Your answers are wrong.
A can solve the task in 2 hours (120 mins) working solo.
B can solve the task in 80 mins working solo.
C can solve the task in 4 hours (240 mins) solo.
As we are dealing with work being done in parallel, we need to work with reciprocal quantities, expressing the 3 task results as follows, using decimal hours throughout:
1) 1/A + 1/B = 5/4
2) 1/A + 1/C = 3/4
3) 1/B + 1/C = 1
Rearranging 3): BC = B + C => B = C / (C - 1) (call this equation X).
Subtracting 2) from 1): 1/B - 1/C = 2/4
=> C - B = 2BC/4
=> B = C - 2BC/4 => B(1 + 2C/4) = C
=> B = C / (1 + 2C/4) (call this equation Y).
Substituting eq X) in eq Y): C / (C - 1) = C / (1 + 2C/4)
Cross multiplying and cancelling: C - 1 = 1 + 2C/4
Gathering terms: 2 = C - 2C/4
=> 2 = C/2
=> C = 4 This is the first result, C takes 4 hours to complete the task.
Substituting this result in eq 3): 1/B + 1/4 = 1
=> 1/B = 3/4 => B = 4/3 hours = 80 minutes, our second result.
Substituting C = 4 in eq 2): 1/A + 1/4 = 3/4
=> 1/A = 2/4 => A = 2 hours, or 120 minutes.
So there is the complete solution, using just basic maths. I just wonder why anyone would need to work this out, and what relevance it has to commercial aviation, especially as it cannot be worked out quickly on the back of a fag packet whilst flying!
It is also quite worrying that even the published solutions are wrong. Maybe the originator of the question ought to return to being the student rather than the teacher / examiner!!
Last edited by pilotmike; 21st February 2008 at 15:12.