[galactus]

Here's an algebra problem you may like to take a stab at.

"1 man and 2 boys can do a job in the same amount of time as 2 men and 1 boy. How long for 1 man to do the job?".

Sometimes these can be tricky to wrap the ol' cranium around.

[Capuchin]

Haha I hate these. It'll keep me off my math problem that I'm working on (I might post it a bit later)

[Capuchin]

My first thought is 3 times as long, but I must be missing something. Too simple.

In terms of rates:

1m+2b = 2m+1b
2b-1b = 2m-1m
m = b

so 1m+2b = 3m

so 1m is three times slower.

?

[asterisk_man]

I have to agree with Capuchin and I can't see how it would be any more confusing. Unless there's some sly joke about male predisposition for laziness in there :)

[galactus]

Here was my solution to this wacky problem.

Let x=boy time
Let y=man time

... [1]

and

... [2]


Solve [1] for x and get:



Sub into [2] and solve for y and we get:



Easy to solve for y=12

Subbing back into the equations, we see that and equal 0.

But the denominators can not be 0, therefore,



The boys work infinite time. :) ;) :D :p :cool:

Actually, just looking at the problem:

1 man and 2 boys do a job in 12 hours
2 men and 1 boy do a job in 6 hours

Since one man does the job in 12 hours, that's the same as 1man and two boys.
the boys are worthless. They lean on their shovels and watch the men work.

[asterisk_man]

Actually, just looking at the problem:

1 man and 2 boys do a job in 12 hours
2 men and 1 boy do a job in 6 hours

I must have missed this part. It doesn't appear anywhere in your initial question. You said the two combinations did the job in an equal amount of time.

[Capuchin]

The answer could be the same if "an equal amount of time" was an infinite amount of time :p

But yes I was perplexed by your answer too gal, thought maybe it was just below me. :p