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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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manalq8
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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] New post   12 Jan 2012, 07:54
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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 \(P^2 + Q^2 + R^2?\)

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M06-08

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C
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manalq8
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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] New post   16 Jan 2012, 04:39
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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] New post   14 Jul 2014, 22:27
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\(p^2 - qr = 10\) .......... (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] New post   02 Jan 2017, 05:10
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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] 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 [#permalink] New post   20 Apr 2019, 11:08
Bunuel wrote:
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



Hi, can you please guide me on how can I work on developing a thought process, like the one you exhibited in this solution. I know this is difficult and something which comes natural, but I will really appreciate it, if you can give me some tips on honing my skills.
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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] New post   20 Apr 2019, 20:24
subtracting eq. 2 from eq. 3
R^2+PQ-Q^2-PR=0
R^2-Q^2+PQ-PR=0
(R+Q)(R-Q)+P(Q-R)=0
(R-Q)(P-Q-R)=0 (R≠Q)
Hence P=Q+R
From EQ. 1
P^2-QR=10
or (Q+R)^2-QR=10
Q^2+R^2+QR=10
Q^2+R^2+(P^2-10)=10
P^2+Q^2+R^2=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] New post   Updated on: 01 Jun 2020, 12:56
Q^2+PR =10
=>P= (10-Q^2)R

R^2+PQ=10
=> P= (10-R^2)/Q

=> (10-Q^2)/R= (10-R^2)/Q
=> 10Q-Q^3=10R-R^3
=> 10= Q^2+R^2+QR
=> 10= Q^2+R^2-10+P^2
=> P^2+Q^2+R^2= 20

Option C

Originally posted by SSgmat1 on 01 Jun 2020, 12:19.
Last edited by SSgmat1 on 01 Jun 2020, 12:56, edited 1 time in total.
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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] New post   01 Jun 2020, 12:41
SSgmat1 wrote:
Q^2+PR =10
=>P= (10-Q^2)R

R^2+PQ=10
=> P= (10-R^2)/Q

=> (10-Q^2)/R= (10-R^2)/Q
=> 10Q-10Q^3=10R-10R^3
=> 10= Q^2+R^2+QR
=> 10= Q^2+R^2-10+P^2
=> P^2+Q^2+R^2= 20

Option C


SSgmat1
How did you deduce the highlighted portion from the previous line?

I am getting this:-

=> (10-Q^2)/R= (10-R^2)/Q
=> 10Q - Q^3 = 10R - R^3

Could you please help further in understanding your step
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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] New post   01 Jun 2020, 12:55
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rsrighosh wrote:
SSgmat1 wrote:
Q^2+PR =10
=>P= (10-Q^2)R

R^2+PQ=10
=> P= (10-R^2)/Q

=> (10-Q^2)/R= (10-R^2)/Q
=> 10Q-10Q^3=10R-10R^3
=> 10= Q^2+R^2+QR
=> 10= Q^2+R^2-10+P^2
=> P^2+Q^2+R^2= 20

Option C


How did you deduce the highlighted portion from the previous line?

I am getting this:-

=> (10-Q^2)/R= (10-R^2)/Q
=> 10Q - Q^3 = 10R - R^3

Could you please help further in understanding your step


Hi
Please consider the same as deduced by you.
It was a typo error.
From
10Q-Q^3=10R-R^3
= 10(Q-R) = (Q-R)(Q^2+R^2+QR)
= 10= Q^2+R^2+QR

Hope this clarifies. I have edited the original as well.
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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] New post   07 Jul 2020, 19:21
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Assume Q=0, Now P=√10 R=√10. Solved
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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] New post   07 Jul 2020, 23:08
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Quote:
Assume Q=0, Now P=√10 R=√10. Solved


pankulbansal
This is indeed an excellent approach that would solve this question in seconds without digging much into solving the equation
However, I do have a question.... did you take this approach after u saw any certain characteristics in the equation?
Would this approach be applicable if the equations were like this:-

\(P^2+QR=10\) (sign changed in this equation)

\(Q^2+PR=10\)

\(R^2+PQ=10\)

\(and R≠Q\)

Could you please share your thought process behind this approach? Would be of real help.
Thanks
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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] New post   08 Jul 2020, 10:31
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rsrighosh
You raised a valid point. However, I didn't look for any pattern in the equations. I tend to assume values in all questions except in those where the value of a particular variable is sought. So yes, I would have assumed values in your question also (the one with the sign changed)
The only constraint in this question was R not equal to S. I made sure my assumed values conformed to this inequality
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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] New post   12 Mar 2022, 02:15
Asked: If \(P^2 - QR = 10\) , \(Q^2 + PR = 10\) , \(R^2 + PQ = 10\) , and \(R \ne Q\) , what is the value of \(P^2 + Q^2 + R^2?\)

Q^2 - R^2 + PR - PQ = 0
(Q+R)(Q-R) + P(R-Q) = 0
(Q-R)(Q+R-P) = 0
P = R + Q since \(Q - R \neq 0\).

P^2 - QR + Q^2 + PR + R^2 + PQ = 30
P^2 + Q^2 + R^2 + P(R +Q) - QR = 30
P^2 + Q^2 + R^2 + P^2 - QR = 30
P^2 + Q^2 + R^2 + 10 = 30
P^2 + Q^2 + R^2 = 20

IMO 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] New post   21 Aug 2023, 06:39
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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

even we only add q2+pr=10 AND r2+pq=10 and use r+q=p we will get the desired value.
we can ignore completely the first equation.
Please correct me if I miss any thing.
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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]
21 Aug 2023, 06:39
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