Is p > q ? 1) p^2 > q^2 2) p^3 > q^3

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Is p > q ? 1) p^2 > q^2 2) p^3 > q^3

by mariofelixpasku » Wed Feb 19, 2014 1:05 pm
Is p > q ?

1) p^2 > q^2

2) p^3 > q^3

Ans: B

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by [email protected] » Wed Feb 19, 2014 1:14 pm
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

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by mariofelixpasku » Wed Feb 19, 2014 1:17 pm
can you explain why from q^3<p^3 follows q<p ?

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by [email protected] » Wed Feb 19, 2014 1:44 pm
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.

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by mariofelixpasku » Wed Feb 19, 2014 1:46 pm
thanks for sharing this general rule. you saved me a lot of time i would have wasted on testing numbers.

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by Matt@VeritasPrep » Wed Feb 19, 2014 6:52 pm
An algebraic way of explaining this, for those curious:

p² > q² = |p|² > |q|², so

p² > q² implies |p| > |q|, but NOT p > q

(For instance, if p = -3 and q = 2, p² > q² and |p| > |q|, but p is not > q.)

For p³ > q³, it's a little more complicated. We have:

p³ > q³, or
p³ - q³ > 0, or
(p - q) * (p² + pq + q²) > 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 (p² + pq + q²) are both positive OR (p - q) and (p² + pq + q²) are both negative.

Let's assume that it's the second case. If (p - q) and (p² + pq + q²) are both negative, then (p² + pq + q²) < 0. But if (p² + pq + q²) < 0, then (p² + q²) < -pq. Since (p² + q²) 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 p³ > q³ (how can a positive be greater than a negative!?) This contradicts our original statement (that p³ > q³), making it impossible.

Hence (p - q) and (p² + pq + q²) 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 :D

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Re: Is p > q ? 1) p^2 > q^2 2) p^3 > q^3

by Dee34596 » Sun Oct 01, 2023 2:12 pm
what if p = 1 and q = 2? 1 cubed is LESS than 2 cubed and in this case p < q and not p>q?

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Re: Is p > q ? 1) p^2 > q^2 2) p^3 > q^3

by beatthegmat » Tue Oct 03, 2023 9:15 am
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.
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