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

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

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New post Updated on: 14 Dec 2017, 01:10
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

67% (02:51) correct 33% (02:14) wrong based on 57 sessions

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[GMAT math practice question]

Attachment:
12.12.png
12.12.png [ 1.16 KiB | Viewed 749 times ]


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. \(\frac{11}{14}\)

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Originally posted by MathRevolution on 12 Dec 2017, 00:42.
Last edited by MathRevolution on 14 Dec 2017, 01:10, edited 1 time in total.
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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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New post 12 Dec 2017, 02:52
MathRevolution wrote:
[GMAT math practice question]

Attachment:
12.12.png


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. \(\frac{11}{14}\)


Let BC = x, so, AB = x/5
Also, AC = BD
AB + BC = BC + CD
AB = CD = x/5

AD = AB + BC + CD = x/ 5 + x+ x/5 = 7x/5
Also, AD = 2/3 - 1/2 = 1/6
7x /5 = 1/6
x = 5/42

C = A + AB + BC = 1/2 + x/5 +x = 1/2 + 6x/5 = 1/2 + (6/5)*(5/42) = 1/2 + 1/7 = 9/14

Answer C
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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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New post 12 Dec 2017, 05:38
shashankism wrote:
MathRevolution wrote:
[GMAT math practice question]

Attachment:
12.12.png


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. \(\frac{11}{14}\)


Let BC = x, so, AB = x/5
Also, AC = BD
AB + BC = BC + CD
AB = CD = x/5

AD = AB + BC + CD = x/ 5 + x+ x/5 = 7x/5
Also, AD = 2/3 - 1/2 = 1/6
7x /5 = 1/6
x = 5/42

C = A + AB + BC = 1/2 + x/5 +x = 1/2 + 6x/5 = 1/2 + (6/5)*(5/42) = 1/2 + 1/7 = 9/14

Answer C


I worked out C too. But why E?
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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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New post 12 Dec 2017, 06:01
I got 9/14 as well. Anyone got the OE?
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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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New post 12 Dec 2017, 07:14
MathRevolution wrote:
[GMAT math practice question]

Attachment:
12.12.png


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. \(\frac{11}{14}\)



Hi MathRevolution,

Is the OA correct?? Getting C = 9/14
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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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New post 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
line.png [ 11.07 KiB | Viewed 604 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 \(AD = (\frac{2}{3} - \frac{1}{2}) = \frac{1}{6}\)

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

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.
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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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New post 14 Dec 2017, 01:13
=>

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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Re: Points A, B, C, and D lie on the number line as shown in th   [#permalink] 14 Dec 2017, 01:13
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