At 12:00 noon, Anna and Bill left home and walked in the same direction. Anna walked at $4 \mathrm{~km} / \mathrm{h}$ and Bill walked at $3 \mathrm{~km} / \mathrm{h}$. At 12:15 their dog Dexter, running at $6 \mathrm{~km} / \mathrm{h}$, left home to run after them. The dog ran until it caught up to Anna, then it ran back to Bill. (In his excitement, Dexter lost no time in turning around once he reached Anna.) At what time did Bill meet Dexter on Dexter's way back?
(A) 1:00 p.m.
(B) 1:15 p.m.
(C) 12:45 p.m.
(D) 1:05 p.m.
(E) 12:50 p.m.
We solve this problem by systematically keeping track of the distance from home of each of Anna, Bill and Dexter. At 12:00 noon, each is $0 \mathrm{~km}$ from home.

At 12:15:

Dexter is $0 \mathrm{~km}$ from home (since he hasn't started running)

Anna is $\frac{1}{4} \times 4=1 \mathrm{~km}$ from home (since she has walked at $4 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{4}$ of an hour)

Bill is $\frac{1}{4} \times 3=\frac{3}{4} \mathrm{~km}$ from home (since he has walked at $3 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{4}$ of an hour)

At 12:15, Dexter leaves and runs until he catches up to Anna.

How long does this take? Since Anna walks at $4 \mathrm{~km} / \mathrm{h}$ and Dexter runs at $6 \mathrm{~km} / \mathrm{h}$ in the same direction, then Dexter gains $2 \mathrm{~km}$ on Anna every hour. Since Anna starts $1 \mathrm{~km}$ ahead of Dexter, then it takes Dexter $\frac{1}{2}$ hour to catch Anna, so he catches her at 12:45.

At 12:45:

Dexter is $3 \mathrm{~km}$ from home (since he has run at $6 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{2}$ hour)

Anna is $\frac{3}{4} \times 4=3 \mathrm{~km}$ from home (since she has walked at $4 \mathrm{~km} / \mathrm{h}$ for $\frac{3}{4}$ of an hour)

Bill is $\frac{3}{4} \times 3=\frac{9}{4} \mathrm{~km}$ from home (since he has walked at $3 \mathrm{~km} / \mathrm{h}$ for $\frac{3}{4}$ of an hour)

At 12:45, Dexter turns around instantaneously and runs back to Bill from Anna.

How long does this take? Since Bill walks at $3 \mathrm{~km} / \mathrm{h}$ and Dexter runs at $6 \mathrm{~km} / \mathrm{h}$ in the opposite direction, then Dexter and Bill are getting closer at a rate of $9 \mathrm{~km} / \mathrm{h}$.

Since Dexter and Bill start $3-\frac{9}{4}=\frac{3}{4} \mathrm{~km}$ apart, then it takes Dexter $\frac{1}{9} \times \frac{3}{4}=\frac{1}{12}$ hour (or 5 minutes) to meet Bill.

Therefore, Bill and Dexter meet at 12:50 p.m.

ANswer: (E)

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