Question

Please answer this question ASAP.

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Answer 1

Using quadratic equation
Let the police catch the thief in ‘n’ minutes
Since the thief ran 1 min before the police start running.
Time taken by the thief before he was caught = (n + 1) min
Distance travelled by the thief in (n+1) min = 100(n+1)m
Given speed of policeman increased by 10m per minute.
Speed of the police in 1st min = 100m/minute
". ". ". ". "2nd min = 110m/minute
" ". ". ". ". 3rd min = 120m/minute
Now ,
100, 110, 120 ....are in AP
TOTAL distance travelled by the police inn minutes =
[n/2]{2a + (n-1)d}.

= (n/2)[2 x 100 +(n – 1)10]
Distance travelled by the police = distance travelled by the theif
100(n+1)=(n/2)[2 x 100 +(n – 1)10]
200(n + 1) = n[200 + 10n – 10]
200n + 200= 200n + n(10n – 10 )
200 = n(n – 1)10
n(n – 1) – 20 = 0
n2 – n– 20 = 0
From this quad eq. you will get n=5 or n = -4
Hence , time can't be negative .
Therefore we consider n=5 minutes

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Answer 2

Answer:

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