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?

Difficulty: Medium

20 h
60 h
15 h
25 h
None of these

Correct Answer: 20 h

Explanation:

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.

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?

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