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If x > y^2 > z^4, which of the following statements could be true?

I. x>y>z II. z>y>x III. x>z>y

A. I only B. I and II only C. I and III only D. II and III only E. I, II and II

As this is a COULD be true question then even one set of numbers proving that statement holds true is enough to say that this statement should be part of correct answer choice.

Given: \(x > y^2 > z^4\).

1. \(x>y>z\) --> the easiest one: if \(x=100\), \(y=2\) and \(z=1\) --> this set satisfies \(x > y^2 > z^4\) as well as given statement \(x>y>z\). So 1 COULD be true.

2. \(z>y>x\) --> we have reverse order than in stem (\(x > y^2 > z^4\)), so let's try fractions: if \(x=\frac{1}{5}\), \(y=\frac{1}{4}\) and \(z=\frac{1}{3}\) then again the stem and this statement hold true. So 2 also COULD be true.

3. \(x>z>y\) --> let's make \(x\) some big number, let's say 1,000. Next, let's try the fractions for \(z\) and \(y\) for the same reason as above (reverse order of \(y\) and \(z\)): \(y=\frac{1}{3}\) and \(z=\frac{1}{2}\). The stem and this statement hold true for this set of numbers. So 3 also COULD be true.

A) I Only B) I and II Only C) I and III Only D) II and III Only E) I, II and III

I think the actual question is: Which of the following could be true?

Plugging in numbers work best for such questions. The only thing to keep in mind is that you should plug in the right numbers. How do you know the right numbers? When I see \(x > y^2 > z^4\), I think that \(y^2\) and \(z^4\) are non negative. Since \(y^2 > z^4\), \(y^2\) cannot be 0. Only z can be 0. x has to be positive. Also, I have to take into account two ranges: 0 to 1 and 1 to infinity. The powers behave differently in these two ranges. I will consider negative numbers only if I have to since with powers, they get confusing to deal with.

The question says: "Which of the following could be true?" We have to find examples where each relation holds.

I. x > y > z This is the most intuitive of course. z = 0, y = 1 and x = 2 \(2 > 1^2 > 0^4\)

II. z > y > x Let me consider the 0 to 1 range here. Say z = 1/2, y = 1/3 and x = 1/4 \(1/4 > 1/9 > 1/16\)

III. x > z > y Let's stick to 0 to 1 range. z > y as in case II above but x has to be greater than both of them. Say z = 1/2, y = 1/3 and x = 1 \(1>1/9 > 1/16\)

So all three statements could be true.
_________________

Hi Bunuel/Karishma, Thanks for the earlier response.. I think, I am very weak in Inequalities.. Could you please post how to go about this question in algebraic way.. ...also if you could let me know how do you make sure about the "Range of the values", that will also work.. Thanks H

Plug-in method is really the best way to handle such kind of questions. No need to look for some kind of textbook or algebraic ways.

Notice that there are are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes). Many difficult inequality problems will often require some sort of plug-ins, as part of your technique or else you'll spend too much time solving them with algebra. Which means that you MUST make plug-in methods part of your arsenal if you want to get a decent score.

P.S. I'm not sure understood the following part of your post: "how do you make sure about the "Range of the values", that will also work.. "
_________________

If x > y^2 >z^4, which of the following statements could be true?

I. x > y > z

x=10000 y=10; y^2=100 z=1; z^4=1 x>y^2>z^4

II. z > y > x z=0.5; z^4=0.0625 y=0.4; y^2=0.16 x=0.3 x>y^2>z^4

III. x > z > y x=0.5 z=0.2; z^4=0.0016 y=0.1; y^2=0.01 x>y^2>z^4

a. I only b. I and II only c. I and III only d. II and III only e. I, II and III

We just need to remember that 1. the number decreases in value with increment in the power of the number if 0< number< 1; if x=0.1; x>x^2>x^3>x^(100) because x is between 0 and 1.

2. the number increases in value with increment in the power of the number if number>1 if x=2; x<x^2<x^(100) because x is more than 1.

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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13 Mar 2012, 00:05

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So, Karishma, can it be said that any configuration of x, y and z COULD be true? Because we can always find fractions or integers which will satisfy.

