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M06 #08

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Re: M06 #08 [#permalink] New post 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.
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Re: M06 #08 [#permalink] New post 26 Aug 2011, 23:11
l really like SVP's method. its short and smart. 8-)
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Re: M06 #08 [#permalink] New post 05 Apr 2012, 20:44
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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
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Re: M06 #08 [#permalink] New post 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
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Re: M06 #08 [#permalink] New post 08 Apr 2013, 06:14
Thanks for the question.....this was very hard
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Re: M06 #08 [#permalink] New post 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.
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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Re: M06 #08 [#permalink] New post 26 Mar 2014, 11:59
Can someone please elaborate on the "intelligent substitution" method? I didn't follow that approach
Re: M06 #08   [#permalink] 26 Mar 2014, 11:59
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