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If x and y are positive and x + y = 20, is y > x?

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chetan2u
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If x and y are positive and x + y = 20, is y > x? [#permalink] New post   02 Jun 2018, 21:39
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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) \(4x^2+4xy+y^2=961\)


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D
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pushpitkc
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Re: If x and y are positive and x + y = 20, is y > x? [#permalink] New post   02 Jun 2018, 22:48
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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


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)
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Re: If x and y are positive and x + y = 20, is y > x? [#permalink] New post   02 Jun 2018, 23:09
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.
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Re: If x and y are positive and x + y = 20, is y > x? [#permalink] New post   03 Jun 2018, 00:21
I think D.

A) Since Y has 3 different Factors, Y has only two possible choices i.e 4(4,2,1) and 9( 9,3,1).
If we take y as 4 , x becomes 16. y<x i.e true
If we take y as 9 , x becomes 11. Y<x again i.e true . Thus A is sufficient.

B) 4x^2 + 4xy + y^2 = 961
=> 4x^2+2xy+2xy+y^2=961
=>2x(2x+y)+y(2x+y)=961
=>(y+2x)^2=31^2
=>y+2x=+-31
Y+2x cant be equal to -31 as both are positive.
=> y+2x=31 and from the question we know x+y=20
We can easily solve for both x and y => Sufficient

Hence : D
Kudos to you. A good question
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Re: If x and y are positive and x + y = 20, is y > x? [#permalink]
03 Jun 2018, 00:21
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