Two water taps together can fill a tank in 3133 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.
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Step-by-Step Solution
Step 1: Define variables and rates
Let tS 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−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/tS and the rate of the larger tap is 1/(tS−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 3133 hours, which is 42/13 hours. Therefore, the combined rate is 13/42. 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=0.
Step 4: Solve the quadratic equation for tS
We use the quadratic formula x=2a−b±b2−4ac to solve for tS. Substitute the coefficients a=13, b=−149, and c=210 into the formula. Calculate the discriminant and then find the two possible values for tS.
Step 5: Determine the valid time for each tap
We have two possible values for tS. Since time cannot be negative, we must discard the solution that leads to a negative time for the larger tap. If tS=1.65 hours, then tL=1.65−5=−3.35 hours, which is not physically possible. Thus, tS=9.81 hours is the correct time for the smaller tap, and tL=9.81−5=4.81 hours for the larger tap.