This is the fifth post of the Solving Mixture Problems Series on PH Civil Service Review. In this post, we are going to solve a problem in which only the total amount of mixture is given.
A chemist creates a mixture with 5% boric acid and combined it with another mixture containing 40% boric acid to obtain a 800 ml of mixture with 12% boric acid. How much of each mixture did he use?
Solution and Explanation
x = mixture with 5% boric acid
800 – x = mixture with 40% boric acid.
Note that if these two mixtures are combined, we will produce a mixture with 800 ml of solution with 12% boric acid. As we have learned in the previous tutorials, the amount of boric acid in the first mixture added to the amount of boric acid in the second mixture is equal to the amount of boric acid in the combined mixtures. That is,
(5%)(x) + (40%)(800-x) = (12%)(800).
Take note that x, 800 – x, and 800 are amount of the mixture and if these are multiplied by the percentage of boric acid, then we will get the exact amount of pure boric acid.
Converting percent to decimals, we have
(0.05)(x) + (0.4)(800-x) = (0.12)(800)
0.05x + 320 – 0.4x = 96.
Simplifying, we have
-0.35x = -224
That means that we need 640 ml of mixture with 5% boric acid and 800 – 640 = 160 ml of mixture with 40% boric acid.
Amount of boric acid in mixture with 5% boric acid: (640 ml)(0.05) = 32ml
Amount of boric acid in mixture with 40% boric acid: (160 ml)(0.4) = 64ml
Amount of boric acid in mixture with 12% boric acid: (800 ml)(0.12) = 96ml
As we can see, 32 ml + 64 ml = 96 ml which means that we are correct.