Two pipes are connected to the same tank. When working together, they can fill the tank in 8 hrs. The larger pipe, working alone, can fill the tank in 12 hrs less time than the smaller one. How long would the smaller one take, working alone, to fill the tank?
Accepted Solution
A:
Answer:24 hoursStep-by-step explanation:The small hose + the large hose = 8 hours of workIf the small hose can get the job done in x hours, then 1/x of the job will get done in 1 hourIf the large hose can get the job done in x-12 hours, then 1/(x-12) of the job will get done in 1 hour.If the whole job can get done in 8 hours, then 1/8 of the job will get done in an hour. Going back to the first entry, we know we are adding the small to the large and setting it equal to the total:[tex]\frac{1}{x}+\frac{1}{x-12}=\frac{1}{8}[/tex]In order to solve for x, we need to find the LCD and multiply it through the whole equation. The LCD is x(x-12)(8):[tex](8)(x)(x-12)[\frac{1}{x}+\frac{1}{x-12}=\frac{1}{8}][/tex]Canceling out between each denominator and the LCD leaves us with:8x - 96(1) + 8x(1) = x² - 12xSimplify through to get8x - 96 + 8x = x² - 12xSince this is a second degree polynomial, we expect to get 2 solutions for x. We will factor to solve for these, then decide which one to use. First get everything on one side of the equals sign, combining like terms as we go:x² - 28x + 96 = 0Factor this however you have been taught (the quadratic formula is the best!) to getx = 4 and x = 24We know it can't be x = 4, because if we fit 4 into the expression for the larger hose, x - 12, we would get a negative time, which is not possible. Therefore, x = 24, the time it would take for the small hose to do the job alone.