Question

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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?



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Answer 1

$\bf{\Huge{\boxed{\sf{\blue{ANSWER\::}}}}}$

$\bf{\Large{\underline{\tt{Given\::}}}}}}$

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.

$\bf{\Large{\underline{\tt{To\:Find\::}}}}}}$

The Policeman take will minutes to catch the thief.

$\bf{\Large{\underline{\bf{\green{Explanation\::}}}}}$

Let the time taken by Policeman to catch the thief be R minutes.

Let the time taken by thief be (R+1) minutes.

The speed of police are form of sum of an A.P are, 100, 110,120,130.................

We know that formula of the sum of arithmetic progression;

$\leadsto\sf{Sn\:=\:\frac{n}{2} [2a+(n-1)d]}$

• First term,(a) = 100
• Common difference,(d) = 10.

&

We know that formula of the Distance;

$\leadsto\sf{\red{Distance\:=\:Speed*Time}}$

According to the question:

$\longmapsto\sf{S*T=\frac{n}{2} [2a+(n-1)d]}$

$\longmapsto\sf{100*(n+1)=\frac{n}{2} [2*100+(n-1)10]}$

$\longmapsto\sf{100*(n+1)=\frac{n}{2} [200+10n-10]}$

$\longmapsto\sf{100n+100=\frac{200n+10n^{2}-10n }{2} }$

$\longmapsto\sf{\cancel{200n}+200=\cancel{200n}+10n^{2} -10n}$

$\longmapsto\sf{10n^{2} -10n-200=0}$

$\longmapsto\sf{10(n^{2} -n-20)=0}$

$\longmapsto\sf{n^{2} -n-20=\cancel{\frac{0}{10} }}$

$\longmapsto\sf{n^{2} -n-20=0}$

$\longmapsto\sf{n^{2} -5n+4n-20=0}$

$\longmapsto\sf{n(n-5)+4(n-5)=0}$

$\longmapsto\sf{(n-5)(n+4)=0}$

$\longmapsto\sf{n-5=0\:\:\:or\:\:\:n+4=0}$

$\longmapsto\sf{n=5\:\:\:or\:\:\:n=-4}$

We know that negative value is not acceptable.

$\longmapsto\sf{\orange{n=5}}$

Thus,

$\bf{\large{\boxed{\tt{\blue{The\:Policeman\:taken\:time\:to\:catch\:the\:thief\:is\:5\:minutes.}}}}}}}$

Answer 2

Hi friend here is your answer

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Let's consider that the thief gets caught by the police after n minutes

From the start the distance travelled by the police every min is then given by

$100 + 110 ......... + (100 + (n - 1)10)$

Distance travelled by the thief is equal to the sum of Distance of above AP

$n \div 2(200 + (n - 1)10) =(n + 1)100$

On solving the equation further we get

${n}^{2} - n - 20 = 0$

On solving this quadratic equation we get

$n = 5 \: \: and \: n = - 4$

But time is not negative so we discard - 4

Hence 5min is the answer

After 5min the policeman will catch the thief

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Hope it helps you

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