25. Thomas has constant speeds for both running and walking. When a down-escalator is moving, Thomas can run down it in 15 seconds or walk down it in 30 seconds. One day, when the escalator was broken (and stationary), it took Thomas 20 seconds to run down it.
How long, in seconds, would it take Thomas to walk down the broken escalator?
A 30
B 40
C 45
D 50
E 60
25. E Let the speed, in $\mathrm{m} / \mathrm{s}$, of the escalator when it is working be $v$. Let the length of the escalator be $l \mathrm{~m}$ and let Thomas's running and walking speeds, in $\mathrm{m} / \mathrm{s}$, be $r$ and $w$ respectively.
As Thomas can run down the moving escalator in 15 seconds and walk down it in 30 seconds, $\frac{l}{v+r}=15 \ldots[1] ; \frac{l}{v+w}=30 \ldots$
Also, as Thomas runs down the broken escalator in 20 seconds, $\frac{l}{r}=20 \ldots$ [3].
From [1] $v+r=\frac{l}{15} \ldots$ [4]; from [2] $v+w=\frac{l}{30} \ldots$ [5]; from [3] $r=\frac{l}{20} \ldots$
Adding [5] and [6] and subtracting [4] gives $v+w+r-v-r=\frac{l}{30}+\frac{l}{20}-\frac{l}{15}$. Therefore $w=\frac{l(2+3-4)}{60}=\frac{l}{60}$. So $\frac{l}{w}=60$.
Hence it would take Thomas 60 seconds to walk down the broken escalator.