A alone can do a work in 22 days. B alone can do the same work in 44 days. C alone can do the same work in 66 days. They work in the following manner :
Day 1 : A and B work
Day 2 : B and C work
Day 3 : C and A work
Day 4 : A and B work and so on.
In how many days will 81 \(\frac{9}{11}\)% of the work be completed?
Answer & explanation
Correct answer: option 2
Let the Total work = 132 (LCM of times 22, 44, 66)
So efficiency of A, B and C is 6 w/d, 3w/d, 2w/d respectively.
Now,
\(9\frac{1}{11}\)% = \(\frac{100}{11}\)% = \(\frac{1}{11}\) and
\(81\frac{9}{11}\)% = \(\frac{9}{11}\)
Work to be done = 81\(\frac{9}{11}\)% of 132 = \(\frac{9}{11}\) × 132 = 108
ATQ:
Work done⇒
1st day (A + B) = 6 + 3 = 9
2nd day (B + C) = 3 + 2 = 5
3rd day (C + A) = 2 + 6 = 8
So in 3 days = 9 + 5 + 8 = 22
accordingly,
Work done in (3 × 4 =) 12 days = 22 × 4 = 88
Now,
13th day work (A + B) = 9
14th day work (B + C) = 5
So, in 14 days work done = 88 + 9 + 5 = 102
Remaining work = 108 - 102 = 6
on 15th day (C + A) does the remaining work in = \(\frac{6}{8}\) = \(\frac{3}{4}\) days.
So the required work complete in = 14 + \(\frac{3}{4}\) = \(14\frac{3}{4}\) days.