Imperfectly opened taps (reduced discharge) Two taps A and B nominally fill a tank in 25 min and 20 min. Due to partial opening, A allows only 5/6 of its normal flow and B allows only 2/3 of its normal flow. How long will they take together to fill.

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:When taps are partially opened, multiply each nominal rate by its effective fraction to get the reduced rate. Then add reduced rates as usual. Given Data / Assumptions: Nominal A: 25 min ⇒ 1/25 per min; effective = (5/6)*(1/25). Nominal B: 20 min ⇒ 1/20 per min; effective = (2/3)*(1/20). Tank initially empty; both run together. Concept / Approach:Compute effective rates and add. Time is the inverse of the total effective rate. Step-by-Step Solution: Rate(A_eff) = (5/6)*(1/25) = 1/30Rate(B_eff) = (2/3)*(1/20) = 1/30Total effective = 1/30 + 1/30 = 1/15 per minTime = 15 min Verification / Alternative check:In 15 min, A contributes 15*(1/30)=1/2 and B contributes 1/2; sum 1 tank. Why Other Options Are Wrong:14, 13, 12 imply total effective rates greater than 1/15, which are not supported by the given fractions. Common Pitfalls:Reducing the times instead of the rates (that yields wrong arithmetic). Final Answer:15 min

Imperfectly opened taps (reduced discharge) Two taps A and B nominally fill a tank in 25 min and 20 min. Due to partial opening, A allows only 5/6 of its normal flow and B allows only 2/3 of its normal flow. How long will they take together to fill the tank?

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