Mixture problems approach

[doclkk]

What is your approach to mixture problems.

I still don't understand.

Something as simple as

Sulfuric acid (H2SO4) can be used to remove water from organic materials leaving behind a residue of black carbon. If beaker A contains a 15% solution of sulfuric acid and beaker B contains a 20% solution of sulfuric acid, how much from each beaker should be used to make 6 liters of a 18% solution of sulfuric acid?

And then

How many ounces of a solution that is 30 % salt must be added to a 50 ounce solution that is 10% salt so that the resulting solution is 20% salt.

The first question doesn't use whole numbers but the latter does. I just need an approach. If someone could dumb it down that would be great. MGMAT book doesn't dumb it down enough for me =/

[jba05d]

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.

[doclkk]

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)

[jba05d]

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.