Quite interesting question. I took lot of time to solve it. Did it by mathematical approach.
Let the efficiency of A, B and C be denoted by a, b and c
\(a = (1+\frac{1}{5})*b => a = \frac{6}{5} b => \frac{a}{b} = \frac{6}{5}\) <----- Also means (Rate of A) : (Rate of B) = 6:5
Let Rate of A = 6x and Rate of B = 5x. Since both do same work, let it be denoted by 1. Make an RTW chart with this data (See attached) to keep track of all the information.
A takes 40 days less than B. From RTW chart we get
\(\frac{1}{5x} - \frac{1}{6x} = 40\)
Solving for x we get \(x = \frac{1}{1200}\)
Put this back in RTW chart
B is 50% more efficient than C. b = \((1+\frac{1}{2}) c => b = \frac{3}{2} c => c = b*\frac{2}{3}\)
Substuting the Rate of B (Already got above) we get Rate of \(C = \frac{1}{360}\)
We now have all the rates.
A works for 50 days. Means A does 1/4 of the work.
B works for 120 days. Means B does 1/2 of the work. Total work completed by A and B = 1/4 + 1/2 = 3/4. Remaining work = 1/4
Remaining work done by C. This means C works for 90 days to complete the work.
Final answer = A
I hope this helps!
theperfectgentleman wrote:
A is 20% more efficient than B and takes 40 days less than B to complete a job. B is 50% more efficient than C. On this job, A works for 50 days and quits. Then B takes over and works for 120 days & quits. C completed the remaining work. The no. of days worked by C is
A) 90
B) 120
C) 240
D) 150
E) 360
Attachments
GMC RTW QUESTION.png [ 33.85 KiB | Viewed 10904 times ]
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