2015 A-level H2 Mathematics (9740) Paper 1 Question 8 Suggested Solutions

November 5, 2015

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. :/

A-level H2 Mathematics MF15 Pure Mathematics Suggested Solution Year 2015

About

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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