SOLUTION: At the same time, but different speed, Mark and Nancy left (driving) from different towns, A and B, and drove toward each other. They first met at the point that was 60 miles from

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Question 682048: At the same time, but different speed, Mark and Nancy left (driving) from different towns, A and B, and drove toward each other. They first met at the point that was 60 miles from Town A. When Mark and Nancy reached their destinations (B and A, respectively) they drove back without stop. 3 hours later after they originally left the towns (the initial staring points), they met again. This time, the meeting point was 40 miles from town A. Suppose each of them was moving all the time at a constant speed, what were the speeds of Mark and Nancy. (note, they have different speeds).
Thanks for your help! It is much appreciated!!

Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
At the same time, but different speed, Mark and Nancy left (driving) from different towns, A and B, and drove toward each other.
They first met at the point that was 60 miles from Town A.
When Mark and Nancy reached their destinations (B and A, respectively) they drove back without stop.
3 hours later after they originally left the towns (the initial staring points), they met again.
This time, the meeting point was 40 miles from town A. Suppose each of them was moving all the time at a constant speed, what were the speeds of Mark and Nancy.
A diagram will help
A------60------*-----d------B; first meeting
A----40----*----(d+20)------B: 2nd meeting
:
Let d = distance from the 1st meeting point to B, therefore
Initially, M travels 60 mi while N travels d mi
:
To the 2nd meeting:
M travels: d + (d+20) = (2d+20) mi
while
N travels: 60 + 40 = 100 mi
:
The M:N distance relationships are the same for each part of the trip
:
1st meeting = 2nd meeting
60%2Fd = %282d%2B20%29%2F100
Cross multiply
d(2d+20) = 60 * 100
2d^2 + 20d = 6000
A quadratic equation
2d^2 + 20d - 6000 = 0
simplify, divide by 2
d^2 + 10d - 3000 = 0
This will factor to
(d+60)(d-50) = 0
the positive solution
d = 50 miles from the 1st meeting point to B
then
60 + 2(50) + 20 = 180 miles traveled by M in 3 hrs, therefore
180%2F3 = 60 mph is M's speed
And
50 + 60 + 40 = 150 miles traveled by N in 3 hrs, therefore
150%2F3 = 50 mph is N's speed

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the distance between towns A and B = +d+
Mark's total trip:
+d+%2B+d+-+40+
Nancy's total trip:
+d+%2B+40+
They both traveled these distances in +3+ hrs
So, their rates are
Mark:
+%28+2d+-+40+%29+%2F+3+ mi/hr
Nancy:
+%28+d+%2B+40+%29+%2F+3+ mi/hr
================
For the 1st meeting, the ratio of their speed is directly
proportional to the distances they cover, since
+d%5B1%5D+=+r%5B1%5D%2At+
+r%5B1%5D+=+d%5B1%5D+%2F+t+
and
+d%5B2%5D+=+r%5B2%5D%2At+
+r%5B2%5D+=+d%5B2%5D+%2F+t+
and
+r%5B1%5D+%2F+r%5B2%5D+=+d%5B1%5D+%2F+d%5B2%5D+
( Mark's rate ) / ( Nancy's rate ) = +60+%2F+%28+d+-+60+%29+
-------------------
Now I can say:
+60+%2F+%28+d+-+60+%29+=+%28+2d+-+40+%29+%2F+%28+d+%2B+40+%29+ ( t 's cancel )
+60%2A%28+d+%2B+40+%29+=+%28+2d+-+40+%29%2A%28+d+-+60+%29+
+60d+%2B+2400+=+2d%5E2+-+40d+-+120d+%2B+2400+
+60d+=+2d%5E2+-+160d+
+2d%5E2+-+220d+=+0+
+d%5E2+-+110d+=+0+
+d%2A%28+d+-+110+%29+=+0+
+d+=+110+ mi
Mark:
+%28+2d+-+40+%29+%2F+3+ mi/hr
+%28+2%2A110+-+40+%29+%2F+3+
+220+-+40%29+%2F+3+
+180%2F3+
+60+
Nancy:
+%28+d+%2B+40+%29+%2F+3+ mi/hr
+%28+110+%2B+40+%29+%2F+3+
+150%2F3+
+50+
-----------
Mark's speed is 60 mi/hr
Nancy's speed is 50 mi/hr
check:
( Mark's rate ) / ( Nancy's rate ) = +60+%2F+%28+d+-+60+%29+
+60+%2F+%28+110+-+60+%29+=+60%2F50+