Three grades of milk are 1%, 2%, 3% fat by volume

[Guest]

All- This doesn't seem like a hard problem but I have a hard time trying to figure out how to solve quickly. Please help.

Three grades of milk are 1%, 2%, 3% fat by volume. If x gallons of the 1% grade, y gallons of the 2% grade, and z gallons of the 3% grade are mixed to give x+y+z gallons of a 1.5 % grade, what is x in terms of y and z?

1) y+3z
2) (y+z)/4
3) 2y + 3z
4) 3y+z
5) 3y+4.5z

Thanks. Answer is A, but I can't figure out quick way of doing this problem.

[StaceyKoprince]

When you have variables in the answer choices, you can plug in real numbers and do the problem arithmetically instead of algebraically. When you're struggling with the algebra, then the arithmetic approach is often more efficient.

Some rules to follow:
- pick for the variables that show up in the answers.
- follow any constraints given or implied by the problem (eg, we can't choose negative numbers for this one)
- generally avoid choosing zero and one
- generally avoid choosing numbers that show up in most or all of the answer choices (in this case, I'd avoid 3)
- choose different numbers if there are multiple variables
- after all of that, pick numbers that make your life easy!

So we need to pick for y and z. Let's make y = 40 and z = 20. Now I can set up arithmetic to solve for x:

(x)(1/100) + (40)(2/100) + (20)(3/100) = (x+40+20)(1.5/100)
That is, the individual amounts of each at its respective percentage will add to the total of the three at 1.5%.

x(1/100) + 0.8 + 0.6 = (x+60)(1.5/100)
x(1/100) + 1.4 = x(1.5/100) + 0.9
0.5 = x(1.5/100 - 1/100)
0.5 = x(0.5/100)
0.5(100/0.5) = x
100 = x

Now plug y = 40 and z = 20 into your answer choices and see which one gives you 100.
A) 40 + 3(20) = 100 works
B) (40+20)/4 = too small (note: don't actually do the math - if you can tell it won't work, then stop)
C) 2(40) + 3(20) = too big
D) 3(40) + stop right here - too big
E) 3(40) + stop here again - too big

A is the only one that works, so it's the right answer!

[Guest]

Thanks Stacey.

[guest612]

thanks for the explanation. i just came across this problem in the exam and decided after a bit to guess and move on. i almost get it now but have a question. can you please explain this second half (the right side of the equation):

x(1/100)+40(2/100)+20(3/100)=(x+40+20)(1.5/100)
x(1/100) + 0.8 + 0.6 = (x+60)(1.5/100)

i understand the left side is the weighted average. it looks like on the right side you're adding across the numerator and multiplying by the 1.5% result but i'm not quite sure i get the concept behind the arithmetic here.


i'm taking the exam this coming weekend!

thank you!

[Raj]

Hello,

(x/100) + (2y/100) + (3z/100) = 1.5 (x+y+z)/100

this equation is based on amount of milk in each.

This gives x + 2y + 3 z = 1.5x + 1.5y +1.5z, which in turn gives x = y+3z

Hope that helps.
-Raj.

thanks for the explanation. i just came across this problem in the exam and decided after a bit to guess and move on. i almost get it now but have a question. can you please explain this second half (the right side of the equation):

x(1/100)+40(2/100)+20(3/100)=(x+40+20)(1.5/100)
x(1/100) + 0.8 + 0.6 = (x+60)(1.5/100)

i understand the left side is the weighted average. it looks like on the right side you're adding across the numerator and multiplying by the 1.5% result but i'm not quite sure i get the concept behind the arithmetic here.


i'm taking the exam this coming weekend!

thank you!

[Raj]

Sorry..

I should have said " the equation is based on amount of fat in each"...

Hello,

(x/100) + (2y/100) + (3z/100) = 1.5 (x+y+z)/100

this equation is based on amount of milk in each.

This gives x + 2y + 3 z = 1.5x + 1.5y +1.5z, which in turn gives x = y+3z

Hope that helps.
-Raj.

thanks for the explanation. i just came across this problem in the exam and decided after a bit to guess and move on. i almost get it now but have a question. can you please explain this second half (the right side of the equation):

x(1/100)+40(2/100)+20(3/100)=(x+40+20)(1.5/100)
x(1/100) + 0.8 + 0.6 = (x+60)(1.5/100)

i understand the left side is the weighted average. it looks like on the right side you're adding across the numerator and multiplying by the 1.5% result but i'm not quite sure i get the concept behind the arithmetic here.


i'm taking the exam this coming weekend!

thank you!

[guest612]

raj, brilliant! you make it look so easy.

[raj.prakarsh]

If we assume z = 0, then we know that x and y have to be equal to give us 1.5% of the resultant. (Since 1.5 is the mean of 1 and 2)

Then we can assume y=2 (or anything else) and z=0 and test out the answer choices to see which of the answer choices give us 2 (Or whatever you assumed y to be) as the value of x.

As simple as that.

[RonPurewal]

raj.prakarsh --

I'm going to guess that you're trying to ask some kind of question here. (No one has posted on this thread in over 5 years, so I don't think you are just leaving a random comment.)

Unfortunately, I can't tell what you are asking. Please clarify, thanks.

[raj.prakarsh]

raj.prakarsh --

I'm going to guess that you're trying to ask some kind of question here. (No one has posted on this thread in over 5 years, so I don't think you are just leaving a random comment.)

Unfortunately, I can't tell what you are asking. Please clarify, thanks.

I posted a simpler approach to the problem. May be, it helps other readers of this forum.

[tim]

Thanks for sharing. I like that approach!

[Dienekes]

If we assume z = 0, then we know that x and y have to be equal to give us 1.5% of the resultant. (Since 1.5 is the mean of 1 and 2)

Then we can assume y=2 (or anything else) and z=0 and test out the answer choices to see which of the answer choices give us 2 (Or whatever you assumed y to be) as the value of x.

As simple as that.

My question is how does one get to the stage of pre-thinking to come up with such quick/smart approaches at time of giving the test?

I mean I did this question wrong in gmatprep lately and in fact while solving this, during pre-thinking a solution i wasn't able to think of a simple approach, the only thing that struck me was that it's a weighted average problem and thus i applied the equation to add products of each quantity's differential and weight to be equal to zero.

Thanks.

[RonPurewal]

For just about anything that's not your "usual" approach...

• Identify the technique you're not currently using (or aren't using enough, at least).
• Do a drill on that technique.
The drill:
– Go through a large number of official problems.
– Use nothing except the method you're drilling.
– Skip problems to which the method doesn't apply.
– When the method does apply, use it.

Basically, just "identify a skill that needs practice, and practice it".

If you do this sort of thing—and don't use other, more familiar methods at all during the entire course of the drill—you'll find that you've become much better at recognizing opportunities by the time it's all said and done.

[RonPurewal]

(^^ in your case, you'd want to do a drill for "plugging in numbers")

[RonPurewal]

Also, don't worry too much about picking "nice" numbers, or numbers that work "smoothly".
In writing a forum post, it's easy to make it all seem so simple and predictable—as though you should somehow always come up with the "magic" numbers in each situation.
... but this is merely an artifact of the fact that forum posts are written in retrospect. If I've been toying with a problem for a long, long time and only post the best method I've found, then, sure, it's going to look a lot easier than it really was.

As I've said many times before, there are 2 keys to success in situations like this:
1/ Do stuff.
2/ Don't not do stuff.

If the only numbers you can come up with are "ugly" numbers, then just pick up the proverbial shovel and start digging. That's how life works.