Question:

Stepwise Solution:

Let policeman run for $n$ minutes. Thief runs for $(n + 1)$ minutes. [1.0 Mark]

Distance covered by thief in $(n + 1)$ minutes $= 100(n + 1)$ metres. [1.0 Mark]

Distance covered by policeman in $n$ minutes is an AP with $a = 100, d = 10$:
$S_n = \dfrac{n}{2}[2(100) + (n - 1)(10)] = \dfrac{n}{2}[200 + 10n - 10] = \dfrac{n}{2}[190 + 10n] = 5n^2 + 95n$. [1.5 Marks]

Equating distances: $5n^2 + 95n = 100(n + 1) = 100n + 100 \Rightarrow 5n^2 - 5n - 100 = 0$. [1.0 Mark]

Divide by 5: $n^2 - n - 20 = 0 \Rightarrow (n - 5)(n + 4) = 0 \Rightarrow n = 5$ (since time $n > 0$).
So the policeman catches the thief after $5$ minutes. [0.5 Mark]

Marking Scheme:

• Setting thief's running time as $n+1$ and distance $= 100(n+1)$: 1.0 Mark
• Expressing policeman's distance as AP sum $S_n = 5n^2 + 95n$: 1.5 Marks
• Forming quadratic $5n^2 - 5n - 100 = 0$: 1.0 Mark
• Solving quadratic for $n = 5$: 1.0 Mark
• Final statement: Policeman catches thief after 5 minutes: 0.5 Mark