Question : A alone can do $\frac{2}{5}$ of a work in 12 days. B is 25% more efficient than A. C alone can do the same work in 12 days less than B. D is 25% less efficient than C. If they all work together, then the work will be completed in how many days?
Option 1: $\frac{240}{53}$ days
Option 2: $\frac{180}{43}$ days
Option 3: $\frac{200}{51}$ days
Option 4: $\frac{300}{47}$ days
Correct Answer: $\frac{240}{53}$ days
Solution :
Given: A alone can do $\frac{2}{5}$ of a work in 12 days.
Total work = Efficiency × Number of days
A will finish one work in $12 \times \frac{5}{2}=30$ days.
A's work in a day is $\frac{1}{30}$ of the total work.
Since B is 25% more productive than A.
B will finish the work in $\frac{100}{125}\times 30 =\frac{4}{5}\times 30=24$ days.
One day's work for B equals $\frac{1}{24}$ of the total work.
C requires 12 fewer days than B = 24 – 12 = 12 days.
C's one day work $\frac{1}{12}$ of the total work.
Currently, D is 25% less productive than C.
So, D will finish the task in $\frac{100}{75}\times 12 = 16$ days.
D's one day work $\frac{1}{16}$ of the total work.
Together one day work $=\frac{1}{30}+\frac{1}{24}+\frac{1}{12}+\frac{1}{16}$.
$=\frac{8+10+20+15}{240}=\frac{53}{240}$ part of work.
The time taken by all together to complete the work = $\frac{240}{53}$ days.
Hence, the correct answer is $\frac{240}{53}$ days.
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