View Full Version : Math question

24th Oct 2007, 07:06

I'm practising my math for the dlr test.

I've a question and i don't how to solve. I'm sure it will be easy but i don't know how to find it:

Person A and B are working together for a task and it takes 48' if they work togethr.

Person A and C are doing the same task and it takes 80 min.

Person B and C are doing the same task and it takes 60 min.

How long are A B C working?

(answer is 48min, 80min, 60min)

Can somebody tell me what calculations you have to make?


24th Oct 2007, 07:29
48 +2c=140 So,c=(140-48)/2 c=46
a=80-c a=34
b=60-c b=14


24th Oct 2007, 08:05
My mistake, wrong anwsers

the right answers are:

person a is 120' working
person b is 240' working
person c is 80' working

but i still don't know how to obtain this answers

anyone can help??

24th Oct 2007, 08:40
your answers are wrong!

24th Oct 2007, 10:30
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!!

24th Oct 2007, 13:53
Captain djaffar,

the answers I gave above are not out of logic....It is that you didn't hear so far about the reductibility of equations....very simple!

24th Oct 2007, 16:47
i didnt notice the hour unit.had the problem too quickly.
that's why i did not take it in consideration.:ugh::ugh::ugh:
but based on a purely equational assumption like i did as well as A13 the results 34,14,46 are perfect.
but accounting for the hour factor 'complicates' the whole matter...and your question ask to take it into account...


24th Oct 2007, 17:02
I have seen simpler questions on aptitude tests along the lines of:

If John completes 5 tasks in 27 minutes and Bob completes 5 tasks in 54 minutes, how long will it take them to complete 5 tasks if they work together?

But never anything so complex as the question above!


The answer to the above is 18 (apparently!)

24th Oct 2007, 17:16
Again, Captain_djaffar is mistaken. It has nothing to do with having 1 hour wrong either - that is simply a red herring, of no relevance.

Trying to convince us that the results 34,14,46 are perfect is simply pointless. The results are simply WRONG, and therefore far from perfect.

You simply cannot say that if two people, A and B take 48 minutes to complete a task, then each of them take LESS time to complete it individually. (airman13 would like us to believe that A takes 34 minutes and B takes 14 minutes to do the task individually, therefore they take a total of 48 minutes when working together). Life just doesn't work like that! Take a closer at my earlier dripping tap example to prove it.

This really doesn't warrant further discussion, as you are not accepting the principle that working individually takes longer than working together, as in the tap example.

I notice that your earlier post with incorrect calculations has mysterioulsy disappeared from this thread now!

Simply, inner asked to be shown the calculation needed, and after receiving some wild guesses which were wrong, I offered him the correct and complete solution in post #6, which your deleted post now has made post #5.


24th Oct 2007, 18:06
i understand ur point pilotmickey...i said the answers are perfect if they represent a simple equation and not involving the factor of time...which i did by mistake....which by the way doesnt solve inner's problem!Perfect only for another problem with mere units rather than hours.

your answer is perfectly descriptive of the real problem and gives the real solution.:D


really it was good explanation and indeed rather tough for an aptitude test.
maybe more suitable for a cambridge Thinking skill 3th paper....

24th Oct 2007, 18:34
it is a harder version of the basic simultaneous equations/system of equation, done at O/standard grade maths level.

i hope the big red sentence at the bottom of the page isnt taking effect here as many pilots/engineers/ numerically orientated people on here may end up getting a bit embaressed.

24th Oct 2007, 19:05
There are some interesting interpretations in this string.

But I think that we need to RTFQ.

The question did not ask how long it would take each individual to complete the task when working alone.

It asked "How long are A B C working?"

Nothing in the question suggests that any individual works less than the full specified time when part of a specified pair.

A works with B for 48 minutes and with C for 80 minutes. So A works for a total of 48 + 80 = 128 minutes.

B works with A for 48 minutes and with C for 60 minutes. So B works for 48 + 60 = 108 minutes.

C works with B for 60 minutes and with A for 80 minutes. So C works for 60 + 80 = 140 minutes.

