M approach to mixture problems is like this: Suppose we work

M approach to mixture problems is like this:

Suppose we work in a science lab and need a mixture of acid of 20%. The we look in the supply closet and find 10% and 35% mixtures of the same acid. We're smart people, WE ARE SCIENTISTS after all, so we decide to mix up our own 20% mixture with what we have.

Lets say we have acid mixed with water to make our mixture. If we have a 10% mixture and we have 10 litres of that mixture. That means there is 1 litres of acid and 9 liters of water.

There are 2 approaches. One is slow, but easier to follow and less abstract, and the other involves the formulas.

I do not recommend the slow way. It's just that slow, and no good on the GMAT.

Equation Method
Let a = total liters of 10% mixture
Let b = total liters of 35% mixture

Total liters of of pure acid will be .1a and .35b.
(Note: the ---- is just there to keep spacing.)
It helps to organize this into a table
-----------Liters Solution-------Percent Acid-------Pure Acid in terms of variable
10% Sol.-------x--------------------10%---------------------.1x
35% sol.-------y--------------------35%---------------------.35y
Desired--------10-------------------20%-----------------(0.2)(10) = 2

We have 2 variables. We need to solve for one of them and then that value will lead to the overall answer.

x + y = 10, then x = 10 - y. Substitute 10-y in for x.

-----------Liters Solution-------Percent Acid-------Pure Acid in terms of variable
10% Sol.-----(10-y)-----------------10%-------------------.1(10-y)
35% sol.-------y--------------------35%----------------------.35y
Desired--------10-------------------20%------------------(0.2)(10) = 2

The last column is what you'll use to set up your equation:

.1(10-y) + .35y = 2
1 - .1y +.35y = 2
1 + 0.25y = 2
.25y = 1
y = 4

This tells us that when we have a 10% mixture of acid added to a 30% mixture of acid to get 10 liters, we need 4 liters of y (the 35% mixture) and 6 liters of x (10-y; the 10%) solution.

I highly suggest the table method as it helps you separate out and label everything. Organization is key to a problem like this because when you get to y = 4, you need to know what that really means because you're not done with the problem!

you also might see the question in DS form:

Do you have enough 10% and 35% mixture to make 10 gallons of 20% mixture?
1) You have 6 gallons of 10% mixture
2) You have half as much 35% mixture as 10% mixture.

The same process applies, but your answer will be in a different form. Yes/No.

EDIT: I just realized I provided the easiest DS question in the history of DS quesions. If you have 6 gallons, you don't know how much 35% you have. It's clear that alone the statements are insufficient. Together, it shows that you have 9 gallons! You don't need to do any calculations. If you're asked if you have enough to make 10 gallons of mixture and all you have it a total of 9....who needs to compute % ? LOL