The answer
(a) shown (\(p = 2.5\))
(b) see comparison
O-Level E-Math 2024 Paper 1 Question 21 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 21 of the O-Level E-Math 2024 Paper 1. It tests mean of a data set with an unknown, in the Central tendency / Dispersion area. It is worth 5 marks: (a) 3, (b) 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) The sum of the eight values is \(12 + 24 + 8 + 21 + 28 + 17 + 2p + 4p^2 = 110 + 2p + 4p^2\). The mean is \(17.5\), so the sum is \(8 \times 17.5 = 140\): \[110 + 2p + 4p^2 = 140 \;\Rightarrow\; 4p^2 + 2p - 30 = 0 \;\Rightarrow\; 2p^2 + p - 15 = 0 \;\Rightarrow\; (2p - 5)(p + 3) = 0.\] So \(p = 2.5\) or \(p = -3\). A count of messages cannot give a negative term, so \(p = 2.5\). (shown)
(b) (1) Means: Li's mean (\(17.5\)) is less than his sister's (\(20\)), so on average his sister received more text messages per day. (2) Standard deviations: Li's standard deviation (\(7.89\)) is smaller than his sister's (\(9.51\)), so Li's daily numbers are more consistent (less spread out) than his sister's.
Answer: (a) shown (\(p = 2.5\))
(b) see comparison
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 mean of a data set with an unknown question from Central tendency / Dispersion, worth 5 marks: (a) 3, (b) 2.
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