Question

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Answer 1

hope it help you meet

Answer 2

HEY MATE!

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HERE IS YOUR ANSWER:

STEP BY STEP EXPLANATION:

We , are given that police runs after thief after 1 minute

Speed of the thief is given as 100km/min

Speed of the police is given in A.P i.e 100, 110,120,130..

Now, as the police catches the thief at a common place,

So,Let Time taken by the thief be x+1 min

and time taken by police to reach that place will be x (as police is 1 min late)

Now, speer of police= 100+110+120+130.....till t minutes

Sum of the speed of police = distance of police=n/2[2a+(n-1)d)]

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

St=t/2[200+(t-1)10]

St=t/2[10(20+t-1]

St=10t/2(19+t)

St=5t(19+t)

Now, distance of police=distance of thief (as thief was caught by police at a distance)

distance of thief=100(t+1)

Now, 5t(19+t)=100(t+1)

t(10+t)=20(t+1)

19t+t^2=20t+20

0=t^2+19t-20t-20

t^2-t-20=0

t^2-5t+4t-20=0

t(t-5)+4(t-5)=0

(t-5)(t+4)=0

t=5,-4

Time cannot be negative , hence the time taken to catch the thief is 5 min.

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HOPE IT HELPED ^_^