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If a man walks at the rate of 5 kmph, he misses a train by 7 minutes.

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Bunuel
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If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   10 Jan 2020, 05:34
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If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the departure of the train. Find the distance to the station.

A. 2 km
B. 3 km
C. 4 km
D. 5 km
E. 6 km
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E
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cantaffordname
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Re: If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   10 Jan 2020, 05:50
There's a 12 min difference in the times it takes him to walk the distance d at the 2 different speeds, so

d/5 – d/6 = 12/60 = 1/5 ............ multiply by 30

6d - 5d = 6

d = 6 km.
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Re: If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   10 Jan 2020, 07:28
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Let the distance to station be d kms

5 kms in 60 min
d kms in = 60/5*d = 12d min

6 kms in 60 min
d kms in = 60/10*d = 10d min

12d-10d = Difference in times = 12
(the 12 comes from 7 minutes late and 5 minutes early, if time to reach is t min, then t+5-(t-7) = 12)
2d = 12
d=6 kms

Ans: E
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If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   10 Jan 2020, 14:54
I used the options:

Difference between the times must be 12 minutes.

Starting always with C.

C) 3km/5kmph = 36 min 1/2 = 30 min --- Discard

D) 2km/5kmph = 24 min 1/3 = 20 min --- Discard

E) 6km/5kmph = 72 min 6/6 = 60 min --- Correct one

E
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Re: If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   20 May 2021, 23:49
Bunuel wrote:
If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the departure of the train. Find the distance to the station.

A. 2 km
B. 3 km
C. 4 km
D. 5 km
E. 6 km


We know, \(D=\frac{Δt×S1×S2}{S1–S2}\)

So, \(D=\frac{12×5×6}{60×1}\)
=6 km
Answer is E

Posted from my mobile device
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Re: If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   03 Jun 2021, 03:02
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60 mins ------------- 5 km
1 min ---------------- 5/60 = 1/12 km/min

60 mins ------------- 6 kms
1 min --------------- 6/60 = 1/10 km/min

speed x time = distance

(t+7)/12 = d

(t-5)/10 = d

(t+7)/12 = (t-7)/10

t=65 mins

d = (t+7)/12 = 72/12 = 6 kms (E)
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If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink] New post   16 Mar 2024, 15:28
IMO E


\(­d = t*s\)


below, x is the distance av. time needed to reach the station

(1)  \(d = 5 ( x + \frac{7}{60} ) \)                      (First person walks at a speed of 5 kmph , and arrives 7 minutes late, which is 7/60 hours)

(2)  \(d = 6 ( x - \frac{5}{60} )     \)                   (Second person walks at a speed of 6kmph , and arrives 5 minutes early , which is 5/60 hours) 



(1) = (2)

\(5x + \frac{7}{12} = 6x - \frac{5}{10}\\
\\
x = \frac{65}{60} \)


Replace \(x=\frac{65}{60}\) in (1) or (2) and get d = 6 mins which is option E 
 ­This was my solution, hope you liked it.
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If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. [#permalink]
16 Mar 2024, 15:28
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