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
Answer: C
Source: GMAT Club Tests
If \(P^2-QR=10,\) \(Q^2+PR=10,\) \(R^2+PQ=10,\) and \(R\ne Q,\) what is the value of \(P^2+Q^2+R^2?\)
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You're free to choose values for P,Q,and R that satisfy the equalities with the exception of R $$\ne$$ Q.
So, let P=Q
P^2 -PR = 10
P^2 + PR =10
from the first two equalities, therefore
P^2 = 10
this means that R must =0
Since P=Q, Q^2 also =10
So P^2+Q^2+R^2 =20,C
So, let P=Q
P^2 -PR = 10
P^2 + PR =10
from the first two equalities, therefore
P^2 = 10
this means that R must =0
Since P=Q, Q^2 also =10
So P^2+Q^2+R^2 =20,C