A thief runs with a uniform speed of 100m/min. After one min a policeman runs after the thief to catch him. He goes with a speed of 100m/min in first min and increase his speed by 10m/min every succeeding min. After how many min the policeman will catch the thief.
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Answers
Answer:
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 1
st
min
=100m/min
Speed of police in 2
nd
min=110m/min
Speed of police in3
rd
min =120m/min. and so on
∴100,110,120,... this forms an AP
Total distance travelled by the police in n min =
2
n
(2×100+(n−1)10)
On catching the thief by police,distance traveled by thief= distance travelled by the police
⇒100(n+1)=
2
n
(2×100+(n−1)10)
⇒100n+100=100n+
2
n
(n−1)10
⇒100=n(n−1)5
⇒n
2
−n−20=0
⇒(n−5)(n+4)=0
⇒n−5=0,n+4=0
⇒n=5 OR n=−4(but this is not possible)
so, n=5
Time taken by the policeman to catch the thief=5min
Answer :
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