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If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q

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If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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New post 12 Jan 2012, 07:54
1
5
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

51% (02:57) correct 49% (03:10) wrong based on 116 sessions

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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 \(P^2 + Q^2 + R^2?\)

10
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25
30

M06-08

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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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New post 16 Jan 2012, 04:39
2
Start by subtracting equation 2 from equation 3.
R^2+PQ-Q^2-PR=0

R^2-Q^2-P(R-Q)=0

(R+Q)(R-Q)=P(R-Q)

(R+Q)=P

Now, sum all the equations together:
P^2+q^2+r^2-QR+PQ+PR=30

P^2+Q^2+R^2-QR+P(Q+R)=30

Since , we can substitute and get
P^2+Q^2+R^2+P^2-QR=30

Since P^2-QR=10 , we can substitute these as well:

P^2+Q^2+R^2=20

The correct answer is C.

Hope this helps
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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New post 14 Jul 2014, 22:27
2
\(p^2 - qr = 10\) .......... (1)

\(q^2 + pr = 10\) ............. (2)

\(r^2 + pq = 10\) .......... (3)

Equation (1) - (2)

\(p^2 - q^2 - r(p+q) = 0\)

\((p+q)(p-q) = r(p+q)\)

p-q = r ................. (4)
OR
p = q+r

Adding (1) (2) & (3)

\(p^2 + q^2 + r^2 = 30 + qr - pr - pq\)

\(p^2 + q^2 + r^2 = 30 + qr - p(r + q)\)

\(p^2 + q^2 + r^2 = 30 + qr - p^2\)

\(p^2 + q^2 + r^2 = 30 - (p^2 - qr)\)

Substitute value from (1)

\(p^2 + q^2 + r^2 = 30 - 10 = 20\)

Answer = 20
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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New post 02 Jan 2017, 05:10
manalq8 wrote:
If \(P^2 - QR = 10\) , \(Q^2 + PR = 10\) , \(R^2 + PQ = 10\) , and \(R \ne Q\) , what is the value of \(P^2 + Q^2 + R^2?\)

10
15
20
25
30


Start by subtracting equation 2 from equation 3.
\(R^2 + PQ - Q^2 - PR = 0\)
\(R^2 - Q^2 - P(R - Q) = 0\)
\((R + Q) (R - Q) = P(R - Q)\)
\(R + Q = P\)

Now, sum all the equations together:
\(P^2 + Q^2 + R^2 - QR + PQ + PR = 30\)
\(P^2 + Q^2 + R^2 - QR + P(R + Q) = 30\)

Since \(P = R + Q\), we can substitute and get

\(P^2 + Q^2 + R^2 + P^2 - QR = 30\)

Since \(P^2 - QR = 10\), we can substitute these as well:

\(P^2 + Q^2 + R^2 = 20\)


Answer: C
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q  [#permalink]

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New post 01 Apr 2018, 07:17
PareshGmat wrote:
\(p^2 - qr = 10\) .......... (1)

\(q^2 + pr = 10\) ............. (2)

\(r^2 + pq = 10\) .......... (3)

Equation (1) - (2)

\(p^2 - q^2 - r(p+q) = 0\)

\((p+q)(p-q) = r(p+q)\)

p-q = r ................. (4)
OR
p = q+r

Adding (1) (2) & (3)

\(p^2 + q^2 + r^2 = 30 + qr - pr - pq\)

\(p^2 + q^2 + r^2 = 30 + qr - p(r + q)\)

\(p^2 + q^2 + r^2 = 30 + qr - p^2\)

\(p^2 + q^2 + r^2 = 30 - (p^2 - qr)\)

Substitute value from (1)

\(p^2 + q^2 + r^2 = 30 - 10 = 20\)

Answer = 20


In this explanation, we could have P+Q=0 as well which is not considered
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Re: If P^2-QR=10, Q^2+PR=10, R^2+PQ=10 and R does not equal to Q &nbs [#permalink] 01 Apr 2018, 07:17
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