Two water taps together can fill a tank in 33133\frac{3}{13} hours. The tap of larger diameter takes 5 hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Answer: The smaller tap takes approximately 9.819.81 hours and the larger tap takes approximately 4.814.81 hours to fill the tank separately.

Step-by-step solution

Step 1: Define variables and rates

Let tSt_S be the time taken by the smaller tap to fill the tank alone. Since the larger tap takes 5 hours less, its time will be tS−5t_S - 5. The rate at which a tap fills the tank is the reciprocal of the time it takes. So, the rate of the smaller tap is 1/tS1/t_S and the rate of the larger tap is 1/(tS−5)1/(t_S - 5).

Step 2: Formulate the combined rate equation

When both taps work together, their rates add up. The combined time to fill the tank is given as 33133\frac{3}{13} hours, which is 42/1342/13 hours. Therefore, the combined rate is 13/4213/42. We set up the equation by summing the individual rates and equating it to the combined rate.

Step 3: Simplify the equation to a quadratic form

First, combine the fractions on the left side of the equation by finding a common denominator. Then, cross-multiply to eliminate the denominators. Expand and rearrange the terms to form a standard quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0.

Step 4: Solve the quadratic equation for tSt_S

We use the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} to solve for tSt_S. Substitute the coefficients a=13a=13, b=−149b=-149, and c=210c=210 into the formula. Calculate the discriminant and then find the two possible values for tSt_S.

Step 5: Determine the valid time for each tap

We have two possible values for tSt_S. Since time cannot be negative, we must discard the solution that leads to a negative time for the larger tap. If tS=1.65t_S = 1.65 hours, then tL=1.65−5=−3.35t_L = 1.65 - 5 = -3.35 hours, which is not physically possible. Thus, tS=9.81t_S = 9.81 hours is the correct time for the smaller tap, and tL=9.81−5=4.81t_L = 9.81 - 5 = 4.81 hours for the larger tap.

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