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If $9$ men working $7\frac{1}{2}$ hours a day can finish a work in $20$ days, then how many days will be taken by $12$ men working $6$ hours a day to finish the work, it being given that $3$ men of former type work as much as $2$ men of the latter type in the same time ?

  1. $12\frac{1}{2}$
  2. $13$
  3. $9\frac{1}{2}$
  4. $11$
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$9$ men working $7\frac{1}{2}$ hours a day can finish a work in $20$ days,

Total men hours required $=9\times \frac{15}{2} \times 20 = 9\times15 \times 10$ hours.

$3$ men of former type work as much as $2$ men of the latter type,

$\therefore$ Total men hours required for latter type = $\frac{2}{3}$ of total men hours for the former type

Let they require $x$ days to finish the work,

$\require{cancel}  \begin{align} 12\times 6\times x &= \frac{2}{\cancel3} \times \cancelto{3}{9}\times 15\times 10 \\ \implies x &=\frac{ \cancel2 \times \cancel3 \times \cancelto{5}{15} \times \cancelto{5}{10}}{\cancelto{2}{12}\times \cancel6} \\ \implies x &= \frac{25}{2} = 12\frac{1}{2} \end{align} $

Option A is correct.

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