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# Question 58. A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

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One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.

In this question it is given that there is a thief who commits a theft amd  runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute.

Now we have to find that after how many minutes, the policeman will catch the thief.

In this question we have solve the question by using the arithmetic progression properties.

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 58

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1. Suppose n minutes is the time taken by the policeman to catch the thief.

Since policeman started running 2 minutes later, the thief ran for (n + 2) minutes.

Therefore, distance travelled by thief = Speed × Time = 50(n + 2) meter.

Speed of the police after every minute are: 60, 65, 70,. . . .

These form an A.P. with first term(a) = 60 and common difference(d) = 65 – 60 = 5

Total distance travelled by police in n minutes = n[2(60) + (n – 1)5] / 2

= n[120 + 5n – 5] / 2

= n[115 + 5n] / 2

Now, according to the question,

Distance travelled by thief in (n + 2) minutes = Distance travelled by police in n minutes

=> 50(n + 2) = n[115 + 5n] / 2

=> 100(n + 2) = 115n +5n2

=> 5n2 + 15n – 200 = 0

=> n2 + 3n + 40 = 0

=> n2 + 8n – 5n + 40 = 0

=> n(n + 8) – 5(n + 8)

=> n = 5 or n = – 8

Ignoring n = – 8 as time cannot be negative. So, we get n = 5.

Hence, after 5 minutes the policeman will catch the thief.

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