How much 60% concentrated cordial must be added to 90% concentrated cordial to make 6L of 80% cordial?
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
Please walk it through for me
Thank you so much!
To find out how much 60% concentrated cordial must be added to 90% concentrated cordial to make 6L of 80% cordial, we need to use the concept of mixing solutions.
Let x represent the amount of 60% concentrated cordial needed in liters.
To solve the problem, we can use the following equation based on the principle of mixtures:
(60% * x) + (90% * (6 - x)) = 80% * 6
Let's break it down:
60% * x represents the amount of cordial (in liters) from the 60% concentrated solution.
90% * (6 - x) represents the amount of cordial (in liters) from the 90% concentrated solution (since the total volume is 6 liters).
80% * 6 represents the desired concentration of the final mixture (80% cordial in 6 liters).
Now, let's solve the equation:
(0.6x) + (0.9(6 - x)) = 0.8(6)
0.6x + 5.4 - 0.9x = 4.8
Combine like terms:
0.6x - 0.9x = 4.8 - 5.4
-0.3x = -0.6
Divide by -0.3 to isolate x:
x = (-0.6) / (-0.3)
x = 2
Therefore, you will need to add 2 liters of 60% concentrated cordial to 4 liters of 90% concentrated cordial to make 6 liters of 80% cordial.