Arithmetic basic mixture and percentage problem help.

Question

6 pints of a 20% solution of methanol in water are mixed with 4 pints of a 10% methanol in water solution. The percentage of methanol in the new solution is?

Answer 1

6(20 + 4(10) = 10(X)

120 + 40 = 10X

10X = 160

X = 16% methanol

If it were 5 pints each of 20% and 10% the mixture would be 10 pints of 15%, the average of 20 and 10

but it's 6 and 4 pints with more of the 20% than the 10%, so the mixture is a little closer to 20 than to 10.

Answer 2

Mixture problems are handled typically by determining a function or expression for each condition and solving simultaneous functions. But this problem is more straightforward because you are given the AMOUNTS of each initial mixture and their CONCENTRATIONS;

6 pints of 20% methanol in water and 4 pints of 10% methanol in water.

You should understand that:

(Amount of Mix 1)(Concentration of Mix 1) PLUS (Amount of Mix 2)(Concentration of Mix 2) IS EQUAL TO THE (Amount of FINAL Mix)(Concentration of FINAL Mix);

6(0.2) + 4(0.1) = 10(x) where x is the concentration of the FINAL mixture AND the FINAL mixture will be 10 pints because 6 + 4 = 10

This problem is straightforward...just solve for x

6(0.2) + 4 (0.1) = 10x

1.2 + 0.4 = 10x

1.6 = 10x

1.6 / 10 = x

x = 0.16 or 16% methanol is water. That is the FINAL concentration of the mixed solution.

Hope this helps... :)