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M06-08

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Re M06-08  [#permalink]

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New post 30 Dec 2016, 23:07
I think this is a high-quality question and I agree with explanation. real good question
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Re M06-08  [#permalink]

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New post 30 Oct 2017, 22:57
I think this is a high-quality question and I agree with explanation.
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Re: M06-08  [#permalink]

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New post 16 Nov 2017, 04:58
Harley1980 wrote:
mejia401 wrote:
The clue helped, but I still couldn't bridge how to eliminate the other variables with what I had. The logic which I'm to substitute or combine is kind of ambiguous now, and this problem exploited that weakness. I found the article below, but any body have other guidance on how to step through expressions in a logical manner?

http://gmatclub.com/blog/2010/11/should-i-use-combination-or-substitution/

Thanks,


Hello mejia401

This is quite painful topic for me too. And I still struggle with it occasionally but I'll try to help to you with this one:

firstly we have three equations:
\(P^2 - QR = 10\)
\(Q^2 + PR = 10\)
\(R^2 + PQ = 10\)

We see that each pair of them have three pair of the same values. Let's take for example another pair (not which Bunuel describe in his solution)
The method will be same:

\(P^2 - QR = 10\)
\(Q^2 + PR = 10\)

our goal is to get rid of squares and received clear values
we can make it by using formula \(a^2-b^2=(a-b)(a+b)\)

here is a lot about algebra and exactly about these Algebraic Identities
http://gmatclub.com/forum/math-algebra-101576.html (at the end of the article in subsection Algebraic Identities)

So we need to make from these two equations
\(P^2 - QR = 10\)
\(Q^2 + PR = 10\)

one equation without exponents
and we can see that by subtraction we will receive \(Q^2-P^2\) so we can try to check this idea:
\(Q^2 + PR - P^2 + QR = 0\)
\(Q^2 - P^2 + PR + QR = 0\)
\((Q - P)(Q + P) + PR + QR = 0\)

Now we need to receive \(Q - P\) or \(Q + P\) for cancellation this equation
from this part \(PR + QR\) we can receive \(R(Q+P)\)
So we have:
\((Q - P)(Q + P) - R(Q+P) = 0\) --> \((Q - P)(Q + P) = -R(Q+P)\) --> cancel \((Q + P)\) from both sides --> \((Q - P) = -R\) or \(P = R + Q\)

And now we should see the way to eliminate these obnoxious parts \(PR\) and \(PQ\)
Again we should take a look on all three equations and try to see what we can eliminate to receive needed result.
\(P^2 - QR = 10\)
\(Q^2 + PR = 10\)
\(R^2 + PQ = 10\)

We have part \(Q+R = P\) so ideally will be to find parts in which we have the same disguised expression
These two equations have it
\(Q^2 + PR = 10\)
\(R^2 + PQ = 10\)

Let's sum them: \(Q^2 + R^2 + PR + PQ = 20\) --> undisguise \(P + Q\) --> \(Q^2 + R^2 + P(R + Q) = 20\)
And now we have expression from our question \(Q^2 + R^2 + P^2 = 20\)

____

For solving such tasks you should know Algebraic identites by heart and do a lot of such tasks with thinking about why author of solution pick exactly these values to cancel or to transform in another view

The best way of practice is to start solving all tasks from section Algebra.
http://gmatclub.com/forum/gmat-ps-quest ... 27957.html

I think after 50-70 solved and dissected tasks you will obtain "feeling" of "where is hide identities which will help to solve the task" and this process even can became boring for 600 lvl tasks ;)
Good luck with your preparation.


When you move the highlighted expression over to the other side the expression moves from a negative to a positive. Therefore \(Q - P = R\) Which would change your solution.
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Re M06-08  [#permalink]

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New post 01 Jul 2018, 16:44
I think this is a high-quality question and I agree with explanation. Excellent question.
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Re: M06-08  [#permalink]

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New post 22 Jul 2018, 00:34
Nawfal, read the ques again, your equation is incorrect and hence the solution too.
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Re: M06-08  [#permalink]

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New post 08 Dec 2018, 22:31
\(P^2-QR=10----1\)
\(R^2+PQ=10----2\)
Subtract 2 from 1
\(P^2-R^2-QR-PQ=0\)
\(P^2-R^2=Q(R+P)\)
(P-R)(P+R)=Q(R+P)
P-R=Q
===>P-Q=R
Also,
\(P^2+Q^2+PR-QR=20\)
\(P^2+Q^2+R(P-Q)=20\)
\(P^2+Q^2+R^2=20\)
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Re: M06-08 &nbs [#permalink] 08 Dec 2018, 22:31

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