Working together, it takes two different sized hoses 35 minutes to fill a small swimming pool. If it takes 60 minutes for the larger hose to fill the swimming pool by itself, how long will it take the smaller hose to fill the pool on its own? Do not do any rounding.

L = rate of the large hoseS = rate of the small hosewe know the sum of their rates deliver the swimming pool filled up in 35 minutes, thus L + S = 35.now, the large hose can do it in 60 minutes, that means that in 1 minute, the large hose has only done 1/60 th of the work.the small hose can do the whole thing in S minutes, that means in 1 minute it has only done 1/S th of the whole work.and since both of them working together can do it in 35 minutes, then in 1 minute they both have done 1/35 th of the whole job.[tex]\bf \stackrel{\textit{large's rate in 1 minute}}{\cfrac{1}{L}}+\stackrel{\textit{small's rate in 1 minute}}{\cfrac{1}{S}}=\stackrel{\textit{done in 1 minute}}{\cfrac{1}{35}} \\\\\\ \cfrac{1}{60}+\cfrac{1}{S}=\cfrac{1}{35}\implies \cfrac{S+60}{60S}=\cfrac{1}{35}\implies 35S+2100=60S\implies 2100=25S \\\\\\ \cfrac{2100}{25}=S\implies 84=S[/tex]