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Points A, B, C, and D are on the number line above, and AB = CD = 1/3
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Bunuel
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Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink] New post  01 May 2018, 03:55
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A
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Points A, B, C, and D are on the number line above, and \(AB = CD = \frac{1}{3}BC\) What is the coordinate of C ?


A. \(\frac{13}{30}\)

B. \(\frac{9}{20}\)

C. \(\frac{11}{24}\)

D. \(\frac{7}{15}\)

E. \(\frac{29}{60}\)


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RohitGoyal1123
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Re: Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink] New post  01 May 2018, 04:28
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Bunuel wrote:
Image
Points A, B, C, and D are on the number line above, and \(AB = CD = \frac{1}{3}BC\) What is the coordinate of C ?


A. \(\frac{13}{30}\)

B. \(\frac{9}{20}\)

C. \(\frac{11}{24}\)

D. \(\frac{7}{15}\)

E. \(\frac{29}{60}\)



Show Spoiler
Attachment:
test.jpg



Assume Length of AB = x
so AB = CD = x and BC = 3x

so distance between AD = \(5x = \frac{1}{3}-\frac{1}{2} = \frac{1}{6}\)
or \(x= \frac{1}{30}\)
now AC = \(4x+\frac{1}{3} = \frac{4}{30} + \frac{1}{3} = \frac{14}{30} or \frac{7}{15}\)

Hence Option D
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push12345
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Re: Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink] New post  18 May 2018, 00:34
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Let B coordinate be x and C be y

So AB=CD=BC/3

x-(1/3)=(1/2)-y=(y-x)/3

x-(1/3)=(y-x)/3 --------(1)

(1/2)-y=(y-x)/3---------(2)

Two equations above

Compare and solve for y
y comes out to be 7/15

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LeoN88
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Re: Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink] New post  25 May 2019, 06:15
Bunuel wrote:
Image
Points A, B, C, and D are on the number line above, and \(AB = CD = \frac{1}{3}BC\) What is the coordinate of C ?


A. \(\frac{13}{30}\)

B. \(\frac{9}{20}\)

C. \(\frac{11}{24}\)

D. \(\frac{7}{15}\)

E. \(\frac{29}{60}\)




Show Spoiler
Attachment:
test.jpg


Let BC is x
Then AB = CD = x/3

Total distance between AD= 1/2 - 1/3 = 5x/3, solve for x.

Now C is D - x/3 = 14/30 Option D 7/15
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BrentGMATPrepNow
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Re: Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink] New post  29 Jul 2019, 09:34
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Bunuel wrote:
Image
Points A, B, C, and D are on the number line above, and \(AB = CD = \frac{1}{3}BC\) What is the coordinate of C ?


A. \(\frac{13}{30}\)

B. \(\frac{9}{20}\)

C. \(\frac{11}{24}\)

D. \(\frac{7}{15}\)

E. \(\frac{29}{60}\)
Show Spoiler
Attachment:
test.jpg


Distance from A to D = 1/2 - 1/3
= 3/6 - 2/6
= 1/6

Let x = the distance from A to B
This means that 3x = the distance from B to C
And x = the distance from C to D
So, the TOTAL DISTANCE from A to D = x + 3x + x = 5x

So, we can conclude that 5x = 1/6
Divide both sides by 5 to get: x = 1/30

We want the coordinate of C.
C is 1/30 LESS THAN 1/2 (the coordinate of D)
1/2 - 1/30 = 15/30 - 1/30
= 14/30
= 7/15

Answer: D

Cheers,
Brent
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jquest68
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Re: Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink] New post  16 Jun 2021, 11:27
I am not really understanding this. How are you getting 3X and then a 5X in the equation?
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Re: Points A, B, C, and D are on the number line above, and AB = CD = 1/3 [#permalink]
16 Jun 2021, 11:27
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