Thread: basic algebra

If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?
is there a rule for this??!

"If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?
is there a rule for this??!"

Well Steven is mixing 20/5= 4 drinks per minute.
Sue is mixing 20/10=2 drinks per minute.
Jack is mixing 20/15=4/3 drinks per minute.

So in 1 minute (60 seconds), the 3 of them can mix 4+2+(4/3)=(22/3) drinks

So:

60 seconds - 22/3 drinks
90/11 seconds - 1 drink
(90/11)*20 seconds = 20 drinks

So the answer is 163.6 seconds.

There is probably a quicker way though

Since the three of them mix drinks at a rate of 22/3 drinks per minute . . .

....20 drinks
--------------.....= 30/11 minutes or 2.73 minutes or 163.6 seconds
(22/3 drinks/min)

If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?
is there a rule for this??!

Working together problems use this format:

Time together / time alone + time tog/ time alone = 1 [full job]
x/5 + x/10 + x/15 = 1 Multiply by common denom (30) to clear the fractions -->
6x + 3x + 2x = 30
11x = 30
x = 30/11

Here's another approach: If they all work until the slowest one is finished (15 mins), how many drinks will they have?
60 + 30 + 20 = 110 in 15 mins
But we only want 20 drinks, so
110/ 20 = 15/x
110x = 20*15
x = 30/11