The answer
(a) \(510\) seconds
(b) shown (11/hour)
(c) Chen \(\approx \$44\,342\) (no); Zhu \(\approx \$49\,134\) (yes)
O-Level E-Math 2022 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 2022 Paper 2. It tests rate (items per time), in the Ratio, rate & proportion / Equations area. It is worth 11 marks: 1 + 2 + 8. 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) One hour \(= 3600\) s for 7 pairs, so one pair takes \(\dfrac{3600}{7} = 514\) s \(= 510\) seconds (nearest 10 s).
(b) Basic pay for 9 hours \(= 9 \times 9.80 = \$88.20\). Bracelet pay \(= 231.75 - 88.20 = \$143.55\). Number of bracelets \(= \dfrac{143.55}{1.45} = 99\), i.e. \(\dfrac{99}{9} = 11\) bracelets per hour. (shown)
(c) Let Chen take \(x\) seconds per necklace, so Zhu takes \((x - 80)\) s. Each works \(8\) h \(= 28\,800\) s, and together they make 132: \[\dfrac{28\,800}{x} + \dfrac{28\,800}{x - 80} = 132 \Rightarrow 11x^2 - 5680x + 192\,000 = 0 \Rightarrow (11x - 400)(x - 480) = 0.\] So \(x = 480\) (rejecting \(x = \tfrac{400}{11}\)). Chen makes \(\dfrac{28\,800}{480} = 60\) necklaces/day; Zhu makes \(\dfrac{28\,800}{400} = 72\) necklaces/day. Take a year as 52 weeks \(\times\) 5 days \(= 260\) working days, of which 18 are holiday (basic pay only) and 242 are item-making days. - Basic pay (all 260 days): \(9.80 \times 8 \times 260 = \$20\,384\). - Chen's necklace pay: \(60 \times 1.65 \times 242 = \$23\,958\). **Chen's annual income \(= 20\,384 + 23\,958 = \$44\,342\), which is below $48 000**, Chen cannot expect the minimum. - Zhu's necklace pay: \(72 \times 1.65 \times 242 = \$28\,749.60\). **Zhu's annual income \(= 20\,384 + 28\,749.60 = \$49\,133.60\), which is above $48 000**, Zhu can expect the minimum.
So they cannot both reach the advertised minimum: only Zhu can. (The exact figures depend on the stated assumption of 52 weeks / 242 working days; the conclusion that Zhu clears $48 000 and Chen does not follows from 72 vs 60 necklaces per day.)
Answer: (a) \(510\) seconds
(b) shown (11/hour)
(c) Chen \(\approx \$44\,342\) (no); Zhu \(\approx \$49\,134\) (yes)
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 rate (items per time) question from Ratio, rate & proportion / Equations, worth 11 marks: 1 + 2 + 8.
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