Points A, B, C, and D lie on the number line as shown in th

I got 9/14 as well. Anyone got the OE?

Hi MathRevolution,

Is the OA correct?? Getting C = 9/14

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 \(AD = (\frac{2}{3} - \frac{1}{2}) = \frac{1}{6}\)

\(7x = \frac{1}{6}\)
\(x = \frac{1}{42}\)

Point \(C = D - \frac{1}{42}\)

\(C = (\frac{2}{3} - \frac{1}{42}) = (\frac{28}{42} - \frac{1}{42}) = (\frac{27}{42}) = \frac{9}{14}\)

Answer C

Bunuel , would you please check the OA? Thanks in advance.

=>

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
\(C = \frac{2}{3} – \frac{1}{42} = \frac{28}{42} – \frac{1}{42} = \frac{27}{42} = \frac{9}{14}.\)

Therefore, the answer is C.

Answer: C

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