weird rate problem that I can't solve

Question

Two teams compete in a relay race, they're composed of 4 members, each of whom runs 100 meters. All of the members of team B runs at $b$ meters per second (mps). The first 3 members of team A can only run at 5$b$/6 (mps) while the 4th member runs at $a$ mps. Find $a$/$b$ if the race is a tie.

Answer 1

The total time it takes team A to run the race is $$T_A=\frac{300 \text{ m}}{5b/6 \text{ ms}^{-1}}+\frac{100\text{ m}}{a \text{ ms}^{-1}}$$ The time it takes team B to run the race is $$T_B=\frac{400\text{ m}}{b\text{ ms}^{-1}}$$ Therefore, $$T_A=T_B \implies \frac{a}{b}=\frac{5}{2}$$ EDIT: My workings - $$\frac{300}{5b/6} + \frac{100}{a} = \frac{400}{b}$$ $$\frac{18}{5b} + \frac{1}{a} = \frac{4}{b}$$ $$\frac{18}{b} + \frac{5}{a}=\frac{20}{b}$$ $$\frac{2}{b}=\frac{5}{a}$$ It's straightforward from here.

Answer 2

The total time for the tie race is

$$ \frac Sb = \frac {\frac {3S}4}{\frac56b}+ \frac{\frac S4}a $$

where $S$ is the total distance that cancels. Solve to obtain $\frac ab = \frac52$.