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

This is a really tough problem. Here is my video explanation:
https://gmatquantum.com/gmatprep-algebr ... tatements/

Dabral

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.


Ans: "E"

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.

Inequality questions to practice.
DS: search.php?search_id=tag&tag_id=184
PS: search.php?search_id=tag&tag_id=189

Hope it helps.

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.. "

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

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.

Hope it's clear.

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.

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.

Dabral

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

If we CAN find a set of values that satisfies a statement AND yields values such that x > y² > z⁴, then we'll keep that statement.

Statement I. x > y > z
If x = 2, y = 1, and z = 0, then x > y² > z⁴
KEEP statement I

Statement II. z > y > x
If x = 1/4, y = 1/3, and z = 1/2, then x > y² > z⁴
KEEP statement II

Statement III. x > z > y
If x = 2, y = -1, and z = 0, then x > y² > z⁴
KEEP statement III

Answer: E

Cheers,
Brent

VeritasKarishma
Bunuel
Please help me understand where I am going wrong.
If you choose x = 1/3 and y = 1/2 then y>x but y^​2<x
So how can we choose the options that include 1 - which is x>y>z. Because that might not be true.
Thank you in advance!!

Stmnt I will be true for some values of x, y and z.
Note that only one of the three statements can be true for a given set of values of x, y and z. The three statements hold for different values of x, y and z as shown in my post above: https://gmatclub.com/forum/if-x-y-2-z-4 ... l#p1032823

Basically, the question should be: Which of the following "could be" true for some values of x, y and z?

I. x > y > z
would be true when
z = 0, y = 1 and x = 2

Its given \(x > y^2 > z^4\)
Since its a "Could be true " question , we assume values for x,y and z and try to prove that each statement could be true or not.


1. x > y > z

Assume x = 1, y = 1/2 , z = 1/4

Then, \(1 > (1/2)^2 > (1/4)^4\)

Also 1 > 1/2 > 1/4

Hence x > y > z could be true

2. z > y > x
Assume x= 1/8, y= 1/4 ,z = 1/3

Then \(1/8 > (1/4)^2 > (1/3)^4 \) >>> \(x > y^2 > z^4\)

Also 1/3 > 1/4> 1/8
Hence z > y > x could be true

3. x > z >y

Assume x= 1, y= -1/2 , z = 1/2

Then \(x > y^2 > z^4\)
Also x > z > y
This statement also could be true.

Since all the Statements could be true, Answer is OPTION E

Thanks,
Clifin J Francis,
GMAT SME

Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Bunuel, I face issues in DS questions like these. How do you come up with cases?

Hello AADINAATH,

This is not a DS question, but I understand what your doubt and concern is regarding. It gets difficult, especially under a time-controlled exam environment, to think of cases or examples which would satisfy or not satisfy different conditions.

Through this post, I will help you identify certain ranges in which all numbers follow some fixed patterns. Knowing these ranges and the patterns within them will help you easily think of examples for every condition in such questions.

Let’s get started!

1. (1, ∞)
   
   This is the range of all numbers greater than 1. For example, 20 and 10.
   
   • PATTERN 1: Observe that 10 < 10\(^2\) < 10\(^3\) < 10\(^4\) < …. and 20 < 20\(^2\) < 20\(^3\) < 204 < ….
     
     • Using this observation, we can generalize that if x > 1, then x < x\(^2\) < x\(^3\) < x\(^4\) < ……
     • That is, as power (n) increases, the value of x\(^n\) also increases for all n > 0.
   • PATTERN 2: Observe that 20 > 10; (20)2 > (10)2, (20)3 > (10)3, and so on.
     
     • Using this observation, we can generalize that if two numbers x and y belong to the range (1, ∞), then if x > y, then x\(^n\) > y\(^n\) for all n > 0.
   
   
2. (0, 1)
   
   
   This is the range of all positive numbers less than 1. For example, 0.4 and 0.5.
   
   • PATTERN 1: Observe that 0.4 > 0.4\(^2\) > 0.4\(^3\) > 0.4\(^4\) > …. and 0.5 > 0.5\(^2\) > 0.5\(^3\) > 0.5\(^4\) >….
     
