12 Days of Christmas GMAT Competition - Day 10: If x > y > z, which : Problem Solving (PS)

Asked: If , which of the following may be true?

I.
If x = 3 > y=2 > z = 1; x^4 = 81 > y^4 = 16 > z^4 = 1
MAY BE TRUE

II.
If x=-1 > y=-2 > z>=-3; z^4 = 81 > y^4 = 16 > x^4 = 1
MAY BE TRUE

III.
Since x>z & x^4 > z^4; x & z are positive; x>y>z>0
But It is not possible that y^4 > x^4 when x > y > 0
NOT POSSIBLE

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

IMO C

1. For any positive no's x>y>z this will be true. eg. x=2, y=1,z=0 so b out

Try negative no's for 2 and 3

2. x= -0.1 y= -0.2 z= -0.3 this will be true so a d out

3. Not possible. Try different set of positive and negative no's including 0. e out

Ans: C

If x>y>z, which of the following may be true?

I.

II.

III.

Let's see... powers of 4. This means it will always be positive (even powers). Let's notice the combination of signs (+/-).

1. x(+) > y(+) > z(+)
-->

2. x(+) > y(-) > z(-)
--> (x = 2, y = -3, z = -10), or
--> (x = 12, y = -3, z = -10)

3. x(+) > y(+) > z(-)
--> (x = 4, y = 3, z = -1)
--> (x = 4, y = 3, z = -10)

4. x(-) > y(-) > z(-)
--> Order reverses.

I and II are possible. Answer.

If we assume all value is positive the x,y,z will satisfy the 1st equation

If we assume all value negative then x,y,z will satisfy the 2nd equation

Here is no group value of x,y,z which can satisfy 3rd equation

So Ans-C

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Bunuel wrote:12 Days of Christmas GMAT Competition with Lots of Fun

If , which of the following may be true?

I.
II.
III.

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

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for all positive integers
if x > y > z
then x^4 > y^4 > z^4

For all negative integers
if x > y > z
then z^4 > y^4 > x^4

For all numbers between 0 and 1 statement I will apply.

So answer is option C

given
If , which of the following may be true
I.
2>1>0 ; true
II.
(-8 )^4>(-6)^4>(-2)^4 ; true

III.
this relation cannot be true because for any -ve value of z and +ve value of x , z>x when raised to power of 4
also when +ve x>y so y^4>x^4 is not possible

OPTION C is correct I and II only

Bunuel wrote:12 Days of Christmas GMAT Competition with Lots of Fun

If , which of the following may be true?

I.
II.
III.

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

This question was provided by Experts'Global for the 12 Days of Christmas Competition Win $30,000 in prizes: Courses, Tests & more

 

Bunuel wrote:12 Days of Christmas GMAT Competition with Lots of Fun

If , which of the following may be true?

I.
II.
III.

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

This question was provided by Experts'Global for the 12 Days of Christmas Competition Win $30,000 in prizes: Courses, Tests & more

 

Solution:
Given, .

If x=4, y=3, z=2
or x=1/2, y=1/3, z=1/4
then . I is possible.

If x=-2, y=-3, z=-4
or x=-1/4, y=-1/3, z=-1/2
then . II is possible.

No other case is possible even after taking either of them as 0.

ANSWER C.

taking fraction, negative values and positive values
only 1 & 2 can be true

If x>y>zx>y>z, which of the following may be true?

I. x4>y4>z4x4>y4>z4
9^4>8^4>7^4 Correct
II. z4>y4>x4
(-10)^4>(-6)^4>5^4 Correct

IMO C

If all are +ve, say x=3, y=2, z=1 then I will hold true
If all are -ve, say x=-1, y=-2, z=-3 then II will hold true
III will not hold true in any scenario even when we consider any variable as 0 or fraction/decimal.

Hence, answer is C.

I. x4>y4>z4x4>y4>z4; x=3, y=2, z=1 yes
II. z4>y4>x4z4>y4>x4; x=-1/4, y=-1/3, z=-1/2
III. y4>x4>z4--Not possible because z^4 will always be more

Ans C

Answer is C
If x>y>z, which of the following may be true?

I. x^4>y^4>z^4 when x > y > z > 0. this is true
II. z^4>y^4>x^4 When 0 >x > y > z this is true
III. y^4>x^4>z^4 this cannot be true

If x>y>z, which of the following may be true?

I. x4>y4>z4
II. z4>y4>x4
III. y4>x4>z4

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

Given x>y>z and all options have even power 4 for all varaiables
which means sign will change if there is a negative number
If all x,y,z are positive 1 will hold true

if z is negative , sign will change for z and will retain it magnitude
for eg if x=2 y=1 z= -5 and given
x>y>z
z^4 =625 which is highest so 2 will hold true

If y is negative, it cannot have the highest magnitude
as x>y>z has always has to be true
for eg x=8 y=-5 and z=-6
so 3 cannot be true


So C is the correct answer

For any positive number greater than 1,
x^4>y^4>z^4 is true.

For example- z=2, y=3, x=4.

Now,
if z=-5, y=3 and x=4,
z^4>y^4>x^4 is true.

Can't think of any combination where statement III would be true. So, statement I and II are true.

Hence, IMO C.

3 > 2> 1

So, 81 > 16> 1

is true



-1> -2> -3

but, 1< 16< 81

II. is true

Option c

If x>y>z which of the following may be true?

I. x4>y4>z4x4>y4>z4
suppose x=3,y=2,z=1 hence 81>16>1 therefore I is true
II. z4>y4>x4
suppose x=3,y=-4,z=-5 hence 625>256>81 Therefore II is also true

hence only I and II are correct