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GMAT 5/18
 
 

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

by GMAT 5/18 Wed May 02, 2007 12:26 pm

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the 1 percent 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 percent grade, what is x in terms of y and z?

a. y + 3z
b. y+z / 4
c. 2y + 3z
d. 3y + z
e. 3y + 4.5z
dbernst
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by dbernst Fri May 04, 2007 10:50 am

To solve this problem quickly, I "broke" several established rules of plugging in numbers. When plugging in numbers on algebraic problems (VICs), it is advisable not to choose 1, 0, or the same number more than once. Doing so will not eliminate the correct answer; however, it will potentially leave you with more than one "correct" answer choice. At first glance, I could not readily find suitable numbers for x,y and z, so I used some GMAT resourcefulness. I knew that the same amount of 1% and 2% milk would provide me with a 1.5% grade, so I chose x=5, y=5, and z=0 (Thus receiving a ruler to the knuckles for breaking "the rules").

However, when I tested each answer choice (solving for x=5, my "target" answer), only choice A gave me the correct value.

Another (slower) option would have been to skip the "plugging in" step altogether, and simply use the answer choices to determine which one would leave me with a 1.5% grade. For example, answer choice a. says y + 3z. If I were to randomly choose numbers for choice a, such as y=2 and z=3, , then x = 11. Now, simply test whether this leaves you with 1.5% milk. With these numbers, I have 11 + 2 + 3 = 16 gallons of milk, and I have 11(1) + 2(2) + 3(3) "parts" fat, = 11 + 4 + 9 = 24. Since 16(1.5) = 24, A is the correct answer.

I will confer with my MGMAT colleagues and one of us will respond with the more advisable way to handle this problem. However, I wanted to post my solution to reinforce the potential to use secondary approaches, even on problems that initially confound you.

-dan
GMAT 5/18
 
 

by GMAT 5/18 Fri May 04, 2007 11:08 am

Thanks, Dan.

I used the first method you illustrated below and came up with answer A as well. Like you, I too broke the rules as I used:

x = 4 gallons
y = 1 gallon
z = 1 gallon

So, that gave me 9%/6gallons = 1.5%.

The only answer choices that left me with were a. and d., and for once, I think being short on time on the GMAT helped me here! :) As a. worked, that is the answer choice I selected!

Thanks for the 2nd method. I like it, and although slower, it has taught me to look for more ways to solve "tricky" questions.
Guest
 
 

Averages

by Guest Tue May 15, 2007 8:30 am

This problem is also relatively straightforward to solve algebraically. You know the weighted average of the milk fat is 1.5%, so take the total milk fat and divide by the total number of gallons of milk:

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

x+2y+3z = 1.5x+1.5y+1.5z

.5x = .5y+1.5z

x = y+3z


Hope this is helpful.
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by christiancryan Wed May 16, 2007 10:03 am

Thanks, Guest! Weighted averages are a great topic: the GMAT considers them difficult (for good reason -- lots of people get them wrong), but you should learn to solve them algebraically as well. I find a quick table useful (to get it to space right, I have to put in underlines, sorry!):

_________A_____B_____C_____Total
Fat
Gallons
Fat %

And then I make sure that the columns divide (Fat/Gallons = Fat %, or Fat % * Gallons = Fat).
Also, the Fat sums across, and the Gallons sum across.

_________A_____B_____C_____Total
Fat______.01x___.02y__.03z____.01x + .02y + .03z
Gallons___x_____y_____z______x + y + z
Fat %____.01____.02___.03____.015

So we get (.01x + .02y + .03z)/(x + y + z) = .015

Multiplying through by 100, we get (x + 2y + 3z)/(x + y + z) = 1.5

(This last step is why you can ignore the conversion of a percentage to decimal, as Guest did -- it's a good shortcut, once you understand it.)

And then the rest of the math is as Guest did it.

You should also develop an intuition for weighted averages. Imagine a 2-foot seesaw marked as follows:

1----1.5----2----2.5----3

We want to put weight at the "1" mark, at the "2" mark, and at the "3" mark to make the seesaw balance exactly at the "1.5" mark. (Again, I have to put in underlines for spacing.)

X________Y_______Z
1----1.5----2----2.5----3


Leave off the X and Y for a minute. What must go at the "1" mark to balance the Z?

Z
Z
Z________________Z
1----1.5----2----2.5----3

Why 3Z? Because the "lever arm" (the distance from 3 to 1.5) for the original Z is 3 times as far as the lever arm from the original X to the balance point (1 to 1.5).

Similarly, to balance the Y at mark 2, you need to put a Y at mark 1.

Together, we get this:

Y
Z
Z
Z________Y_______Z
1----1.5----2----2.5----3

Therefore X = Y + 3Z.
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Re: Three grades of milk are 1%, 2%, 3% fat by volume

by rkafc81 Tue Aug 07, 2012 5:01 pm

hi

i had a lot of trouble picking numbers on this one... it just didn't seem to work.

when can I not pick numbers then on a VICS problem?

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

by tim Thu Aug 09, 2012 3:06 pm

you can always pick numbers on a VICs problem, it's just harder for some problems than for others. in addition, sometimes you don't have complete freedom to pick all the variables because some of them are expressed in terms of others..
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Re: Three grades of milk are 1%, 2%, 3% fat by volume

by WendyY887 Tue Sep 19, 2017 7:50 am

Is it possible to solve this problem using the teeter-totter method?

I've drawn out the below but got stuck at how to execute:

x ------------ y ------------ z
x = 1
y = 2
z = 3
Total = 1.5

So seems like the ratio is 1.5x to 0.5(y+z).

Thereafter, I wasn't sure how to continue or if this method even works with three different percentages. Someone help?
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Re: Three grades of milk are 1%, 2%, 3% fat by volume

by Sage Pearce-Higgins Sun Oct 08, 2017 7:04 am

Yes, it is possible to solve this using the teeter-totter method. Draw out the teeter-totter and add the information that we're given. 1.5%, the strength of the fat in the milk, should be on the balance point. x, at the 1% point, is to the left of the balance, and y, at 2%, and z, at 3%, are to the right.

The next bit is the tricky stage, but you can use this to make an equation. Since the scale "balances", the left side "force" is equal to the right side. The distance of the x liters from the pivot point is 0.5, so 0.5x is on the left side of the equation. On the right side, you have 0.5y (as y liters is a distance of 0.5% points from the pivot) added to 1.5z. So, in conclusion 0.5x = 0.5y + 1.5z.