The answer
(a)(i) \(\approx 26\) min
(ii) \(\approx 22\) min
(iii) \(\approx \dfrac{1}{10}\)
(b) 60 children
(c)(i) \(k = 54\)
(ii) \(37.7^{\circ}\)
O-Level E-Math 2025 Paper 2 Question 2 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 2 of the O-Level E-Math 2025 Paper 2. It tests cumulative frequency (median, in the Data analysis / Probability area. It is worth 8 marks: (a) 1 + 2 + 2, (b) 1, (c) 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)(i) Median is the time at cumulative frequency \(\tfrac12(80) = 40\). Reading across from \(40\) on the curve gives a median of about \(26\) minutes.
(ii) Lower quartile is at CF \(= 20\) (\(\approx 19\) min) and upper quartile at CF \(= 60\) (\(\approx 41\) min). IQR \(\approx 41 - 19 \approx 22\) minutes (reading from the curve).
(iii) At \(46\) minutes the cumulative frequency is about \(72\), so the number taking longer than \(46\) min is \(80 - 72 = 8\). Probability \(= \dfrac{8}{80} = \dfrac{1}{10}\).
(b) On a box-and-whisker plot, \(38\) minutes is the upper quartile (the right-hand edge of the box). The upper quartile marks the value below which \(75\%\) of the data lie, so the number of children who took at most \(38\) min is \(75\% \times 80 = 60\) children.
(c)(i) Range \(= \text{largest} - \text{smallest} = 5.43 \times 10^7 - 1.276 \times 10^5 = 54\,172\,400 = 54.1724\) million. To the nearest integer, \(k = 54\).
(ii) Total visitors \(= 8.76\times10^5 + 5.43\times10^7 + 1.276\times10^5 + 9.78\times10^6 + 2.84\times10^7 = 93\,503\,600\). Angle for D \(= \dfrac{9.78\times10^6}{93\,503\,600} \times 360^{\circ} = 37.66^{\circ} = 37.7^{\circ}\) (1 d.p.).
Answer: (a)(i) \(\approx 26\) min
(ii) \(\approx 22\) min
(iii) \(\approx \dfrac{1}{10}\)
(b) 60 children
(c)(i) \(k = 54\)
(ii) \(37.7^{\circ}\)
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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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 cumulative frequency (median question from Data analysis / Probability, worth 8 marks: (a) 1 + 2 + 2, (b) 1, (c) 1 + 1.
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