SOLUTION: Each day a man meets his wife at the commuter train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an e

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Question:
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Each day a man meets his wife at the commuter train station after work, and
 then she drives him home. She always arrives exactly on time to pick him up.
 One day he catches an earlier train and arrives at the station an hour early.
 He immediately begins walking home along the same route the wife drives.
 Eventually his wife sees him on her way to the station and drives him the
 rest of the way home. When they arrive home, the man notices that they
 arrived 20 minutes earlier than usual. How much time did the man spend
 walking?


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Establish variables to be used and context for use:
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T = Time index he normally arrives at station
w = walk-time for man to walk until intercepted
c = car-time for man and woman to normally drive from station to home
x = time walking-man and driving-woman in car after meet&pickup&ride home


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Create equations using given problem info and established variables:
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T - 60         = Time index when he arrives early at station (60 min early)
T - 60 + w     = Time index to when man is picked up in car
T + c          = Time index for when they normally get home
T - 60 + w + x = Time index for when man arrives home with woman after pickup while walking
T + c -20      = Time index arriving home 20 minutes earlier than usual after walk+ride (arrive home 20 minutes early)
T - c          = Time index when she leaves home to go to station

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Assumptions: (not all assumptions included -- just critical assumptions)
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1) c is same time to drive from station to home as home to station


2) x is same to drive from meet-point to home as home to meet-point


3) zero-time is needed at meet point to switch from "walk" to "drive-from-meet-point"


4) c, w, x, are all elements of "Real" numbers and greater-than-or-equal-to zero. (Travelling back in time is not allowed :-)

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If we assume time "c" to drive to station is same time to drive back, then:
T - c          = Time index when she leaves their house to go to station


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If we assume time "x" to drive home from point of intersection/meet is the same
time for her to drive car from home to intersection point, then:
T - c + x      = time index for interception


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"Real" work begins:
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Given from above:
T + c - 20     = Time index when man and woman arrived home early
and
T - 60 + w + x = Time index for when man arrives home with woman after pickup while walking

But these are the same indexed time, therefore they are equivalent:
T - 60 + w + x = T + c - 20
Subtract "T" from both sides:
-T              -T
  - 60 + w + x =     c - 20
Add 20 to both sides:
   +20                 + 20
  - 40 + w + x =     c

Solve for c:

c = w + x - 40


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Given from above:
T - c + x      = time index for interception
and:
T - 60 + w          = Time index to when man is picked up in car

But these are both indexed times to the same event, so are equivalent.

Therefore:
T - c + x = T - 60 + w

But: (from above work)
c = w + x - 40

Therefore: (substitute equation for c instead of c)
T - ( w + x - 40 ) + x = T - 60 + w 
Multiply all in the parens by "-1" and drop parens:
T - w - x + 40 + x     = T - 60 + w
move the "x" and "-x" to make "x" elimination obvious:
T - w - x + x + 40     = T - 60 + w
Showing x - x is 0, remove:
T - w         + 40     = T - 60 + w
Subtract T from both sides:
-T                      -T
Add "w" and "60" to both sides:
   -w    + 40          =   - 60 + w
   +w    + 60              + 60 + w
Gives us:
       +100             =    2w
Divide by 2:
       (100)/2          =   (2w)/2
Gives us:
        50=w
Solve for w:
w=50


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Sanity Check:
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T - 60 + w  + x  = T + c - 20

Requirements: x and c must be positive

Let w=50
T - 60 + 50 + x  = T + c - 20
Subtract T from both sides:
-T               -T
  - 60 + 50 + x  =   + c - 20
Add 20 to both sides:
       + 20             + 20
20 - 60 + 50 + x =   + c
          10 + x =   + c


Solve for c:
c = x + 10

Test 1:

Let x = 100
c = 100 + 10
c = 110

T - 60 + 50 + x   ?= T + c   - 20
T - 60 + 50 + 100 ?= T + 110 - 20
T           + 90  ?= T + 90

Yes:  T + 90 = T + 90

Let X = 1000
c = 1000 + 10
c = 1010
T - 60 + 50 + x    ?= T + c    - 20
T - 60 + 50 + 1000 ?= T + 1010 - 20
T           + 990   ?= T + 990
Yes:  T + 990 = T + 990

Is "c" always greater than "x" (as we would expect) ?
c = x + 10
Yes.

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Answer:
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w = walk-time for man to walk until intercepted
w = 50 minutes

"50 minutes."

Q.E.D.

How many Internets do I win?
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