Is a > c?

[factor26]

Is a > c?

(1) b > d

(2) ab^2 - b > b^2c - d

Source: Manhattan GMAT Practice Test

Answer: E

Can someone please show me some pointers/hints on this question? Thanks!

[VivianKerr]

For a question like this, I always write down 1) the question type, 2) the given info, and 3) what info is needed for sufficiency.

1) This is a Y/N question. In order to be sufficient, a statement must allow us to say 100% yes or 100% no.

2) "a" and "c" are numbers, but we don't know if they are positive, negative, zero, integers, fractions, etc. We're given no info about them.

3) In order to determine sufficiency, we need to know the possible a's and c's.

Statement 1 tells us nothing about "a" and "c," so it is obviously insufficient.

Statement 2 tells us that AB^2 - B > B^2C - D.

Let's choose a set of numbers such that yes A > C that makes this statement's inequality true. If A = 1 and C = 0, then we get (1)B^2 - B > B^2(0) - D. This becomes B^2 - B > - D. If B = 1 and D = 1, for example, then 0 > -1, and we could answer "yes" to "A > C?"

But we could also choose a set of numbers such that no, A < C. If A = 0, and C = 1, then we get (0)B^2 - B > B^2(1) - D. This becomes - B > B^2 - D. If B = -1 for example, and D = 1, then we get 1 > 0, and we could answer "no" to "A > C?"

Therefore Statement 2 is insufficient. Even combined, we could still choose numbers that gave us a "yes" and a "no." Basically, for DS involving variables, picking numbers is always the way to go.

[Mike@Magoosh]

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

[GMATGuruNY]

Is a > c?

(1) b > d

(2) ab^2 - b > b^2c - d

Source: Manhattan GMAT Practice Test

Answer: E

Can someone please show me some pointers/hints on this question? Thanks!

Statement 1: b>d
No information about a or c.
INSUFFICIENT.

Statement 2: ab² - b > b²c - d
If b=0, we get:
a(0²) - 0 > 0²c - d
0 > -d
0 < d.
Thus, if b=0, the only requirement is that d>0, implying that a and c can be ANY VALUES.
Thus, it's possible that a<c, a=c, or a>c.
INSUFFICIENT.

Statements 1 and 2 combined:
Adding together b>d and ab² - b > b²c - d, we get:
b + ab² - b + b > d + b²c - d
ab² > b²c.

We can divide by b² if we can prove that b≠0 when the 2 statements are combined.
If b=0 in statement 2, we saw above that d>0.
But if b=0 and d>0, then d>b, which contradicts the requirement in statement 1 that d<b.
Thus, when the 2 statements are combined, it is not possible that b=0, and we can safely divide by b².
Even better, since the square of a value cannot be negative, we know b²>0 -- implying that there is no need to change the direction of the inequality.

Dividing ab² > b²c by b², we get:
a > c.
SUFFICIENT.

The correct answer is C.

[LeoBen]

is a > c?

1. b > d ==> no clue on a or c given --so insuff

2. ab^2 - b > b^2 c - d ==> no clue of b & d -- so insuff

eliminated AD, B. contest for C

ab^2 - b > b^2 c - d ==> rearrange b sq to left and b to right --

ab^2 - b^2c > b - d

b^2 (a-c) > b - d

now we know b > d hence right is + ve. Therefore left side should be positive.

b2 needs to be positive as imaginary nos are OOS.

therefore a-c must equate a positive outcome. the upshot, thus, is a > c and C is the answer.

Hope the above makes sense.

Are you sure the answer is E? (spoiler says E). If so -- pls post the OE. Thanks.