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Thief has a lead of 50*2 = 100m
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
10, 15, 20 ... is an AP whose sum is 100
a= 10 , d = 5
Sum = 100
100 = n/2(2a + (n-1)d) = n/2(20 +(n-1)5 )
100= n/2(20 + 5n -5)
200 = n(15 + 5n)
divide by 5
40 = n(n + 3)
n^2 + 8n - 5n - 40 =0
n(n+8) - 5(n+8)= 0
>> n = 5 as it cant be negative
Therefore cop catches after 5 min + 2min (when cop was at rest) = 7min
Hope you help it
plz mark as brainlist
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
10, 15, 20 ... is an AP whose sum is 100
a= 10 , d = 5
Sum = 100
100 = n/2(2a + (n-1)d) = n/2(20 +(n-1)5 )
100= n/2(20 + 5n -5)
200 = n(15 + 5n)
divide by 5
40 = n(n + 3)
n^2 + 8n - 5n - 40 =0
n(n+8) - 5(n+8)= 0
>> n = 5 as it cant be negative
Therefore cop catches after 5 min + 2min (when cop was at rest) = 7min
Hope you help it
plz mark as brainlist
Priyanshu31072002:
Correct
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