Answer:
A. 605 mL of the 35% solution
B. 55 mL of the 95% solution
Step-by-step explanation:
It is given that, you need 660 mL of a 40% alcohol solution. On hand, you have 35% alcohol mixture. You also have a 95% alcohol mixture.
Let x and y be the amount of 35% alcohol solution and 95% alcohol solution (in mL) respectively.
So,
Amount equation: [tex]x+y=660[/tex] ...(1)
Alcohol equation : [tex]0.35x+0.95y=0.40(660)\Rightarrow 0.35x+0.95y=264[/tex] ...(2)
On solving (1) and (2), we get
[tex]0.35(660-y)+0.95y=264[/tex]
[tex]231-0.35y+0.95y=264[/tex]
[tex]0.6y=264-231[/tex]
[tex]y=\dfrac{33}{0.6}[/tex]
[tex]y=55[/tex]
Now, substitute y=55 in (1).
[tex]x+55=660[/tex]
[tex]x=660-55[/tex]
[tex]x=605[/tex]
Therefore, we need 605 mL of the 35% solution and 55 mL of the 95% solution.
