Two fillers together take 12 h; one is 10 h faster If two pipes running together fill a tank in 12 hours, and one pipe is 10 hours faster than the other, how long does the faster pipe take by itself?

Question

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

Accepted Answer

Introduction / Context:Difference-of-times problems with a known joint time reduce to a quadratic in the individual times when working in rate form. Choose the physically valid root. Given Data / Assumptions: Together time = 12 h. Let slower = x h; faster = x − 10 h. Rates add: 1/x + 1/(x − 10) = 1/12. Concept / Approach:Clear denominators, solve the quadratic, and discard negative or non-sensible times (x − 10 > 0). Step-by-Step Solution: (1/x) + (1/(x − 10)) = 1/12(2x − 10) / (x(x − 10)) = 1/1212(2x − 10) = x^2 − 10x ⇒ 24x − 120 = x^2 − 10xx^2 − 34x + 120 = 0 ⇒ x = (34 ± 26)/2 ⇒ x = 30 (valid) or 4 (invalid)Slower = 30 h; Faster = 20 h Verification / Alternative check:1/30 + 1/20 = (2 + 3)/60 = 5/60 = 1/12 as required. Why Other Options Are Wrong:60, 15, 25 mismatch the quadratic solution. Common Pitfalls:Using time differences directly rather than rates, leading to linear (but wrong) equations. Final Answer:20 h

Two fillers together take 12 h; one is 10 h faster If two pipes running together fill a tank in 12 hours, and one pipe is 10 hours faster than the other, how long does the faster pipe take by itself?

More Questions from Pipes and Cistern

Discussion & Comments