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(D) Let the time taken by $A, B, C$ working together be $t$ hours.Then $A$ takes $(t+6)$ hours,$B$ takes $(t+1)$ hours,and $C$ takes $2t$ hours.The co

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$4/3$

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(D) Let the time taken by $A, B, C$ working together be $t$ hours.Then $A$ takes $(t+6)$ hours,$B$ takes $(t+1)$ hours,and $C$ takes $2t$ hours.The combined work rate is $\frac{1}{t+6} + \frac{1}{t+1} + \frac{1}{2t} = \frac{1}{t}$.Subtracting $\frac{1}{2t}$ from both sides: $\frac{1}{t+6} + \frac{1}{t+1} = \frac{1}{2t}$.$\frac{(t+1) + (t+6)}{(t+6)(t+1)} = \frac{1}{2t} \Rightarrow \frac{2t+7}{t^2+7t+6} = \frac{1}{2t}$.$4t^2 + 14t = t^2 + 7t + 6 \Rightarrow 3t^2 + 7t - 6 = 0$.$(3t-2)(t+3) = 0$. Since time must be positive,$t = 2/3$ hours.Now,$A$ takes $2/3 + 6 = 20/3$ hours and $B$ takes $2/3 + 1 = 5/3$ hours.Combined rate of $A$ and $B = \frac{3}{20} + \frac{3}{5} = \frac{3+12}{20} = \frac{15}{20} = 3/4$ work per hour.Time taken by $A$ and $B$ together = $4/3$ hours.

Three men $A, B, C$ working together can do a job in $6$ hours less time than $A$ alone,in $1$ hour less time than $B$ alone,and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in (in hours):

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