A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?


Given:

Speed of the thief $=50$ m/minute.

After 2 minutes, a policeman runs to catch him. He goes 60 m in the first minute and increases his speed by 5m/minute every succeeding minute.

To do:

We have to find the time taken by the policeman to catch the thief.
Solution:

Let the time taken by the policeman to catch the thief be $x$ minutes after he starts running.

According to the question,

The distance covered by the policeman in each successive minute (in metres) is $60, 65, 70, ......, [60+(x-1)5]$

This series is in A.P., here,

$a=60, d=65-60=5$

We know that,

Sum of $n$ terms in an A.P $S_n=\frac{n}{2}[2a+(n-1)d]$

Therefore,

The total distance covered by the policeman $=\frac{x}{2}[2(60)+(x-1)5]$......(i)

The distance covered by the thief in $x+2$ minutes $=$ The total distance covered by the policeman

We know that,

Distance $=$ Speed $\times$ Time

The distance covered by the thief in $x+2$ minutes $=50\times(x+2)$....(ii)

From (i) and (ii)

$\frac{x}{2}[2(60)+(x-1)5]=50(x+2)$

$x(120+5x-5)=2(50x+100)$

$115x+5x^2=100x+200$

$5(x^2+23x)=5(20x+40)$

$x^2+23x-20x-40=0$

$x^2+3x-40=0$

$x^2+8x-5x-40=0$

$x(x+8)-5(x+8)=0$

$(x-5)(x+8)=0$

$x-5=0$ or $x+8=0$

$x=5$ or $x=-8$ which is not possible as time cannot be negative

Therefore, the policeman will catch the thief after 5 minutes.

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Updated on: 10-Oct-2022

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