PS - Word rate problem

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

\(t = \frac{1}{1} = 1\)