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O-Level E-Math · 2020 · P2 Q10 Everyday mathematics (money & graphs) · Reading a time-series graph 10 marks: (a)(i) 1, (a)(ii) 3, (b) 6 · number & algebra (everyday maths: money, graph reading) difficulty 4 of 5

O-Level E-Math 2020 Paper 2, Question 10: Reading a time-series graph

The answer

(a)(i) \(\approx \$1.825\)
(a)(ii) \(\approx \$14.11\)
(b) a defensible figure is \(\approx \$47\,000\)-\(\$48\,000\) (worked model below)

O-Level E-Math 2020 Paper 2 Question 10 · Verified worked solution by the Genius Plus Academy teaching team

What this question tests

This is Question 10 of the O-Level E-Math 2020 Paper 2. It tests reading a time-series graph, in the Everyday mathematics (money & graphs) area. It is worth 10 marks: (a)(i) 1, (a)(ii) 3, (b) 6. 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.

Step-by-step solution

(a)(i) Reading the curve at date \(15\): \(\pounds 1 \approx \$1.825\) (graph read to 3 d.p.).

(a)(ii) Greatest rate \(\approx 1.86\), least rate \(\approx 1.795\). Difference in cost \(= 217 \times (1.860 - 1.795) = 217 \times 0.065 \approx \$14.11\) (graph read; values within the usual tolerance).

(b) One defensible model (the mark is for stated assumptions and a consistent method). Weekly costs \(= 217 + 60 + 57 + 25 + 15 = \pounds 374\). For a year (\(52\) weeks): \(374 \times 52 = \pounds 19\,448\). Travel (\(12\) months): \(56.90 \times 12 = \pounds 682.80\). Total \(\approx \pounds 20\,130.80\). To be sure of having enough, convert at the least favourable rate on the two-year graph, the smallest pounds-per-dollar value, \(\approx \pounds 0.49\) per $ (which needs the most dollars): \(\dfrac{20\,130.80}{0.49} \approx \$41\,083\). Adding the \(15\%\) buffer: \(41\,083 \times 1.15 \approx \$47\,246\). So a sensible allowance is about \(\$47\,000\)-\(\$48\,000\) (round up for safety).

Answer: (a)(i) \(\approx \$1.825\)
(a)(ii) \(\approx \$14.11\)
(b) a defensible figure is \(\approx \$47\,000\)-\(\$48\,000\) (worked model below)

Same structure, different numbers

A question is hard because of its structure, not its surface.

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.

Questions students ask

What does O-Level E-Math 2020 Paper 2 Question 10 test?

It is a reading a time-series graph question from Everyday mathematics (money & graphs), worth 10 marks: (a)(i) 1, (a)(ii) 3, (b) 6.

Is this the same as IP Math?

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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