The answer
(a) \(14\) hours
(b) \(\approx 30.6\%\)
O-Level E-Math 2018 Paper 1 Question 12 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 12 of the O-Level E-Math 2018 Paper 1. It tests inverse proportion (product constant), in the Direct & inverse proportion area. It is worth 3 marks: 1 + 2. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(a) Time \(\times\) workers is constant (inverse proportion): \(4 \times 17.5 = 70\) worker-hours. For 5 workers, time \(= \dfrac{70}{5} = 14\) hours.
(b) \(F = \dfrac{k}{d^2}\). Increasing \(d\) by 20% gives a new distance \(1.2d\), so the new force is \(\dfrac{k}{(1.2d)^2} = \dfrac{k}{1.44\,d^2} = \dfrac{F}{1.44}\). The force becomes \(\dfrac{1}{1.44} = \dfrac{100}{144} = \dfrac{25}{36}\) of the original, so the reduction is \(1 - \dfrac{25}{36} = \dfrac{11}{36} = 0.3056 = 30.6\%\) (3 s.f.).
Answer: (a) \(14\) hours
(b) \(\approx 30.6\%\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a inverse proportion (product constant) question from Direct & inverse proportion, worth 3 marks: 1 + 2.
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