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22 Sep 2023, 02:39
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
51% (02:52) correct
49%
(02:36)
wrong
based on 147
sessions
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If m and n are positive numbers and m/(m + n) = n/(m - n), what is the value of m/n?
A. \(1 - \sqrt{2}\)
B. \(\sqrt{2} - 1\)
C. 1
D. \(\sqrt{2}\)
E. \(1 + \sqrt{2}\)
A. \(1 - \sqrt{2}\)
B. \(\sqrt{2} - 1\)
C. 1
D. \(\sqrt{2}\)
E. \(1 + \sqrt{2}\)
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22 Sep 2023, 11:37
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Bunuel
A great question!
\(\frac{m}{(m + n)} = \frac{n}{(m - n)}\)
Cross multiplying we get
\(m^2 - mn = mn + n^2\)
\(m^2 - mn - mn - n^2 = 0\)
\(m^2 - 2mn - n^2 = 0\)
\(m^2 - 2mn + n^2 - 2n^2 = 0\)
\((m-n)^2 - 2n^2 = 0\)
\((m-n)^2 - (\sqrt{2}n)^2 = 0\)
\([(m-n)-\sqrt{2}n][(m-n)+\sqrt{2}n] = 0\)
So either \((m-n)-\sqrt{2}n = 0\) or \((m-n)+\sqrt{2}n = 0\)
Case 1:
\((m-n)-\sqrt{2}n = 0\)
\(m-n-\sqrt{2}n = 0\)
\(m = n + \sqrt{2}n\)
\(m = n(1 + \sqrt{2})\)
\(\frac{m}{n} = 1 + \sqrt{2}\)
Case 2:
\((m-n)+\sqrt{2}n\)
\(m-n+\sqrt{2}n\)
\(m = n-\sqrt{2}n\)
\(m = n (1 -\sqrt{2})\)
\(\frac{m}{n} = 1 -\sqrt{2}\)
Case 2 is however invalid because \(1 -\sqrt{2}\) will yield a negative number while both m and n are positive.
Hence, only case 1 is valid.
Option E
16 Sep 2024, 00:53
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Here's a bit simpler solution:
m/n = (m+n) / (m-n)
just multiply RHS by n/n, we get
m/n = [ (m/n + 1) / (m/n - 1)]
Now assume m/n as x
x = (x+1)/(x-1)
x2 - 2x -1 = 0 and solve this quadratic
We get; x = m/n = 1−root(2) and 1 + root(2)
since both m and n are positive, simply reject the one solution that is a negative, i.e., 1 - 1.4142.. it will be a negative number.
Ans will be the option E, 1 + root(2)
m/n = (m+n) / (m-n)
just multiply RHS by n/n, we get
m/n = [ (m/n + 1) / (m/n - 1)]
Now assume m/n as x
x = (x+1)/(x-1)
x2 - 2x -1 = 0 and solve this quadratic
We get; x = m/n = 1−root(2) and 1 + root(2)
since both m and n are positive, simply reject the one solution that is a negative, i.e., 1 - 1.4142.. it will be a negative number.
Ans will be the option E, 1 + root(2)
