The answer
(a) \(360 = 2^3 \times 3^2 \times 5\)
(b) \(p = 3\), \(q = 5\)
O-Level E-Math 2019 Paper 1 Question 4 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 4 of the O-Level E-Math 2019 Paper 1. It tests prime factorisation, in the Numbers (primes, powers, roots) area. It is worth 2 marks: 1 + 1. 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) \(360 = 2^3 \times 3^2 \times 5\).
(b) For a perfect cube every prime index must be a multiple of 3. In \(2^3 \times 3^2 \times 5^1\), the \(2^3\) is already fine; the \(3^2\) needs one more factor of 3 (so multiply by \(p = 3\) to give \(3^3\)), and the lone \(5\) must be removed (so divide by \(q = 5\)). Then \(\dfrac{360 \times 3}{5} = 216 = 6^3\). So \(p = 3\) and \(q = 5\).
Answer: (a) \(360 = 2^3 \times 3^2 \times 5\)
(b) \(p = 3\), \(q = 5\)
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 prime factorisation question from Numbers (primes, powers, roots), worth 2 marks: 1 + 1.
Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.
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