The answer
(a) 12 cm
(b) \(\approx 6.35\) cm
O-Level E-Math 2019 Paper 1 Question 20 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 20 of the O-Level E-Math 2019 Paper 1. It tests linear scale factor for lengths, in the Mensuration (similar figures & solids) area. It is worth 4 marks: 2 + 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) Linear scale factor (small : large) \(= \dfrac{4}{10} = \dfrac{2}{5}\). Smaller height \(= 30 \times \dfrac{2}{5} = 12\) cm.
(b) Volume ratio (new : small) \(= \dfrac{500}{125} = 4\), so the linear scale factor is \(\sqrt[3]{4}\). New base diameter \(= 4 \times \sqrt[3]{4} = 4 \times 1.5874 = 6.35\) cm (3 s.f.).
Answer: (a) 12 cm
(b) \(\approx 6.35\) cm
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 linear scale factor for lengths question from Mensuration (similar figures & solids), worth 4 marks: 2 + 2.
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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