Notice that we don't have much information about the 3 variables a, b and c
Since b + c < 0, we know that at least one variable is negative (which means a must be negative), but that's pretty much it.
The missing part here is that we don't really know whether c is positive or negative.

Given the absences of solid information about c, we might solve this question by testing values

Case i: a = -3, b = -2, and c = 1 (these values satisfy the given information). We get:
Quantity A: ac = (-3)(1) = -3
Quantity B: 0
In this case, Quantity B is greater

Case ii: a = -3, b = -2, and c = -1 (these values satisfy the given information). We get:
Quantity A: ac = (-3)(-1) = 3
Quantity B: 0
In this case, Quantity A is greater

Answer: D


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if b+c<0, and a<b<c, then we know that A and B MUST be negative numbers and C can be a number ranging all the way from a negative up to a huge positive (just as long as the negative value of B is a larger negative than C is a positive such that their sum would still be less than 0). Quantity A being AC therefore could either be a negative*negative=positive (in which case quantity A>B) or a negative*positive =negative (in which case quantity A<B). Since we don't have a definitive answer, the correct choice is D.

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.

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.