Help me out in solving this

This is a problem that I am finding hard to solve it. Two men(A& take 24hrs to complete a particular work.If A alone does it , it takes 40minutes to complete the same work.How much time will it take if B alone does the same work.If U can solve it please tell me how U have solved the same

A takes 40 minutes by himself, but with B standing there distracting him he takes 24 hours? It seems most unlikely that B can do the work at all.

Let's assume you meant minutes when you said hours.

Suppose the work can be divided into 40 units. A does one unit per minute. So when A and B work together, A does 24 units in the 24 minutes; so B must do 16 units in those 24 minutes. So B is working at 2/3 of the rate of A. So if B does the work alone, he will take 60 minutes.

Or, one could read 'The Mythical Man Month' and realize that adding folks to projects usually slows down progress. So maybe the 40 minutes/24 hours thing was real.

This makes little sense I assume you mean A takes 40hrs not minutes Then B takes 56 hours since the average must be 48man hours.

i.e. A&B together take 24 hours = 2x24 man hours = 48 hours To do the task separately twice would take 96 hours therefore, multiplying by 2 B = 96 - A = 96 - 40 = 56hrs

This is invalid, because 'a man hour' (when A is the man) is more valuable than 'a man hour' (when B is the man), so you can't just do a straight average of 40 'A hours' with 56 'B hours' to get 48 'generic hours'.

The correct method is the one used by Nick Wedd:

A and B together finish in 24 hours, so they do 1/24 of the job every hour.

A by himself finishes in 40 hours, so he does 1/40 of the job every hour, and B must do (1/24 - 1/40) of the job every hour. This works out to (5/120 - 3/120 = 2/120 = 1/60), so B by himself would finish in 60 hours.

Here's another way to look at that invalid method:

Suppose A and B start at the same time, intending for each to do half the job. After 20 hours, A has finished his half; B has 10 hours left on his.

Suppose A quits, and C shows up and starts helping B with the remaining portion, and C works at the same speed as B. Obviously they will cut that 10 hours in half, making it 5 hours. (Averaging 'B hours' and 'C hours' *is* valid, because they are equally valuable.)

Now suppose that the same things happen, except that C works as the same speed as A (and faster than . They will cut down the 10 hours by *more* than half; it turns out that they will cut it down to 4 hours.

Answer to revised problem = 60 minutes. As stated , the answer is negative 24/35 hours. I guess this means that B screws stuff up at this rate, which has to be undone by A , working alone. Actually he screws stuff up faster than the other guy can fix it if they are working alone!!

RJ Pease

actually, use the formula portion of job done by each = 1 job thus 1/40 + 1/B = 1/24

RJ Pease