Question : Two men and 7 women can complete a work in 28 days, whereas 6 men and 16 women can do the same work in 11 days. In how many days will 5 men and 4 women, working together, complete the same work?
Option 1: 20
Option 2: 18
Option 3: 14
Option 4: 22
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Correct Answer: 22
Solution :
Given,
2 men and 7 women can complete a work in 28 days
6 men and 16 women can do the same work in 11 days
Use the following formula to solve the question:
$\frac{M_{1}\times D_{1}}{W_{1}}=\frac{M_{2}\times D_{2}}{W_{2}}$
Where $M_1$ and $M_2$ are men, $D_1$ and $D_2$ are days and $W_1$ and $W_2$ are work done.
Let the efficiency of 1 man be M.
And the efficiency of 1 woman is W.
According to the question,
2 men and 7 women complete a work in 28 days = 6 men and 16 women complete the same work in 11 days
⇒ (2M + 7W) × 28 = (6M + 16W) × 11
⇒ 56M + 196W = 66M + 176W
⇒ 10M = 20W
⇒ M = 2W
2 men and 7 women = 2M + 7W
⇒ 2 men and 7 women = 4W + 7W
⇒ 2 men and 7 women = 11W
5 men and 4 women = 5M + 4W
⇒ 5 men and 4 women = 10W + 4W
⇒ 5 men and 4 women = 14W
$\frac{M_1D_1}{W_1}=\frac{M_2D_2}{W_2}$
⇒ 11W × 28 = 14W × D
2
⇒ D
2
= 22 days
Hence, the correct answer is 22.
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