Question

A thief runs with the uniform speed of $100$m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of $100$ m/minute in the first minute and increases his speed by $10$ m/minute. After how many minutes the policeman will catch the thief?

Answer 1

Let the police catch the thief in $n$ min

As the thief ran $1$ min before the police

Time taken by the thief before being caught $=(n+1)$ min

Distance travelled by the thief in $(n+1)$ min

$=100(n+1)$m

Speed of police in $1^{st}$ min

$=100$m/min

Speed of police in $2^{nd}$ min$=110$m/min

Speed of police in$3^{rd}$ min $=120$m/min. and so on

$\therefore 100,110,120,...$ this forms an AP

Total distance travelled by the police in $n$ min $=\frac{n}{2}(2\times 100+(n-1)10)$

On catching the thief by police,distance traveled by thief$=$ distance travelled by the police

$\Rightarrow 100(n+1)=\frac{n}{2}(2\times 100+(n-1)10)$

$\Rightarrow 100n+100=100n+\frac{n}{2}(n-1)10$

$\Rightarrow 100=n(n-1)5$

$\Rightarrow n^2-n-20=0$

$\Rightarrow (n-5)(n+4)=0$

$\Rightarrow n-5=0,n+4=0$

$\Rightarrow n=5$ OR $n=-4$(but this is not possible)

so, $n=5$

Time taken by the policeman to catch the thief$=5$min