}=40\) hours: since \(42 \gt 40\), the budget is not enough time. [1 mark]
(d) Multiplying the time needed by the hourly cost:
\[42\times1.5\times10^{1}=42\times15=630=6.3\times10^{2} \text{ pounds}\] [2 marks]Each part builds on the previous one's answer (width from the area, time from the area again, cost from the time), so an early slip, such as an incorrect width in part (a), would carry through and change every later value; that is why re-deriving each quantity from the original given data, as shown above, guards against that.
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