If 3 men or 6 boys can do a piece of work in 10 days, working 7 hours a day; how many days will it take to compete a piece of work twice as large with 6 men and 2 boys working together for 8 hours a day ?

A) 6

B) $7\frac{1}{2}$

C) $8\frac{1}{2}$

D) 9


Option B


3 men = 6 boys = (6 men + 2 boys) = 14 boys.

More work, More days (Direct Proportion)

More boys, Less days (Indirect Proportion)

More hours per day, Less days      (Indirect Proportion)

$\left\{\begin{array}{c}Work\quad\quad\quad\quad\quad\quad 1:2\\ Boys\quad\quad\quad\quad\quad\quad 14:6\\Hours/Day\quad\quad\quad\quad 8:7\end{array}\right\}::10:x$

$\therefore (1\times 14\times 8\times x)$ $=(2\times 6\times 7\times 10)$  

$\Leftrightarrow x=\frac{840}{112}$ $=7\frac{1}{2}$