SOLUTION: Bev takes a train home at 4:00, arriving at the station at 6:00. Every day, driving at the same rate, her husband meets her at the station. One day, she takes the train an hour ear

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Bev takes a train home at 4:00, arriving at the station at 6:00. Every day, driving at the same rate, her husband meets her at the station. One day, she takes the train an hour ear      Log On

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Question 1062445: Bev takes a train home at 4:00, arriving at the station at 6:00. Every day, driving at the same rate, her husband meets her at the station. One day, she takes the train an hour early and arrives at 5:00. Her husband leaves home to meet her at the usual time, so Bev begins to walk home. He meets her on the way and they arrive home 20 min earlier than usual. How many minutes did Bev walk?
Here is how I solved the problem: I think the time of 2 hours spent traveling between the Work and Train Station is irrelevant to the problem. The key to the problem is the second situation, when her husband picks her up on a way from work, when he found her walking home, because she arrived early. So TIME they meet is critical to find. But how do we do that? I think the distance husband spends driving until he finds her walking and the distance she is walking equals the usual distance that picks her up all the time (that would be the distance between home and train station). So, if he arrives home with her at the usual time at t=x hours, the time it will take him to arrive home with her 20 min earlier is t=(x-0.333) hours. But how do I find the time husband meets wife walking home from work to find the times she spent walking towards her husband. I figured, they walk TOWARDS EACH OTHER, so I was going to use the same reasoning than when any two people walk towards each other or travel on the same line, while walking towards each other. That's why I set the first equation to d(H)+d(B)=usual distance spent driving to the train station from home, where d(H) is distance of husband driving to the train station from home and where d(B) is the distance wife walks home from the train station. But in the first situation and (thus, this first equation), the wife just waits at the train station for the husband to arrive, so she is not walking, because she knows he is coming to pick her up at 6:00. So we can't really use this first equation, because her distance walking would be zero and the equation will thus reduce to express only the distance that husband travels and that is the total distance traveled one way, meaning half the loop (NOT whole loop, or NOT on a first round), so we need to double that distance to make it WHOLE LOOP (FULL LOOP). SO HUSBAND TRAVELS d(H)=whatever that full distance is that he drives to the train station and then home with the wife after picking her up at the train station. So that distance from Home to Train Station is a certain unknown number, but this number is the same than the distance when husband travels to meet wife at the train station when she arrived early and began walking home. So, it's the same distance in both situations. But how do we figure out, when do they meet? What is the t(meet)? And then what is the time that she spends walking home before husband picks her up? So, in this situation, they are both heading TOWARDS EACH OTHER, so we can use the second equation: d(H)+d(A)=total distance between home and train station. So, if I can figure out what the total distance or time is in a first situation (from home to train station) and what husband's time is in second situation, I can subtract the two, and get the time she spent walking home before husband picked her up. I can set up the second equation and just say: Since Bev is walking at y miles/hour and meets husband in (x-0.333 hours), we have Y miles/hour TIMES (x-0.333) hours to find her distance she walks; we also have her husband's speed at z miles/hour and his time traveling till he meets her, which is (x-0.333) hours, so husband's distance is z miles/hour TIMES (x-0.333) hours. When adding the husband's distance and Bev's distance, we get a distance from train station to home. But what is that distance? Can I somehow get it from the first equation, if all I know is that those distances equal?
Thank you very much and Merry Christmas to you!

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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When you read my solution to the end, you will laugh loudly:

          SO     SIMPLE   IT   IS   !

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What we need to do is to extract some (or all) logical consequences from the condition.

1.  Since the husband returned home with Bev 20 minutes earlier than usual, he spent 10 minutes less time in each direction than usual. 


2.  It means that the husband met Bev not at 6:00 pm at the station as usual, but he met her walking from the station at 5:50 pm.


3.  Hence, Bev was walking from 5:00 pm till 5:50 pm, i.e. 50 minutes.

Answer. How many minutes did Bev walk? - 50 minutes. From 5:00 pm till 5:50 pm.


          S  O  L  V  E  D  !


Thanks  for  sending  so  enjoyable  problem  !
Surely,  it is not a Math,  but is a brilliant training exercise for a brain  !

Brilliant problem for the day of April, 1.