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M06-08

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16 Sep 2014, 00:27
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Difficulty:

95% (hard)

Question Stats:

54% (02:42) correct 46% (02:41) wrong based on 98 sessions

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If $$P^2 - QR = 10$$, $$Q^2 + PR = 10$$, $$R^2 + PQ = 10$$, and $$R \ne Q$$, what is the value of $$P^2 + Q^2 + R^2?$$

A. 10
B. 15
C. 20
D. 25
E. 30

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16 Sep 2014, 00:27
Official Solution:

If $$P^2 - QR = 10$$, $$Q^2 + PR = 10$$, $$R^2 + PQ = 10$$, and $$R \ne Q$$, what is the value of $$P^2 + Q^2 + R^2?$$

A. 10
B. 15
C. 20
D. 25
E. 30

Start by subtracting equation 2 from equation 3.
$$R^2 + PQ - Q^2 - PR = 0$$
$$R^2 - Q^2 - P(R - Q) = 0$$
$$(R + Q) (R - Q) = P(R - Q)$$
$$R + Q = P$$

Now, sum all the equations together:
$$P^2 + Q^2 + R^2 - QR + PQ + PR = 30$$
$$P^2 + Q^2 + R^2 - QR + P(R + Q) = 30$$

Since $$P = R + Q$$, we can substitute and get

$$P^2 + Q^2 + R^2 + P^2 - QR = 30$$

Since $$P^2 - QR = 10$$, we can substitute these as well:

$$P^2 + Q^2 + R^2 = 20$$

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14 Nov 2014, 09:17
1
What a brilliant question; R different from Q is a great hint. R-Q different from zero --> denominator of a rational number cannot be zero + I like difference of squares = subtract equations and simplify. Had I noticed that small nuance when I was solving, I would not have gotten it wrong. Alas, It will come in handy next time.
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24 Feb 2015, 14:52
Bunuel, could you explain the mental steps you took to get to the different steps in your solution? I would not have even known to get to some of those steps.
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19 Jul 2015, 14:10
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,
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19 Jul 2015, 15:53
3
1
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
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.
gmat-ps-question-directory-by-topic-difficulty-127957.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
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19 Jul 2015, 23:43
Oversaw hint, wasted too much time and finally got wrong answer.

Great question and requires great deal of attention.

Definitely one for the book..
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10 Oct 2015, 23:03
3
Oh man! I saw the explanation and finally understood, I can nowhere reach close to the answer in 2 mins.

A shortcut trick helped me. I made a blind guess of plugging in numbers and voila that worked.

P=10^0.5
Q=0
R=10^0.5

I agree this trick cannot be generalized. Just some practice might help.
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22 Oct 2015, 09:42
Bunuel wrote:
If $$P^2 - QR = 10$$, $$Q^2 + PR = 10$$, $$R^2 + PQ = 10$$, and $$R \ne Q$$, what is the value of $$P^2 + Q^2 + R^2?$$

A. 10
B. 15
C. 20
D. 25
E. 30

Very Good question. Got it right .
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22 Oct 2015, 12:06
Seemed very simple but a tricky question. Took a lot of time
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18 Nov 2015, 05:21
is this question representative of official gmat? there is no any shortcut to solution, ever.
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30 Jan 2016, 22:50
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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30 Jan 2016, 23:46
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate.

i'll try and explain in a shorter method, but it wou;ld also mean playing with the Equations..

Quote:
If P2−QR=10, Q2+PR=10, R2+PQ=10, and R≠Q, what is the value of P2+Q2+R2?

A. 10
B. 15
C. 20
D. 25
E. 30

lets pick two equations containing R^2 and Q^2 terms, as it is given R≠Q..
Q^2+PR=10,
R^2+PQ=10..
subtract two ..
Q^2+PR-( R^2+PQ)=10-10..
Q^2-R^2+PR-PQ=0..
(Q-R)(Q+R)-P(Q-R)=0..
(Q+R-P)(Q-R)=0
As it is given R≠Q,
so (Q+R-P)=0..
so P=R+Q..(i)

P^2−QR+ Q^2+PR+R^2+PQ=30,
P^2+ Q^2+R^2+PR+PQ−QR=30,
(P^2+ Q^2+R^2)+PR+PQ−QR=30,
(P^2+ Q^2+R^2)+PR+PQ−QR=30,
(P^2+ Q^2+R^2)+P(R+Q)−QR=30,..(ii)

from (i), we get P=R+Q.. substitute in (ii)
(P^2+ Q^2+R^2)+P*P−QR=30..
also it is given P^2−QR=10..
so (P^2+ Q^2+R^2)+10=30..
or (P^2+ Q^2+R^2)=20..
C..

