Question

a theif run with a uniform speed of 10m/minute after 10 minute policeman run after the theif to catch him he goes with speed of 100m/min in first minute and increase his speed by 10m/min every succeeding mintue after how many mintues the policemen will catch theif

if answer u will get 75 points pls help​

Answer 1

Answer:

5 minutes

Hey guy, your question must be a theif run with a uniform speed of 100m/minute after 10 minute instead of a theif run with a uniform speed of 10m/minute after 10 minute

Step by step explaination:

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

Speed of police in3rd min =120m/min. and so on

∴100,110,120,... this forms an AP

Total distance travelled by the police in n min =

n/2 (2×100+(n−1)10)

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

⇒100(n+1)= n/2 (2×100+(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+4=0

⇒n=5 OR n=−4(but this is not possible) because nobody can run in negative speed.

so, n=5

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

hope it helps.

Answer 2

Answer:

☆ Question:

A thief run with uniform speed of 10 m/min after 10 minute police run after the thief to catch him he goes with speed of 100m/min in first minute and increase his speed by 10m/min every succeeding mintue after how many mintues the policemen will catch theif .

☆ Given:

$\textsf{Uniform speed of the theif = 10 m/min}$

$\textsf{Time after policeman start chasing = 10 min}$

$\textsf{Uniform speed of the police man = 100 m/min}$

$\textsf{Speed increases by the policeman = 10 m/min}$

☆ To find:

$\textsf{Time taken to catch the theif by policeman}$

☆ Concept:

First , find the distance covered by theif in 10 minutes.

Then , find time taken by the policeman to catch the theif .

☆ Taken:

Speed of the police at 1st min = 100 m/min

Speed of the police after 1 min = 110 m/min

(The speed of the police is increasing per min by 10 m/min .)

Speed of the police after 2 min = 120 m/min

Speed of the police after 3 min = 130 m/min

Speed of the police after 4 min = 140 m/min

So , now 100,110,120,130,140....... m/min are in AP.

And 100 m/min and the 110 m/min are the first term of it .

So let the first term be :

y=100 m/min

z=110 m/min-100 = 10

Let sum of the x terms of AP

= x/2(2y+(x-1)z)

(RHS part)

= 100(x+1)

(LHS part)

☆ Solution:

Now ,

>> 100(x+1)=x/2(2y+(x-1)z)

>> 100x+100=x/2(2×100+(x-1)10)

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

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

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

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

>> 10x^2-10n-200=0

>> x^2-x-20=0

>> x^2-5x+4x-20=0

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

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

>> So , the x = 5 and x = -4

And the x can't be in the negative.

So , the time taken by the policeman to catch the theif was 5 mins .

☆ Answer:

So , 5 mins taken by the policeman to catch the theif .

☆ Extra information:

$\bold{▪︎Speed=\frac{Distance}{Time}}$

$\bold{▪︎Distance=Speed×Time}$

$\bold{▪︎Time=\frac{Distance}{Speed}}$

HOPE IT HELPS YOU