Question 621888: How many L of a 50% sugar solution must be mixed with 6.6 L of a 80% sugar solution to make a 61% solution? Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website! Solution A
Amount = x
Concentration =50% =0.5
Solution B
Amount = 6.6L
Concentration =80% = 0.8
Resultant Solution
Amount =(6.6+x)L
Concentration =61%=0.61
[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(x)(0.5)+(6.6)(0.8)=(6.6+x)(0.61)
0.5x+5.28=4.026+0.61x
0.5x-0.61x=4.026-5.28
0.5x-0.61x=-1.254
-0.11x=-1.254
-0.11x/-0.11=-1.254/-0.11
x=11.4
11.4 L of a 50% sugar solution must be mixed with 6.6 L of a 80% sugar solution to make a 61% solution of 18L?
