Question

Heya ❤ A thief runs with a uniform speed of 100m/min.After one min. A policeman runs after a thief to catch him.He goes with a speed of 10m/min. In the first minute and increases his speed by 10m/minute every succeding minute.After how many minutes the policemen will catch thief.Kindly solve :)

Answer 1

≡QUESTION≡

A thief runs with a uniform speed of 10 m/minute . After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/min every succeeding minute. After how many minutes the police man will catch the thief ?

                                                         

║⊕ANSWER⊕║

Let us assume total time be n minutes

Also, total distance covered by the thief in n minutes

=Speed × Time

= 100 × n

= 100n m

Distance travelled by the thief in (n+1) min = 100(n+1)m

Speed of police in 1st min= 100 m/min

Speed of police in 2nd min = 110 m/min

3rd min = 120m/min.........so on

100,110,120,............ this forms an AP

And, Total distance covered by policeman = 1st min. + 2nd min + …… (n – 1) terms

Here, we have:

a = 100

d = 110 – 100

   = 10

‘n’ = n – 1

⇒200n = (n-1)[200 + (n-2)10]

⇒200n = (n – 1) [200 + 10n – 20]

⇒10n2 + 180n – 10n – 180 – 200n = 0

⇒10n2 – 30n – 180 = 0

⇒10 (n2 – 3n – 18) = 0

⇒n2 – 3n – 18 = 0

⇒n (n – 6) + 3 (n – 6) =0

⇒(n + 3) (n – 6) = 0

∴ n + 3 = 0

n = - 3

Or n – 6 = 0

n = 6

As the value of n i.e., time cannot be negative therefore n = 6

∴ Time taken by the police to catch the thief = n – 1

                                                                              = 6 – 1

                                                                              = 5 minutes

                                                             

Answer 2

Answer:

Let the police catch the thief in n min

As the thief ran 1 min before the police...time taken by the thief before being caught = (n+1) min

Distance travelled by the thief in (n+1) min = 100(n+1)m

Speed of police in 1st min=100m/min

Speed of police in 2nd min=110m/min

...3rd min = 120m/min.........so on

100,110,120,............ this forms an AP

Total distance travelled by the police in n min = n/2(2 x 100 +(n-1)10)

On catching the thief by police,distance traveled by thief= distance travelled by the police

100(n+1)= n/2(2 x 100 + (n-1)10)

100n + 100= 100n +n/2(n-1)10

100=n(n-1)5

n2 -n-20 = 0

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

n-5 = 0

n= 5    OR       n= -4...(but this is not possible)

so, n= 5

Time taken by the policeman to catch the thief = 5min

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