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In a rectangular coordinate system, the point (u, v) lies in the sec
[#permalink]
15 Sep 2026, 03:26
15 Sep 2026, 03:26
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Question Stats:
40% (02:53) correct
60% (01:54) wrong based on 5 sessions
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In a rectangular coordinate system, the point (u, v) lies in the second quadrant, and the point (x, y) lies in the fourth quadrant such that \( |u|>|x|\) and \(|v|<|y| \).
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x y-u v\) |
0 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
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: In a rectangular coordinate system, the point (u, v) lies in the sec
[#permalink]
01 Oct 2026, 05:01
01 Oct 2026, 05:01
Expert Reply
Let
\(u=-a, \quad v=b, \quad x=c, \quad y=-d\)
where a, b, c, d>0 .
The conditions give:
\(a>c, \quad b<d\)
Now,
\(x y=c(-d)=-c d \)
\(u v=(-a)(b)=-a b\)
So Quantity A is:
\(x y-u v=-c d-(-a b)=a b-c d\)
We know a>c and $b<d$, but the product a b could be greater than, less than, or equal to c d .
If a=10, c=1, b=1, d=2 , then a b=10, c d=2 , so Quantity A > 0.
If a=2, c=1, b=1, d=10 , then a b=2, c d=10 , so Quantity A < 0.
Thus the comparison cannot be determined.
\(u=-a, \quad v=b, \quad x=c, \quad y=-d\)
where a, b, c, d>0 .
The conditions give:
\(a>c, \quad b<d\)
Now,
\(x y=c(-d)=-c d \)
\(u v=(-a)(b)=-a b\)
So Quantity A is:
\(x y-u v=-c d-(-a b)=a b-c d\)
We know a>c and $b<d$, but the product a b could be greater than, less than, or equal to c d .
If a=10, c=1, b=1, d=2 , then a b=10, c d=2 , so Quantity A > 0.
If a=2, c=1, b=1, d=10 , then a b=2, c d=10 , so Quantity A < 0.
Thus the comparison cannot be determined.
gmatclubot
Re: In a rectangular coordinate system, the point (u, v) lies in the sec [#permalink]
01 Oct 2026, 05:01
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