Question

Can someone please help me on this problem? I tried setting it up but kept getting the wrong answer... Please help me!

Answer 1

Answer:

• 240 mL of pure alcohol
• 360 mL of 80% solution

Step-by-step explanation:

Let the variable (x) represent the amount of 100% alcohol that must be added. You want to find the amount of that such that the final solution will be 80% alcohol. We assume that we use the entire amount on hand of 40% mixture (see comment below).

  100%·x + 40%·120 = 80%·(x + 120)

  0.20x = 48 . . . . . . . simplify, subtract 48+0.80x

  48/0.20 = x = 240 . . . . . divide by the coefficient of x

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You will need

240 mL of pure alcohol

to obtain

360 mL of the desired solution.

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Comment on the problem

The problem statement does not tell you how much solution is needed or how much of the 40% stock you're supposed to use. As a consequence, you can make any amount of desired 80% solution up to 360 mL. You could make 123 mL of 80% mix, for example. This means any pair of numbers will do for an answer as long as the second one is 3/2 times the first one. For 123 mL of mix, you would need 82 mL of pure alcohol to go with the 41 mL of 40% solution you would use.

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Comment on mixture problem setup

As a rule, the setup for a mixture problem is to add the contributions of the constituents and set that equal to the total amount in the mix. In this problem, each solution contributes some amount of alcohol. In "candy and nut" problems, each ingredient contributes some amount of cost to the total.

For this problem, the quantity of one of the ingredients is known. In some problems, you're to find the two quantities that result in some total. For those problems, you can set them up using two variables, with the second equation being the one that says the total of the two quantities is the specified amount.