6 configurations are possible, x>y>z, x>z>y, y>x>z, y>z>x, z>x>y and z>y>x. The answer choices in the question are the first, second and the last but I think any of these 6 configurations COULD be true.
_________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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13 Mar 2012, 01:14

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Hi,

Agree its a super tricky problem. I was so focused on the 0-1 range for x,y & z that forgot that for statement 3 you could have larger numbers
_________________

Giving +1 kudos is a better way of saying 'Thank You'.

If x > y^2 > z^4, which of the following statements could be true?

I. x>y>z II. z>y>x III. x>z>y

A. I only B. I and II only C. I and III only D. II and III only E. I, II and II

As this is a COULD be true question then even one set of numbers proving that statement holds true is enough to say that this statement should be part of correct answer choice.

Given: \(x > y^2 > z^4\).

1. \(x>y>z\) --> the easiest one: if \(x=100\), \(y=2\) and \(z=1\) --> this set satisfies \(x > y^2 > z^4\) as well as given statement \(x>y>z\). So 1 COULD be true.

2. \(z>y>x\) --> we have reverse order than in stem (\(x > y^2 > z^4\)), so let's try fractions: if \(x=\frac{1}{5}\), \(y=\frac{1}{4}\) and \(z=\frac{1}{3}\) then again the stem and this statement hold true. So 2 also COULD be true.

3. \(x>z>y\) --> let's make \(x\) some big number, let's say 1,000. Next, let's try the fractions for \(z\) and \(y\) for the same reason as above (reverse order of \(y\) and \(z\)): \(y=\frac{1}{3}\) and \(z=\frac{1}{2}\). The stem and this statement hold true for this set of numbers. So 3 also COULD be true.

Answer: E.

Isn't it stated in the exam that assume all numbers are integers? We can't try fractions unless they say they are not integers.

No that's not true at all. All numbers on the test represent real numbers: Integers, Fractions and Irrational Numbers. You cannot assume a variable is integer if you are not explicitly told so.
_________________

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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07 Aug 2015, 14:45

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Algebraic solution:

In the question we are given: x>y2>z4, hence from concepts of inequalities we break it into 2 parts: x>y2 and y2>z4. 1. x>y2 means -x(1/2)<y<x(1/2) 2. y2>z4 means -y<z2<y, but a square cannot be negative so 0<z2<y, this implies -y(1/2)<z<y(1/2).

Now we plot all these points on number line with the intersection of there ranges. But before that we need to understand that we will only be taking x,y,z as positive since if we take y as negative for example(easiest one) the value z(2) becomes negative, whereas a square can never be negative.

hence we plot all of them on the positive x-axis. From above 1 & 2 point we get a general range as such 0<z<y(1/2)<y<x(1/2)<x. Now, we need to see that we haven't in reality considered various values of x,y,z but have come up with a general idea of how they look on the number line. Now we define the ranges, since we know about a^x graph varies for values 0<a<1 and a>1, we also take such cases for all three of them. 1. 0<x<1 and x>1 2. 0<y<1 and y>1 3. 0<z<1 and z>1

Hence looking at the combinations we find we have 8 possibilities (2*2*2). taking the 2 general ones: 1. x>1 y>1 z>1. In the general formula we simply put x,y,z and get x>y>z. (Would have figured initially). 2. 0<x<1, 0<y<1 and 0<z<1. In this possibility put x as 1/x, y as 1/y and z as 1/z in general formula we get z>y>x. 3. x>1 0<y<1 and 0<z<1. In this put y as 1/y and z as 1/z. Keep x as x in general formula, we see x>1/y>1/z. since only 1/y>1/z are in reciprocal hence z>y by inequalities. thus x>z>y.

Therefore we can get 8 possibilities and the fact is all of them are correct.

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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22 Aug 2015, 22:59

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Hi vinnisatija,

Yes, if you see a "could be" question, then just one example that satisfies the given condition is sufficient. In case of "must be" questions, the required condition must hold true under all circumstances along with whatever additional constraint is given in the problem.

