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.

Updated On: Sep 13, 2024
  • 22341\frac{223}{41} days
  • 22441\frac{224}{41} days
  • 22541\frac{225}{41} days
  • 22641\frac{226}{41} days
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The Correct Option is B

Solution and Explanation

The correct option is (B): 22441\frac{224}{41} days.
Let total work be (LCM of 10,15,25 and 30) = 150
Efficiency of X = (15010\frac{150}{10}) = 15
Efficiency of Y = (15015\frac{150}{15}) = 10
Efficiency of Z = (15025\frac{150}{25}) = 6
Efficiency of W = (15030\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 =18641 \frac{186}{41}
Required difference = 10 - 18641 \frac{186}{41} = 22441\frac{224}{41} days.
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