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If y < x, is x^2 < y^2 ? (1) y > 0 (2) x > 0 01 Nov 2018, 03:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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35% (medium)

Question Stats:

71% (01:31) correct 29% (01:28) wrong based on 1215 sessions

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If \(y < x\), is \(x^2 < y^2\)?

(1) y > 0
(2) x > 0
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A
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If y < x, is x^2 < y^2 ? (1) y > 0 (2) x > 0 01 Nov 2018, 05:02
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If \(y < x\), is \(x^2 < y^2\)?

(1) y > 0
(2) x > 0
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if y > 0

then x > y > 0

then it is not true that x^2 < y^2 sufficient.

x > 0 then we don't know about y

if x = 2 and y = -3

then it is not true.

if x = 2 and y = -1 then is it true.

Not sufficient.

Answer choice A
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If y < x, is x^2 < y^2 ? (1) y > 0 (2) x > 0 19 Apr 2020, 04:15
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If \(y < x\), is \(x^2 < y^2\)?

(1) y > 0
(2) x > 0
Show more

MODIFY the question stem..


is \(x^2 < y^2\)? Means is |x|<|y|? Or is \(x^2-y^2<0.....(x-y)(x+y)<0\)
Now it is given that \(y<x......x-y>0\)
So 'is \((x-y)(x+y)<0\) ?' becomes 'is x+y<0?'.
(a) if both x and y are positive ..NO
(b) if both x and y are negative...YES
(c) If x>0 and y<0, we will require to know the exact value of x and y to answer

(1) y > 0
Now x>y, so x>y>0..
Case (a) above....NO...Suff

(2) x > 0
So both case (a) and case (c) possible
Insuff

A
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If y < x, is x^2 < y^2 ? (1) y > 0 (2) x > 0 28 Jul 2021, 07:32
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chetan2u
carcass
If \(y < x\), is \(x^2 < y^2\)?

(1) y > 0
(2) x > 0

MODIFY the question stem..


is \(x^2 < y^2\)? Means is |x|<|y|? Or is \(x^2-y^2<0.....(x-y)(x+y)<0\)
Now it is given that \(y<x......x-y>0\)
So 'is \((x-y)(x+y)<0\) ?' becomes 'is x+y<0?'.
(a) if both x and y are positive ..NO
(b) if both x and y are negative...YES
(c) If x>0 and y<0, we will require to know the exact value of x and y to answer

(1) y > 0
Now x>y, so x>y>0..
Case (a) above....NO...Suff

(2) x > 0
So both case (a) and case (c) possible
Insuff

A
Show more

If x=1/2 and y=1/4 then statement 1 is not sufficient. No where the question mentions that x and y are integers.
Please help!
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If y < x, is x^2 < y^2 ? (1) y > 0 (2) x > 0 28 Jul 2021, 07:49
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PrernaTalreja
chetan2u
carcass
If \(y < x\), is \(x^2 < y^2\)?

(1) y > 0
(2) x > 0

MODIFY the question stem..


is \(x^2 < y^2\)? Means is |x|<|y|? Or is \(x^2-y^2<0.....(x-y)(x+y)<0\)
Now it is given that \(y<x......x-y>0\)
So 'is \((x-y)(x+y)<0\) ?' becomes 'is x+y<0?'.
(a) if both x and y are positive ..NO
(b) if both x and y are negative...YES
(c) If x>0 and y<0, we will require to know the exact value of x and y to answer

(1) y > 0
Now x>y, so x>y>0..
Case (a) above....NO...Suff

(2) x > 0
So both case (a) and case (c) possible
Insuff

A

If x=1/2 and y=1/4 then statement 1 is not sufficient. No where the question mentions that x and y are integers.
Please help!
Show more


statement I gives answer as NO always, so statement I is sufficient.
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If y < x, is x^2 < y^2 ? (1) y > 0 (2) x > 0 22 Jan 2026, 04:34
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Your Example: "If x = 1/2 and y = 1/4 then statement 1 is not sufficient."

The Critical Error: Your example doesn't satisfy the given constraint y < xy < x!

Let's check with x = 1/2, y = 1/4:
- Is y < xy? Is 1/4 < (1/2)(1/4) = 1/8?
- NO! 1/4 > 1/8

Your example violates the given constraint, so it's not a valid test case!

The Key Insight: When both x and y are positive, the constraint y < xy < x forces:
- x > 1 (from y < xy)
- 0 < y < 1 (from xy < x)

Since no positive values with x < 1 can satisfy the constraint, your example fails.

PrernaTalreja


If x=1/2 and y=1/4 then statement 1 is not sufficient. No where the question mentions that x and y are integers.
Please help!
Show more
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