Brain Teasers
Remember That Train Scene In 'Stand By Me'?
A man is standing on a railway bridge joining locations A and B. He is 3/8 the distance across, walking from A to B. He hears a train approaching behind him (from A's side); it is traveling 80 kph (that's kilometers per hour for you non-metric fans!) If he were to run back to A, he would meet the train at A. If he were to run to B, the train would overtake him at B.
How fast can the man run?
How fast can the man run?
Hint
You will never know exact distances in this problem, nor will you actually need to. Remember distance equals rate times time, or D = RT. I recommend drawing a picture to map out other distances, such as the man to A and B, respectively.Answer
Start by assigning some variables. We have three - the distance from the train to A, which I will call x; the distance from A to B, which I will call y; and the rate of the running man, in kph, which I will call z.The man is 3/8 the distance of A to B (which is y), so he is 3/8 y from A and 5/8 y from B (please note that the 'y' belongs in the numerator - the top of the fraction - not the bottom, so to be truly technical it is (3y)/8 and (5y)/8).
The time it takes the train to reach A is its rate divided by the distance, or 80/x. The time it takes the man to reach A would then be z/(3/8 y). Since they meet there, 80/x = z / (3/8 y). Solve for z, and simplify, to get z = 30y /x.
The time it takes the train to reach B is 80/(x + y), and the time it takes the man is z/(5/8 y). Again, since they meet there, 80/(x + y) = z/(5/8 y), and we get z = 50y /(x + y).
Set these two z-values equal to each other, so that 30y /x = 50y / (x + y), and cross-multiply. You then get 30xy + 30 y^2 = 50xy. Regroup to get 30y^2 = 20xy. Divide by 30y to get y = 2/3 x. Don't worry about y being 0 and making this pointless - y is a distance, and distances *can't* be 0, right?
So what does that do for us? Well, y is 2/3 of x. That means that the distance from A to B is 2/3 the distance from the train to A. Now, the man himself is 3/8 the distance from A to B (y), so he is 3/8 of 2/3 x away from A, or just 1/4 x away from A. So, in the time it takes the train to travel x, the man travels 1/4 x. So the man is running at 1/4 the speed of the train, or 20kph.
Feel like checking? OK. The train travels x and 2/3 x to get to B, or just 5/3 x. The man travels 5/8 y, or 5/8 (2/3 x), which is 5/12 x. And 5/3 divided by 5/12 is 1/4 again, so there you go.
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I solved this a little differently. I said the speed with which the man and the train approach each other is either (80 + M) if the man runs toward the train, or (80 - M) if he runs away. We know that running three "units" at the faster speed leads to a collision, as would running 5 units at the slower speed. So the equation I set up was: 3 * (80 + M) = 5 * (80 - M). Solving for M, you get M = 20 kph, just as the answer stated. Great question.
I just reasoned that when the train gets to A, the man will have moved 3/8 of the bridge, to either point A or to 6/8 of the way across. This means that the train will travel 8/8 of the bridge in the time it takes him to run 2/8 of the bridge. The train is 4 times faster.
Of course, to solve this we have to assume the train doesn't slow down when the engineer sees the man on the bridge.
Gizzer.. that's actually completely negligible... because, the scary things about trains is, they take over a mile to come to a halt. So basically, if your car stalls in front of one... GET OUT, and don't try to start it!!!! I mean, if you've got a few hundred feet, be smart.. get out and push it (and, don't be dumb and forget to put it in neutral )
The *real* answer, though, is quite obvious. "Not fast enough." Hehe..
Oh, and another comment... he's FAT! What kinda fatso can only run 20kph?!? I could run 12MPH (19.3kph) when i was 6 years old.. lol.
The *real* answer, though, is quite obvious. "Not fast enough." Hehe..
Oh, and another comment... he's FAT! What kinda fatso can only run 20kph?!? I could run 12MPH (19.3kph) when i was 6 years old.. lol.
too.......much........algebra.........
I mean i just did my homework!
I mean i just did my homework!
too.......much........algebra.........
I mean i just did my homework!
I mean i just did my homework!
Frist of all the formula used for the velocity (speed) in the explanation of this puzzel is not correct.
The correct formula is
Speed=Distance / Time;
Now consider: when the man hear the train, it (train) is at distance Ω from the point A. The total length of the bridge is X. The speed of man is S.
Let T1 be the time when train overtakes the man at point B
Hence we get two equations:
T1 =5*X / 8*S
T1 = (X+Ω) / 80
Let T2 be the time when man catches the train at point A
Hence we get two equations:
T2 = 3*X / S
T2 = Ω / 80
Solve this four simultaneous equations to obtain the value for variable S
and you will find S = 20
Therefore the speed of the man is 20 KPL.
The correct formula is
Speed=Distance / Time;
Now consider: when the man hear the train, it (train) is at distance Ω from the point A. The total length of the bridge is X. The speed of man is S.
Let T1 be the time when train overtakes the man at point B
Hence we get two equations:
T1 =5*X / 8*S
T1 = (X+Ω) / 80
Let T2 be the time when man catches the train at point A
Hence we get two equations:
T2 = 3*X / S
T2 = Ω / 80
Solve this four simultaneous equations to obtain the value for variable S
and you will find S = 20
Therefore the speed of the man is 20 KPL.
Ummm.. you just proved his point. He got the same answer. You can solve for velocity using velocity/distance, or distance/time. Either way.
I got it eventually, but was confused at first because on a quick reading of the teaser (first mistake!) I thought the bridge went across the railway - not many railway bridges run in the same direction as the railway! Anyway, nice teaser, even if the given asnwer is 3 times longer than necessary - no algebra really required, as earlier comments have shown.
i love hard teasers .......but come ON! THAT! i took fourtylenol this morning but now i have a killer headache
This problem was on a PA Math League test!
I_am_the_Omega: 20 kph is actually not TOO slow.
20 kph = 20 000 meters/h = 5.555 meters/second.
That's about 18 seconds for the 100 meter run.
Actually that is kind of slow.
20 kph = 20 000 meters/h = 5.555 meters/second.
That's about 18 seconds for the 100 meter run.
Actually that is kind of slow.
All this heavy algebra is not needed. Assume the bridge is 80 km long (That's kilometres not kilometers. A meter is an instrument for measuring things like a barometer or a thermometer whereas a metre is a unit of length and a litre is a unit of capacity). Where was I? If he runs back 30 kilometres the train is at the beginning of the bridge and so it is if he runs 30 km forward. Thus he has 20 km to run (forward) in the time the train takes to cover the whole bridge which is 80 km. Thus he is one quarter as fast as the train = 20 km/h.
same as realhair and jimbo. great teaser though.
Why do people feel compelled to make these problems so difficult? As Realhair points out, it takes him the same time to run 2/8 of the bridge as it takes the train to cross the whole bridge, so the man is running 2/8 the speed of the train.
Sheesh. Had the answer by the time I finished reading the question.
Sheesh. Had the answer by the time I finished reading the question.
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