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Ginny and Jenna are 20 miles from home. They have one pair of roller skates. Jenna walks 4mph and skates 9mph. Ginny walks 3 mph and skates 8mph. They start for home at the same time.

First, Ginny has the skates and Jenna walks. Ginny skates for a while, then takes the roller skates off and starts walking.When Jenna reaches the roller blades, she puts them on and starts skating home.

If they both start at 4:00 and arrive home at the same time, what time is it when they get home?

My solution,

I assumed that the time that Ginny skates is $a$ hours and walks for $b$ hours. And hence Jenna skates for $b$ hours and walks for $a$ hours. And since total distance covered is 20 for both, I got the following 2 equations.

$ \begin{align} 8a + 3b &= 20 \\ 4a + 9b &= 20 \end{align} $

I solved this system of equations by elimination to get $b = \dfrac{4}{3}$ and $a = 2$, and $a + b = \dfrac{10}{3}$. This doesn't check out with the required solution which is $4$ and arriving at $8$ pm.

I have checked the simultaneous equation, so I have probably made a logic error. Any ideas where I went wrong. Thanks.

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    You’ve got Jenna walking for $a$ hours, but in fact she must walk longer than that: after $a$ hours she still has to cover the distance to the point at which Ginny took off the skates.2012-03-04

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Suppose, as you did, that Ginny skates for $a$ hours and walks for $b$ hours; that gives us one relationship between $a$ and $b$, namely, $8a+3b=20\;.\tag{1}$

After $a$ hours Ginny has covered $8a$ miles, and Jenna, walking at $4$ mph, has covered $4a$ miles. It will take Jenna another $a$ hours to reach the skates. Ginny covered the remaining distance in $b$ hours, walking at $3$ mph, so the remaining distance is $3b$ miles. Skating at $9$ mph, Jenna will need $\frac{3b}9=\frac{b}3$ hours to cover this distance. Thus, Jenna’s total elapsed time must be $2a+\frac{b}3$. But we know that they took the same total amount of time, so $2a+\frac{b}3=a+b\;,$ or $a=\frac23b\;;\tag{2}$ this gives us a second relationship between $a$ and $b$. Now just solve the system consisting of equations $(1)$ and $(2)$.

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    I saw your initial comment, realized my mistake went back to redo my work. Came back to see you have also answered the question after the hint. Very well explained, Thank you!2012-03-04