AP lap times against GP lap times · source reference

Printed question

A marathon runner and a sprinter run laps on a 400m track. The sprinter takes 56 seconds for lap 1, then 6 seconds more each subsequent lap. The marathon runner takes 1 minute for lap 1, then 4% more than the previous lap.

(i) Show the sprinter’s time for $n$ laps is $3n^2+53n$. [2]

(ii) Find the marathon runner’s time for $n$ laps in terms of $n$. [2]

(iii) How many complete laps does it take for the marathon runner to first exceed 40 minutes total? [3]

They decide to do a 12km race with the same starting point.

(iv) Determine who finishes first. [3]

(v) Determine which lap the winner is on when first overtaking the other runner. [3]

Printed working

(i) Sprinter: $56+62+68+\cdots+[56+(n-1)6]$; AP first term 56, difference 6.

$$\frac n2[2(56)+(n-1)6]=n(56+3n-3)=3n^2+53n.$$

(ii) Marathon: $60+60(1.04)+60(1.04)^2+\cdots+60(1.04)^{n-1}$; GP first 60, ratio 1.04.

$$\frac{60(1-1.04^n)}{1-1.04}=\frac{60(1.04^n-1)}{1.04-1}=1500(1.04^n-1).$$

(iii) $1500(1.04^n-1)>2400$.

Method 1: $1.04^n-1>1.6\Rightarrow1.04^n>2.6\Rightarrow n>\lg2.6/\lg1.04>24.362$. Needs 25 laps.

Method 2: use the table below; the first value above 2400 is at 25 laps.

(iv) $n=12000/400=30$. Sprinter time $3(30)^2+53(30)=4290$s; marathon time $1500(1.04^{30}-1)=3365.1$s (5 s.f.). Marathon runner finishes first.

(v) Consider $1500(1.04^n-1)<3n^2+53n$. Method 1 uses the table below.

Method 2 sketches $Y_1=1500(1.04^x-1)$, $Y_2=3x^2+53x$. Their first positive intersection is $x\approx3.2471963,y\approx203.73425$; source rounds $n\ge3.25$ (3 s.f.), hence $n=4$. Marathon runner first overtakes on lap 4.

Source tables and calculator displays

$n$	Marathon total seconds
24	2345 <2400
25	2498.8 >2400
26	2658.7 >2400

$n$	$1500(1.04^n-1)$	$3n^2+53n$
3	187.3	186
4	254.79	260
5	324.98	340

Source diagrams redrawn

Printed answer key

(ii) $1500(1.04^n-1)$; (iii) 25 laps; (iv) marathon runner; (v) 4th lap

Mathematical review and qualifications

(i) $T_S(n)=\frac n2[112+6(n-1)]=3n^2+53n$.

(ii) $T_M(n)=60(1.04^n-1)/0.04=1500(1.04^n-1)$ seconds.

(iii) Solve $1500(1.04^n-1)>2400$, giving $n>\ln2.6/\ln1.04\approx24.36$. First complete count is 25 (24 gives about 2345 s; 25 gives 2499 s).

(iv) For 30 laps, $T_S=4290$ s and $T_M\approx3365.10$ s, so the marathon runner wins.

(v) At three laps, $T_S=186<T_M=187.296$; at four, $T_M=254.788<T_S=260$. The marathon runner first overtakes during lap 4 under the usual constant-speed-per-lap model. The first three lap endpoint comparisons retain the sprinter lead.

Lap totals establish a lead change between laps 3 and 4. The precise within-lap crossing time needs a speed model; the source gives only lap durations.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2023/JPJC/Prelim/I/8] Question PDF pages 4; solution PDF pages 18, 19. The original files remain private QA records outside this native content.