sandy wrote:
\(x > y\)
\(xy \neq 0\)
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
_________________
If you found this post useful, please let me know by pressing the Kudos ButtonRules for PostingGot 20 Kudos? You can get Free GRE Prep Club TestsGRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months
GRE Prep Club tests
23 Jun 2018, 12:33
44% (01:43) correct
close
