The Train And The Cyclist
A railway track runs parallel to a road until a bend brings the road to a level crossing. A cyclist rides along the road everyday at a constant speed of 12 miles per hour. He normally meets a train that travels in the same direction at the crossing.
One day, the cyclist was late by 25 minutes and met the train 6 miles in front of the level crossing. What is the speed of the train (in miles per hour)?
[This problem is not clearly phrased.]
The answer is 72.
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6 solutions
There is some mistake in the question. The train is supposed to be late by 25 minutes, rather than the cyclist being late.
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The question is perfectly fine and without proper visualization you are bound to make a mistake. Check the solution of Ben. I hope some one can make an illustrated diagram.
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No, the question is wrong. Say the train is meant to reach the crossing at 8:00 am. That means that at 8:00 am, both the cyclist and the train are at the crossing. However, if the cyclist is 25 minutes late, that means that he reaches the crossing at 8:25. at 8:25, the train is now (its speed * 25/60) miles ahead of the crossing. If the cyclist is able to catch up to the train and meet it -after- the crossing, then he is going faster than the train. If you solve for the train's speed assuming they meet 6 miles after, like the problem says, you get 6.5454 mph. The question is meant to say that cyclist meets it 6 miles -before- the crossing. Indeed, if you look at some of the solutions in here, they all make that assumption.
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@Connor Lemp – In normal circumstances the cyclist meets the train at the crossing, so when he is late how can he meet the train before crossing?? I guess the question should read "train was late by 25 minutes".
@Connor Lemp – The question does say that the cyclist meets the train before the crossing. "One day, the cyclist was late by 25 minutes and met the train 6 miles ahead of the level crossing." This makes sense if you picture the train catching up to the biker earlier when the biker is running late.
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@Ben Resnick – The way you stated does allow for a reasonable solution, but it is not what is implied by the problem. At least in my experience, whens someone says location blab is 3 miles AHEAD of location zoob, that means I should get to zoob and go 3 miles further to reach blab. If you interpret the problem this way you get the wrong answer. If you mentally swap "ahead" for "before", you can get a sensible answer, but it's a little ridiculous to expect solvers to have to interpret and correct the wording of problems as they solve them.
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@Connor Lemp – I understand your point, and it could be worded clearer. Ahead can mean different things in different contexts; I thought it was ok based on this context and definition of "ahead of" in the dictionary: "in front or advance of." synonyms include "before". http://www.merriam-webster.com/dictionary/ahead%20of In a different context, such as "he is 2 miles ahead of me on the run," the term means farther along.
@Ben Resnick – Question is definitely wrong. There is clearly said "cyclist was late". If the train is faster and cyclist is late, they would never meet as they are going to same direction.
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@Humaid Ashraf – Nope it's not wrong, the question says that the cyclist and the train meet 6 miles BEFORE the crossing, which means the train passes the cyclist and arrives at the crossing 5 minutes after that.
@Humaid Ashraf – Let me ask you this. Suppose they normally meet at the crossing at 8am. If the biker is running 25 minutes late, that means he will reach the crossing at 8:25 am. Where is he at 8 am when the train is at the crossing? And what time is it when the train passes the bike 6 miles before the crossing?
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@Ben Resnick – If you say anything that someone doesn't understand, it doesn't make you brilliant. Say something that makes sense. Like "Question is wrong" that's it.
@Ben Resnick – The way problem states, The biker would be 5 miles before the crossing at 8 am. and they would meet at 8:55 am in 6 miles ahead of the crossing.
as Connor pointed - You are saying "before" in place of "ahead" - but that was not stated in the problem.
How a late cyclist could meet the faster train ahead of the usual place? wrong question made this problem rating high.
im probably wrong somewhere but from what i understand of the sum the cyclist is faster than the train as they meet after the crossing not before my answer goes as follows
givens distance (train)= 6m time (train)= (y+25)/60 speed = ? d(cyclist) = 6 m t(cyclist) = y/60 s(cyclist)= 12 m/hr t (c) = d(c)/ s y/60 = 6/12 y= 30 mins therefore t(train)= 25+30/60 s(t) = 6 x 60/55 and we get an irrational number
The question is definitely worded wrong. If it were worded correctly you should have gotten something like (30+25)/60= .916. Then take the distance (6 miles) divided by the time (.916 hours) to get the rate (6.55 mph) roughly.
