The answer
(a) \(0.825\) kWh
(b) \(\approx 4.44\) hours
(c) open-ended, a worked, justified model below (e.g. Type D, 7 panels, total area \(\approx 12.4\) m²)
O-Level E-Math 2023 Paper 2 Question 9 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 9 of the O-Level E-Math 2023 Paper 2. It tests energy = power × time, in the Problem-solving in real-world contexts / Mensuration area. It is worth 10 marks: 1 + 2 + 7. 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) Energy \(= \) power \(\times\) time \(= 165 \text{ W} \times 5 \text{ h} = 825\) Wh \(= 0.825\) kWh. *(Or \(0.165 \text{ kW} \times 5 \text{ h}\).)*
(b) Mean \(= \dfrac{4.7 + 4.9 + 5.0 + 4.7 + 4.3 + 4.2 + 4.3 + 4.4 + 4.5 + 4.4 + 3.9 + 4.0}{12} = \dfrac{53.3}{12} = 4.44\) hours (3 s.f.).
(c) Open-ended; a defensible model with clearly stated decisions. - Daily electricity used. Use the most recent (highest) year. The top curve reaches about 3000 kWh by December, so the household uses about \(3000\) kWh in a year, i.e. \(\dfrac{3000}{365} = 8.22\) kWh \(= 8220\) Wh per day. - Peak sunlight hours. Use the yearly average from part (b): \(4.44\) hours per day. - Choose the panel type. To use the least area, pick the type with the greatest daily output per square metre. Daily output \(=\) watts \(\times 4.44 \times 0.75 =\) watts \(\times 3.33\) Wh; area \(=\) length \(\times\) width. - A: \(300 \times 3.33 = 999\) Wh over \(1.46 \times 1.05 = 1.533\) m² \(\Rightarrow 652\) Wh/m² - B: \(370 \times 3.33 = 1232\) Wh over \(1.96\) m² \(\Rightarrow 629\) Wh/m² - C: \(390 \times 3.33 = 1299\) Wh over \(1.98\) m² \(\Rightarrow 656\) Wh/m² - D: \(400 \times 3.33 = 1332\) Wh over \(1.69 \times 1.05 = 1.7745\) m² \(\Rightarrow 751\) Wh/m² ← best Choose Type D. - Number of panels. \(\dfrac{8220}{1332} = 6.17\), so round up to 7 panels (to meet the daily demand). - Total area \(= 7 \times 1.7745 = 12.4\) m² (3 s.f.).
So about 12.4 m² of Type D panels is a reasonable estimate. (Any answer is acceptable if the daily usage, sunlight hours, chosen panel and the panel-count rounding are stated and the area is computed consistently; the figure varies with the assumptions made.)
Answer: (a) \(0.825\) kWh
(b) \(\approx 4.44\) hours
(c) open-ended, a worked, justified model below (e.g. Type D, 7 panels, total area \(\approx 12.4\) m²)
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 energy = power × time question from Problem-solving in real-world contexts / Mensuration, worth 10 marks: 1 + 2 + 7.
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