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# M06 #08

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04 Nov 2008, 00:59
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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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The hint in the question is R <>Q.

Hence, we need to form an equation where (R-Q) or (Q-R) becomes a factor.

Subtracting equation 3 from 2,
Q^2 - R^2 = PQ - PR
or, (Q+R)(Q-R) = P(Q-R)
and since, Q <>R,
hence, P = Q + R.

Now, equation 1 is still left out, hence, let us use this equation now.
P^2 - QR = 10
or, P^2 - (P-R)R = 10
or, P^2 + R^2 = 10 + PR.
or, P^2 + R^2 = 10 + 10 - Q^2
or, P^2 + Q^2 + R^2 = 20.
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04 Nov 2008, 08:05
Scthakur

I'm new to the forum and just started preparing. I have seen your quite a few replies and I'm impressed with your crisp logic. It's short and sweet. How did you learn? What books etc?

Thx
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04 Nov 2008, 08:28
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HG wrote:
Scthakur

I'm new to the forum and just started preparing. I have seen your quite a few replies and I'm impressed with your crisp logic. It's short and sweet. How did you learn? What books etc?

Thx

Hi HG,

To be frank, it is the active participation in the forum that has helped me learn the logic.
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11 May 2009, 12:17
If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

could someone please tell me how am i suppose to id the fastest way to solve this problem?

thanks,
millhouse
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13 May 2009, 03:12
Didn't you like the Official Explanation for this question?
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13 May 2009, 03:24
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millhouse wrote:
If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

could someone please tell me how am i suppose to id the fastest way to solve this problem?

thanks,
millhouse

I will solve this with intelligent substitution.

say p=sqrt(10) q= sqrt(10) r=0

clearly satisfies the all equations.

Ans = 10+10+0 =20
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13 May 2009, 05:48
x2suresh wrote:
millhouse wrote:
If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

could someone please tell me how am i suppose to id the fastest way to solve this problem?

thanks,
millhouse

I will solve this with intelligent substitution.

say p=sqrt(10) q= sqrt(10) r=0

clearly satisfies the all equations.

Ans = 10+10+0 =20

Can you please elaborate? I didn't understand how you calculated.

I'm confused with this one, I dont know how to proceed...besides is this Q really GMAT equivalent?
The least I could do,

p^2 = 10+QR
q^2 = 10-PR
r^2 = 10-PQ
or
p^2-QR = q^2+PR, r^2 +PQ = 10

dont know where to go from here!
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14 May 2009, 20:15
dzyubam wrote:
Didn't you like the Official Explanation for this question?

explanation is fine. My problem is that I wouldnt instinctively tackle the problem in that way. And the ways I did led me no where, even without a 2 minute time limit.

I think ill stick with the intelligent substitution idea.

thanks x2suresh, +1.
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26 Jan 2010, 03:10
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The OE says to start from subtracting equation 2 from equation 3.
I started from adding all the equations in hope to reduce some variables later but came to nowhere. It was too difficult and time consuming for me to play with 3 equations until I got to the right solution.
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26 Jan 2010, 03:33
The question is tough. x2suresh proposed a really nice alternative solution for it. While his approach won't necessarily work for all similar questions, it's very good for this one. Equation 2 was subtracted from equation 3 because further simplifications were possible after doing so.
Igor010 wrote:
The OE says to start from subtracting equation 2 from equation 3.
I started from adding all the equations in hope to reduce some variables later but came to nowhere. It was too difficult and time consuming for me to play with 3 equations until I got to the right solution.

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26 Jan 2010, 04:32
dzyubam wrote:
The question is tough. x2suresh proposed a really nice alternative solution for it. While his approach won't necessarily work for all similar questions, it's very good for this one. Equation 2 was subtracted from equation 3 because further simplifications were possible after doing so.
Igor010 wrote:
The OE says to start from subtracting equation 2 from equation 3.
I started from adding all the equations in hope to reduce some variables later but came to nowhere. It was too difficult and time consuming for me to play with 3 equations until I got to the right solution.

Thank you dzyubam!
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30 Mar 2010, 08:44
Awesome Question.
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30 Mar 2010, 12:03
I was totally stumped looking at the question. Thanks for posting this.
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30 Mar 2010, 19:25
i am also in total blank state, spent more then 25 min and use complex logic but

GMAT like simplicity. great Question.
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28 May 2010, 05:24
I second this approach. Even I tried to add all the three equations and tried to figure out, but no use.

I found that except eq (1), rest are having addition of variables and Q!= R. So, I took P = R. Adding all the equations and using P=R and using some numbers, I hit on the C.

x2suresh wrote:
millhouse wrote:
If $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$ , what is the value of $$P^2 + Q^2 + R^2?$$

could someone please tell me how am i suppose to id the fastest way to solve this problem?

thanks,
millhouse

I will solve this with intelligent substitution.

say p=sqrt(10) q= sqrt(10) r=0

clearly satisfies the all equations.

Ans = 10+10+0 =20

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12 Jun 2010, 22:12
I totally support the number substitution method on this qtn. I attempted to do it through simplification, and it took almost 15 minutes to get to the answer. I'm guessing if I'm doing far too elaborate calculations, chances are there's a faster way of solving it within 2 mins, otherwise it wouldn't be on the GMAT.

Thanks to Suresh for showing a fast way!
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01 Apr 2011, 04:02
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P^2 - QR = Q^2 + PR

=> -PR -QR = Q^2 - P^2

=> -R(Q + P) = (Q+P)(Q-P)

=> Q - P = -R

Now P^2 + Q^2 + R^2 - QR + PR + PQ = 30

=> P^2 + Q^2 + R^2 - R(Q - P) + PQ = 30

=> P^2 + Q^2 + R^2 + R^2 + 10 - R^2 = 30

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

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01 Apr 2011, 17:52
Good question. Time consuming though.
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02 Apr 2011, 01:44
Let P^2-QR=10-->equ. 1
Q^2+PR=10-->equ. 2
R^2+PQ=10-->-->equ. 3

Equs. 1+2+3=>P^2+Q^2+R^2-QR+P(R+Q)=30---->equ. 4

Equating LHS of equs. 1 & 2
P^2-QR=Q^2+PR
P=R+Q-->equ. 5

Put equ. 5 in equ. 4 =>P^2+Q^2+R^2-QR+P^2=30-->-->equ. 6
Put equ. 1 in equ. 6 =>P^2+Q^2+R^2=20
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08 May 2011, 17:04
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$$P^2-QR = 10$$----equation 1
$$Q^2+PR = 10$$----equation 2
$$R^2+PQ = 10$$----equation 3

Q#R

so pick equations whereever we see q , r alone and solve them. this happens to be equations 2 and 3

$$Q^2+PR = 10$$ = $$R^2+PQ = 10$$
=>$$Q^2-R^2 = P(Q-R)$$
=>P =Q+R-- equation 4

now substituting 4 in 1 we have

$$Q^2+R^2+QR = 10$$

adding P^2 on both sides we have

$$P^2+Q^2+R^2+QR = P^2+10$$

=>$$P^2+Q^2+R^2 = P^2+10-QR$$

using 1 RHS deduces to 10+10
=>$$P^2+Q^2+R^2 = 20$$

Re: M06 #08   [#permalink] 08 May 2011, 17:04

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# M06 #08

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