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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15 Feb 2012, 11:11

I agree with E. Last one was tricky but once you realize that y and z could be fractions it becomes clear. took less than a minute but still a good question to make sure your reasoning is solid

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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28 Feb 2012, 01:31

Hi Bunuel/Karishma, Thanks for the earlier response.. I think, I am very weak in Inequalities.. Could you please post how to go about this question in algebraic way.. ...also if you could let me know how do you make sure about the "Range of the values", that will also work.. Thanks H
_________________

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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28 Feb 2012, 03:01

Thanks Bunuel for your response.. This is what I mean when I said range - Red Part in Karishma's response"

VeritasPrepKarishma wrote:

arps wrote:

1) x > y2 > z4

which of the following is true:

I x>y>z II z>y>x III x>z>y

A) I Only B) I and II Only C) I and III Only D) II and III Only E) I, II and III

I think the actual question is: Which of the following could be true?

Plugging in numbers work best for such questions. The only thing to keep in mind is that you should plug in the right numbers. How do you know the right numbers? When I see \(x > y^2 > z^4\), I think that \(y^2\) and \(z^4\) are non negative. Since \(y^2 > z^4\), \(y^2\) cannot be 0. Only z can be 0. x has to be positive. Also, I have to take into account two ranges: 0 to 1 and 1 to infinity. The powers behave differently in these two ranges. I will consider negative numbers only if I have to since with powers, they get confusing to deal with.

The question says: "Which of the following could be true?" We have to find examples where each relation holds.

I. x > y > z This is the most intuitive of course. z = 0, y = 1 and x = 2 \(2 > 1^2 > 0^4\)

II. z > y > x Let me consider the 0 to 1 range here. Say z = 1/2, y = 1/3 and x = 1/4 \(1/4 > 1/9 > 1/16\)

III. x > z > y Let's stick to 0 to 1 range. z > y as in case II above but x has to be greater than both of them. Say z = 1/2, y = 1/3 and x = 1 \(1>1/9 > 1/16\)

Thanks Bunuel for your response.. This is what I mean when I said range - Red Part in Karishma's response"

VeritasPrepKarishma wrote:

First notice that since x>z^4 (x is greater than some nonnegative value) then x>0.

Now, as Karishma correctly noted, numbers in powers behave differently in the range {0. 1} and {1. +infinity}. For example:

If 0<a<1 then a, a^2 and a^4 will be ordered as follows: 0--(a^4)--(a^2)--(a)--1

If a>1 then a, a^2 and a^4 will be ordered as follows: 1--(a)--(a^2)--(a^4)--

So, we should take the above difference in ordering into account when picking numbers for x, y, and z, since we need to find the values which satisfy 3 different statements.

Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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20 Jan 2013, 11:48

Sorry to open up a new thread!

Bunuel, I was impressed from your geometric way to solve inequalities. I was trying this one but could not really figure out how? Would you mind giving us a geometric way to solve this problem?

A) I Only B) I and II Only C) I and III Only D) II and III Only E) I, II and III

I think the actual question is: Which of the following could be true?

Plugging in numbers work best for such questions. The only thing to keep in mind is that you should plug in the right numbers. How do you know the right numbers? When I see \(x > y^2 > z^4\), I think that \(y^2\) and \(z^4\) are non negative. Since \(y^2 > z^4\), \(y^2\) cannot be 0. Only z can be 0. x has to be positive. Also, I have to take into account two ranges: 0 to 1 and 1 to infinity. The powers behave differently in these two ranges. I will consider negative numbers only if I have to since with powers, they get confusing to deal with.

The question says: "Which of the following could be true?" We have to find examples where each relation holds.

I. x > y > z This is the most intuitive of course. z = 0, y = 1 and x = 2 \(2 > 1^2 > 0^4\)

II. z > y > x Let me consider the 0 to 1 range here. Say z = 1/2, y = 1/3 and x = 1/4 \(1/4 > 1/9 > 1/16\)

III. x > z > y Let's stick to 0 to 1 range. z > y as in case II above but x has to be greater than both of them. Say z = 1/2, y = 1/3 and x = 1 \(1>1/9 > 1/16\)

So all three statements could be true.

Hey Karishma,

Can you please confirm if this type of solving is justified

As x>y^2>z^4. It can be inferred that x>y^2 and is true for both x>y and x<y (As we don't have any limitations for the three variables) Also, y^2>z^4 which means y>z^2 and the inequality is valid for both y>z and y<z Additionally, x>z^4, similar to prior cases x>z and z<x

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