Is p > q ?
1) p^2 > q^2
2) p^3 > q^3
Ans: B
Is p > q ? 1) p^2 > q^2 2) p^3 > q^3
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 mariofelixpasku
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Hi mariofelixpasku,
This DS question is built around some Number Properties (how numbers "relate" to one another).
Is P > Q?
This is a YES/NO question; we know nothing about P nor Q.
Fact 1: P^2 > Q^2
The Number Properties here:
(Positive)^2 = Positive
(Negative)^2 = Positive
(Zero)^2 = 0
Since P^2 > Q^2, that means that P^2 MUST be positive. However, P could be positive OR negative. There's no way to know whether P is bigger than Q or not. As an example....
If P = 1, Q = 0, then the answer to the question is YES
If P = 1, Q = 0, then the answer to the question is NO
Fact 1 is INSUFFICIENT
Fact 2: P^3 > Q^3
The Number Properties here:
(Positive)^3 = Positive
(Negative)^3 = Negative
(Zero)^3 = 0
Here, since P^3 > Q^3, we know that the relationship between P and Q will be the same (P MUST be greater than Q). The answer to the question will ALWAYS be YES.
Fact 2 is SUFFICIENT
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
This DS question is built around some Number Properties (how numbers "relate" to one another).
Is P > Q?
This is a YES/NO question; we know nothing about P nor Q.
Fact 1: P^2 > Q^2
The Number Properties here:
(Positive)^2 = Positive
(Negative)^2 = Positive
(Zero)^2 = 0
Since P^2 > Q^2, that means that P^2 MUST be positive. However, P could be positive OR negative. There's no way to know whether P is bigger than Q or not. As an example....
If P = 1, Q = 0, then the answer to the question is YES
If P = 1, Q = 0, then the answer to the question is NO
Fact 1 is INSUFFICIENT
Fact 2: P^3 > Q^3
The Number Properties here:
(Positive)^3 = Positive
(Negative)^3 = Negative
(Zero)^3 = 0
Here, since P^3 > Q^3, we know that the relationship between P and Q will be the same (P MUST be greater than Q). The answer to the question will ALWAYS be YES.
Fact 2 is SUFFICIENT
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
 mariofelixpasku
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Hi marioflexipasku,
Here are a series of examples that will prove that when P^3 > Q^3 that P > Q:
Both Positive Integers
P=2, Q =1
P^3 = 8
Q^3 = 1
P^3 > Q^3 and P > Q
Both Negative Integers
P = 1
Q = 2
P^3 = 1
Q^3 = 8
P^3 > Q^3 and P > Q
One Positive, One 0
P = 1
Q = 0
P^3 = 1
Q^3 = 0
P^3 > Q^3 and P > Q
Etc. There are other examples (0, Negative) and (Positive, Negative), but they'd lead to the same, consistent result.
This process of TESTING Values is a great way to handle many DS questions and ultimately prove whether a consistent result occurs or not.
GMAT assassins aren't born, they're made,
Rich
Here are a series of examples that will prove that when P^3 > Q^3 that P > Q:
Both Positive Integers
P=2, Q =1
P^3 = 8
Q^3 = 1
P^3 > Q^3 and P > Q
Both Negative Integers
P = 1
Q = 2
P^3 = 1
Q^3 = 8
P^3 > Q^3 and P > Q
One Positive, One 0
P = 1
Q = 0
P^3 = 1
Q^3 = 0
P^3 > Q^3 and P > Q
Etc. There are other examples (0, Negative) and (Positive, Negative), but they'd lead to the same, consistent result.
This process of TESTING Values is a great way to handle many DS questions and ultimately prove whether a consistent result occurs or not.
GMAT assassins aren't born, they're made,
Rich
 mariofelixpasku
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thanks for sharing this general rule. you saved me a lot of time i would have wasted on testing numbers.

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An algebraic way of explaining this, for those curious:
pÂ² > qÂ² = pÂ² > qÂ², so
pÂ² > qÂ² implies p > q, but NOT p > q
(For instance, if p = 3 and q = 2, pÂ² > qÂ² and p > q, but p is not > q.)
For pÂ³ > qÂ³, it's a little more complicated. We have:
pÂ³ > qÂ³, or
pÂ³  qÂ³ > 0, or
(p  q) * (pÂ² + pq + qÂ²) > 0
(that's how you factor the difference of cubes ... don't worry about it on the GMAT, I haven't seen it tested there yet)
This tells us that either (p  q) and (pÂ² + pq + qÂ²) are both positive OR (p  q) and (pÂ² + pq + qÂ²) are both negative.
Let's assume that it's the second case. If (p  q) and (pÂ² + pq + qÂ²) are both negative, then (pÂ² + pq + qÂ²) < 0. But if (pÂ² + pq + qÂ²) < 0, then (pÂ² + qÂ²) < pq. Since (pÂ² + qÂ²) CAN'T be negative, pq must be positive, which means that exactly one of p and q is negative.
Since we're assuming (p  q) < 0, we have p < q, so p is the negative one and q the positive. But if q is positive and p is negative, we can't have pÂ³ > qÂ³ (how can a positive be greater than a negative!?) This contradicts our original statement (that pÂ³ > qÂ³), making it impossible.
Hence (p  q) and (pÂ² + pq + qÂ²) are both positive. Since (p  q) is positive, p > q.
Woof: too much work! This is why you want to try numbers on test day
pÂ² > qÂ² = pÂ² > qÂ², so
pÂ² > qÂ² implies p > q, but NOT p > q
(For instance, if p = 3 and q = 2, pÂ² > qÂ² and p > q, but p is not > q.)
For pÂ³ > qÂ³, it's a little more complicated. We have:
pÂ³ > qÂ³, or
pÂ³  qÂ³ > 0, or
(p  q) * (pÂ² + pq + qÂ²) > 0
(that's how you factor the difference of cubes ... don't worry about it on the GMAT, I haven't seen it tested there yet)
This tells us that either (p  q) and (pÂ² + pq + qÂ²) are both positive OR (p  q) and (pÂ² + pq + qÂ²) are both negative.
Let's assume that it's the second case. If (p  q) and (pÂ² + pq + qÂ²) are both negative, then (pÂ² + pq + qÂ²) < 0. But if (pÂ² + pq + qÂ²) < 0, then (pÂ² + qÂ²) < pq. Since (pÂ² + qÂ²) CAN'T be negative, pq must be positive, which means that exactly one of p and q is negative.
Since we're assuming (p  q) < 0, we have p < q, so p is the negative one and q the positive. But if q is positive and p is negative, we can't have pÂ³ > qÂ³ (how can a positive be greater than a negative!?) This contradicts our original statement (that pÂ³ > qÂ³), making it impossible.
Hence (p  q) and (pÂ² + pq + qÂ²) are both positive. Since (p  q) is positive, p > q.
Woof: too much work! This is why you want to try numbers on test day
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 beatthegmat
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Hello,
If you are going to pick numbers for the values of p and q, then the numbers have to match up with the information in the statements. Statement 1 says that p^2 > q^2 and statement 2 says that P^3 > q^3. In neither of those statements is p=1,q=2 a valid option. If you try again with numbers that match up then you will get the correct answer.
If you are going to pick numbers for the values of p and q, then the numbers have to match up with the information in the statements. Statement 1 says that p^2 > q^2 and statement 2 says that P^3 > q^3. In neither of those statements is p=1,q=2 a valid option. If you try again with numbers that match up then you will get the correct answer.
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