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x>0 and y<0
[#permalink]
27 Aug 2026, 13:09
27 Aug 2026, 13:09
Expert Reply
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Question Stats:
50% (02:06) correct
50% (01:29) wrong based on 4 sessions
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\( x>0 \) and \(y<0 \) such that \(x+y \neq 0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\( (x+y)^3 - (x^3+y^3)\) |
\(x+y\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Re: x>0 and y<0
[#permalink]
30 Aug 2026, 15:49
30 Aug 2026, 15:49
1
Quantity A
( x + y )^3 - ( x^3 + y^3)
(x + y)( x^2 + 2xy + y^2) - ( x + y ) (x^2 - xy + y^2 )
Factorise
( x + y )( x^2 + 2xy + y^2 - x^2 + xy - y^2)
( x + y )3xy
If y has a greater numerical value than x
Quantity A will be positive and Quantity B negative
If x has a greater numerical value
A will be negative and B positive so answer is D
( x + y )^3 - ( x^3 + y^3)
(x + y)( x^2 + 2xy + y^2) - ( x + y ) (x^2 - xy + y^2 )
Factorise
( x + y )( x^2 + 2xy + y^2 - x^2 + xy - y^2)
( x + y )3xy
If y has a greater numerical value than x
Quantity A will be positive and Quantity B negative
If x has a greater numerical value
A will be negative and B positive so answer is D
gmatclubot
Re: x>0 and y<0 [#permalink]
30 Aug 2026, 15:49
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