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

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

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12 Dec 2017, 00:42
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Question Stats:

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[GMAT math practice question]

Attachment:

12.12.png [ 1.16 KiB | Viewed 387 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. $$\frac{11}{14}$$
[Reveal] Spoiler: OA

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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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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

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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, 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

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

$$7x = \frac{1}{6}$$
$$x = \frac{1}{42}$$

Point $$C = D - \frac{1}{42}$$

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

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

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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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