chetan2u wrote:
If x and y are positive and x+y=20, is y>x?
(1) y is an integer having three different factors.
(2) \(4x^2+4xy+y^2=961\)
New tricky question
Self made
OA: D
Given: x and y are positive and x+y=20
(1) y is an integer having three different factors.
Square of prime number has 3 different factors.
so y can be 4 (factors: 1,2,4) or 9 (factors: 1,3,9).
in both cases , x>y
Answer for question
is y>x? is No.
So Statement 1 alone is sufficient.
(2) \(4x^2+4xy+y^2=961\)
\((2x)^2+2*2x*y+y^2=961\)
\((2x+y)^2-(31)^2=0\)
\((2x+y-31)(2x+y+31)=0\)
\(2x+y+31\neq{0}\) as x and y are positive.
\(2x+y-31 =0\)
Using \(x+y=20\) from question stem, we get \(x =11\) and \(y =9\)
Answer for question
is y>x? is No.
So Statement 2 alone is sufficient.