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A thief runs with a uniform speed of 100 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/minute every succeeding minute. - Algebra

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A thief runs with a uniform speed of 100 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/minute every succeeding minute. After how many minutes the policeman will catch the thief.

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Solution

Suppose the policeman catches the thief after t minutes.

Uniform speed of the thief = 100 metres/minute

∴ Distance covered by thief in (t + 1) minutes = 100 metres/minute × (t + 1) minutes = 100 (t + 1) metres

Distance covered by policeman in t minutes = Sum of t terms of an AP with first term 100 and common difference 10

=`t/2`[2×100+(t1)×10] m

=t[100+5(t1)]

=t(5t+95)

=5t2+95t

When the policeman catches the thief, 

5t2+95t=100(t+1)

5t2+95t=100t+100

5t25t100=0

t2t20=0

⇒ t = − 4 and t = 5

t = 5  (As t cannot be negative)

Thus, the policeman catches the thief after 5 minutes.

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method
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