I think the time taken by the train should be 55 (30+25) minutes, instead of 5 minutes. So speed of train will be 6*60/55 miles/hour.
[If train is faster than the cyclist, then it is not possible for him to catch the train after getting late].
Please tell me if I have gone wrong somewhere.
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I found it is easy to make a mistake without visualizing it properly.
We want to know the time and position at 2 points for the train to calculate speed.
Where is the biker at the time the train is at the crossing?
25/60 hours x 12mph = 5 miles behind the crossing.
How long before this did they meet 6 miles behind the crossing?
1 mile / 12mph = 1/12 hours
Thus the train went 6 miles in 1/12 hours, 6*12=72
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Can you explain how 25/60 came
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@Kavita Parmar – It's a unit conversion from minutes to hours. 25 minutes = 25/60 hours.
The question should say like the train was late ...else its same direction and late cyclist catching up with train means that its slower than the cycle ..so the question has to be corrected
wow the ques was wrong -_-
the cyclist is late.....not the train.....
Suppose that the train and the cyclist meet everyday at the crossing at 8:00A.M. i. e; starts at 7:00A.M Since the cyclist is late by 25 minutes, he starts at 7:25A.M As his speed is 12 miles per hour, he reaches the crossing at 7:25A.M + 1 Hour = 8:25A.M By 8:30A.M the train is 6 miles ahead of the cyclist The difference between their timings = 8:30A.M – 8:25A.M = 5 Minutes The difference between their distances = 6 Miles Therefore,the train travels 6 miles in 5 minutes In 1 minute it travels ---------------------? = [(1 * 6) /5] * 60 = 72 Miles/hour
ha ha ha ,comedy explanation that is ...
Call S is distance from bicycle's student's home to cross bend, V is velocity of the train. So have: 1 2 S − 6 × V + 6 + 1 2 5 × V = 1 2 S × V => V=72.
Suppose normally the cyclist meet train on the crossing at 6.00 PM. By the speed of cyclist , he is travelling 1 mile in 5 minutes and he was 25 minutes late , means at 6.00 PM he was 5 miles away from crossing. But he met train at 6 miles ahead , so he met train at 5.55 PM. The train is right time, means it will cover 6 miles in 5 minutes to reach the crossing at 6.00 PM. Which gives the speed of 12*6= 72 miles per hour.
It made sense for me when I used a concrete number for the time. Also, I assumed that the meaning of ahead was before or "in advance of".
Let's say the train and bike meet @ 1 o'clock today, 6 miles before/ahead of the railroad crossing.
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Because the bike is traveling @6mph, we know that the cyclist will get to the crossing @1:30.
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Because the cyclist is 25 minutes late, we know he normally will hit the crossing @1:05.
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We know then that the train will cross @1:05.
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Since the train is going to travel 6 miles in 5 minutes, we find it is traveling @72mph.
Let the distance between starting point of Train and Cyclist be x. In the first case, let the time taken to reach the crossing=t and by equating the distances, we get: 12 t +x = v t (where v is the velocity of the Train)...eqn 1 In the second case, let the time taken by the train be t' and equating the distances again, we get: 12 t+x-6=v t'....eqn 2 Time taken by cyclist in second case is t'-25/60 or t'-5/12. Again equating the distances traversed, we get: 12(t'-5/12)+x=v*t'...eqn 3 Solving the above 3 equations gives us v=72
Yeah, the word ahead is misleading but a bit of drawing helps...I wish I could have posted this one with a drawing!
cyclist covers 6 miles in 30 min.as time lag is 25 min,the train covers 6 miles in 5 min. =>speed of train is 6/(5/60)=72 miles/hour