Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?
A. 19
B. 25
C. 26
D. 190
E. 196
20 machines working at a different constant rates. Let the rates be r1, r2, r3, r4...........r18,r19,r20.
Each day exactly two people work = 20 C 2 = (20*19)/(1*2) = 190 cases.
Each rate repeats for 19 times. If Machine 1 is chosen, then machine 1 can pair with exactly 19 other machines.
So the summation of rates is 19*(r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20) = 19* A
Where, r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20 = Rate = A
Given, 20 machines work for another 6 days.
6* (r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20) = 6*A
Total work = 19*A + 6*A = 25*A
The number of days when all 20 people work
= (25*A ) / (Total rates )
= (25*A) / (r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20)
= (25*A)/ (A)
=
25 Option B