Question

X, Y, Z and W can do a certain piece of work in 10, 15, 25 and 30 days respectively. All of them started the work together but X left the job 2 days before the completion of work, Z left the work 1 day before completion of work, W worked with double of his efficiency. Find total time taken by them to complete the work is how much more or less than the total time taken by X alone to complete the work.

• A. $\frac{223}{41}$ days
• B. $\frac{224}{41}$ days
• C. $\frac{225}{41}$ days
• D. $\frac{226}{41}$ days

Answer 1

Answer: (B)

$\frac{224}{41}$ days

Answer 2

The correct option is (B): $\frac{224}{41}$ days.
Let total work be (LCM of 10,15,25 and 30) = 150
Efficiency of X = ($\frac{150}{10}$) = 15
Efficiency of Y = ($\frac{150}{15}$) = 10
Efficiency of Z = ($\frac{150}{25}$) = 6
Efficiency of W = ($\frac{150}{30}$) = 5
Let work get completed in z days.
[15*(z-2) +6*(z-1) +10z + 52*z] = 150
15z - 30 + 6z - 6 + 10z + 10z = 150
41z - 36 = 150
41z = 186
z =$\frac{186}{41}$
Required difference = 10 - $\frac{186}{41}$ = $\frac{224}{41}$ days.