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B) p=2q is not sufficient to answer to our question, because p and q can be negatice or positive; hence q > p or p > q.

C) p>q but in this case we can have for instance 2 > 3 that is false. But if we have for example -1/2 > -1/3 or -2 > -3 are true. So is unclear if p > q.

For D ) and E ) I have applied the similar reasoning.

Re: In this exponent problem P > Q ???? [#permalink]
15 Feb 2012, 10:32

Expert's post

2

This post was BOOKMARKED

In which one of the following choices must p be greater than q?

Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

A. 0.9^p = 0.9^q --> p=q. Discard; B. 0.9^p = 0.9^2q --> p=2q --> if p=-2 and q=-1 then p<q. Discard; C. 0.9^p > 0.9^q --> if p=1 and q=2 then then p<q. Discard; D. 9^p < 9^q --> 9^{p-q}<1 --> p-q<0 --> p<q. Discard; E. 9^p > 9^q --> 9^{p-q}>1 --> p-q>0 --> p>q. Correct.

Re: In this exponent problem P > Q ???? [#permalink]
15 Feb 2012, 11:04

Expert's post

Bunuel wrote:

In which one of the following choices must p be greater than q?

Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

A. 0.9^p = 0.9^q --> p=q. Discard; B. 0.9^p = 0.9^2q --> p=2q --> if p=-2 and q=-1 then p<q. Discard; C. 0.9^p > 0.9^q --> if p=1 and q=2 then then p<q. Discard; D. 9^p < 9^q --> 9^{p-q}<1 --> p-q<0 --> p<q. Discard; E. 9^p > 9^q --> 9^{p-q}>1 --> p-q>0 --> p>q. Correct.

Answer: E.

Bunuel thanks for the explanation very clear. This was a tricky problem and sometimes I have the propensity to over calculate: I'm wrong ??' do you have a suggestion ?? of course I guess to have idea in which direction I have to go but is not enough to beat problems at high level..... _________________

B) p=2q is not sufficient to answer to our question, because p and q can be negatice or positive; hence q > p or p > q.

C) p>q but in this case we can have for instance 2 > 3 that is false. But if we have for example -1/2 > -1/3 or -2 > -3 are true. So is unclear if p > q.

For D ) and E ) I have applied the similar reasoning.

Thanks GMAT club.

This question is tricky. You might end up losing lots of valuable time trying to plug in numbers in each option and getting lost mid-way. The trick is to use exponents theory. I like to use number plugging only where it is apparent that it is useful (e.g. in options A and B above to rule them out in 10 sec) or where I can't think of anything else. The specific theory which helps in dealing with this and similar questions:

If the base is positive, no matter what power you give to it, it will not become negative. e.g. 5^n must be positive no matter what the value of n is. The reason is that 5^n is just 5 multiplied by itself n number of times, whatever n is. So this number can never be negative.

Also, If n is positive, 5^n > 1 If n = 0, 5^n = 1 If n is negative, 0 < 5^n < 1

Another thing, the positive base, a, can lie between one of two ranges - 'between 0 and 1' and 'greater than 1'. We saw what happens if the base is greater than 1 above (in 5^n example)

Let's see what happens when the base lies between 0 and 1: If n is positive, a^n < 1 If n = 0, a^n = 1 If n is negative, a^n > 1

These relations hold the other way around too. If a is between 0 and 1, and a^n > 1, n must be negative etc. Plug in values to understand them.

Now look at this question:

In these three options, each term is positive so we can take the term on the right to left hand side so that we have to deal with a single base.

(C)0.9^p > 0.9^q (0.9)^{p-q} > 1 Here, p-q must be negative i.e. p-q < 0 or p<q (from red above)

(D) 9^p < 9^q 9^{p-q} < 1 p-q < 0 (from blue above) or p < q

(E)9^p > 9^q 9^{p-q} > 1 So p-q > 0 (from green above) or p > q _________________

Re: In this exponent problem P > Q ???? [#permalink]
22 Feb 2012, 22:07

if x^2 >x^3 : we write x^2-x^3 >0 ie we subtract and dont divide in this case we are dividinge is that right?

Bunuel wrote:

In which one of the following choices must p be greater than q?

Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

A. 0.9^p = 0.9^q --> p=q. Discard; B. 0.9^p = 0.9^2q --> p=2q --> if p=-2 and q=-1 then p<q. Discard; C. 0.9^p > 0.9^q --> if p=1 and q=2 then then p<q. Discard; D. 9^p < 9^q --> 9^{p-q}<1 --> p-q<0 --> p<q. Discard; E. 9^p > 9^q --> 9^{p-q}>1 --> p-q>0 --> p>q. Correct.

Re: In this exponent problem P > Q ???? [#permalink]
22 Feb 2012, 22:30

Expert's post

devinawilliam83 wrote:

if x^2 >x^3 : we write x^2-x^3 >0 ie we subtract and dont divide in this case we are dividinge is that right?

Bunuel wrote:

In which one of the following choices must p be greater than q?

Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

A. 0.9^p = 0.9^q --> p=q. Discard; B. 0.9^p = 0.9^2q --> p=2q --> if p=-2 and q=-1 then p<q. Discard; C. 0.9^p > 0.9^q --> if p=1 and q=2 then then p<q. Discard; D. 9^p < 9^q --> 9^{p-q}<1 --> p-q<0 --> p<q. Discard; E. 9^p > 9^q --> 9^{p-q}>1 --> p-q>0 --> p>q. Correct.

Answer: E.

