Mary wants to make 20 gallons of a 52% acid solution by mixing together a 84% acid solution and a 4% acid solution. how much of each solution must she use?

Step-by-step explanation:

I like to use an "X-diagram" to solve mixture problems. On the left side are the constituents of the mix; in the middle is the result of the mix; and on the right side are the differences between the numbers on each diagonal. These differences are the ratio numbers for the mix.

Here, that means the ratio of 84% solution to 4% solution is ...

  48 : 32 = 12 : 8

Note that the last two "ratio numbers" were chosen so their sum is 20, hence they represent the number of gallons of the corresponding constituent in the mix. (The sum of the first two ratio numbers is 48+32=80, so to get a sum of 20, we divide each by 4.)

Mary must use ...

You may note that this solution takes much longer to explain than to do. The math here can all be done without a calculator.

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Check

12 × 84% + 8 × 4% = 10.40 = 20 × 52%

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Usual Solution

A more conventional approach would be to assign x to the amount of 84% solution needed. Then the number of gallons of acid in the mix is ...

  0.84x + 0.04(20 -x) = 0.52(20)

  0.80x + 0.8 = 10.4 . . . . simplify

  0.80x = 9.6 . . . . . . . . . . subtract 0.8; next, divide by 0.8

  x = 9.6/0.8 = 12 . . . . gallons of 84% acid

  20-x = 8 . . . . . . . . . . gallons of 4% acid

333.74 ! should be your answer for this question