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If x and y are integers and x + y = 21, is x > 19? (1) 2y > 19 + y

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If x and y are integers and x + y = 21, is x > 19? (1) 2y > 19 + y [#permalink]

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Kudos [?]: 139678 [1], given: 12794

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Re: If x and y are integers and x + y = 21, is x > 19? (1) 2y > 19 + y [#permalink]

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New post 15 Mar 2017, 04:53
x +y = 21

St 1: 2y>19+y or y>19. hence x<2. therefore xis not >19 . ANSWER

St 2: 3x/y >51x
or x/y>17x
or x(1-17y)/y>0
therefore x>0, 0<y<1/17. for this x>19

or x<0 , y<0 or y>1/17. for this we cannot say. INSUFFICIENT

Option A

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Re: If x and y are integers and x + y = 21, is x > 19? (1) 2y > 19 + y [#permalink]

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New post 13 Dec 2017, 07:24
Bunuel wrote:
If x and y are integers and x + y = 21, is x > 19?

(1) 2y > 19 + y
(2) 3x/y > 51x


Hi Bunuel,

Are you sure that answer is A? How do you prove that Statement 2 is insufficient?

My answer as follows:

2) 3x/y > 51x............x/y>17x

If y>0.... then x>17xy.....x-17xy>0....x(1-17y)>0

As y is positive... (1-17y) is negative........x must be negative in order to have x(1-17y)>0

x is NOT greater than 19

If y<0.....then x<17xy.....x-17xy<0.....x(1-17y)<0

As y is negative... (1-17y) is positive........x must be negative in order to have x(1-17y)<0

x is NOT greater than 19

So it should be Sufficient .

Another proof for statement 2:

Statement 2:
3x/y > 51x
x/y > 17x

In the resulting inequality, x can be positive only if y is also positive.
There is ONLY ONE SOLUTION for x+y = 21 such that x>19 and x and y are both positive:
x=20, y=1.
If this solution does not satisfy x/y > 17x, then it is not possible that x>19.
Plugging x=20 and y=1 into x/y > 17x, we get:
20/1 > 17*20
20 > 340.
Doesn't work.
Thus, it is not possible that x>19, implying that the answer to the question stem is NO.
SUFFICIENT.

Thanks

Kudos [?]: 324 [0], given: 170

Re: If x and y are integers and x + y = 21, is x > 19? (1) 2y > 19 + y   [#permalink] 13 Dec 2017, 07:24
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