If its given that x^2<x^3 then since x^2>0 then we can reduce by it and write x>1;

But if its given that x>x^3 then since we don't know the sign of x then we can not reduce by it and should rewrite: x(x^2-1)<0 --> (x+1)x(x-1)<0 --> x<-1 or 0<x<1.

Re: In this exponent problem P > Q ???? [#permalink]
23 Feb 2012, 01:48

Expert's post

devinawilliam83 wrote:

if x^2 >x^3 : we write x^2-x^3 >0 ie we subtract and dont divide in this case we are dividinge is that right?

If the variable is given to be non zero, you can divide in an equation and if you also know the sign of the variable, you can divide in an inequality too. (and adjust the inequality according to the sign of the variable)

e.g. x^3 = x. What values can x take? x can be 0 so we don't divide. x^3 - x = 0 x(x^2 - 1) = 0 x = 0, 1, -1

Here, if you forget about 0 and divide by x, you get x^2 = 1 so x = +1 or -1. You lose out one solution: x = 0

But in case you have x^3 = x. What values can x take if x is non zero? Now you can divide and you will get x = 1 or -1

Similarly, say you have an inequality: x^2 > x^3 If you divide by x^2 thinking it's non negative, you get the range x < 1 which is incorrect since you have to specify that 'x cannot be 0'. It holds for every value less than 1 other than 0. x^2 > x^3 does not hold for x = 0.

So instead, you can choose to solve in two ways:

1. Take cases and use division: x^2 > x^3 If x \neq 0, we can divide. We get x < 1. If x = 0, the inequality does not hold. Hence, x \neq 0 Answer: x < 1, x \neq 0

2. Subtract: x^3 - x^2 < 0 x^2 (x - 1) < 0 One of the terms must be negative and the other positive. Since x^2 cannot be negative, x - 1 must be negative (so x < 1) and x^2 must be positive so x \neq 0 Answer: x < 1, x \neq 0

Either way, you will get the same answer (obviously!)

In this question, as Bunuel said, 9^q must be positive. It cannot take value 0 or negative for any value of q so there are no complications at all. You can easily divide. _________________

Re: In this exponent problem P > Q ???? [#permalink]
23 Feb 2012, 01:54

Expert's post

VeritasPrepKarishma wrote:

devinawilliam83 wrote:

if x^2 >x^3 : we write x^2-x^3 >0 ie we subtract and dont divide in this case we are dividinge is that right?

If the variable is given to be non zero, you can divide in an equation and if you also know the sign of the variable, you can divide in an inequality too. (and adjust the inequality according to the sign of the variable)

e.g. x^3 = x. What values can x take? x can be 0 so we don't divide. x^3 - x = 0 x(x^2 - 1) = 0 x = 0, 1, -1

Here, if you forget about 0 and divide by x, you get x^2 = 1 so x = +1 or -1. You lose out one solution: x = 0

But in case you have x^3 = x. What values can x take if x is non zero? Now you can divide and you will get x = 1 or -1

Similarly, say you have an inequality: x^2 > x^3 If you divide by x^2 thinking it's non negative, you get the range x < 1 which is incorrect since you have to specify that 'x cannot be 0'. It holds for every value less than 1 other than 0. x^2 > x^3 does not hold for x = 0.

So instead, you can choose to solve in two ways:

1. Take cases and use division: x^2 > x^3 If x \neq 0, we can divide. We get x < 1. If x = 0, the inequality does not hold. Hence, x \neq 0 Answer: x < 1, x \neq 0

2. Subtract: x^3 - x^2 < 0 x^2 (x - 1) < 0 One of the terms must be negative and the other positive. Since x^2 cannot be negative, x - 1 must be negative (so x < 1) and x^2 must be positive so x \neq 0 Answer: x < 1, x \neq 0

Either way, you will get the same answer (obviously!)

In this question, as Bunuel said, 9^q must be positive. It cannot take value 0 or negative for any value of q so there are no complications at all. You can easily divide.

One little thing Karishma.

If its given that x^2<x^3 then it's clear that x\neq{0} thus x^2>0 and you can safely reduce by it and write x>1. _________________

Re: In this exponent problem P > Q ???? [#permalink]
23 Feb 2012, 03:38

Expert's post

Bunuel wrote:

One little thing Karishma.

If its given that x^2>x^3 then it's clear that x\neq{0} thus x^2>0 and you can safely reduce by it and write x>1.

Sure Bunuel, I understand that one can see that when x^2>x^3, x cannot be 0. But since here, you get x < 1, a range that includes x = 0, we need to specify that x can take any value less than 1 except 0. The point I am making here is that if you only reduce it and write 'x<1', it's not correct. I think you got a typo 'x > 1' above. _________________

Re: In this exponent problem P > Q ???? [#permalink]
23 Feb 2012, 03:49

Expert's post

VeritasPrepKarishma wrote:

Bunuel wrote:

One little thing Karishma.

If its given that x^2>x^3 then it's clear that x\neq{0} thus x^2>0 and you can safely reduce by it and write x>1.

Sure Bunuel, I understand that one can see that when x^2>x^3, x cannot be 0. But since here, you get x < 1, a range that includes x = 0, we need to specify that x can take any value less than 1 except 0. The point I am making here is that if you only reduce it and write 'x<1', it's not correct. I think you got a typo 'x > 1' above.

Actually I do have a typo but not the one you pointed above: in 4 posts above I considered x^2<x^3 (not x^2>x^3) which when reduced gives x>1.

So, it should read: if its given that x^2<x^3 then it's clear that x\neq{0} thus x^2>0 and you can safely reduce by it and write x>1. _________________

Re: In which one of the following choices must p be greater than [#permalink]
29 Jan 2014, 13:11

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