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

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

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New post 08 Sep 2013, 08:58
Bunuel wrote:
chetan86 wrote:
Restrictions are not provided on the variables so I planned to check different values I and used x=y=z=1/2.
As if I take 1/2 for each variables, its given condition would be satisfied and it will become 1/2>1/4>1/8.

So according to me none of the conditions are satisfied.
Am I doing anything wrong here?


Notice that the question asks "which of the following statements could be true" NOT "which of the following statements must be true"



Thanks Bunuel for identifying my mistake.
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Re: x > y^2 > z^4 [#permalink]

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New post 07 Nov 2013, 02:10
Bunuel wrote:
Orange08 wrote:
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.
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Re: x > y^2 > z^4 [#permalink]

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New post 07 Nov 2013, 02:19
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SaramiR wrote:
Bunuel wrote:
Orange08 wrote:
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.
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Re: x > y^2 > z^4 [#permalink]

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New post 19 Nov 2013, 11:52
Bunuel wrote:
Orange08 wrote:
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.



Hi Bunuel,

Can we not use negative integers.
For e.g.: x =5, y=-2,z=-1 then the first inequality would be 5>(-2)^2>(-1)^4. In this case x>z>y and y is not greater than z.

Am i missing something?
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Re: x > y^2 > z^4 [#permalink]

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New post 19 Nov 2013, 15:21
chitrasekar2k5 wrote:
Bunuel wrote:
Orange08 wrote:
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.



Hi Bunuel,

Can we not use negative integers.
For e.g.: x =5, y=-2,z=-1 then the first inequality would be 5>(-2)^2>(-1)^4. In this case x>z>y and y is not greater than z.

Am i missing something?


The questions asks which of the following COULD be true not MUST be true. As shown above each option COULD be true for certain numbers.
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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New post 04 Apr 2014, 22:06
I and II are pretty much evident.. However III is no, it is better to split the third option in to two parts and focus just on Z>Y which is quite possible.. So the answer is all three..
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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

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New post 07 Aug 2015, 15:45
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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.
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Re: If x > y^2 >z^4 [#permalink]

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New post 19 Aug 2015, 04:01
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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New post 19 Aug 2015, 08:19
For statement 3: x>z>y
Question: x > y^2 > z^4

It translates to, z>y & y^2 > z^4 (x is a big value and feature in both equalities)

This is same as statement 2 now: z>y
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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New post 20 Aug 2015, 07:13
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
\(1>1/9 > 1/16\)

So all three statements could be true.



Karishma...In "could be true" scenarios, do we need just a one situation that fits in to answer the question ?
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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New post 22 Aug 2015, 23:59
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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.

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

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New post 30 Sep 2015, 21:15
Bunuel wrote:
Orange08 wrote:
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.


Bunuel actually for option 3, since the sign stays the same z>y as in option 2, there is no need to reverse the value of z and y from option 2 right?
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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New post 01 Oct 2015, 00:46
SamuelWitwicky

Yes, there is no need to change the examples that Bunuel used for z and y in II, they work in III as well.

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

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New post 01 Oct 2015, 01:02
dabral ok cool thanks. I thought maybe I was missing something
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If x > y^2 > z^4, which of the following statements could be [#permalink]

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New post 10 Feb 2016, 22:18
since we have no constraints all the inequalities could be true because we can play with +ve and -ve and integer (tiny and huge) and fractions (proper and improper) in any way to satisfy inequalities. if needed we can assume any number to be a fraction and nothing prevents us from considering this fraction to be as small as possible and maximize other numbers
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Re: If x > y^2 > z^4, which of the following statements could be [#permalink]

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

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New post 10 Mar 2017, 10:41
Quote:
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


We are given that x > y^2 > z^4 and need to determine which statements must be true. Let’s test each Roman Numeral.

I. x > y > z

Notice that the order of arrangement of x, y, and z in the inequality x > y > z is the same as the order of arrangement of x, y^2, and z^4 in the inequality x > y^2 > z^4, so we want to test positive integers in this case.

x = 10

y = 3

z = 1

Notice that 10 > 3 > 1 for x > y > z AND 10 > 9 > 1 for x > y^2 > z^4.

We see that I could be true.

II. z > y > x

Notice that the order of arrangement of x, y, and z in the inequality z > y > x differs from the order of arrangement of x, y^2, and z^4 in the inequality x > y^2 > z^4, so we want to test positive proper fractions in this case. This is because we need to decrease the value of y and z to make them work within the given inequality.

x = 1/5

y = 1/3

z = 1/2

Notice that 1/2 > 1/3 > 1/5 for z > y > x AND 1/5 > 1/9 > 1/16 for x > y^2 > z^4.

We see that II could be true.

III. x > z > y

Notice that the order of arrangement of y and z in the inequality x > z > y differs from the order of arrangement of y^2 and z^4 in the inequality x > y^2 > z^4, so we once again want to test positive proper fractions. This is because we need to decrease the value of z to make it work within the given inequality (that is, we want to swap the order of z4 and y2 even if z > y).

x = 1/2

y = 1/4

z = 1/3

Notice that 1/2 > 1/3 > 1/4 for x > z > y AND 1/2 > 1/16 > 1/81 for x > y^2 > z^4.

We see that III could be true.

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

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New post 11 May 2017, 05:35
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

Solution:
To answer questions like this, use ZONEF.
Z = zero
O = one
N = negative integers
E = Extreme Integers (Read as Positive Integers > 1)
F = Fractions (think both positive and negative)

For x > y^2 > z^4, the following numbers work: x = 10, y = 3, z = 1. So , I is true.
Eliminate (D).

The easiest way to prove II true is to multiply the above numbers by -1, but as you are dealing with even exponents, negative numbers will not work.
That leaves us with F(Fractions).

Let z = 0.9, y = 0.7 and x = 0.1
These numbers keep both x > y^2 > z^4 and z > y > x true.
Eliminate (A) and (C).

Now, proving the third is easy. Just take x = any big positive number, say 10
Now, you have x = 10, y = 0.7 and z = 0.9
These values keep both x > y^2 > z^4 and x > z > y true.

So, all three are true.
The answer is (E).
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Re: If x > y^2 > z^4, which of the following statements could be   [#permalink] 11 May 2017, 05:35

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