Respuesta :
well, first off, I don't ever recall anyone named "birth" or "phone", so let's just call them B and P.
so B is slower than P, say P can do the whole job in "p" minutes, and since B is three times as slow, then B can do it in "3*p" or "3p" minutes.
we know that they together can do it in 90 minutes, so in 1 minute alone, both of their rates added up, have only done 1/90 of the job, whilst P has done 1/p of the work and B has done 1/3p of the work.
[tex]\bf \stackrel{B's~rate}{\cfrac{1}{3p}}+\stackrel{P's~rate}{\cfrac{1}{p}}~~=~~\cfrac{1}{90}\impliedby \begin{array}{llll} \textit{let's multiply both sides by }\stackrel{LCD}{90p}\\ \textit{to do away with the denominators} \end{array} \\\\\\ 90p\left( \cfrac{1}{3p}+\cfrac{1}{p} \right)~~=~~90p\left( \cfrac{1}{90} \right)\implies 30+90=p\implies 120=p[/tex]
so it takes 120 minutes or 2 hours for P to do it alone, and 3p to B to do it alone.
so B is slower than P, say P can do the whole job in "p" minutes, and since B is three times as slow, then B can do it in "3*p" or "3p" minutes.
we know that they together can do it in 90 minutes, so in 1 minute alone, both of their rates added up, have only done 1/90 of the job, whilst P has done 1/p of the work and B has done 1/3p of the work.
[tex]\bf \stackrel{B's~rate}{\cfrac{1}{3p}}+\stackrel{P's~rate}{\cfrac{1}{p}}~~=~~\cfrac{1}{90}\impliedby \begin{array}{llll} \textit{let's multiply both sides by }\stackrel{LCD}{90p}\\ \textit{to do away with the denominators} \end{array} \\\\\\ 90p\left( \cfrac{1}{3p}+\cfrac{1}{p} \right)~~=~~90p\left( \cfrac{1}{90} \right)\implies 30+90=p\implies 120=p[/tex]
so it takes 120 minutes or 2 hours for P to do it alone, and 3p to B to do it alone.
