(1) -x < x^3
(2) x < x^2
can anyone please explain why the answer is B (statement 2 alone is sufficient?)
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DS: Is x between 0 and 1?
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Hi 750+,
We're asked if X is between 0 and 1. This is a YES/NO question.
1) -X < X^3
With this Fact, we know that X MUST be POSITIVE.
IF X is between 0 and 1, then the answer to the question is YES.
IF X is greater than 1, then the answer to the question is NO.
Fact 1 is INSUFFICIENT
2) X < X^2
With the Fact, X could be positive or negative. HOWEVER, X cannot be between 0 and 1 (since squaring a positive fraction makes it SMALLER). Thus, whatever X is, it's NOT between 0 and 1, so the answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT.
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
We're asked if X is between 0 and 1. This is a YES/NO question.
1) -X < X^3
With this Fact, we know that X MUST be POSITIVE.
IF X is between 0 and 1, then the answer to the question is YES.
IF X is greater than 1, then the answer to the question is NO.
Fact 1 is INSUFFICIENT
2) X < X^2
With the Fact, X could be positive or negative. HOWEVER, X cannot be between 0 and 1 (since squaring a positive fraction makes it SMALLER). Thus, whatever X is, it's NOT between 0 and 1, so the answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT.
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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The first inequality would be true for any positive x, meaning x could be between 0 and 1 (or it could be greater than one).750+ wrote:(1) -x < x^3
(2) x < x^2
can anyone please explain why the answer is B (statement 2 alone is sufficient?)
The second inequality is true for all values of x, except those in the interval between 0 and 1, inclusive. Hence, we can say with certainty that x is not in that interval.
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Matt@VeritasPrep
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S1
x³ + x > 0
x * (x² + 1) > 0
so either x and x² + 1 are both positive OR x and x² + 1 are both negative. In the first case, we'd have
x > 0
x² > -1
which is true for any positive x. In the second, we'd have
0 > x
-1 > x²
which isn't true for ANY real x. So this only tells us x > 0; NOT SUFFICIENT.
S2:
x² > x
Since x² is positive (it can't be 0 here, as 0² isn't > 0), we can divide both sides by x², giving
1 > 1/x
If x is negative, this is always true, so 0 > x works. If x is positive, multiply both sides by x, giving
x > 1
So we have two solution sets: 0 > x and x > 1. In neither case is 1 > x > 0, so we can say that x is NOT between 0 and 1, and we're done.
x³ + x > 0
x * (x² + 1) > 0
so either x and x² + 1 are both positive OR x and x² + 1 are both negative. In the first case, we'd have
x > 0
x² > -1
which is true for any positive x. In the second, we'd have
0 > x
-1 > x²
which isn't true for ANY real x. So this only tells us x > 0; NOT SUFFICIENT.
S2:
x² > x
Since x² is positive (it can't be 0 here, as 0² isn't > 0), we can divide both sides by x², giving
1 > 1/x
If x is negative, this is always true, so 0 > x works. If x is positive, multiply both sides by x, giving
x > 1
So we have two solution sets: 0 > x and x > 1. In neither case is 1 > x > 0, so we can say that x is NOT between 0 and 1, and we're done.
