The answer
(a) \(3\) h \(31\) min
(b) \(336\) km
(c) total time \(\approx 10\) h \(43\) min (\(< 12\) h) and total cost \(= \$155.84\) (\(< \$165\)), so Erin is correct on both counts
(d) the journey time will be longer
O-Level E-Math 2025 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 2025 Paper 2. It tests speed = distance/time, in the Rates / Real-world problem solving area. It is worth 9 marks: (a) 1, (b) 1, (c) 6, (d) 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) York to Edinburgh \(= 317\) km at \(90\) km/h: time \(= \dfrac{317}{90} = 3.522\) h \(= 3\) h \(31\) min (nearest minute).
(b) Range at \(80\%\) \(= 0.80 \times 420 = 336\) km.
(c) Erin starts fully charged (\(420\) km of range). Route legs (driving at \(90\) km/h):
London → Manchester (via Oxford): distance \(= 95 + 260 = 355\) km. Driving time \(= \dfrac{355}{90} = 3.944\) h \(= 3\) h \(57\) min. After this leg the remaining range is \(420 - 355 = 65\) km, so to reach \(80\%\) (\(336\) km) she must add \(336 - 65 = 271\) km of range. Rapid charger: \(193\) km per \(30\) min, so waiting time \(= \dfrac{271}{193}\times 30 = 42.1 \approx 42\) min. Cost \(= 271 \times \$0.45 = \$121.95\).
Manchester → York: distance \(= 135\) km. Driving time \(= \dfrac{135}{90} = 1.5\) h \(= 1\) h \(30\) min. After driving she has \(336 - 135 = 201\) km left; she recharges back to \(336\) km, adding \(135\) km of range. Fast charger: \(64\) km per \(30\) min, so waiting time \(= \dfrac{135}{64}\times 30 = 63.3 \approx 63\) min. Cost \(= 135 \times \$0.251 = \$33.89\).
York → Edinburgh: distance \(= 317\) km. Driving time \(= 3\) h \(31\) min (from part (a)). After this leg she still has \(336 - 317 = 19\) km of range, enough to finish, so no charge needed.
Total journey time \(= 3\text{ h }57 + 42\text{ min} + 1\text{ h }30 + 63\text{ min} + 3\text{ h }31 = 10\) h \(43\) min, which is **less than \(12\) hours**. ✓ Total charging cost \(= \$121.95 + \$33.89 = \$155.84\), which is less than $165. ✓ So under Erin's (constant-rate) assumption, both of her statements are correct.
(d) The graph shows the real charging rate is lower than the assumed constant maximum rate for much of the charge (especially as the battery fills). A lower charging rate means the battery takes longer to charge, so the waiting times at the chargers are longer. Therefore Erin's total journey time will be longer than she calculated.
Answer: (a) \(3\) h \(31\) min
(b) \(336\) km
(c) total time \(\approx 10\) h \(43\) min (\(< 12\) h) and total cost \(= \$155.84\) (\(< \$165\)), so Erin is correct on both counts
(d) the journey time will be longer
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 speed = distance/time question from Rates / Real-world problem solving, worth 9 marks: (a) 1, (b) 1, (c) 6, (d) 1.
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