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23 Dec 2022, 07:00
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C
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Difficulty:
25%
(medium)
Question Stats:
74% (01:40) correct
26%
(01:27)
wrong
based on 260
sessions
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12 Days of Christmas GMAT Competition with Lots of Fun
If \(x > y > z\), which of the following may be true?
I. \(x^4 > y^4 > z^4\)
II. \(z^4 > y^4 > x^4\)
III. \(y^4 > x^4 > z^4\)
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
If \(x > y > z\), which of the following may be true?
I. \(x^4 > y^4 > z^4\)
II. \(z^4 > y^4 > x^4\)
III. \(y^4 > x^4 > z^4\)
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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23 Dec 2022, 07:25
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Asked: If \(x > y > z\), which of the following may be true?
I. \(x^4 > y^4 > z^4\)
If x = 3 > y=2 > z = 1; x^4 = 81 > y^4 = 16 > z^4 = 1
MAY BE TRUE
II. \(z^4 > y^4 > x^4\)
If x=-1 > y=-2 > z>=-3; z^4 = 81 > y^4 = 16 > x^4 = 1
MAY BE TRUE
III. \(y^4 > x^4 > z^4\)
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
I. \(x^4 > y^4 > z^4\)
If x = 3 > y=2 > z = 1; x^4 = 81 > y^4 = 16 > z^4 = 1
MAY BE TRUE
II. \(z^4 > y^4 > x^4\)
If x=-1 > y=-2 > z>=-3; z^4 = 81 > y^4 = 16 > x^4 = 1
MAY BE TRUE
III. \(y^4 > x^4 > z^4\)
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
23 Dec 2022, 07:56
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If x>y>z, which of the following may be true?
I. \(x^4>y^4>z^4\)
II. \(z^4>y^4>x^4\)
III. \(y^4>x^4>z^4\)
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(+)
--> \(x^4>y^4>z^4\)
2. x(+) > y(-) > z(-)
--> \(z^4>y^4>x^4\) (x = 2, y = -3, z = -10), or
--> \(x^4>z^4>y^4\) (x = 12, y = -3, z = -10)
3. x(+) > y(+) > z(-)
--> \(x^4>y^4>z^4\) (x = 4, y = 3, z = -1)
--> \(z^4>x^4>y^4\) (x = 4, y = 3, z = -10)
4. x(-) > y(-) > z(-)
--> Order reverses. \(z^4>y^4>x^4\)
I and II are possible. Answer.
I. \(x^4>y^4>z^4\)
II. \(z^4>y^4>x^4\)
III. \(y^4>x^4>z^4\)
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(+)
--> \(x^4>y^4>z^4\)
2. x(+) > y(-) > z(-)
--> \(z^4>y^4>x^4\) (x = 2, y = -3, z = -10), or
--> \(x^4>z^4>y^4\) (x = 12, y = -3, z = -10)
3. x(+) > y(+) > z(-)
--> \(x^4>y^4>z^4\) (x = 4, y = 3, z = -1)
--> \(z^4>x^4>y^4\) (x = 4, y = 3, z = -10)
4. x(-) > y(-) > z(-)
--> Order reverses. \(z^4>y^4>x^4\)
I and II are possible. Answer.
