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x > y xy 0
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Updated on: 30 Oct 2024, 03:25
Updated on: 30 Oct 2024, 03:25
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Question Stats:
52% (01:52) correct
47% (02:15) wrong based on 106 sessions
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\(x > y\)
\(xy \neq 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(xy \neq 0\)
Quantity A |
Quantity B |
\(\dfrac{x^2}{(y+\frac{1}{y})}\) |
\(\dfrac{y^2 }{(x+\frac{1}{x})}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: x > y xy 0
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26 Jun 2018, 07:45
26 Jun 2018, 07:45
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sandy wrote:
\(x > y\)
\(xy \neq 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(xy \neq 0\)
Quantity A |
Quantity B |
\(x^2 \div (y+\frac{1}{y})\) |
\(y^2 \div (x+\frac{1}{x})\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Here since no value are give let us take
both x and y be positive integer such that x= 3 and y =2
From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{18}{5}\) = 3.6 (putting the value of x and y in the equation)
From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{12}{10}\)= 1.2 (putting the value of x and y in the equation)
From above we get statement 1 > Statement 2
Now if we consider negative integer such that x= -2 and y = -3 (since x>y)
From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{- 12}{10}\)= -1.2 (putting the value of x and y in the equation)
From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{- 18}{5}\) = -3.6(putting the value of x and y in the equation)
From above we have Statement 1 > Statement 2
But we have to look into every possibilities
Now if we consider negative integer such that x= 2 and y = -2 (since x>y)
From statement 1: \(x^2 \div (y+\frac{1}{y})\) = \(\frac{- 4}{5}\)= -0.8 (putting the value of x and y in the equation)
From Statement 2 : \(y^2 \div (x+\frac{1}{x})\) = \(\frac{4}{5}\) = 0.8(putting the value of x and y in the equation)
From above we have Statement 2 > Statement 1.
Hence the option is D
