Question

You need 660 mL of a 40% alcohol solution. On hand, you have 35% alcohol mixture. You also have a 95% alcohol mixture. How much of each mixture will you need to add to obtain the desired solution?
You will need...
A. ___ mL of the 35% solution
B. ___ mL of the 95% solution

Answer 1

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: $x+y=660$        ...(1)

Alcohol equation : $0.35x+0.95y=0.40(660)\Rightarrow 0.35x+0.95y=264$     ...(2)

On solving (1) and (2), we get

$0.35(660-y)+0.95y=264$

$231-0.35y+0.95y=264$

$0.6y=264-231$

$y=\dfrac{33}{0.6}$

$y=55$

Now, substitute y=55 in (1).

$x+55=660$  

$x=660-55$  

$x=605$  

Therefore, we need 605 mL of the 35% solution  and 55 mL of the 95% solution.