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705-805 (Hard)|   Algebra|      
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Bunuel
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If q^2 + pr = 10 and r^2 + pq = 10, and r q, what is the value of 12 Jan 2012, 07:54
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C
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Difficulty:

85% (hard)

Question Stats:

58% (02:48) correct 42% (03:01) wrong based on 606 sessions

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If \(q^2 + pr = 10\) and \(r^2 + pq = 10\), 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


M06-08

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C
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Bunuel
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If q^2 + pr = 10 and r^2 + pq = 10, and r q, what is the value of 02 Jan 2017, 05:10
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Official Solution:

If \(q^2 + pr = 10\) and \(r^2 + pq = 10\), 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

Given:

(1): \(q^2 + pr = 10\);
(2): \(r^2 + pq = 10\)

Subtract (2) from (1):

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

Since it is given that \(q - r \ne 0\), we can safely divide the equation by it to obtain: \(q+r - p = 0\), which can be rewritten as \(p = q+r\)

Now, sum (1) and (2):

\(q^2 + pr + r^2 + pq= 20\)
\(q^2 + r^2 + p(r + q) = 20\)

Since we found that \(p = q+ r\), then:

\(r^2 + q^2 + p(r + q) = 20\)
\(r^2 + q^2 + p^2= 20\)

Answer: C
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If q^2 + pr = 10 and r^2 + pq = 10, and r q, what is the value of 07 Jul 2020, 19:21
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Assume Q=0, Now P=√10 R=√10. Solved
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If q^2 + pr = 10 and r^2 + pq = 10, and r q, what is the value of 06 Nov 2024, 20:22
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How are we supposed to know to add, then to subtract the statements? Is this a strategy we can use for other questions with two statements but three variables?
Bunuel
Official Solution:

If \(q^2 + pr = 10\) and \(r^2 + pq = 10\), 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

Given:


(1): \(q^2 + pr = 10\);
(2): \(r^2 + pq = 10\)

Subtract (2) from (1):


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

Since it is given that \(q - r \ne 0\), we can safely divide the equation by it to obtain: \(q+r - p = 0\), which can be rewritten as \(p = q+r\)

Now, sum (1) and (2):


\(q^2 + pr + r^2 + pq= 20\)
\(q^2 + r^2 + p(r + q) = 20\)

Since we found that \(p = q+ r\), then:


\(r^2 + q^2 + p(r + q) = 20\)
\(r^2 + q^2 + p^2= 20\)

Answer: C
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Bunuel
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If q^2 + pr = 10 and r^2 + pq = 10, and r q, what is the value of 07 Nov 2024, 00:09
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mellapeche
How are we supposed to know to add, then to subtract the statements? Is this a strategy we can use for other questions with two statements but three variables?
Bunuel
Official Solution:

If \(q^2 + pr = 10\) and \(r^2 + pq = 10\), 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

Given:


(1): \(q^2 + pr = 10\);
(2): \(r^2 + pq = 10\)

Subtract (2) from (1):


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

Since it is given that \(q - r \ne 0\), we can safely divide the equation by it to obtain: \(q+r - p = 0\), which can be rewritten as \(p = q+r\)

Now, sum (1) and (2):


\(q^2 + pr + r^2 + pq= 20\)
\(q^2 + r^2 + p(r + q) = 20\)

Since we found that \(p = q+ r\), then:


\(r^2 + q^2 + p(r + q) = 20\)
\(r^2 + q^2 + p^2= 20\)

Answer: C
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Adding and subtracting equations, multiplying and dividing equations, as well as expressing one variable in terms of others, are standard techniques when working with several equations and unknowns. For this question, adding and subtracting turned out to be the best approach. These techniques typically come with practice.
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If q^2 + pr = 10 and r^2 + pq = 10, and r q, what is the value of 12 Dec 2025, 00:18
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If \(q^2 + pr = 10\) and \(r^2 + pq = 10\), and \(r \ne q\), what is the value of \(p^2 + q^2 + r^2?\)

\(q^2 + pr = 10\)
\(r^2 + pq = 10\)
\(r \ne q\)

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

\(q^2 + pr + r^2 + pq = 20\)
\(q^2 + r^2 + p(r+q) = 20\)
\(p^2 + q^2 + r^2 = 20\)

IMO C
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