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24 Jun 2009, 04:04
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N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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24 Jun 2009, 10:08
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The answer is E.
\(\text{rate} \times \text{time} = \text{work}\)
Given
\(work = 1\) (filling 1 tank)
For A+B,
\(\text{work} = 1\)
\(\text{time} = \frac{6}{5}\)
\(\text{rate} = \frac{\text{work}}{\text{time}} = \frac{1}{\frac{6}{5}} = \frac{5}{6}\)
For A+C,
\(\text{work} = 1\)
\(\text{time} = \frac{3}{2}\)
\(\text{rate} = \frac{\text{work}}{\text{time}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}\)
For B+C,
\(\text{work} = 1\)
\(\text{time} = 2\)
\(\text{rate} = \frac{\text{work}}{\text{time}} = \frac{1}{2}\)
Adding all three rates
\(\Rightarrow \text{(A+B)+(A+C)+(B+C)} = \frac{5}{6} + \frac{2}{3} + \frac{1}{2} = \frac{5}{6} + \frac{4}{6} + \frac{3}{6} = \frac{12}{6}\)
\(\Rightarrow \text{2A+2B+2C} = \frac{12}{6} = 2\)
\(\Rightarrow \text{2(A+B+C)} = 2\)
\(\Rightarrow \text{(A+B+C)} = \frac{2}{2} = \frac{1 Tank}{hour}\)
again
\(r \times t = w\)
\(r(A+B+C) = 1\) -- from above and \(w = 1\)
Hence
\(t = \frac{1}{1} = 1\)
\(\text{rate} \times \text{time} = \text{work}\)
Given
\(work = 1\) (filling 1 tank)
For A+B,
\(\text{work} = 1\)
\(\text{time} = \frac{6}{5}\)
\(\text{rate} = \frac{\text{work}}{\text{time}} = \frac{1}{\frac{6}{5}} = \frac{5}{6}\)
For A+C,
\(\text{work} = 1\)
\(\text{time} = \frac{3}{2}\)
\(\text{rate} = \frac{\text{work}}{\text{time}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}\)
For B+C,
\(\text{work} = 1\)
\(\text{time} = 2\)
\(\text{rate} = \frac{\text{work}}{\text{time}} = \frac{1}{2}\)
Adding all three rates
\(\Rightarrow \text{(A+B)+(A+C)+(B+C)} = \frac{5}{6} + \frac{2}{3} + \frac{1}{2} = \frac{5}{6} + \frac{4}{6} + \frac{3}{6} = \frac{12}{6}\)
\(\Rightarrow \text{2A+2B+2C} = \frac{12}{6} = 2\)
\(\Rightarrow \text{2(A+B+C)} = 2\)
\(\Rightarrow \text{(A+B+C)} = \frac{2}{2} = \frac{1 Tank}{hour}\)
again
\(r \times t = w\)
\(r(A+B+C) = 1\) -- from above and \(w = 1\)
Hence
\(t = \frac{1}{1} = 1\)
24 Jun 2009, 10:28
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Right solution. It is 1.
