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Re: a < b < c [#permalink]
1
The only way to get b + c < 0 is

B.........C
"-"......."-"
"-"......."+" (you know that b < c)

When "B" is "-", "A" should be "-"

Since the question is about a.c,
"A" can be only "-"
but,
"C" can be "+" or "-".

Therefore, D is the answer.
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a < b < c [#permalink]
1
a<b<c

b+c<0

There are three possible cases where b+c < 0 is satisfied

Case 1 -> b (negative) and c (positive)
We know that a<b so value of a must be negative. Therefore QA (ac) < QB (0)

Case 2 -> b (positive) and c (negative) -> This is not valid case since in question it is given that b<c.

Case 3 -> b (negative) and c (negative)
We know that a<b so value of a must be negative. Therefore QA (ac) > QB (0)

Based on case 1 and case 3 we can select answer D.
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Re: a < b < c [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: a < b < c [#permalink]
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