Hi, there. I'm happy to give my 2¢ on this one.
The prompt is very simple: Is a > c?
Remember, no restriction is placed on a and c, so they could be any numbers --- positive integers, negative integers, fractions, etc.
Statement #1: b > d
This statement, by itself, tell us zilch about a & c. On its own, completely
insufficient.
Statement #2: ab^2 - b > b^2c - d
OK, so let's test a bit: we will plug numbers in one pair where a > c, another pair where a < c, and see if they both would be allowed.
Try a > c: for example, try a = 3, c = 1 (keeping it simple!) Then 3b^2 - b > b^2 - d. Well, 3b^2 is clearly going to be bigger than b^2. Let's say that b = 5. Then 3(5^2) - 5 = 70, and we can certainly find a value of d such that (25 - d) is less than 70. So, a pair where a > c would be consistent with this equation.
Now, try a < c: for example, a = 1, b = 3. Then b^2 - b > 3b^2 - d. Subtract b^2 from both sides, to get -b > 2b^2 - d. Well, we know b^2 is going to be positive, so if -b be is going to be greater, then -b must be positive: that means, b is negative. Let's say b = -5. Then, the inequality becomes 5 > 2(5^2) - d, or 5 > 50 - d. Well, clearly for small values of d, this is not true, but in this statement, there's no restriction on d. We could pick d = 700 if we want: when we pick a large value of d, it makes the inequality true. Thus, a pair where a < c also would be consistent with this equation.
Since both a > c and a < c are consistent with this statement, this statement by itself does not help us at all to answer the question: Is a > c? Statement #2, by itself, is
insufficient.
Combined Statements #1 and #2:
(1) b > d
(2) ab^2 - b > b^2c - d
This is really the crux of the problem: when we combine useless-by-itself Statement #1 with Statement #2, can we determine anything?
First, I am going to re-arrange Statement #2:
d - b > b^2c - ab^2
or
d - b > b^2(c - a) ***
From Statement #1, we know b > d, which is equivalent to 0 > d - b. Thus, the left side of the starred inequality
must be negative. The only way the starred can be true is if the right side is even more negative. Well, the factor b^2 must be positive, so the only way the right side can be negative is if (c - a) < 0, which is equivalent to c < a. If we combine the statements, we can definitively say: Yes, a > c. Combined, the statements are
sufficient.
Answer =
C.
I'm sorry, but I disagree with the answer you posted. Perhaps you miscopied it from your source.
Does all of this make sense? Please let me know if you have any questions on what I've said.
Mike
