Is a > bc? (1) a/c > b (2) c > 3

Is a > bc?
is a -bc >0 ?


(1) a/c > b
\(\frac{a-bc}{c} > 0\)
Insufficient. as both numerator and denominator can be +ive OR both can be -ive.

(2) c > 3
not sufficient.

Together:

c is +ive so a-bc has to be positive.

Answer C.

1) a/c > b

=> a > bc ( if c>0)

or

=> a < bc ( if c < 0)

Not sufficient

2) c >3

Not sufficient

1) & 2)

a > bc ( as c=3 >0)

Choice C

Hi All,

This question can be solved with Number Properties and/or by TESTing VALUES.

We're asked if A > BC. This is a YES/NO question.

Fact 1: A/C > B

While it might be tempting to try to cross-multiply this inequality, we don't know if C is positive or negative. THAT value would effect the inequality. If C is positive, then A > BC. If C is negative, then A < BC. The following TESTs prove the insufficiency.

IF....
A = 2
B = 1
C = 1
2/1 > 1
2 > (1)(1) and the answer to the question is YES

IF....
A = -2
B = 1
C = -1
-2/-1 > 1
-2 is NOT > (1)(-1) and the answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: C > 3

This tells us nothing about the values of A and B.
Fact 2 is INSUFFICIENT

Combined, we know:
A/C > B
C > 3

Since we know that C is POSITIVE, we can cross-multiply the first inequality, which gives us...

A > BC
The answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

VERITAS PREP OFFICIAL SOLUTION

Playing “middle school hearts”, many test-takers will run through this progression:

Step one: If you multiply both sides by c, you get a > bc so this looks sufficient*. The answer, then, would be A or D.

Step two: Forget everything you learned about statement 1 since you’ve already made your decision about it. Statement 2 is clearly insufficient on its own, so the answer must be A*.

(*we know the math here is slightly flawed; demonstration purposes only!)

But here’s how you’d play the game as an adult, or as a 700-level test-taker:

Step one: Same thing – if you multiply both sides by c you’ll get a > bc, so this one looks sufficient.

Step two: Wait a second – statement 2 is absolutely worthless. And statement one wasn’t *that* hard or interesting. Maybe the author of this question is “shooting the moon”…

Step three: Look at both statements together, reconsidering statement 1 by asking myself if statement 2 matters. If statement 2 is true and c is, say, 10, then a/10 > b would mean that a > 10b, so this still holds. But what if c is -10, and statement 2 is not true. a/(-10) > b would mean that when I multiply both sides by -10 I have to flip the sign, leaving a < -10b. This time it’s not true. Statement 2 *seems* worthless but in actuality it’s essential. Statement 1 is not sufficient alone; as it turns out I need statement 2.

What’s the difference between the two methodologies?

The 500-level, “middle school hearts” approach – NEVER consider the statements together unless they’re each insufficient alone – leaves you vulnerable to the author’s bait. On hard questions, authors love to shoot the moon…that’s their best chance of tricking savvy test-takers.

The 700-level, “playing hearts with grownups” approach seems counterintuitive, much like saving your king of hearts and knowingly accepting points in a hearts game would seem strange to a seventh-grader. But it’s important because it saves you from that bait. On a question like this, it’s easy to think that statement 1 is sufficient; abstract algebra is great at getting your mind away from numbers like negatives, zero, fractions… But statement 2′s worthlessness (ALONE) functions two ways: it’s a trap for the unsuspecting 500-level types, and it’s a reward for those who know how to play the game. That worthless statement 2 is akin to the author leading a high heart early in the game – the novice player sees it as a freebie; the expert considers “why did she do that?” and re-examines statement 1 by asking specifically “what if statement 2 weren’t true; would that change anything?”.

Remember, when you’re taking the GMAT you’re playing against other very-intelligent adults, and so the authors of these questions have a responsibility to “shoot the moon”. While the rules of the game dictate that you don’t want to consider the statements together until you’ve eliminated A, B, and D, there’s a caveat – if you have reason to believe that the author of the question is trying to trick you (which is very frequently the case on 600+ level questions), you have to consider what one statement might tell you about the other; you have to play the game.

Statement 1 is a big trap as it simplifies exactly like the question stem.

As per st 1: if c > 0 then a > bc
if c < 0 then a < bc (Insufficient)

As per st 2: c > 3 : we don't know about a and b. Insufficient

Combining 1 & 2 tells us that c > 0
So a > bc Sufficient (C)

The question asks, is a>bc

1) a/c > b

if c= +ve, a> bc
c= -ve, a<-bc . INSUFFICIENT

2) c>3, c= +ve. Thus a>bc. SUFFICIENT

I used your approach in the below question. if apply your approch in the below question, then i have to go for A for the question above.


But now you are coming with another approch. Which one is the right one? The one in the below question or the one in this question? Both question seem quite the same.

https://gmatclub.com/forum/is-yz-x-1-y- ... l#p1514140

Thanks in Advance!

You refer to Veritas Prep solutions but I'd offer you mine and maybe thy help.

Is a > bc?

(1) a/c > b

Here we cannot multiple both sides by c and write a > bc because we don't know the sign of c. If c > 0, then yes, when multiplying by positive number we'd keep the sign of the inequality and get a > bc BUT if c < 0, the when multiplying by negative number we'd flip the sign of the inequality and get a < bc. So:

If c > 0, then a > bc, which gives an YES answer to the question.
If c < 0, then a < bc, which gives a NO answer to the question.

Two different answers. Not sufficient.

(2) c > 3. This one is clearly insufficient.

(1)+(2) Since from (2) c > 0, then we have the first case from (1): a > bc. Sufficient.

Answer: C.

Hope it's clear.

P.S. Solution for another question is here: https://gmatclub.com/forum/is-yz-x-1-y- ... l#p3135409

Bunuel, thank you once again, you really make a difference! You are the man!

It makes sense to know the sign of C, whether it is + or -, because of the rules of inequality(when we multiply or divide inequality by a negative number, the inequality sign flips).

In this case if C is negative and we multiply both sides by C, then we have to flip the inequality sign. But if C is positive, we donot have to flip the inequality sign. I overlooked this rule for this question.

Thanks a million!

Greeting from Holland.

Rebaz