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I am not sure how you arrived at the solution a > b > c or a < b < c, but you can try to plug in few values for a, b and c.

If a = 1, b = 1 and c = 1 then b is the median.

If a = 1, b = -1 and c = 1 then b is not the median.

AyeMon
I am confused with St.1.
The question asked whether b is the median or not ... so I understand whether b is the middle value, bigger than one number and lower than another number.
So; either a>b>c or c>b>a

State 1: b/a = c/b
if a > b, b must be > c ... then a>b>c
if a < b, b must be < c ... then a<b<c

So ... Statement 1 seems sufficient to me :'(
Someone please help me to clarify this

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Bunuel
Is b the median of 3 numbers a, b, and c?

(1) b/a = c/b
(2) ab < 0

Analyzing the question:
We would need to prove either \(a >= b >= c\) or \(a <= b <= c\). Note it is easier to prove this sufficient by answering "no" to the question. For example if we are given that \(b\) is the biggest, it would be sufficient since \(b\) cannot be in the middle.

(1) \(b^2 = a*c\). We can have \(b\) between \(a\) and \(c\) if a, b, c are positive, or we can have negative \(b\) with positive \(a\) and \(c\). Therefore this is insufficient.

(2) Nothing on \(c\), insufficient. This tells us \(a\) and \(b\) have opposite signs however.

(3) Combining with \(b^2 = a*c\), \(a\) and \(c\) must have same signs in order to have \(a*c >= 0\). Then since \(b\) always has opposite signs from \(a\) and \(c\), \(b\) is never in the middle. Therefore this is sufficient.

Answer: C
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AyeMon
I am confused with St.1.
The question asked whether b is the median or not ... so I understand whether b is the middle value, bigger than one number and lower than another number.
So; either a>b>c or c>b>a

State 1: b/a = c/b
if a > b, b must be > c ... then a>b>c
if a < b, b must be < c ... then a<b<c

So ... Statement 1 seems sufficient to me :'(
Someone please help me to clarify this

IMO, when dealing with most fractions, it's always clearer to make one variable the subject.

(1) \(\frac{b}{a }= \frac{c}{b}\) --> \(b^2 = ac\)
Now you can more easily test cases. If b is any negative number, you can make a and c any positive number for b not to be the median.
If b = 2, a = 1 and c = 4, then b is the median. Therefore, insufficient.
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A provides the information b^2 =ac however it could 6,6,6 so answer yes, 12,6,3 answer no .
B provides a<0 or b<0 clearly insufficient
when combined B^2 = ac which means a>o , C>0 , b could be positive or negative however second option provides B is negative (ab<0) therefore b is not the median yes the information are IMO C
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Bunuel
Is b the median of 3 numbers a, b, and c?

(1) b/a = c/b
(2) ab < 0

There are good solutions above. I looked at things slightly differently:

For Statement 1 alone, the letters can all equal 1, and then b is the median. But if b = -1 and a = c = 1, then b is not the median, so Statement 1 is not sufficient. Statement 2 is clearly not sufficient alone.

Using both Statements:

From Statement 2, we know ab is negative. If ab is negative, then b/a is automatically negative. But Statement 1 tells us b/a equals c/b, so c/b must be negative too, which means bc is negative.

So we know ab is negative and bc is negative. If b is positive, then a and c must both be negative, and b is the largest thing in the set. If b is negative, then a and c must both be positive, and b is the smallest thing in the set. So it is impossible for b to be the median, and the answer is C.

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