All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let A_{50} and B_{50} denote the total time A & B take respectively, in seconds.

A_{50} = \frac{50}{2}[2T+49(2)] = 50T + 2450

1.5H = 5400s

1.75H = 6300s

5400 \le 50T +2450 \le 6300

59 \le T \le 77

\therefore, \{ T\in \mathbb{R} | 59 \le T \le 77\}

(ii)
B_{50} = \frac{t(1.02^{50}-1)}{1.02-1} = 50t(1.02^{50}-1)

5400 \le 50t(1.02^{50}-1) \le 6300

63.84 \le t \le 74.483

63.9 \le t \le 74.4 (3 SF)

\therefore, \{ t\in \mathbb{R} | 63.9 \le t \le 74.4\}

(iii)
For A: Time taken = 59 + 49(2) = 157
For B: Time taken = 63.84533(1.02)^{49} = 168.475

Difference = 168.475 - 157 \approx 11 \mathrm{seconds} (nearest seconds)

Back to 2015 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

KS Comments:

  • Be careful of the units, its in seconds.
  • Be careful to write in sets.
  • Be careful when you’re rounding off (ii) as you need to round off according to the range of values. If you rounded down to 63.8, this means you accept 63.81, which is out of range and will cause him to arrive before 1.5H. I reckon this part will get many students who are less careful. :/

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