This girl we know was interviewing for a CFO position last week, and she was given a problem that we've been arguing about all weekend. According to her interviewer, this can be solved with a one variable equation. I think the guy is full of crap, so I'm seeing if anyone here can figure it out. Here is the Q: A guy is walking across a bridge that is 150 feet long. He is a third of the way down the bridge when a train is heading towards him from behind. If he walks at the train, he'll nail it at the beginning of the bridge. If he runs away from it, he'll get nailed at the end of the bridge. How far away is the train?
Train (x) - Beginning of bridge - Guy (y) - end of bridge x---x_distance---|----y----------| What's the speed at which the guy's running? Remember, you can have a two-variable equation that can be transformed into a ONE-variable equation: TWO VARIABLES to get N: x+y=n x = (n-y) Turns into ONE variable to get N: (n - y ) + y = n
I take that back, J-bone. I don't even know the speed of the train, so it could be a constant on both sides: the speed of the guy, the speed of the train. Yet confusing'er.
Why doesn't he just jump off the bridge? Even if it's over land, he'll probably have a better chance of surviving than getting hit by a train. Better yet, hang off the side of the bridge, then reach over and grab the wooden planks in the middle and hang from those. If the train isn't too long, he should be able the hang until it passes.
I suspect the answer is 150 feet, here is my thinking: Let the distance between the train and the beginning of the bridge be: d Let the the speed of the train be: s Now, suppose the time it takes for the unlucky guy to get to the beginning of the bridge is: t seconds, then it would take 2t seconds for the guy to get to the end of the bridge. We also know that from where the training is right now, it takes the train t seconds to get to the beginning of the bridge, and we also know that it takes the train 2t seconds to get to the end of the bridge, since distance/speed = time, we now have 2 equations: d (1) --------- = t s (2) d+ 150 -------------- = 2t s Now, if we divide equation (1) by equation (2), we get: d (3) ------------ = 1/2 d + 150 Equation 3 only as a single variable, and solving it gives you d = 150. Does this make sense?
Oh, looks like my formatting was screwed up, here are the 3 equations again: (1) (d/s) = t (2) ((d+150)/s) = 2t (3) = (1)/(2) => (d/(d+150)) = 1/2 => d = 150 This is a math problem, not a brain teaser.
Well, when you see a training coming at you, you will run at your top speed, no matter in which direction. Yes, you are right in pointing out that I assumed that the man's speed and the train's speed are constant. If they're not, then the equations have an infinite number of solutions.
The person is correct. It is one variable equation. Z = distance train is from bridge T = speed of train X = speed of man 50/x = Z/T 100/x = (Z+150)/T so 2Z/T = (Z+150)/T 2Z = Z+150 Z = 150 Guess someone already solved it earlier. That's what you get for not refresh the page!
Somebody hasn't seen Stand By Me... O'Connell trips bc he's the fat character, then Cory Feldman and River Phoenix have to save his ass.
newplayer, what is the speed of the man (a) walking, or (b) running? Dude. First of all, she's in 3rd grade. Second, I can... but I just want to teach her a lesson: everyone knows math. Fourth, I never thought you'd bring my personal sh*t into a thread . Fifth, see if I ever email you through the board to hire you again. "Don't I just feel like a f*ck*ng a-hole?" - macalu/jessip I kid, freaker. She's doing a lot better.
LOL...b*stard. had me going there for a minute. but if there was a single bit of truth to my insensitivity, my apologies.
