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Lisa gets picked up from work by her husband at exactly 4 pm every day. One day Lisa finished work half an hour earlier and started walking to meet up with her husband. She met her husband on the way, and they drove home together. When they arrived home Lisa noticed that they got back 10 minutes earlier than usual.

For how long did Lisa walk?

A few notes:

  • Her husband drove from home
  • The route is symmetric with regards to speed/distance
  • The car can take an instantaneous U-turn
  • Lisa didn't tell her husband that she left early
  • It's neither more nor less traffic than usual.
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    $\begingroup$ Stewie Griffin asking a question about Lisa (Simpson?). Another crossover episode here? $\endgroup$
    – CodeNewbie
    Sep 21, 2015 at 12:48
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    $\begingroup$ For what it's worth, I think the many variables is a non-issue. There are more puzzles here that seemingly have missing variables. But I think there is one important rule when it comes to puzzles: If you believe information is missing just assume the most straightforward scenario. Making up some variables can have a direct influence on the solution, for example, say it was stated "picking up Lisa and turning the car around takes X minutes". But just the fact that it isn't stated you may always assume that this takes no time, except when a question is tagged lateral-thinking of course. $\endgroup$
    – Ivo
    Sep 21, 2015 at 13:39
  • $\begingroup$ @StewieGriffin - Sorry if I am being a wet-blanket. It's not personal. I think we have to minimally say whether a question belongs to a fantasy world or to the real world. $\endgroup$
    – chasly - supports Monica
    Sep 21, 2015 at 13:40
  • $\begingroup$ @IvoBeckers - I understand your point of view. I've been caught out on this site with puzzles that I set that weren't sufficiently specified. I've gradually come to change my view. I now believe that, because this is a specialist puzzle site, the standards need to be especially high. I'm fully aware that opinions differ on this and I'm sure my downvote is likely to be cancelled by one or more sympathetic upvotes. $\endgroup$
    – chasly - supports Monica
    Sep 21, 2015 at 13:46
  • $\begingroup$ Does Lisa always work until she gets picked up by her husband? $\endgroup$
    – HelloGoodbye
    Sep 21, 2015 at 19:35

4 Answers 4

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Lisa walked

25 minutes.

Because:

To arrive home 10 minutes earlier than usual, her husband has to drive 5 minutes less on each way (bidirectional). So he picked her up at 3:55 pm. Since she started walking at 3:30 pm, she had to have been walking for 25 minutes in order for the above condition to be true.

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    $\begingroup$ What's funny is that this also means that Lisa is only 1/5th slower than the car, so Lisa walks pretty fast I think or the car is slow $\endgroup$
    – Ivo
    Sep 21, 2015 at 11:28
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    $\begingroup$ I think Lisa is pretty athletic and her husband is in the rush hour ;) $\endgroup$
    – Wa Kai
    Sep 21, 2015 at 11:30
  • $\begingroup$ Sorry that I down-voted but I agree with the people in the pub. This answer makes too many unwarranted assumptions. $\endgroup$
    – chasly - supports Monica
    Sep 21, 2015 at 13:20
  • $\begingroup$ @WaKai I would definitely delete the math section...it just confuses the issue, and is unneeded. In fact, your math requires the same inferences that you made in the logic section just to get it to work out, making the math irrelevant. See my answer for some mathematical rigor. Ultimately, unless you make your logic deductions, it's impossible to solve. $\endgroup$
    – dberm22
    Sep 21, 2015 at 13:54
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User Wa Kai was much faster, but I cannot make myself delete this, having spent so much time figuring it out.

To arrive 10 minutes earlier, the husband has to drive 5 minutes less each way. Thus, they met 3:55, after Lisa walked 25 minutes.

However, this only works under the assumption that her husband departs from their home, which was, for me, not clear from the text.

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Solution:

This two-dimensional graph shows time vertically and position along the route horizontally. The car's position normally follows the green line. Today Lisa's walk is light blue and the return trip home is blue.

a) when husband normally leaves home. i) when he normally picks her up. e) when they normally get home. d) when they got home today.

edgi is a parallelogram: home and work do not move, so these lines are both vertical and therefore parallel. eg and ei are likewise parallel as the rate of traffic doesn't differ (given). de is 10 minutes, so gi is 10 minutes.

Since the rate doesn't differ, the angle of dgi and aig the same. This makes ghj and hij symmetric, so gh and hi are the same length. gh and hi are both 5 minutes.

The required distance fh is therefore 30-5=25 minutes.

enter image description here

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So, the answers above say that he met her at

15:55?

How could that be if he did not know that she was leaving early.

Seems to me like she would be walking at least

30 minutes until it's 16:00 (4 PM) and first then she has a chance of meeting her husband along the road -- a road which is now a bit shorter.

For example:

NORMAL 16:00 work---60---home => 90km/h = 40mins (=16:40). TODAY 15:30 work---15---pickup---45---home => 30km/h (30 mins walk) + 90km/h (30 mins drive) (=16:30). (fast walking, I know)

But now, we can certainly tweak those variables, and end up with other answers:

NORMAL 16:00 work---40---home => 40km => 40km/h = 60mins (=17:00). TODAY 15:30 work---10---pickup---30---work = 35 minute walk (17km/h) and 45 mins drive for 30 km with 40km/h, and we arrive at 16:50, so, she got picked up at 16:05, and has walked for 35 minutes (with an insane speed of 17km/h).

I think the answer is that there is no answer until we get more variables (her walking speed, total distance, car walking speed).

(I'm sorry for/if not following SE conventions here. Personal first post.)

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    $\begingroup$ The husband usually arrives at exactly 4 pm. To do this, he must leave home before 4, so it is possible for them to meet on the road before 4. $\endgroup$
    – f''
    Sep 21, 2015 at 20:38
  • $\begingroup$ Oh, drat. And that does constrain the answer. Never mind, move along, nothing to see here. $\endgroup$
    – Sedulus
    Sep 21, 2015 at 20:59

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