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

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:        ...(1)

Alcohol equation :     ...(2)

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

Now, substitute y=55 in (1).

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

no

explanation:

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