Decent algebra-based DS problem

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Decent algebra-based DS problem

by Brian@VeritasPrep » Tue Sep 28, 2010 3:07 pm
Hello, everyone:

I mentioned to a few posters that I'd throw up a few original problems this week - here's another:



Is 10 - 6x < 0?


(1) 5x^2 > 3x^3


(2) 4 > 3/x
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by aimscore » Tue Sep 28, 2010 4:24 pm
My answer : A

It is a Y/N question
With Statement 1, 5x^2 > 3x^3
This can be possible when x = 1 or x = any -ve no
hence x = { 1, -1, -2, -3...}
Substituting in the equn 10- 6x
I will get either 4 (when x=1) or
I will get 16,22, 28 when x= -1, -2 , -3 resp.
In either case I get 10 - 6x > 0
Hence A is sufficient.

Narrowing down to A,D

Statement 2
4 > 3/x is the same as x >3/4

Hence the values of the equn 10 - 6x can be positive when I plug in values like x =1 and negative when I plug in values like x =2. Hence Statement 2 is insufficient.

Hence my answer : A

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by Rahul@gurome » Tue Sep 28, 2010 9:30 pm
Consider first statement (1) alone.
It says 5x^2 > 3x^3.
Or 5x^2 - 3x^3 > 0.
Or x^2(5 - 3x) > 0.
Now x^2 is always positive.
So 5-3x > 0.
This means 3x < 5.
Or 6x < 10.
Or 10 - 6x > 0.
So (1) alone is sufficient to answer the question.

Next consider (2) alone.
It says 4 > 3/x.
Now we cannot directly say that 4x > 3 because the sign of x will decide the inequality.
If x > 0, we have 4x > 3 or x > ¾.
If x < 0, we have 4x < 3 or x < ¾.
From this we cannot say definitely whether 10 - 6x < 0 or not.
So (2) alone is not sufficient.

The correct answer is (A).
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by hi.itz.mani » Tue Sep 28, 2010 9:39 pm
Statement 1: 5x^2 > 3 x^3
since x^2 will always be positive , we can divide both sides if x != 0

case 1: x = 0 case 2: x != 0
10 - 6x > 0 5 > 3x => 10 - 6x > 0
or 10 > 0

since in both the cases we get 10 - 6x > 0 , hence Statement 1 is sufficient

Statement 2: 4 > 3 / x

case 1 x is negative case 2 x is positive
then 4x < 3 4x > 3

since case 1 and case 2 will end up in either < or > , this is not suffiecient

Hence answer = A

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by selango » Tue Sep 28, 2010 10:48 pm
Question can be rephrased as,

10-6x<0

10<6x or x>5/3

stmt1,

5x^2 > 3x^3

x<5/3

suff

stmt2,

4>3/x

x>3/4

We don't know value of x. x may or may not be greater than 5/3

Insuff

Pick A
--Anand--

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by Brian@VeritasPrep » Wed Sep 29, 2010 11:43 am
Nice work, everyone - what I think is the biggest takeaway here is kind of the inverse of a pretty critical concept. When you see questions that involve inequalities and variables, you have to be very careful about multiplying/dividing by a variable if you don't know the sign of the variable (because then you won't know the direction of the inequality).

In this case, though, we know that x^2 will be greater than 0 (it can't be negative, and statement 1 proves that it isn't 0), so because we know that it will be positive, we can divide both sides by x^2 in statement 1 to get:

5>3x
5/3 > x

So, therefore, x is less than 5/3 and the answer to the overall question (which you can manipulate algebraically to solve for x: Is 10 < 6x, so is x >5/3?) is NO.


This question is a good example of what I guess I'll call "GMAT counterintelligence" - the authors of the GMAT know that examinees constantly make mistakes by dividing by variables in inequalities. If, say, 60% of test-takers are susceptible to that mistake, the authors of the test know that maybe 15% of those who are not bound to make that mistake have simply taught themselves "never divide by a variable in an inequality" - but that's not entirely true. If you know the sign of the variable you can divide by it. So the GMAT can take this application to weed out that next 15% so that they can correctly identify those at the top of the food chain who understand all sides of this concept. Beware of the counterintelligence - the authors know that people love the quick-fix...if they've been burned by a variable, they'll teach themselves quickly to not divide by a variable, but even that correction isn't 100%, so it's important to look at how the authors can use your own momentum against you.
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by alivapriyada » Thu Sep 30, 2010 3:31 am
Rahul@gurome wrote:Consider first statement (1) alone.
It says 5x^2 > 3x^3.
Or 5x^2 - 3x^3 > 0.
Or x^2(5 - 3x) > 0.
Now x^2 is always positive.
So 5-3x > 0.
This means 3x < 5.
Or 6x < 10.
Or 10 - 6x > 0.
So (1) alone is sufficient to answer the question.

Next consider (2) alone.
It says 4 > 3/x.
Now we cannot directly say that 4x > 3 because the sign of x will decide the inequality.
If x > 0, we have 4x > 3 or x > ¾.
If x < 0, we have 4x < 3 or x < ¾.
From this we cannot say definitely whether 10 - 6x < 0 or not.
So (2) alone is not sufficient.

The correct answer is (A).
Brilliant Explanation!!!!
one more vote for A