how do i solve this?

[nangeley]

If Steffon can mix 20 drinks in 5 minutes, Dwayne can mix 20 drinks in 10 minutes, and Jacob can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?

 

[tkhunny]

Several ways to go about it. One way would be to determine a least common multiple.

We have 5 min, 10 min, and 15 min. It appears that 30 min is a common multiple.

Converting

Steffon can mix 120 drinks in 30 minutes, Dwayne can mix 60 drinks in 30 minutes, and Jacob can mix 40 drinks in 30 minutes

Adding 220 drinks in 30 minutes

Scaling

Adding 220/11 drinks in 30/11 minutes

 

[JeffM]

Several ways to go about it. One way would be to determine a least common multiple.

We have 5 min, 10 min, and 15 min. It appears that 30 min is a common multiple.

Converting

Steffon can mix 120 drinks in 30 minutes, Dwayne can mix 60 drinks in 30 minutes, and Jacob can mix 40 drinks in 30 minutes

Adding 220 drinks in 30 minutes

Scaling

Adding 220/11 drinks in 30/11 minutes This step is of course correct, but its rationale may be obscure. Where does the 11 come from? The combined rate of drink making is 220 drinks / 30 minutes (see above) but we want the minutes m for making 20 drinks. Well the rate is assumed to be independent of the time so 20 drinks / m minutes = 220 drinks /30 minutes. Now one way to solve this is by cross-multiplying m = 20 * 30 / 220 = 2.73 minutes approximately.
If however you see that 220 / 11 = 20, then 220/ 30 = (220/11) / (30/11) = 20 / (11/30) = 20/2.73 approximately. Both ways are right and so give the same answer, but, at least for me, the cross-multiplication method requires a little less intuition than tkhunny's scaling method. So now you have two methods for solving this class of problem

 

[mmm4444bot]

Here's another of the "several ways".

One could determine how many drinks each person makes per minute.

Then one could add these results to discover the total number of drinks made each minute when all three work together.

After that, I think that it's a division problem.