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M16-34
[#permalink]
16 Sep 2014, 01:00
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74% (01:48) correct
If \(a\), \(b\), \(c\), and \(d\) are non-zero numbers such that \(\frac{a}{b} = \frac{c}{d}\) and \(\frac{a}{d} = \frac{b}{c}\), which of the following must be true?
A. \(|a| = |c|\)
B. \(|b| = |d|\)
C. \(|a| = |d|\)
D. \(|b| = |a|\)
E. \(|b| = |c|\)
A. \(|a| = |c|\)
B. \(|b| = |d|\)
C. \(|a| = |d|\)
D. \(|b| = |a|\)
E. \(|b| = |c|\)
Re M16-34
[#permalink]
16 Sep 2014, 01:00
Expert Reply
Official Solution:
If \(a\), \(b\), \(c\), and \(d\) are non-zero numbers such that \(\frac{a}{b} = \frac{c}{d}\) and \(\frac{a}{d} = \frac{b}{c}\), which of the following must be true?
A. \(|a| = |c|\)
B. \(|b| = |d|\)
C. \(|a| = |d|\)
D. \(|b| = |a|\)
E. \(|b| = |c|\)
First, cross-multiply each equation to obtain: \(ad = bc\) and \(ac = bd\).
Next, multiply these equations together, resulting in: \(a^2dc = b^2dc\).
Given that the unknowns are non-zero, we can safely cancel out \(dc\) from both sides, yielding: \(a^2 = b^2\).
By taking the square root of both sides, we arrive at: \(|a| = |b|\).
Answer: D
If \(a\), \(b\), \(c\), and \(d\) are non-zero numbers such that \(\frac{a}{b} = \frac{c}{d}\) and \(\frac{a}{d} = \frac{b}{c}\), which of the following must be true?
A. \(|a| = |c|\)
B. \(|b| = |d|\)
C. \(|a| = |d|\)
D. \(|b| = |a|\)
E. \(|b| = |c|\)
First, cross-multiply each equation to obtain: \(ad = bc\) and \(ac = bd\).
Next, multiply these equations together, resulting in: \(a^2dc = b^2dc\).
Given that the unknowns are non-zero, we can safely cancel out \(dc\) from both sides, yielding: \(a^2 = b^2\).
By taking the square root of both sides, we arrive at: \(|a| = |b|\).
Answer: D
Re: M16-34
[#permalink]
31 Oct 2018, 23:04
is cross multiplication allowed? We don't know the sign of any number. Please suggest.