Answer to Question #322468 in Linear Algebra for Kenth Smith

Question

Two pipes empty a vat in 1hr and 12 min. When one pipe is used to fill and one to empty the vat, it is filled in 3 hrs. How long would it take each pipe to empty the vat if only one pipe is open?

Answer 1

x - time of emptying the vat with the first pipe

y - operating time of the second pipe

First pipe alone - (1vat/x minute)(x minute) = 1vat

Second pipe alone - (1vat/y minute)(y minute) = 1vat

Total time 72minute, then 72/x + 72/y = 1 (1)

1/x - amount emptied by the outlet pipe in one hour

1/y - amount filled by the inlot pipe in one hour

1/180 - amount filled by both pipe in one hour (3 hours × 60 = 180 minute)

Then 1/x - 1/y = 1/180 (2)

$\begin{cases} 72/x + 72/y = 1 \\ 1/x - 1/y = 1/180 \end{cases}$ divide by 72 equation (1)

$\begin{cases} 1/x + 1/y = 1/72 \\ 1/x - 1/y = 1/180 \end{cases}$

$2/x = 1/72 + 1/180\\ 2/x = 7/360\\ x = 360×2/7\\ x = 103 minute = 1 hour 43 minute$

Substitute x in the first equation

$72/103 + 72/y = 1\\ 72/y = 1- 72/103\\ 72/y = 31/103\\ y = 72×103/31\\ y = 239 minute = 3 hours 59 minute$

Answer: First pipe - 1 hour 43 minute. Second pipe - 3 hours 59 minute. During this time, each pipe alone will do the work