Working together, both pipes take 12 days to fill the tank. Therefore, the sum of their individual rates equals (1)/(12). Combining the fractions gives (a+b)/(ab) = (1)/(12), which means ab = 12(a+b).

Working together, both pipes take 12 days to fill the tank. Therefore, the sum of their individual rates equals (1)/(12). Combining the fractions gives (a+b)/(ab) = (1)/(12), which means ab = 12(a+b).

Multiplying the equation by 2 gives (b)/(a) + (a)/(b) = (25)/(12). Letting k = (a)/(b), we obtain k + (1)/(k) = (25)/(12), which simplifies to the quadratic 12k² - 25k + 12 = 0. Factoring gives k = (4)/(3) or k = (3)/(4).

Multiplying the equation by 2 gives (b)/(a) + (a)/(b) = (25)/(12). Letting k = (a)/(b), we obtain k + (1)/(k) = (25)/(12), which simplifies to the quadratic 12k² - 25k + 12 = 0. Factoring gives k = (4)/(3) or k = (3)/(4).

Using ab = 12(a+b), if a = (4)/(3)b, dividing both sides by b gives (4)/(3)b = 28, so b = 21 days and a = 28 days. By symmetry, if a = (3)/(4)b, then a = 21 days and b = 28 days. Thus, one pipe takes 20 days and the other takes 28 days.

Using ab = 12(a+b), if a = (4)/(3)b, dividing both sides by b gives (4)/(3)b = 28, so b = 21 days and a = 28 days. By symmetry, if a = (3)/(4)b, then a = 21 days and b = 28 days. Thus, one pipe takes 20 days and the other takes 28 days.

Two pipes, A and B, working together can fill a water tank in exactly 12 days.If pipe A works alone for half of the total time it takes for B alone to fill the tank, and then pipe B works alone for half of the total time it takes for A alone to fill the tank, the tank ends up being exactly $\frac{25}{24}$ full (overflowing slightly).How many days would it take each pipe, working completely alone, to fill the tank?

Answer: One pipe takes 21 days and the other takes 28 days (either Pipe A takes 21 days and Pipe B takes 28 days, or Pipe A takes 28 days and Pipe B takes 21 days).

Step-by-step solution

Step 1: Define variables and rates

Let $a$ represent the number of days taken by pipe A alone to fill the tank, and let $b$ represent the number of days taken by pipe B alone. The work rate per day of pipe A is $\frac{1}{a}$ of the tank, and the work rate per day of pipe B is $\frac{1}{b}$ of the tank.

Step 2: Formulate equation from the combined work

Working together, both pipes take 12 days to fill the tank. Therefore, the sum of their individual rates equals $\frac{1}{12}$ . Combining the fractions gives $\frac{a+b}{ab} = \frac{1}{12}$ , which means $ab = 12(a+b)$ .

Step 3: Formulate equation from the split-time condition

Pipe A works for half the time B takes alone, which is $\frac{b}{2}$ days, filling $\frac{b}{2a}$ of the tank. Then pipe B works for half the time A takes alone, which is $\frac{a}{2}$ days, filling $\frac{a}{2b}$ of the tank. In total, they fill $\frac{25}{24}$ of the tank.

Step 4: Find the ratio between the times

Multiplying the equation by 2 gives $\frac{b}{a} + \frac{a}{b} = \frac{25}{12}$ . Letting $k = \frac{a}{b}$ , we obtain $k + \frac{1}{k} = \frac{25}{12}$ , which simplifies to the quadratic $12k^2 - 25k + 12 = 0$ . Factoring gives $k = \frac{4}{3}$ or $k = \frac{3}{4}$ .

Step 5: Solve for the individual times

Using $ab = 12(a+b)$ , if $a = \frac{4}{3}b$ , dividing both sides by $b$ gives $\frac{4}{3}b = 28$ , so $b = 21$ days and $a = 28$ days. By symmetry, if $a = \frac{3}{4}b$ , then $a = 21$ days and $b = 28$ days. Thus, one pipe takes 20 days and the other takes 28 days.

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