अस्य प्रश्नस्य समाधानं किम् अस्ति?
उत्तरा तत् देहि अतः एव जानामि
thief run at the speed of 100m/min .After 1 min a policeman run behind him to catch him with the speed of 100m/min and increases his speed 10m/min in every succeeding min .
find how much time will be taken by policeman to catch the thief.
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Answers
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5 minutes
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step-by-step explanation :
Let,
the police catch the thief in ‘t’ minutes
Since,
the thief ran 1 min before the police start running,
therefore,
Time taken by the thief before he was caught = (t + 1) min
Now,
we know that,
Distance = speed × time
therefore,
Distance travelled by the thief in (t+1) min
= 100(t+1)m
now,
Given,
speed of policeman increased by 10 m/min
therefore,
Speed of policeman in the 1st min = 100 m/min
again,
Speed of policeman in the 2nd min = 110 m/min
and
Speed of policeman in the 3rd min = 120 m/min
Hence,
on observing the pattern,
we get, an A.P i.e.,
100, 110, 120….............
where,
a = 100
d = 110 - 100 = 10
nOw,
By the formula for sum of n terms of an AP,
we get,
Total distance travelled by the police in t minutes
=(t/2)[2a+(t – 1)d]
Putting the value of a, and d,
we get,
= (t/2)[2 x 100 +(t – 1)10]
now,
After the thief was caught by the police,
Distance traveled by the thief = distance travelled by the police
=> 100(t+1)= (t/2)[2 x 100 +(t-1)10]
=> 200(t + 1) = t[200 + 10t – 10]
=> 200t + 200= 200t + t(10t – 10 )
=> 200 = t(t – 1)10
=> t(t – 1) – 20 = 0
=> – t– 20 = 0
=> – 5t + 4t – 20 = 0
=> t(t – 5) + 4(t – 5) = 0
after factorisation,
we get,
=> (t – 5) (t +4) = 0
therefore,
(t – 5) = 0 or (t + 4) = 0
=> t = 5 or t = -4
but
we know that,
time can't be negative,
therefore,
t = 5
Hence,
the time taken by the policeman to catch the thief = 5 minutes
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____________
5 minutes
____________________
____________
step-by-step explanation :
Let,
the police catch the thief in ‘t’ minutes
Since,
the thief ran 1 min before the police start running,
therefore,
Time taken by the thief before he was caught = (t + 1) min
Now,
we know that,
Distance = speed × time
therefore,
Distance travelled by the thief in (t+1) min
= 100(t+1)m
now,
Given,
speed of policeman increased by 10 m/min
therefore,
Speed of policeman in the 1st min = 100 m/min
again,
Speed of policeman in the 2nd min = 110 m/min
and
Speed of policeman in the 3rd min = 120 m/min
Hence,
on observing the pattern,
we get, an A.P i.e.,
100, 110, 120….............
where,
a = 100
d = 110 - 100 = 10
nOw,
By the formula for sum of n terms of an AP,
we get,
Total distance travelled by the police in t minutes
=(t/2)[2a+(t – 1)d]
Putting the value of a, and d,
we get,
= (t/2)[2 x 100 +(t – 1)10]
now,
After the thief was caught by the police,
Distance traveled by the thief = distance travelled by the police
=> 100(t+1)= (t/2)[2 x 100 +(t-1)10]
=> 200(t + 1) = t[200 + 10t – 10]
=> 200t + 200= 200t + t(10t – 10 )
=> 200 = t(t – 1)10
=> t(t – 1) – 20 = 0
=> – t– 20 = 0
=> – 5t + 4t – 20 = 0
=> t(t – 5) + 4(t – 5) = 0
after factorisation,
we get,
=> (t – 5) (t +4) = 0
therefore,
(t – 5) = 0 or (t + 4) = 0
=> t = 5 or t = -4
but
we know that,
time can't be negative,
therefore,
t = 5
Hence,
the time taken by the policeman to catch the thief = 5 minutes