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# M06 #08

Author Message
Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 190

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

WE 1: SAP consultant-IT 2 years
WE 2: Entrepreneur-family business 2 years

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30 May 2011, 00:19
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
Posts: 1

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

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26 Aug 2011, 23:11
l really like SVP's method. its short and smart.

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Manager
Joined: 27 Apr 2010
Posts: 113

Kudos [?]: 130 [2], given: 61

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05 Apr 2012, 20:44
2
KUDOS
i really like this problem... uses a bunch of substitution to solve.

here's how i did it:

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

Steps:
1) $$P^2 - QR = 10$$ , $$Q^2 + PR = 10$$ , $$R^2 + PQ = 10$$ , and $$R \ne Q$$
2) $$P^2-QR=Q^2+PR$$
3) $$P^2-Q^2=PR+QR$$
4) $$(P-Q)^2=PR+QR$$
5) $$(P+Q)(P-Q)=R(P+Q)$$
6) $$P-Q=R$$ or $$P=R+Q$$ or $$Q=P-R$$
7) $$R^2+(R+Q)Q=10$$
8) $$R^2+RQ+Q^2=10$$
9) $$R^2+Q^2=10-RQ$$
10) $$R^2+Q^2=10+10+P^2$$
11) $$P^2+Q^2+R^2=20$$

Kudos [?]: 130 [2], given: 61

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

Kudos [?]: 90 [0], given: 67

Location: United States (OH)
Concentration: Strategy, Entrepreneurship
GMAT 1: 750 Q50 V42
GPA: 2.9
WE: Consulting (Consulting)

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08 Apr 2013, 05:42
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

Kudos [?]: 90 [0], given: 67

Intern
Joined: 14 Feb 2013
Posts: 1

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

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08 Apr 2013, 06:14
Thanks for the question.....this was very hard

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Intern
Joined: 14 Feb 2013
Posts: 7

Kudos [?]: [0], given: 1

Concentration: Strategy, Entrepreneurship
Schools: IE '15
GPA: 3.51

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28 May 2013, 03:11
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.

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
Posts: 19

Kudos [?]: 7 [0], given: 72

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26 Mar 2014, 11:59
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, 11:59

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# M06 #08

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