$28 \cdot 28$ Jack and Jill start running a $10 \mathrm{~km}$ course from the same starting point, the first $5 \mathrm{~km}$ up to the top of a small hill, and the last $5 \mathrm{~km}$ back down to the starting point. Jack starts 10 minutes earlier, with an uphill speed of $15 \mathrm{~km}$ /h and a downhill speed of $20 \mathrm{~km}$ /h. Jill's uphill and downhill speeds are $16 \mathrm{~km}$ /h and $22 \mathrm{~km}$ /h, respectively. The distance from the meeting point to the top of the hill when they meet head-on is
(A) $\frac{5}{4} \mathrm{~km}$.
(B) $\frac{35}{27} \mathrm{~km}$.
(C) $\frac{27}{20} \mathrm{~km}$.
(D) $\frac{7}{3} \mathrm{~km}$.
(E) $\frac{28}{9} \mathrm{~km}$.
(42nd American High School Mathematics Examination, 1991)
[Solution] Jack's time from departure to the top of the mountain is $\frac{5}{15}=\frac{1}{3}$ (hour), which is 20 minutes. When Jack reaches the top, the distance between Jill and Jack (the top) (Jack starts 10 minutes earlier) is
$$
5-\frac{16}{60} \cdot(20-10)=\frac{7}{3}(\mathrm{~km}) .
$$

The distance from the top when the two meet is
$$
20 \cdot \frac{\frac{7}{3}}{16+20}=\frac{35}{27}(\mathrm{~km}) \text {. }
$$

Therefore, the answer is $(B)$.