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

Question

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

arjtryarjtry · Answer

allen , there is also a simpler procedure..
0.1a+.35b=.2(a+b).
or a/b=1.5/1.
so we clever scientists mix the mixture(i prefer if there was alchol... my stomach would have been an additional container to mix ) :wink:

in the ratio of 1.5:1 or in other words, the same that you have arrived at 4 and 6(since 6/4=1.5)....

durgesh79 · Answer

you can complicate the question by adding "Assume that you have enough water available" :wink:

jallenmorris · Answer

I was originally posting this for leonidas as (s)he had asked for information regarding mixture problems. Thanks for the clarification on the simpler formula. I thought of that, but wasn't sure it would work. I had seen the table method somewhere when I was learning it myself. The table method does help if someone is a very visual person, but the .1a + .35b = .2(a+b) is good to. As long as the person can then apply the ratio to get the needed volume.

1:1.5...2:3=5 4:6 = 10...there it is. It's just a matter of being able to recognize what information is present and what needs to be done to get that information into the correct format for the answer.

1:1.5...2:3=5 4:6 = 10...there it is. It's just a matter of being able to recognize what information is present and what needs to be done to get that information into the correct format for the answer.

leonidas · Answer

Allen and arjtryarjtry. Appreciate your responses to my request.

Allen, I also found this piece (attachment) that follows the table approach. I am positing this for folks who might want to see other examples. This also seperates in terms of dry mixture and chemical mixture problems.

jallenmorris · Answer

I'm guessing you're referring to Purple Math?

I use that website all the time. Whoever writes it is great!

leonidas · Answer

I'm guessing you're referring to Purple Math?

I use that website all the time. Whoever writes it is great!

I got this from one of the docs I downloaded from esnips "petersons.com"

jallenmorris · Answer

Ok, well you might try looking at purple math. (Just do a google search for Purple math)

That website really does a great job of explaning these concepts.

leonidas · Answer

Purplemath website is very good. Mostly it is basic stuff, however, that's the kind of foundation one needs for GMAT. I grew up doing differentiations and integrations (advanced math), but my basic skills used to be very average. I am now re-visiting what I learnt in my 6th grade to 10th grade

nmohindru · Answer

I generally use this

Let x be % mixture in original, 'a' fraction be removed and y % mixture be added for 'a' fraction such that final is z% mixture

(1-a)*x + a*y = z

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

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