Question

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

Answer 1

speed = 50/per min
after two minutes police catch the thief
in first minute he runs 60m
in another min he runs 55m
the police catch him in 115m.

Answer 2

Answer:

7 minutes will the policeman catch the thief.

Step-by-step explanation:

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

To find : After how many minutes will the policeman catch the thief?

Solution :

Thief has a lead of $50\times 2=100$

Relative distance covered by cop in 1st min = 60-50 = 10

Relative distance covered by cop in 2nd min = 65-50 = 15

Relative distance covered by cop in 3rd min = 70 -50 = 20

This form an A.P - 10, 15, 20 ...

Where, first term a= 10, common difference d = 5

Sum = 100

The sum formula, $S_n=\frac{n}{2}(2a+(n-1)d)$

$100=\frac{n}{2}(2(10)+(n-1)5)$

$100=\frac{n}{2}(20+5n-5)$

$200=n(15+5n)$

$\frac{200}{5}=n(3+n)$

$40=3n+n^2$

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

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

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

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

$n=-8,5$

Reject n=-8

The number of term is n=5.

The cop catches after 5 min+2 min (when cop was at rest) = 7 minute

Therefore, 7 minutes will the policeman catch the thief.