Respuesta :
A = larger, faster pipe
B = smaller, slower pipe
let's say B can fill the tank in hmmm 8hrs... so in 1hr, it has only done 1/8 of the job, in 2hrs it has done 2/8 of the job and in 3hrs has done 3/8 of the job and so on, to finish the job, it needs to do 8/8 of the job or 8/8=1 whole
we dunno how long it took the B pipe to do it though, let's say it took "t" hours, so in 1hr it had done 1/t of the job
now, if B took "t" hours to do the whole job, pipe A is faster and thus it did it 5hrs less than that, so, A can do it in "t - 5" hours
so, in 1hr, A had done 1/(t-5) of the job
now, we know the rates in 1hr of each pipe, we know together, they can do the job in 6hrs
so
[tex]\bf \textit{in 1hr, both pipes have done} \\\\\\ \begin{array}{llllll} \cfrac{1}{t}&+&\cfrac{1}{t-5}&=&\cfrac{1}{6}\\ smaller&&larger&&job\ done\\ B&&A \end{array} \\\\\\ \textit{let's add the left-side, our LCD is t(t-5)} \\\\\\ \cfrac{t-5+t}{t(t-5)}=\cfrac{1}{6}\implies \cfrac{2t-5}{t(t-5)}=\cfrac{1}{6}\implies 6(2t-5)=t(t-5) \\\\\\ 12t-30=t^2-5t\implies 0=t^2-17t+30 \\\\\\ 0=(t-15)(t-2)\implies \begin{cases} 0=t-15\implies &\boxed{15=t}\\ 0=t-2\implies &2=t \end{cases}[/tex]
well, clearly, if both pipes take 6hrs, the smaller B can't do it in 2hrs by itsef, thus 15 = t
B = smaller, slower pipe
let's say B can fill the tank in hmmm 8hrs... so in 1hr, it has only done 1/8 of the job, in 2hrs it has done 2/8 of the job and in 3hrs has done 3/8 of the job and so on, to finish the job, it needs to do 8/8 of the job or 8/8=1 whole
we dunno how long it took the B pipe to do it though, let's say it took "t" hours, so in 1hr it had done 1/t of the job
now, if B took "t" hours to do the whole job, pipe A is faster and thus it did it 5hrs less than that, so, A can do it in "t - 5" hours
so, in 1hr, A had done 1/(t-5) of the job
now, we know the rates in 1hr of each pipe, we know together, they can do the job in 6hrs
so
[tex]\bf \textit{in 1hr, both pipes have done} \\\\\\ \begin{array}{llllll} \cfrac{1}{t}&+&\cfrac{1}{t-5}&=&\cfrac{1}{6}\\ smaller&&larger&&job\ done\\ B&&A \end{array} \\\\\\ \textit{let's add the left-side, our LCD is t(t-5)} \\\\\\ \cfrac{t-5+t}{t(t-5)}=\cfrac{1}{6}\implies \cfrac{2t-5}{t(t-5)}=\cfrac{1}{6}\implies 6(2t-5)=t(t-5) \\\\\\ 12t-30=t^2-5t\implies 0=t^2-17t+30 \\\\\\ 0=(t-15)(t-2)\implies \begin{cases} 0=t-15\implies &\boxed{15=t}\\ 0=t-2\implies &2=t \end{cases}[/tex]
well, clearly, if both pipes take 6hrs, the smaller B can't do it in 2hrs by itsef, thus 15 = t
