1. Larry can swim from Harvard to MIT (with the current of the Charles River) in 40 minutes, or back (against the current) in 45 minutes. How long does it take him to row from Harvard to MIT, if he rows the return trip in 15 minutes? (Assume that the speed of the current and Larry's swimming and rowing speeds relative to the current are all constant.) Express your answer in the format mm:ss.
Answer: $14: 24$
Solution: Let the distance between Harvard and MIT be 1, and let $c, s, r$ denote the speeds of the current and Larry's swimming and rowing, respectively. Then we are given
$$
s+c=\frac{1}{40}=\frac{9}{360}, \quad s-c=\frac{1}{45}=\frac{8}{360}, \quad r-c=\frac{1}{15}=\frac{24}{360},
$$
so
$$
r+c=(s+c)-(s-c)+(r-c)=\frac{9-8+24}{360}=\frac{25}{360},
$$
and it takes Larry $360 / 25=14.4$ minutes, or $14: 24$, to row from Harvard to MIT.