Question 1124934
rate * time = quantity of work produced.


the rate of the new machine is 3 times the rate of the old machine.


if the rate of the old machine is x, then the rate of the new machine is 3x.


when both machines work together, their rates are additive.


therefore:


(x + 3x) * time = quantity of work produced.


time is 6 hours and quantity of work produced is 20,000, therefore:


(x + 3x) * 6 = 20,000.


combine like terms to get 4x * 6 = 20,000


\divide both both sides of the equation by 24 to get:


x = 20,000 / 24 = 833 and 1/3.


3x is therefore 3 * (833 + 1/3) = 2500


to see if these rates are accurate, replace x and 3x in the original equation to see if it holds truel.


the original equation is (x + 3x) * 6 = 20,000.


substituting 833 and 1/3 for x and 2500 for 3x, we get:


((833 + 1/3) + 2500) * 6 = 25,000.


this results in 25,000 = 25,000, which is true, therefore you can assume that the rates are accurate.


given that the rate of the new machbine is 2500 aluminum cans per hour, then the formula for the new machine is 2500 * time = 20,000.


this results in time = 20,000 / 2500 = 8 hours.


the new machine would take 8 hours working alone.


breaking down the 6 hours when working together, you get:


(833 + 1/3) * 6 = 5000 aluminum cans in 6 hours.


2500 * 6 = 15000 aluminum cans in 6 hours.


total for 6 hours = 20,000 cans.


the new machine, at 3 times the rate of the old machine, produced 3 times the number of cans in the same 6 hours.


your solution is that the new machine would take 8 hours to make 20,000 aluminum cans, working alone.