If AC=BD and AB=BC/3 in the number line, then what is the va

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--|-----------|--------------------------|----------------------------|---------|-----------
A(1) B C D(2)

If AC=BD and AB=BC/3 in the number line, then what is the value of C?

A. 11/10
B. 19/10
C. 29/15
D. 6/10
E. 9/5

OA is E

I am trying to solve this question but my approaches are not working. The following is one of the approach that i tried

Given AC = BD
Suppose AB = x, BC = y, CD = z
So, AB + BC = BC + CS
x + y = y + z
hence, x = z
Now given AB = BC/3 ---> x = y/3 = z

--|----------------------|--------------------------|------------------------|---------|-----------
A(1) y/3 B y C y/3 D(2)

Now, y/3 + y + y/3 = 1
5y/3 = 1
y = 3/5

As, question asks for C,
so C = (y/3 + y) / (5y/3) = 4/5.

Please let me know where i am making a mistake.
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by Vincen » Mon May 14, 2018 2:46 am
vinni.k wrote:--|-----------|--------------------------|----------------------------|---------|-----------
A(1) B C D(2)

If AC=BD and AB=BC/3 in the number line, then what is the value of C?

A. 11/10
B. 19/10
C. 29/15
D. 6/10
E. 9/5

OA is E

I am trying to solve this question but my approaches are not working. The following is one of the approach that i tried

Given AC = BD
Suppose AB = x, BC = y, CD = z
So, AB + BC = BC + CS
x + y = y + z
hence, x = z
Now given AB = BC/3 ---> x = y/3 = z

--|----------------------|--------------------------|------------------------|---------|-----------
A(1) y/3 B y C y/3 D(2)

Now, y/3 + y + y/3 = 1
5y/3 = 1
y = 3/5

As, question asks for C,
so C = (y/3 + y) / (5y/3) = 4/5.

Please let me know where i am making a mistake.
Hello vinni.k.

Your approach is perfect.

There iare no mistakes, it's just that since A=1 then, you have to add 1 to the value of C that you found. Therefore,

C = 4/5 + 1 = 9/5.

That way you get that the answer is the option E.