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

The Correct Option is B

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) +10*z + 5*2*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.