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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.