     • Using this observation, we can generalize that if 0< x < 1, then x > x\(^2\) > x\(^3\) > x\(^4\) > ……
     • That is, as power (n) increases, the value of x\(^n\) decreases for all n > 0.
   • PATTERN 2: Observe that 0.5 > 0.4; (0.5)\(^2\) > (0.4)\(^2\), (0.5)\(^3\) > (0.4)\(^3\), and so on.
     
     • Using this observation, we can generalize that if two numbers x and y belong to the range (0, 1), then if x > y, then x\(^n\) > y\(^n\) for all n > 0.
   
   
3. (-∞, -1)
   
   This is the range of all numbers less than -1. For example, -10 and -20.
   
   • PATTERN 1: Observe that 0 > -10 > (-10)\(^3\) > (-10)\(^5\) > …. and 0 < (-10)\(^2\) < (-10)\(^4\) < (-10)\(^6\) <….
     
     • Using this observation, we can generalize that if x < -1, then 0 > x > x\(^3\) > x\(^5\) > …. and 0 < x\(^2\) < x\(^4\) < x\(^6\) < …
     • That is, for odd powers n, as power increases, the value of x\(^n\) decreases for all integer n > 0.
     • And for even powers (m), as power increases, the value of x\(^m\) also increases for all integer m > 0.
   • PATTERN 2: Observe that -10 > -20; (-10)\(^3\) > (-20)\(^3\) and (-10)\(^5\) > (-20)\(^5\) and so on but (-10)\(^2\) < (-20)\(^2\) and (-10)\(^4\) < (-20)\(^4\) and so on.
     
     • Using this observation, we can generalize that if two numbers x and y belong to the range (-∞, -1), then if x > y, then x\(^n\) > y\(^n\) for all odd integers n > 0 and x\(^m\) < y\(^m\) for all even integers m > 0.
   
   
4. (-1, 0)
   
   This is the range of all negative numbers greater than -1. For example, -0.5 and -0.4.
   
   PATTERN 1: Observe that -0.5 < (-0.5)\(^3\) < (-0.5)\(^5\) < …. and (-0.5)\(^2\) > (-0.5)\(^4\) > (-0.5)\(^6\) >….
   
   Using this observation, we can generalize that if -1 < x < 0, then x < x\(^3\) < x\(^5\) < …. and x\(^2\) > x\(^4\) > x\(^6\) > …
   That is, for odd powers (n), as power increases, the value of x\(^n\) also increases for all n > 0.
   And for even powers (m), as power increases, the value of x\(^m\) decreases for all n > 0.
   PATTERN 2: Observe that -0.4 > -0.5; (-0.4)\(^3\) > (-0.5)\(^3\) and (-0.4)\(^5\) > (-0.5)\(^5\) and so on but (-0.4)\(^2\) < (0.5)\(^2\) and (0.4)\(^4\) < (0.5)\(^4\) and so on.
   
   Using this observation, we can generalize that if two numbers x and y belong to the range (-1, 0), then if x > y, then x\(^n\) > y\(^n\) for all odd integers n > 0 and x\(^m\) < y\(^m\) for all even integers m > 0.

These are the ranges that you should consider while thinking of cases and examples.
Having these ranges in mind and using a little bit of inference will make it easier for you to solve questions of such kind.

TAKEAWAYS:
c) For positive numbers, if x > y, all positive powers of x will be greater than the same power of y. So, if x > y, x\(^n\) > y\(^n\) where n > 0.
d) For negative numbers, if x < y, x\(^n\) < y\(^n\) where n is odd but x\(^m\) > y\(^m\) where m is even.

Hope this helps!

Best Regards,
Ashish Arora
Quant Expert, e-GMAT

This is more of a semantic problem than a quant one lol. I took x =1/5, y = 1/3 and z = 1/2

These numbers satisfy the condition: x>y^2>z^4

But since the numbers do not satisfy the 1st condition, I rejected the 4 choices containing that condition and selected the one choice that does not contain the 1st condition but alas I was wrong.

KarishmaB For such questions that say COULD BE true, we only need one set of numbers to satisfy a condition to accept it, and for the questions that say MUST BE true, we only need one set of numbers that DO NOT Satisfy the condition to reject it. Correct?