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Bunuel

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




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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Bunuel

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. \(\frac{29}{60}\)
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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I am not really understanding this. How are you getting 3X and then a 5X in the equation?
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we have AB=CD=(1/3)BC.

that means BC=3*AB

so (1/3)+AB+BC+CD=1/2

becomes (1/3)+AB+3AB+AB=(1/2)

5*AB=(1/6)

AB=(1/30)

point C =(1/3)+AB+3*AB

---->C =(1/3)+4*AB

sub AB-----> (1/3)+(3/30)=(14/30)=(7/14)
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