"Words failurise me" ................................. George Dubbya Bush.
"Don't let them failurise you"................... RTFQ.

24th Oct 2007, 21:02
Oh feck! is anyone else reading this realising that maths is a very weak point of theirs like i am. Why cant we all just get jobs on flying skills alone (sigh):bored:

24th Oct 2007, 21:10
If they were French, then the answer is a lot easier.

They weren't working at all but on strike or reading the newspaper, or actually off doing a little job on the 'noir'....and having a friend X clock them in and out.

So the answer for A B C is Zero... (that's if they're French).

24th Oct 2007, 21:32
How does any of this help you to fly a plane ?

25th Oct 2007, 06:48
This question is again about tasks completed in parallel, so reciprocals are required. Working in minutes this time, for convenience, the equation is:

It reminds me when elecrical resistors are in parallel: there you have to use reciprocals (for the same reasons) too.


25th Oct 2007, 09:51
It reminds me when elecrical resistors are in parallel: there you have to use reciprocals (for the same reasons) too.

Precisely. As the voltage across parallel resistors is identical, we are considering the total flow of current through the various resistors, just like multiple taps dripping into a basin, or workers hopefully working alongside each other.
How does any of this help you to fly a plane ?
Exactly the point I made at the end of my initial solution - its of no use to aviation whatsoever! Just a stupid test of irrelevant maths, where there are far more useful aspects, eg mental arithmetic or logic, which could have been tested.

But inner simply wanted help with seeing how to solve the problem, so I obliged.


31st Oct 2007, 23:21
Help me I cant stand It!!!

I sit here every other night or so just quietly checking out what my peers are up to and saying nothing but this is too much.:ugh:

The question is rubbish, its not really a problem, who asked it, I want to kick them in the shins! And the answer is b:mad:s! and if its part of the problem its not clever it just becomes an illogical problem or bigger b:mad:s

Person A works for 48 mins with B and 80 mins with C = 128 mins!
Person B works for 48 mins with A and 60 mins with C = 108 mins!!
Person C works for 80 mins with A and 60 mins with B = 140 mins!!!

Question: How long are A B C working?
A+B+C= 376 mins. or to be really clever:8 6 hrs and 16 mins.

Kieth I salute you.
Pilot mike I salute you also, the force is strong with you but WOOD FOR TREES:ok:

1st Nov 2007, 13:36
Pilot mike I salute you also, the force is strong with you but WOOD FOR TREES
As nobody can twig what the examiner wanted, and we've all branched out to get to the root of the problem, possibly we are all barking up the wrong tree? In which case yew or I could be right fir now, and I wood suggest that we leaf it there rather than stick at it any longer, before everyone gets sycamore of this maths.


1st Nov 2007, 13:46
but the obvious question is, what sort of job are they doing?
Are they UK council workers because if so, there's no way they would work that long and furthermore, there'd be twice as many workers doing half as much work, taking into account the two 'workers' who satnd and watch the proceedings!

PS, I'm s**t at maths.

2nd Nov 2007, 10:18
So ,john completes 5 tasks in 27 min ,and bob completes the same 5 tasks in 54 min .
This means john is 2 times faster than bob.In this case john will complete 10 tasks in 54 min .So in 54 min the both of them will complete 15 tasks together.easily to see that 5 tasks will be complete in54/3 =18 min.

2nd Nov 2007, 10:21
the answer above is for pilotmike especially.

Dick Whittingham
2nd Nov 2007, 11:12
This is lovely! To quote Isaac Newton what we have here is not a mathematical problem. The question is, what is the mathematical problem.

So far as I can see, the most logical interpretation of the question is that there exists a standard task T, and on different occasions three diferent teams, A + B, A + C and B + C complete T in different times. From this we have to work out how long each individual would take to complete T, working on his own.

If you accept this as the mathematical problem then pilotmike has the mathematical answer.

Remember that Godel said that in any formal mathematical system there are theorems that are true that cannot be proved within that system.

and vice versa!


Fg Off Max Stout
2nd Nov 2007, 16:14
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.


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. :ok:

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