Question 1170459: Hi
A train after travelling 50km meets an accident and and after reducing 3/4 of its former speed arrives 25min late. If the accident occurred 24km behind it would have arrived 35 minutes late.
What is the speed and distance travelled by the train.
Thanks
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! A train after travelling 50km meets an accident and and after reducing 3/4 of its former speed arrives 25min late. If the accident occurred 24km behind it would have arrived 35 minutes late.
What is the speed and distance travelled by the train.
let speed of train be x km/h
Let total distance be y km
A train after travelling 50km meets an accident
Time for 50 km
50/x
and after reducing 3/4 of its former speed arrives 25min late.
speed now = 0.25x
Time for remainng journey
(y-50)/0.25x
50/x + (y-50)/0.25x =y/x +5/12
multiply by x
50+ y/0.25 -50/0.25 =y +5x/12
Multiply by 12
600 +48y-2400= 12y +5x
36y-5x =1800................................................(1)
Multiply by x
26/x +(y-26)/0.25x = y/x + 7/12
26 + y/0.25 -26/0.25 = y +7x/12
Multiply by 12
312 +48y-1248 = 12y +7x
36y-7x = 936......................................................(2)
solve 1 & 2
speed = 432 km/h distance = 110 km
Answer by ikleyn(52790)  (Show Source):
You can put this solution on YOUR website! .
Hi
A train after travelling 50km meets an accident and and after reducing 3/4 of its former speed arrives 25min late.
If the accident occurred 24km behind it would have arrived 35 minutes late.
What is the speed and distance travelled by the train.
Thanks
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I write these lines for the future generations of students who will read this post.
It is a classic/standard well known problem, but in the arrived post it is given in mispresented/perverted form.
Its standard/canonic formulation is THIS
A train after traveling for 50 km meets with an accident and then proceeds
at 3/4 of its former speed and arrives at its destination 35 minutes late.
Had the accident occurred 24 km farther, it would have reached the destination
only 25 minutes late. What is the speed of the train?
In this formulation, you can find this problem in many sources in the Internet
https://gmatclub.com/forum/a-train-after-traveling-at-50kmph-meets-with-an-accident-and-then-189660.html
https://www.beatthegmat.com/a-train-after-traveling-for-50-km-meets-with-an-accident-t303201.html
https://www.toppr.com/ask/en-us/question/a-train-after-travelling-50-km-meet-with-an-accident-and-then-proceeds-at-34th/
As you see, the formulation given in the arrived post is DIFFERENT from the classic.
So, it is not surprised that the numbers in the @mananth solution are unrealistic.
Again, for the right formulation and an adequate solution see the referred sources.
It is often happens at this forum that the problems come mutilated beyond recognition.
In that formulation in the arrived post, I would consider this " problem "
as a garbage, or as a unfortunate misunderstanding,
or as a fruit of creativity of unprofessional Math composers,
who are not responsible and can not be responsible for their compositions
(and even do not understand the meaning of these words).
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