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

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