Mixture problems approach

The fastest way to solve mixture problems is to set up a weighted average equation and to then solve for the weights. For example, in the sulfuric acid example you are told that beaker A is 15% sulfuric acid and beaker B is 20% sulfuric acid. When these solutions are added together in a certain ratio they yield 18% sulfuric acid. Therefore, setup an equation were the weighted sum of the concentration of sulfuric acid in the two solutions equals the concrentation of sulfuric acid in the new solution. It should look something like thins.

0.15(X) + 0.20(1-X) = (.18)

The solve for X to determine the weights.

0.15X + 0.20 - 0.20X = 0.18

-0.05X = -0.02

x = 0.02/0.05 = 40%

Therefore, 40% o

Therefore, the weight of beaker A is 40% and the weight of beaker B is 60%. The total solution is 6 liters. Therefore, we need 40% of 6 liters from Beaker A, or 2.4 liters, and 60% of 6 liters from Beaker B, or 3.6liters.

You can setup a similiar equation for the second question. For example,

30%(x) + 10%(1-x) = 20%

0.30X + 0.10 - 0.10X = 0.20

0.20x = 0.10

X = 0.10/0.20 = 50%

Therefore, the weight of each solution is 50%. The 10% solution had 50 ounces (which is half) of the total solution. Therefore, the 30% must be 50 ounces as well.

jba05d wrote:The fastest way to solve mixture problems is to set up a weighted average equation and to then solve for the weights. For example, in the sulfuric acid example you are told that beaker A is 15% sulfuric acid and beaker B is 20% sulfuric acid. When these solutions are added together in a certain ratio they yield 18% sulfuric acid. Therefore, setup an equation were the weighted sum of the concentration of sulfuric acid in the two solutions equals the concrentation of sulfuric acid in the new solution. It should look something like thins.

0.15(X) + 0.20(1-X) = (.18)

The solve for X to determine the weights.

0.15X + 0.20 - 0.20X = 0.18

-0.05X = -0.02

x = 0.02/0.05 = 40%

Therefore, 40% o

Therefore, the weight of beaker A is 40% and the weight of beaker B is 60%. The total solution is 6 liters. Therefore, we need 40% of 6 liters from Beaker A, or 2.4 liters, and 60% of 6 liters from Beaker B, or 3.6liters.

You can setup a similiar equation for the second question. For example,

30%(x) + 10%(1-x) = 20%

0.30X + 0.10 - 0.10X = 0.20

0.20x = 0.10

X = 0.10/0.20 = 50%

Therefore, the weight of each solution is 50%. The 10% solution had 50 ounces (which is half) of the total solution. Therefore, the 30% must be 50 ounces as well.

This is a great explanation - but why is it (1-x)

The value is (1-x) because it represents the other part of the whole. The first part of the whole is x; that is, X represents the weight/percent/or part of the full mixture. Therefore, (1-x) represents that other part of the whole to make the mixture full.

*Note

Keep in mind that this formula would not work if we were adding three mixtures and we did not know the weights of at least one. In this case we could not solve for various different parts of the whole. Consequently, another approach would be necessary. I don't think the GMAT will test these more complicated mixture problems though.