Respuesta :
suppose the larger pump alone can empty the tank in L hours, and the smaller pump can finish the job in S hours, then each hour the large pump empties 1/L portion of the tank, and the small pump empties 1/S per hour
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
let's say the larger pump takes "a" hours to empty the whole lot.
and let's say the smaller pump takes "b" hours to empty it by itself.
but we know that if both team up, they get it done in 3 hours flat.
now, if we know the large pump takes "a" hours, how many hours has it done in say 1 hour? well, it has only done thus far 1/a of the total work.
and we know the small one takes "b" hours, so in 1 hour it has done only thus far 1/b of the whole lot, just that much.
we know, both teaming up can do it in 3 hrs, so in just 1 hr, they have both done only 1/3 of the total work.
[tex]\bf \stackrel{large~pump}{\cfrac{1}{a}}+\stackrel{small~pump}{\cfrac{1}{b}}=\stackrel{total~work}{\cfrac{1}{3}}[/tex]
but we know that, whatever "b" hours is, we know that the large pump can do it 4 hours less than that, so, it can do it in "b - 4", namely, a = b - 4.
[tex]\bf \stackrel{large~pump}{\cfrac{1}{a}}+\stackrel{small~pump}{\cfrac{1}{b}}=\stackrel{total~work}{\cfrac{1}{3}}\qquad \qquad \boxed{a=b-4}\\\\\\ \cfrac{1}{b-4}+\cfrac{1}{b}=\cfrac{1}{3}\implies \cfrac{(b)+(b-4)}{b(b-4)}=\cfrac{1}{3}\implies \cfrac{2b-4}{b^2-4b}=\cfrac{1}{3} \\\\\\ 6b-12=b^2-4b\implies 0=b^2-10b+12[/tex]
[tex]\bf \stackrel{\textit{quadratic formula}}{b=\cfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(12)}}{2(1)}}\implies b=\cfrac{10\pm \sqrt{100-48}}{2} \\\\\\ b=\cfrac{10\pm\sqrt{52}}{2}\implies b=\cfrac{10\pm\sqrt{2^2\cdot 13}}{2}\implies b=\cfrac{10\pm 2\sqrt{13}}{2} \\\\\\ b=5\pm\sqrt{13}\implies b\approx \begin{cases} 8.60555\\ 1.39445 \end{cases}[/tex]
well, it can't be 1.394 because that'd make b - 4 a negative value, so it has to be the other value.
and let's say the smaller pump takes "b" hours to empty it by itself.
but we know that if both team up, they get it done in 3 hours flat.
now, if we know the large pump takes "a" hours, how many hours has it done in say 1 hour? well, it has only done thus far 1/a of the total work.
and we know the small one takes "b" hours, so in 1 hour it has done only thus far 1/b of the whole lot, just that much.
we know, both teaming up can do it in 3 hrs, so in just 1 hr, they have both done only 1/3 of the total work.
[tex]\bf \stackrel{large~pump}{\cfrac{1}{a}}+\stackrel{small~pump}{\cfrac{1}{b}}=\stackrel{total~work}{\cfrac{1}{3}}[/tex]
but we know that, whatever "b" hours is, we know that the large pump can do it 4 hours less than that, so, it can do it in "b - 4", namely, a = b - 4.
[tex]\bf \stackrel{large~pump}{\cfrac{1}{a}}+\stackrel{small~pump}{\cfrac{1}{b}}=\stackrel{total~work}{\cfrac{1}{3}}\qquad \qquad \boxed{a=b-4}\\\\\\ \cfrac{1}{b-4}+\cfrac{1}{b}=\cfrac{1}{3}\implies \cfrac{(b)+(b-4)}{b(b-4)}=\cfrac{1}{3}\implies \cfrac{2b-4}{b^2-4b}=\cfrac{1}{3} \\\\\\ 6b-12=b^2-4b\implies 0=b^2-10b+12[/tex]
[tex]\bf \stackrel{\textit{quadratic formula}}{b=\cfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(12)}}{2(1)}}\implies b=\cfrac{10\pm \sqrt{100-48}}{2} \\\\\\ b=\cfrac{10\pm\sqrt{52}}{2}\implies b=\cfrac{10\pm\sqrt{2^2\cdot 13}}{2}\implies b=\cfrac{10\pm 2\sqrt{13}}{2} \\\\\\ b=5\pm\sqrt{13}\implies b\approx \begin{cases} 8.60555\\ 1.39445 \end{cases}[/tex]
well, it can't be 1.394 because that'd make b - 4 a negative value, so it has to be the other value.
