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16 Sep 2014, 00:27
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C
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Difficulty:
95%
(hard)
Question Stats:
48% (02:42) correct
52%
(02:45)
wrong
based on 692
sessions
History
Date
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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
A. 10
B. 15
C. 20
D. 25
E. 30
16 Sep 2014, 00:27
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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:
Subtract (2) from (1):
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):
Since we found that \(p = q+ r\), then:
Answer: C
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): \(r^2 + pq = 10\)
(2): \(r^2 + pq = 10\)
Subtract (2) from (1):
\(q^2 + pr - r^2 - pq= 0\)
\(q^2 - r^2 + pr - pq= 0\)
\((q - r)(q + r) - p(q - r)= 0\)
\((q-r)(q+ r - p)= 0\)
\(q^2 - r^2 + pr - pq= 0\)
\((q - r)(q + r) - p(q - r)= 0\)
\((q-r)(q+ r - p)= 0\)
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):
\(q^2 + pr + r^2 + pq= 20\)
\(q^2 + r^2 + p(r + q) = 20\)
\(q^2 + r^2 + p(r + q) = 20\)
Since we found that \(p = q+ r\), then:
\(r^2 + q^2 + p(r + q) = 20\)
\(r^2 + q^2 + p^2= 20\)
\(r^2 + q^2 + p^2= 20\)
Answer: C
General Discussion
05 Aug 2016, 03:48
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I think this is a high-quality question and I agree with explanation. Brilliant Algebraic manipulation question!