NOTE..
the moment we see ..
three equations containing single term of square and we have to find sum of these three terms, In moost likelihood we will have to add the three terms...
second point is when we see R≠Q, we can choose two equations where we get a term Q-R..
REST ofcourse is calculation
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11 Mar 2016, 20:35
Bunuel wrote:
Official Solution:

If $$P^2 - QR = 10$$, $$Q^2 + PR = 10$$, $$R^2 + PQ = 10$$, and $$R \ne Q$$, what is the value of $$P^2 + Q^2 + R^2?$$

A. 10
B. 15
C. 20
D. 25
E. 30

Start by subtracting equation 2 from equation 3.
$$R^2 + PQ - Q^2 - PR = 0$$
$$R^2 - Q^2 - P(R - Q) = 0$$
$$(R + Q) (R - Q) = P(R - Q)$$
$$R + Q = P$$

Now, sum all the equations together:
$$P^2 + Q^2 + R^2 - QR + PQ + PR = 30$$
$$P^2 + Q^2 + R^2 - QR + P(R + Q) = 30$$

Since $$P = R + Q$$, we can substitute and get

$$P^2 + Q^2 + R^2 + P^2 - QR = 30$$

Since $$P^2 - QR = 10$$, we can substitute these as well:

$$P^2 + Q^2 + R^2 = 20$$

Why does it become this equation
$$R^2 + PQ - Q^2 - PR = 0$$

and not this one?

$$R^2 + PQ - Q^2+ PR = 0$$
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Posts: 7756

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11 Mar 2016, 20:54
1
intogamer wrote:
Bunuel wrote:
Official Solution:

If $$P^2 - QR = 10$$, $$Q^2 + PR = 10$$, $$R^2 + PQ = 10$$, and $$R \ne Q$$, what is the value of $$P^2 + Q^2 + R^2?$$

A. 10
B. 15
C. 20
D. 25
E. 30

Start by subtracting equation 2 from equation 3.
$$R^2 + PQ - Q^2 - PR = 0$$
$$R^2 - Q^2 - P(R - Q) = 0$$
$$(R + Q) (R - Q) = P(R - Q)$$
$$R + Q = P$$

Now, sum all the equations together:
$$P^2 + Q^2 + R^2 - QR + PQ + PR = 30$$
$$P^2 + Q^2 + R^2 - QR + P(R + Q) = 30$$

Since $$P = R + Q$$, we can substitute and get

$$P^2 + Q^2 + R^2 + P^2 - QR = 30$$

Since $$P^2 - QR = 10$$, we can substitute these as well:

$$P^2 + Q^2 + R^2 = 20$$

Why does it become this equation
$$R^2 + PQ - Q^2 - PR = 0$$

and not this one?

$$R^2 + PQ - Q^2+ PR = 0$$

Hi,
$$Q^2 + PR = 10$$, $$R^2 + PQ = 10$$,
$$R^2 + PQ-(Q^2 + PR) = 10-10$$
$$R^2 + PQ-Q^2 - PR) = 0$$.

Hope it helps
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16 Jul 2016, 12:01
I think this is a high-quality question and I agree with explanation.
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05 Aug 2016, 03:47
I think this is a high-quality question and I agree with explanation.
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05 Aug 2016, 03:48
I think this is a high-quality question and I agree with explanation. Brilliant Algebraic manipulation question!
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05 Aug 2016, 11:56
1
Step-by-step solution

It is a dangerous question! If you manipulate anything incorrectly you are doomed.
>> !!!

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07 Oct 2016, 16:22
5
Alternative solution:
The 3 equations are true for every (P,Q,R) such as R≠Q. So it's also true for P=Q.

In this case
(1)P^2−QR=10 -> P^2-PR=10 (1')
(2)Q^2+PR=10 -> P^2+PR=10 (2')

(1') and (2') -> P^2=10=Q^2 (as P=Q)

(3) R^2+PQ=10-> R^2+P^2=10 -> R^2=0

So P^2+Q^2+R^2=20
Re: M06-08   [#permalink] 07 Oct 2016, 16:22

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