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Re: Points A, B, C, and D lie on the number line as shown in th [#permalink]

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12 Dec 2017, 20:31

MathRevolution wrote:

[GMAT math practice question]

Attachment:

The attachment 12.12.png is no longer available

Points \(A, B, C\), and \(D\) lie on the number line as shown in the figure above. If \(AC=BD\) and \(AB=\frac{BC}{5}\), what is the value of \(C\)?

A. \(\frac{7}{14}\) B. \(\frac{8}{14}\) C. \(\frac{9}{14}\) D. \(\frac{10}{14}\) E. \(\frac{11}{14}\)

Attachment:

line.png [ 11.07 KiB | Viewed 242 times ]

If\(AB = \frac{BC}{5}\), then \(BC\) is 5 times bigger than\(AB\)

Let \(BC = 5x\) Thus \(AB = x\) And \(AC = 6x\)

\(AC = BD = 6x\) They share middle part \(BC = 5x\)

\(CD\)? \(= x\) Both segments above the line are equal They share \(5x\). One has another segment of length x (AB). The other segment must have a segment with length x, too. That's CD. \(CD = AB = x\)

Length of whole segment \(AD = (x + 5x + x) = 7x\) Length of whole segment also \(AD = (\frac{2}{3} - \frac{1}{2}) = \frac{1}{6}\)

Suppose \(d\) is the distance between \(A\) and \(B\). Then \(CD = d\) and \(BC = 5d\). Thus, \(AD = d + 5d + d = 7d\). Since \(A = \frac{1}{2}\) and \(D = \frac{2}{3}\), \(7d = \frac{2}{3} – \frac{1}{2} = \frac{4}{6} – \frac{3}{6} = \frac{1}{6},\) and \(d = \frac{1}{42}.\) So, \(D – C = \frac{1}{42}\), and \(C = \frac{2}{3} – \frac{1}{42} = \frac{28}{42} – \frac{1}{42} = \frac{27}{42} = \frac{9}{14}.\)