2021 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)

2 marks

We model Mrs Wong’s ten times as an arithmetic progression (AP) with first term $a = 40$ , last term $l = 25$ and $n = 10$ terms. The sum of an AP is $S_{n} = \frac{n}{2}(a + l)$ .

S_{10} = \frac{10}{2}(40 + 25)

= 5(65)

= 325

To check your answer: the common difference is

d = \frac{25 - 40}{10 - 1} = - \frac{5}{3}

; the ten times

40,\ 38\frac{1}{3},\ 36\frac{2}{3},\ \ldots,\ 25

sum to

325

. ✓

$\therefore$ the total time to swim 10 lengths is $325\ s$ .

Part (b)

5 marks

Approach

Suzie’s 30-length swim splits into three blocks of 10 lengths, each under a different rule — Mrs Wong’s AP (part (a)), a constant $25\ s$ per length, and Alfie’s GP — so we find the total time as the sum of these three blocks, then divide the total distance by it.

For Alfie’s programme, the times form a geometric progression (GP) with first term $a = 25$ and $10$ th term $ar^{9} = 40$ .

ar^{9} = 40

\Rightarrow r^{9} = \frac{40}{25} = \frac{8}{5}

\Rightarrow r = \left( \frac{8}{5} \right)^{1/9}

By GC, $r \approx 1.05361$ . Using $S_{n} = \frac{a\left( r^{n} - 1 \right)}{r - 1}$ ,

S_{10} = \frac{25\left( r^{10} - 1 \right)}{r - 1}

\approx 319.797\ s\ \text{(by GC)}

Suzie’s total time for the 30 lengths is the sum of the three blocks: Mrs Wong’s programme, the middle 10 lengths at $25\ s$ each, and Alfie’s programme.

\text{Total time} = 325 + 10(25) + 319.797

= 894.797\ s

Her total distance is $30$ lengths of the $35\ m$ pool.

\text{Total distance} = 30 \times 35 = 1050\ m

\text{Average speed} = \frac{1050}{894.797}

\approx 1.17\ \text{m s}^{- 1}\ \left( 3\ \text{s.f.} \right)

To check your answer:

r^{9} \approx {1.05361}^{9} \approx 1.600 = \frac{8}{5}

✓, matching the required ratio; and the average speed of

1.17\ \text{m s}^{- 1}

is consistent with covering

1050\ m

in roughly

15

minutes. ✓

$\therefore$ Suzie’s average speed is $1.17\ \text{m s}^{- 1}$ (3 s.f.).

Part (c)

2 marks

We track the cumulative time at the end of each length. At the end of length $10$ (Mrs Wong’s programme), the cumulative time is $325\ s$ (part (a)). Lengths $11$ to $20$ each take a constant $25\ s$ , so the cumulative time at the end of length $10 + k$ is $325 + 25k$ .

325 + 25k = 480

\Rightarrow k = 6.2

Since $k = 6$ gives $t = 475\ s$ (end of length $16$ ) and $k = 7$ gives $t = 500\ s$ (end of length $17$ ), and $475 < 480 < 500$ , at $t = 480\ s$ Suzie is partway through length $17$ .

To check your answer: cumulative time at the end of length

16

is

325 + 25(6) = 475\ s

and at the end of length

17

is

325 + 25(7) = 500\ s

; since

475 < 480 < 500

,

t = 480\ s

does fall within length

17

. ✓

Note: consecutive lengths alternate direction along the pool, with the first length swum away from the starting point, so odd-numbered lengths are swum away from it and even-numbered lengths towards it. Length $17$ is odd.

$\therefore$ exactly 8 minutes after starting, Suzie is swimming away from her starting point (she is on length $17$ ).

2021 H2 Maths Paper 2, Question 4

Part (a)

Part (b)

Part (c)

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