Question

Four men and three women finish a job in six days. And five men and seven women can do the same job in four days. How long will one man and one woman take to do the work.
A. \[22\dfrac{2}{7}\]
B. \[25\dfrac{1}{2}\]
C. \[5\dfrac{1}{7}\]
D. \[12\dfrac{7}{{22}}\]

Answer 1

Hint: In this question, we will proceed by finding out how many days it will take to complete the work by a single woman and a single man. Then we will find how many days it takes to complete the work by one man and one woman. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given that work done by four men and three women in six days is equal to the work done by five men and seven women in four days.
Let us consider that one man takes \[x\] days and one woman takes \[y\] days to complete the work. So, work done per day by a man is \[\dfrac{1}{x}\] and the work done per day by a woman is \[\dfrac{1}{y}\].
Since 4 men and 3 women can done a work in 6 days, the work done by them per day is given by
\[ \Rightarrow \dfrac{4}{x} + \dfrac{3}{y} = \dfrac{1}{6}...........................\left( 1 \right)\]
Since 5 men and 7 women can done a work in 4 days, the work done by them per day is given by
\[ \Rightarrow \dfrac{5}{x} + \dfrac{7}{y} = \dfrac{1}{4}...........................\left( 2 \right)\]
Now, multiplying equation (1) with 5 we have
\[ \Rightarrow \dfrac{{20}}{x} + \dfrac{{15}}{y} = \dfrac{5}{6}...........................\left( 3 \right)\]
And, multiplying equation (2) with 4 we have
\[ \Rightarrow \dfrac{{20}}{x} + \dfrac{{28}}{y} = 1...........................\left( 4 \right)\]
Subtracting equation (3) from (4), we have
\[
   \Rightarrow \left( {\dfrac{{20}}{x} + \dfrac{{28}}{y}} \right) - \left( {\dfrac{{20}}{x} + \dfrac{{15}}{y}} \right) = 1 - \dfrac{5}{6} \\
   \Rightarrow \dfrac{{20}}{x} - \dfrac{{20}}{x} + \dfrac{{28}}{y} - \dfrac{{15}}{y} = \dfrac{{6 - 5}}{6} \\
   \Rightarrow \dfrac{{28 - 15}}{y} = \dfrac{1}{6} \\
   \Rightarrow \dfrac{{13}}{y} = \dfrac{1}{6} \\
  \therefore y = 13 \times 6 = 78 \\
\]
Substituting \[y = 78\] in equation (1), we have
\[
   \Rightarrow \dfrac{4}{x} + \dfrac{3}{{78}} = \dfrac{1}{6} \\
   \Rightarrow \dfrac{4}{x} = \dfrac{1}{6} - \dfrac{3}{{78}} \\
   \Rightarrow \dfrac{4}{x} = \dfrac{{13 - 3}}{{78}} \\
   \Rightarrow \dfrac{4}{x} = \dfrac{{10}}{{78}} \\
  \therefore x = \dfrac{{78 \times 4}}{{10}} = \dfrac{{156}}{5} \\
\]
Now, time taken to finish the work by 1 man and 1 woman \[ = \dfrac{1}{x} + \dfrac{1}{y}\]
\[
   = \dfrac{1}{{\dfrac{{156}}{5}}} + \dfrac{1}{{78}} \\
   = \dfrac{5}{{156}} + \dfrac{1}{{78}} \\
   = \dfrac{{5 + 2}}{{156}} \\
   = \dfrac{7}{{156}}{\text{ }} \\
\]
Therefore, the number of days taken to complete the work by one man and one woman is equal to \[\dfrac{{156}}{7} = 22\dfrac{2}{7}{\text{ days}}\].
Thus, the correct option is A. \[22\dfrac{2}{7}\].

Note: In questions like these, work done is calculated in fraction, this fraction is actually the reciprocal of the number of days in which work is done. So, the denominator of the fraction stands for the number of days. Here the obtained number of days in which only one man and one woman can finish the work must be less than the number of days taken to finish the work by four men and three women and also the number of days taken to finish the work by five men and seven women. Otherwise our obtained answer is wrong.