How much 60% concentrated cordial must be added to 90% concentrated cordial to make 6L of 80% cordial?

Answer 1

Please walk it through for me

Answer 2

Let the amount of the 90% concentrate be x L
then you will need 6-x L of the 60% concentrate.

.9x + .6(6-x) = .8(6)
multiply by 10 to make the numbers "nicer"
9x + 6(6-x) = 8(6)
9x + 36 - 6x = 48
3x = 48-36
3x = 12
x = 4
So you will need 4 L of the 90% stuff, and 2 L of the 60% stuff

you could have defined it this way:

Let the amount of the 60% concentrate be k L
then you will need 6-k L of the 90% concentrate.

.6k + .9(6-k) = .8(6)
6k + 9(6-k) = 8(6)
6k + 54 - 9k = 48
-3k = -6
k = 2 , then 6-k = 4
notice we get the same answer, just follow my new definition

Answer 3

Thank you so much!