2015 H2 Maths Paper 1 Q8 — Worked Solution

Since each lap is $400\ m$ and $50$ laps make up the $20\ km$ run, the required total time lies between $1\frac{1}{2}$ hours and $1\frac{3}{4}$ hours inclusive, i.e.

5400\ s \leq \text{total time} \leq 6300\ s.

Part (i)

4 marks

Athlete $A$ ’s lap times form an arithmetic progression with first term $T$ and common difference $2$ , so the total time for $50$ laps is

S_{50} = \frac{50}{2}\left( 2T + 49(2) \right)

= 50T + 2450.

Using the required time interval:

5400 \leq 50T + 2450 \leq 6300.

Solving:

2950 \leq 50T \leq 3850

59 \leq T \leq 77.

$\therefore$ $T$ satisfies $59\ s \leq T \leq 77\ s$ .

Part (ii)

4 marks

Approach

“2% more than the time for the previous lap” makes each lap time $1.02$ times the one before, so the lap times form a geometric progression with common ratio $1.02$ , not an arithmetic one.

Athlete $B$ ’s lap times form a geometric progression with first term $t$ and common ratio $1.02$ , so the total time for $50$ laps is

S_{50} = \frac{t\left( {1.02}^{50} - 1 \right)}{1.02 - 1}.

By GC, ${1.02}^{50} = 2.69159\ldots$ , so

S_{50} = 84.5794\ldots\, t.

Using the required time interval:

5400 \leq 84.5794\ldots\, t \leq 6300.

Solving:

63.8 \leq t \leq 74.5\ \ (3\ s.f.).

$\therefore$ $t$ satisfies $63.8\ s \leq t \leq 74.5\ s$ (to 3 s.f.).

Note: $84.5794\ldots\, t$ is increasing in $t$ , so the direction of both inequalities is unchanged when dividing through by it.

Part (iii)

3 marks

Approach

“Completes the run in exactly $1\frac{1}{2}$ hours” fixes $S_{50} = 5400$ for each athlete, which pins down $T$ and $t$ at the lower end of the ranges found in (i) and (ii). We then find each athlete’s 50th lap time directly from their progression.

For athlete $A$ , setting $S_{50} = 5400$ in $50T + 2450 = 5400$ gives $T = 59$ (the lower bound found in (i)). The 50th lap time is the 50th term of the arithmetic progression:

T + 49(2) = 59 + 98 = 157\ s.

For athlete $B$ , setting $S_{50} = 5400$ gives $t = 63.845\ldots$ (the lower bound found in (ii), kept unrounded here). The 50th lap time is the 50th term of the geometric progression:

t \times {1.02}^{49}.

By GC,

63.845\ldots \times {1.02}^{49} = 168.48\ldots\ s.

The difference in the athletes’ 50th lap times is

168.48\ldots - 157 = 11.48\ldots\ s.

$\therefore$ the difference is $11\ s$ (to the nearest second), with athlete $B$ ’s 50th lap taking longer.

To check your answer:

84.5794\ldots \times 63.845\ldots = 5400

, confirming

t

was recovered correctly from

S_{50} = 5400

. ✓

2015 H2 Maths Paper 1, Question 8

Part (i)

Part (ii)

Part (iii)

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