Question

one thief run at the speed of 100m/min .After 1 min a policeman run behind him to catch him with the speed of 100m/min and increases his speed 10m/min in every succeeding min .
find how much time will be taken by policeman to catch the thief.
plz ans....​

Answer 1

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5 minutes

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step-by-step explanation :

Let,

the police catch the thief in ‘t’ minutes

Since,

the thief ran 1 min before the police start running,

therefore,

Time taken by the thief before he was caught = (t + 1) min

Now,

we know that,

Distance = speed × time

therefore,

Distance travelled by the thief in (t+1) min

= 100(t+1)m

now,

Given,

speed of policeman increased by 10 m/min

therefore,

Speed of policeman in the 1st min = 100 m/min

again,

Speed of policeman in the 2nd min = 110 m/min

and

Speed of policeman in the 3rd min = 120 m/min

Hence,

on observing the pattern,

we get, an A.P i.e.,

100, 110, 120….............

where,

a = 100

d = 110 - 100 = 10

nOw,

By the formula for sum of n terms of an AP,

we get,

Total distance travelled by the police in t minutes

=(t/2)[2a+(t – 1)d]

Putting the value of a, and d,

we get,

= (t/2)[2 x 100 +(t – 1)10]

now,

After the thief was caught by the police,

Distance traveled by the thief = distance travelled by the police

=> 100(t+1)= (t/2)[2 x 100 +(t-1)10]

=> 200(t + 1) = t[200 + 10t – 10]

=> 200t + 200= 200t + t(10t – 10 )

=> 200 = t(t – 1)10

=> t(t – 1) – 20 = 0

=> ${t}^{2}$ – t– 20 = 0

=> ${t}^{2}$– 5t + 4t – 20 = 0

=> t(t – 5) + 4(t – 5) = 0

after factorisation,

we get,

=> (t – 5) (t +4) = 0

therefore,

(t – 5) = 0 or (t + 4) = 0

=> t = 5 or t = -4

but

we know that,

time can't be negative,

therefore,

t = 5

Hence,

the time taken by the policeman to catch the thief = 5 minutes

Answer 2

Answer:

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