A radiator contains 25 quarts of a water and antifreeze solution of which 60% is antifreeze. How much of this solution should be drained and replaced with water for the new solution to be 40% antifreeze?
Can somebody please explain how it's done?
A radiator contains 25 quarts of a water and antifreeze solution of which 60% is antifreeze. How much of this solution should be drained and replaced with water for the new solution to be 40% antifreeze?
Can somebody please explain how it's done?
Oh, wait scrap what I said. I think I get it now!!!!!!!! omg THANK YOU.
Err..okay I have another one:
A gardener has 26lb of a mixture of fertilizer and weed killer. If 1 lb of the mixture is replaced by weed killer the result is a mixture that is 5 % weed killer. What percent of the original mixture was weed killer?
There doesn't need to be an answer, just steps on how to get there. Thanks!
Hello, strigy!
Let = percent of weed killer (WK) in original mixture.A gardener has 26lb of a mixture of fertilizer and weed killer.
If 1 lb of the mixture is replaced by weed killer the result is a mixture that is 5% weed killer.
What percent of the original mixture was weed killer?
He starts with 26 lbs of mix which is weed killer.
. . It contains: . lbs of WK.
One pound of mix is removed.
. . It contains: . lbs of WK.
One pound of pure (100%) WK is added.
. . It contains: . lb of WK.
Hence, the final mix contains: . lbs of WK.
But we know that the final mix will be 26 lbs which is 5% WK.
. . So it contains: . lbs of WK.
We just described the amount of WK in the final mix in two ways.
There is our equation! . . .
Solve for