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trail mix

by kushal.adhia » Sun Nov 07, 2010 11:53 pm
Three different kinds of trail mix contain 10%, 15%, and 30% peanuts by weight. If x pounds of the 10% kind, y pounds of the 15% kind, and z pounds of the 30% kind are mixed to make 10 pounds of trail mix that is 20% peanuts by weight, what is y in terms of z?

A. 0.5z
B. 4z
C. 2 - 0.3z
D. 10 - z
E. 20 - 4z

Please explain...

Thanks

Kushal
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by Geva@EconomistGMAT » Mon Nov 08, 2010 12:16 am
Instead of messing around with a three part weighted average, use creative plugging in.
We want a 20% average. If we could just take the average of x (10%) and z (30%) in equal amounts, we'd have our 20% average. The only thing that's bothering us is the y at 15%. Bypass this problem by plugging in y=0, then plug in x=z=5 pounds each to create a 10 pound mixture of exactly 20% average.
the question asks for y in terms of z. What have we found? That if x and z equal to 5, then y can equal zero. At this point all we need to do is plug in z=5 into the answer choices, and eliminate the ones that do not give a result of 0 for that plug in:
A 0.5*z will equal 2.5 for z=5, and not zero. Eliminate.
B 4*z will equal 20 for z=5, and not zero. Eliminate
C 2-0.3z whatever this is, this will not equal zero for z=5. Eliminate
D 10-z will equal 5 for z=5, and not zero. Eliminate.
E 20-4z will equal zero for z=5. Since this is the only answer choice which matched our goal for z=5, this is the right answer, and we needn't worry about why this is so.

That's the beauty of the plugging in approach - the right answer should match your goal for whatever numbers you choose, even 'weird' numbers such as y=0. The other answer choices may match your goal for some values, but not for others. If you manage to eliminate four answer choices because they do not match your goal, then the last one is the correct one by POE.
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by Geva@EconomistGMAT » Mon Nov 08, 2010 12:31 am
kushal.adhia wrote:Three different kinds of trail mix contain 10%, 15%, and 30% peanuts by weight. If x pounds of the 10% kind, y pounds of the 15% kind, and z pounds of the 30% kind are mixed to make 10 pounds of trail mix that is 20% peanuts by weight, what is y in terms of z?

A. 0.5z
B. 4z
C. 2 - 0.3z
D. 10 - z
E. 20 - 4z

Please explain...

Thanks

Kushal
for the sake of completeness, here's the algebraic, schoolbook solution. I Much prefer the plugging in approach to messing around with this, but here goes:

form the following weighted average equation
(group1*weight1 + group2*weight2 + group3*weight3) / total weights = average:
0.1x+0.15y+0.3z / x+y+z = 0.2 /*(x+y+z)

0.1x+0.15y+0.3z = 0.2(x+y+z)
Now, since we know that we end up with 10 pounds of trail mix, we can form the equation x+y+z=10. Plug this into our equation:

0.1x+0.15y+0.3z = 0.2*10 /*10
x+1.5y+3z = 20
Need to get rid of that x, and leave an equation with y and z. recall that x+y+z=10, and rewrite the above equation as
(x+y+z) + 0.5y + 2z = 20
10 + 0.5y + 2z = 20
0.5y = 10-2z /*2
y = 20-4z
Last edited by Geva@EconomistGMAT on Mon Nov 08, 2010 2:17 am, edited 1 time in total.
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by kushal.adhia » Mon Nov 08, 2010 2:04 am
Hi Geva

Thanks for replying.

I have a question.
form the following weighted average equation
(group1*weight1 + group2*weight2 + group3*weight3) / total weights = average:
0.1x+0.15y+0.3z / x+y+z = 0.2 /*(x+y+z)
How do u get the average as 0.2 /*(x+y+z) ?
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by Geva@EconomistGMAT » Mon Nov 08, 2010 2:17 am
kushal.adhia wrote:Hi Geva

Thanks for replying.

I have a question.
form the following weighted average equation
(group1*weight1 + group2*weight2 + group3*weight3) / total weights = average:
0.1x+0.15y+0.3z / x+y+z = 0.2 /*(x+y+z)
How do u get the average as 0.2 /*(x+y+z) ?
Sorry - I meant
0.1x+0.15y+0.3z / x+y+z = 0.2
The "*(x+y+z)" part is just the operation to do to get rid of the denominator on the left side.
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by kushal.adhia » Mon Nov 08, 2010 2:25 am
Ah ok

One more question. Could you please explain this?
x+1.5y+3z = 20
Need to get rid of that x, and leave an equation with y and z. recall that x+y+z=10, and rewrite the above equation as
(x+y+z) + 0.5y + 2z = 20
10 + 0.5y + 2z = 20
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by Geva@EconomistGMAT » Mon Nov 08, 2010 2:44 am
kushal.adhia wrote:Ah ok

One more question. Could you please explain this?
x+1.5y+3z = 20
Need to get rid of that x, and leave an equation with y and z. recall that x+y+z=10, and rewrite the above equation as
(x+y+z) + 0.5y + 2z = 20
10 + 0.5y + 2z = 20
What we're doing here is "extracting" x+y+z from the original equation x+1.5y+3z = 20 by rewriting the equation in a different, yet equivalent form:

Split 1.5y into y + 0.5y
Split 3z into z + 2z

You get x+(y+0.5y)+(z+2z)=20.
We haven't changed anything - just rewrote the equation in a different form.
Now rearrange the order so that x+y+z are next to each other:
x+y+z + the remaining 0.5y+2z = 20
Replace x+y+z with 10 to get
10 + 0.5y + 2z = 20.

Again - this part is conceptually difficult to see - which is why I'm a big fan of the plugging in approach, which allows you to solve the problem without resorting to algebra juggling.
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by kushal.adhia » Mon Nov 08, 2010 2:47 am
Got it :)

Thank you very much.
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