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02 Jun 2018, 21:39
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:43) correct
68%
(02:38)
wrong
based on 104
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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
(1) y is an integer having three different factors.
(2) \(4x^2+4xy+y^2=961\)
New tricky question
Self made
02 Jun 2018, 22:48
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chetan2u
From the question stem, we are given that x+y = 20 and both x and y are positive
1. When y is an integer with 3 different factors - the possibilities are 4(1,2,4) and 9(1,3,9)
When x = 16, y = 4 and when x = 11, y = 9. y > x is FALSE in both the cases (Sufficient)
2. This equation can be simplified as follows : (2x + y)^2 = 31^2 -> 2x + y = 31(as x,y are positive)
Solving for x and y, we will get x = 11 and y = 9. Again, y > x is FALSE (Sufficient - Option D)
02 Jun 2018, 23:09
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chetan2u
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.
