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A train met with an accident 60km away from station A. It co

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A train met with an accident 60km away from station A. It co [#permalink]

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New post 18 Oct 2011, 07:37
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A train met with an accident 60km away from station A. It completed the remaining journey at 5/6th of the original speed and reached station B 1hr 12mins late. Had the accident taken place 60km further, it would have been only 1hr late. what was the original speed of the train?

A. 60 km/hr
B. 55 km/hr
C. 65 km/hr
D. 70 km/hr
E. 48 km/hr
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2013, 14:09, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: train speed [#permalink]

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dreambeliever wrote:
A train met with an accident 60km away from station A. It completed the remaining journey at 5/6th of the original speed and reached station B 1hr 12mins late. Had the accident taken place 60km further, it would have been only 1hr late. what was the original speed of the train?

a. 60 km/hr
b. 55 km/hr
c. 65 km/hr
d. 70 km/hr
e. 48 km/hr



let the original speed be 6x. A/q, traveling 60 km at 5/6th of original speed costs him 12 minutes etc

60/5x =60/6x +12/60 -> x =10 , 6x= 60.

Ans A
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Re: train speed [#permalink]

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New post 18 Oct 2011, 09:10
My solution:
s=original spped, t=time taken in minutes to cover journey usually
(t+72-60/s)(5/6s)+60=(t+60-120/s)(5/6s)+120...........as total distance covered is same in both cases
solving, s=1 km/min=60km/hr
hence ans A
I know its too long.....but taking shortcuts was messing me up....
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Re: train speed [#permalink]

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New post 18 Oct 2011, 12:15
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dreambeliever wrote:
A train met with an accident 60km away from station A. It completed the remaining journey at 5/6th of the original speed and reached station B 1hr 12mins late. Had the accident taken place 60km further, it would have been only 1hr late. what was the original speed of the train?

a. 60 km/hr
b. 55 km/hr
c. 65 km/hr
d. 70 km/hr
e. 48 km/hr


The logical approach here is this:

The accident took place 60 km from A. The train was delayed by 1hr 12 min. If it had instead taken place 120 km from A, it would have been only 1 hr late. What does this 12 min gap in time signify? It is the delay caused by reducing the speed by (1/6)th over a distance of 60 km. The ratio of the speed over this 60 km stretch is 5:6 (speed reduced by a sixth : original speed)
Time taken 6:5 (Time taken with reduced speed : original time taken)
We know this difference of 1 in time taken is given by 12 min. So original time taken is 12*5 = 60 min = 1 hr
Time taken is 1 hr to cover a distance of 60 km so speed = 60 km/hr
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Re: train speed [#permalink]

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Agree with A but took too long to reach the solution:(
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Re: train speed [#permalink]

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New post 22 Oct 2011, 08:26
Let the original speed be v .

In case 2, the 60 kms further which the train traveled was with speed v.
In case 1, these 60 kms (which in case 2 the train traveled further where accident happened) was traveled at a speed of (5v)/6. And this caused the extra 12 mins delay in case 1.

So,

60/v +12/60=60/((5v)/6)
v=60
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Re: train speed [#permalink]

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New post 24 Oct 2011, 09:18
my sol. was just too lengthy..

let s be the original speed and x+60 be total distance between A and B:
60/s + x/(5s/6) = (x+60)/s + 72/60

Also, 120/s + (x-60)/(5s/6) = (x+60)/s + 1

subtracting the 2 equations: 60/s + 60/(5s/6) = 12/60
therefore s = 60.

thanks Everybody!!
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Re: A train met with an accident 60km away from station A. It co [#permalink]

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Re: A train met with an accident 60km away from station A. It co [#permalink]

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New post 28 Jan 2015, 20:03
Hi all,

Can someone please help me think through this problem?

Given:
It arrives 1 hr 12 min late if the accident took 60 km further from A
It arrives 1 hr late if the accident took 120 km further from A

It implies:
the two "accident spots" are separated by 12 min and 60 km

Hence:
R = (60)*(60/12) = 300 kmph
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Re: A train met with an accident 60km away from station A. It co [#permalink]

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New post 28 Jan 2015, 23:51
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thorinoakenshield wrote:
Hi all,

Can someone please help me think through this problem?

Given:
It arrives 1 hr 12 min late if the accident took 60 km further from A
It arrives 1 hr late if the accident took 120 km further from A

It implies:
the two "accident spots" are separated by 12 min and 60 km

Hence:
R = (60)*(60/12) = 300 kmph


You are assuming that the train covered 60 km in 12 mins. That gives you the speed as 300 kmph. But is that really the case? 12 mins is the difference in the time taken to cover 60 kms at two different speeds. It is the delay caused by reducing the speed by (1/6)th over a distance of 60 km. The ratio of the speed over this 60 km stretch is 5:6 (speed reduced by a sixth : original speed)

Time taken 6:5 (Time taken with reduced speed : original time taken)

We know this difference of 1 in time taken is given by 12 min. So original time taken is 12*5 = 60 min = 1 hr

Time taken is 1 hr to cover a distance of 60 km so speed = 60 km/hr
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Re: A train met with an accident 60km away from station A. It co [#permalink]

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New post 05 Dec 2015, 12:40
oh man....took some time to solve it...did it in a very long way...
time the train would have done if no accident happened D/x

time when accident happened 60km away:
60/x + 6*(D-60)/5x
now, the difference is 1h12m or 6/5 hours

60/x + 6*(D-60)/5x - D/x = 6/5
rewrite everything as:
(300+6D-360-5D)/5x = 6/5

or (D-60)/5x = 6/5
cross multiply:
30x = 5D-300


now, if the accident happened 120km away:
time traveled before accident 120/x
time after accident:
6*(D-120)/5x
total time: 120/x + 6*(D-120)/5x

now, in this case, the train arrived 1 hour late:
120/x + 6*(D-120)/5x - D/x = 1
rewrite everything:
(120*5 +6D-720-5D)/5x = 1
or
D-120/5x = 1
or 5x=D-120

ok, now we have two formulas:
30x = 5D-300
and 5x=D-120
multiply the second one by 6
30x=6D-720

substitute 30x in the first equation with 6D-720
6D-720 = 5D-300

D = 420

ok, looks good.

5x=D-120
5x = 420-120
5x = 300
x = 60


A
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A train met with an accident 60km away from station A. It co [#permalink]

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New post 06 Dec 2015, 14:44
t=normal time
s=original speed
d=distance
E1: t+6/5=60/s+(d-60)/(5s/6)
E2: t+1=120/s+(d-120)/(5s/6)
subtracting E2 from E1,
1/5=-60/s+360/5s
s=60 kph
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A train met with an accident 60km away from station A. It co [#permalink]

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New post 06 Dec 2015, 15:59
d=distance
s=original speed
(d-60)/(d-120)=(6/5)/1
d=420 k
(420/s)+1=(120/s)+300/(5s/6)
s=60 kph
A train met with an accident 60km away from station A. It co   [#permalink] 06 Dec 2015, 15:59
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