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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by M7MBA » Fri Mar 12, 2021 10:52 am

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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

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Source: Beat The GMAT — Problem Solving | Fri Mar 12, 2021 10:52 am

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

Re: If $P^2-QR=10,$ $Q^2+PR=10,$ $R^2+PQ=10,$ and $R\ne Q,$ what is the value of $P^2+Q^2+R^2?$

by regor60 » Sun Mar 14, 2021 6:36 am

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

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