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Is a > bc?

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Math Expert
Joined: 02 Sep 2009
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17 Apr 2015, 04:36
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Is a > bc?

(1) a/c > b

(2) c > 3

Kudos for a correct solution.

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Re: Is a > bc?  [#permalink]

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17 Apr 2015, 04:52
2
1
Bunuel wrote:
Is a > bc?

(1) a/c > b

(2) c > 3

Kudos for a correct solution.

1. If c>0 then
a>bc

But, if c<0 then
a<bc

Not Sufficient

2. c>3
No information given about a or b

Not Sufficient

1 and 2:
St 2 tells us that c>0, therefore a>bc

Sufficient

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17 Apr 2015, 06:22
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1
#1 a/c > b
we don't know the sign of c so after multiplying by c we should either change the sign to the opposite if c is negative or remain the same othewrise, thus insufficient
#2 c > 3
since we got no information about a and b, we cannot answer our question, insufficient
#1 +#2, the sign of c is explicitly definied in the #2 so we can safely assume that the inequality sign remains the same as in #1, thus we can answer our question. SUFFICIENT

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Re: Is a > bc?  [#permalink]

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17 Apr 2015, 06:56
1
1
Bunuel wrote:
Is a > bc?

(1) a/c > b

(2) c > 3

Kudos for a correct solution.

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.

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Re: Is a > bc?  [#permalink]

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18 Apr 2015, 14:36
1
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
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Re: Is a > bc?  [#permalink]

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18 Apr 2015, 20:14
1
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

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Posts: 60460
Re: Is a > bc?  [#permalink]

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20 Apr 2015, 05:14
1
Bunuel wrote:
Is a > bc?

(1) a/c > b

(2) c > 3

Kudos for a correct solution.

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.

(*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.
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Re: Is a > bc?  [#permalink]

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13 Jan 2017, 11:09
Bunuel wrote:
Is a > bc?

(1) a/c > b

(2) c > 3

Kudos for a correct solution.

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)
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Re: Is a > bc?  [#permalink]

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26 Nov 2019, 21:06

1) a/c > b

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

2) c>3, c= +ve. Thus a>bc. SUFFICIENT
Re: Is a > bc?   [#permalink] 26 Nov 2019, 21:06
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