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# M06 #08

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Intern
Joined: 07 Sep 2010
Posts: 18

Kudos [?]: 5 [0], given: 3

Re: M06 #08 [#permalink]

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02 Apr 2011, 02:44
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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Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 191

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WE 1: SAP consultant-IT 2 years
WE 2: Entrepreneur-family business 2 years
Re: M06 #08 [#permalink]

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30 May 2011, 01:19
i also liked scthakur's way

actually the trick is to interpret statement r not equal q and then proceed to get(R+Q) in some factor form. only eq 2 and 3 will give us that.

the substitution idea looks great but wonder how many of us have bulbs in out brains that will actually light up to think of the numbers on Gday. Looks great after seeing the numbers though.
_________________

Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

Kudos [?]: 15 [0], given: 12

Intern
Joined: 22 Jul 2010
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Re: M06 #08 [#permalink]

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27 Aug 2011, 00:11
l really like SVP's method. its short and smart.

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Manager
Status: Applying
Joined: 01 Dec 2012
Posts: 91

Kudos [?]: 89 [0], given: 67

Location: United States (OH)
Concentration: Strategy, Entrepreneurship
GMAT 1: 750 Q50 V42
GPA: 2.9
WE: Consulting (Consulting)
Re: M06 #08 [#permalink]

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08 Apr 2013, 06:42
scthakur 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

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.

Great reply!! I usually know if I havent figured out the answer within 1-1.5 minutes that I'm just not "thinking the right way" ; The above logic of subtracting Eq.3 from Eq.2 was crucial - Thank you for the wonderful question & solution!

Regards,
Vishnu
_________________

Regards,
Vishnu

Article --> http://www.topmba.com/blog/could-mba-degree-be-way-forward

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Intern
Joined: 14 Feb 2013
Posts: 1

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Re: M06 #08 [#permalink]

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08 Apr 2013, 07:14
Thanks for the question.....this was very hard

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Intern
Joined: 14 Feb 2013
Posts: 7

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Concentration: Strategy, Entrepreneurship
Schools: IE '15
GPA: 3.51
Re: M06 #08 [#permalink]

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28 May 2013, 04:11
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.
Please help me understand why exactly the 2nd from the 3rd and what logic should I follow when dealing with such problems?

Well, I would say that you should always try to anticipate the type of equations that may be formed when you try to solve algebra.
In this case, as the question clearly gives a hint by R<>Q, we should look for a way to reach a case whereR-Q comes into picture. An apparent way is to subtract eq 3 from eq 2, making (R-Q) a factor.
Way ahead from there is a piece of cake.

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Intern
Joined: 01 Apr 2013
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Re: M06 #08 [#permalink]

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26 Mar 2014, 12:59
Can someone please elaborate on the "intelligent substitution" method? I didn't follow that approach

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Re: M06 #08   [#permalink] 26 Mar 2014, 12:59

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# M06 #08

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