The total area covered by the long-jump sand pits at a school sports day is \(3.2\times10^{1}\) m\(^2\), made up of \(4\) identical rectangular pits, each m...
Question 1 Report
The total area covered by the long-jump sand pits at a school sports day is \(3.2\times10^{1}\) m\(^2\), made up of \(4\) identical rectangular pits, each measuring \(4\) m by \(x\) m.
- Form an equation and find the value of \(x\). (3)
- Next year the school wants \(5\) pits of the same size, but only has \(3.6\times10^{1}\) m\(^2\) of ground available. State, with a reason, whether \(5\) pits of this size will fit. (2)
Answer Details
The total sand-pit area is shared equally among the four identical rectangular pits, so setting up an equation for the total area in terms of \(x\) and solving it finds each pit's unknown dimension; the same per-pit area can then be scaled up to check whether a different number of pits will fit in a new, limited space.
(a) Each pit has area \(4\times x=4x\) m\(^{2}\); four identical pits give a total area of \(4\times4x=16x\) m\(^{2}\). Setting this equal to the given total: \(16x=32\), so \(x=2\). [3 marks]
(b) Each pit's area is \(4\times2=8\) m\(^{2}\), so five pits need \(5\times8=40\) m\(^{2}\) of ground. Since \(40\) m\(^{2}\) is more than the \(36\) m\(^{2}\) available, five pits of this size will NOT fit. [2 marks]
Finding the area of one pit first, as in part (b), makes scaling up to any number of pits straightforward; multiplying the required total directly, rather than trying to compare \(x\)-values, keeps the comparison in the same units throughout.
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