Question

A thief runs away from a police station with a uniform speed of 100m/ minute. After 1 minute a policeman runs behind the thief to catch him. He goes at a speed of 100m/minute in first minute and increases his speed by 10m/minute in each succeeding minute. how many minutes will the policeman take to catch the thief? <br />please answer it . hurry. without the use of Google.

Answer 1

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Let the policeman catch the thief in n minutes.

Given that uniform speed of the thief = 100m/min.

Given that After 1 minute a policeman runs to catch him at the speed of 100 in first minutes.

(n + 1) minutes = 100(n + 1) minutes.

Given that speed of policeman increase by 10m/min.

The speed of police in 1 minute = 100m/min.

The speed of police in 2 minutes = 110m/min.

The speed of police in 3 minutes = 120m/min.

The speed of police in 4 minutes = 130m/min.

Hence 100,110,120,130 are in AP.

Let a be the first term and d be the common difference.

a = 100,d = 110 - 100 = 10.

We know that sum of n terms of an ap = n/2(2a + (n - 1) * d)

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


The distance traveled by the thief = Distance traveled by the police

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

100n + 100 = n/2(200 + 10n - 10)

200n + 200 = 10n^2 + 190n

10n^2 + 190n - 200 = 200n

10n^2 - 10n - 200 = 0

n^2 - n - 20 = 0

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

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

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

n = 5 (or) n = -4.

n cannot be negative.


Therefore the time is taken by the policeman to catch the thief = 5minutes.

Answer 2

ANSWER
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Let the policeman catch the thief in x minutes.

uniform speed of the thief = 100m/min.

After 1 minute a policeman runs to catch him at the speed of 100 in first minutes.

(X + 1) minutes = 100(x + 1) minutes.


speed of policeman increase by 10m/min.

The speed of police in 1 minute = 100m/min.

The speed of police in 2 minutes = 110m/min.

The speed of police in 3 minutes = 120m/min.

The speed of police in 4 minutes = 130m/min.

Hence 100,110,120,130 are in AP.

Let y be the first term and d be the common difference.

Y = 100,d = 110 - 100 = 10.


sum of x terms of an ap = x/2(2y+ (x - 1) * d)

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

The distance traveled by the thief = Distance traveled by the police

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

100x + 100 = x/2(200 + 10x - 10)

200x + 200 = 10x^2 + 190x

10x^2 + 190x - 200 = 200x

10x^2 - 10x - 200 = 0

X^2 - x - 20 = 0

X^2 -5x + 4x. - 20 = 0

X(x- 5) + 4(x - 5) = 0

(X -5)(x + 4) = 0

X= 5 (or) x =. ( -4. ))


X cannot be negative.


Therefore the time is taken by the policeman to catch the thief = 5minutes.