I just knocked this off for 'fun' in about 5 mins without reading beyond the first post. I've now read the rest of the threads and seen that (apart from a transcription error on the part of the question writer), my answers match those in post 3. As there seems to be a lot of misunderstanding, and as I have far more important work to be doing which I need an excuse to avoid, here's the solution.
Although it looks like a straightforward 3-variable simultaneous equation, the trap is that you have to realise that you are dealing with reciprocals. If you do not spot this, you blunder down a path where you consider that the time taken for two people to do the work is the sum of the times taken for each individual to do it alone. Clearly not right, as it will always be quicker for two people to work together.
So...
Three people, A, B & C, who each can work at their own particular speed, which we'll call work rates Ra, Rb, and Rc (work done per hour) respectively. The total amount of work to be done, the size of the job, we'll call W (work).
The three lines of data in the original question, we'll refer to as Equation 1, Eq 2 and Eq 3, concerning A&B, A&C, B&C respectively.
Looking at Eq1:
48mins is 4/5 hr. Individual A, working at a work rate of Ra (work/hr), for 4/5 hr, will perform 4/5 Ra units of work. Similarly B will do 4/5 Rb units of work. The total work done by them both together in a fixed amount of time is the sum of the work done by them individually in that amount of time. Hence the work done by A&B is 4/5 Ra + 4/5 Rb. In this amount of time they have completed the job, which we called the unit amount of work, W. This gives us:
Eq1 4/5 Ra + 4/5 Rb = W
Doing this for all three cases gives us:
Eq1 4/5 Ra + 4/5 Rb = W
Eq2 4/3 Ra + 4/3 Rc = W
Eq3 Rb + Rc = W
=>
Eq1 4 Ra + 4 Rb = 5W
Eq2 4 Ra + 4 Rc = 3W
Eq3 Rb + Rc = W
Now solve as a standard 3-variable simultaneous equation for work
rates.
Subtract Eq2 from Eq 1 to give Eq4:
Eq1 { 4 Ra + 4 Rb = 5W }
Eq2 - { 4 Ra + 4 Rc = 3W } -
----- -------------------------------
Eq4 { 0 Ra + 4 Rb - 4 Rc = 2W }
----- -------------------------------
Eq4 2 Rb - 2 Rc = W
Add Eq4 to 2 x Eq3 to solve for Rb
Eq4 { 2 Rb - 2 Rc = W }
Eq3 + { 2 Rb + 2 Rc = 2W } +
-----------------------------
{ 4 Rb + 0 Rc = 3W }
-----------------------------
=>
Rb = 3/4 W (units of work per hour)
Substitute into Eq3:
3/4 W + Rc = W
=>
Rc = 1/4 W
Substitute into Eq1:
=>
Ra = 1/2 W
We have now solved for the three individuals
work rates. If you use these values in Eq1, Eq 2 & Eq3, they all tally up.
Work rates, R:
Ra = 1/2 W }
Rb = 3/4 W } (units of work done per hour)
Rc = 1/4 W }
To carry out the total work alone, would take an individual... the amount of work, W (units of work) divided by their personal work rate R (units of work done per hour), giving time taken in hours. For the one unit of work, W, the time taken in hours is the reciprocal of the hourly work rate:
Time taken, T:
Ta = 2/1 hrs = 2 hrs = 120 mins
Tb = 4/3 hrs = 1hr 20mins = 80 mins
Tc = 4/1 hrs = 4 hrs = 240 mins
This pretty much matches the textbook answers. I think they have transcibed B and C somewhere.
Additionally you can say that the work rate of the three working together is:
Ra + Rb + Rc = Rabc = 3/2 W
Therefore the time taken would be 2/3 hrs = 40 mins.
Job done. Sorry to be long winded. I don't mean to teach granny to suck eggs but hopefully this eliminates any room for misunderstanding. It was a lot quicker to do the calc than to type this up!
You won't find yourself doing these sort of calculations as a professional pilot, but it is essential that a pilot has an above average working knowledge of maths and good mental arithmetic.
