The work done by a man is double the work done by a woman in the same time. If 10 men can do a piece of work in 8 days, then in how many days that work can be done by 3 men and 4 women?

### Concept & Efficiency Equivalence When the efficiency ratio between two types of workers is given, we can convert a mixed group of workers entirely into one uniform type to easily solve the time and work equation. $$ M_1 D_1 = M_2 D_2 $$ ### Step-by-Step Solution * Given efficiency: Work of 1 man = Work of 2 women (or $1\text{M} = 2\text{W}$). * The problem states $M_1 = 10 \text{ men}$ can do the work in $D_1 = 8 \text{ days}$. * We need to find the time taken by a new group: 3 men and 4 women. * Convert the women into equivalent men to match our $M_1$ unit: Since $2\text{W} = 1\text{M}$, then $4\text{W} = 2\text{M}$. * Total equivalent men in the new group ($M_2$): $3\text{M} + 2\text{M} = 5\text{M}$. * Apply the formula $M_1 D_1 = M_2 D_2$: $$ 10 \times 8 = 5 \times D_2 $$ * Solve for $D_2$: $80 = 5 \times D_2 \Rightarrow D_2 = 16$. ### Exam Strategy & Shortcut Instead of doing algebraic conversions, you can assign unit efficiencies. Let a woman do 1 unit/day, so a man does 2 units/day. Total Work = 10 men $\times$ 2 units $\times$ 8 days = 160 units. New capacity = (3 men $\times$ 2 units) + (4 women $\times$ 1 unit) = 6 + 4 = 10 units/day. Days = 160 units / 10 units/day = 16 days. This mental math avoids fractions and equations entirely. ### Common Pitfall A common mistake is reading "work done is double" and incorrectly writing $2\text{M} = 1\text{W}$ instead of the correct $1\text{M} = 2\text{W}$. Always map the multiplier to the less efficient unit to balance the equation. ### Final Answer Therefore, the correct answer is **16**.