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Inequality DS question

by Mo2men » Wed Mar 15, 2017 9:05 am

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If x and y are integers and x + y = 21, is x > 19?

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

OA: A

Can you expert comment on Algebraic solution below?

y & x are integers but can take any POSITIVE or NEGATIVE signs such

42-21=21 (one is positive and other is negative)

5+16=21 (both are positive)


1) 2y > 19 + y ....... y> 19....y=20...x=1

Sufficient

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 .

Where did I go wrong? any advice will be helpful.

Thanks
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Source: — Data Sufficiency |
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by DavidG@VeritasPrep » Wed Mar 15, 2017 9:43 am

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Mo2men wrote:If x and y are integers and x + y = 21, is x > 19?

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

OA: A

Can you expert comment on Algebraic solution below?

y & x are integers but can take any POSITIVE or NEGATIVE signs such

42-21=21 (one is positive and other is negative)

5+16=21 (both are positive)


1) 2y > 19 + y ....... y> 19....y=20...x=1

Sufficient

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 .

Where did I go wrong? any advice will be helpful.

Thanks
Your reasoning looks good to me. If you transcribed the problem properly, the OA should be D.

What's the source?
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by Mo2men » Wed Mar 15, 2017 9:45 am

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DavidG@VeritasPrep wrote:
Mo2men wrote:If x and y are integers and x + y = 21, is x > 19?

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

OA: A

Can you expert comment on Algebraic solution below?

y & x are integers but can take any POSITIVE or NEGATIVE signs such

42-21=21 (one is positive and other is negative)

5+16=21 (both are positive)


1) 2y > 19 + y ....... y> 19....y=20...x=1

Sufficient

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 .

Where did I go wrong? any advice will be helpful.

Thanks
Your reasoning looks good to me. If you transcribed the problem properly, the OA should be D.

What's the source?
Thanks David
Source below:

https://gmatclub.com/forum/if-x-and-y-a ... l#p1819700
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by DavidG@VeritasPrep » Wed Mar 15, 2017 9:46 am

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1) 2y > 19 + y ....... y> 19....y=20...x=1
Small note. You're right that statement 1 alone is sufficient, but we don't know that x = 1. We just know that 20 is the minimum value for y and thus 1 is the maximum value for x. y could be 21 and x 0, etc. But clearly x is not greater than 19.
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by DavidG@VeritasPrep » Wed Mar 15, 2017 9:50 am

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Mo2men wrote:
DavidG@VeritasPrep wrote:
Mo2men wrote:If x and y are integers and x + y = 21, is x > 19?

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

OA: A

Can you expert comment on Algebraic solution below?

y & x are integers but can take any POSITIVE or NEGATIVE signs such

42-21=21 (one is positive and other is negative)

5+16=21 (both are positive)


1) 2y > 19 + y ....... y> 19....y=20...x=1

Sufficient

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 .

Where did I go wrong? any advice will be helpful.

Thanks
Your reasoning looks good to me. If you transcribed the problem properly, the OA should be D.

What's the source?
Thanks David
Source below:

https://gmatclub.com/forum/if-x-and-y-a ... l#p1819700
Gotcha. Note that the expert who posted the question hasn't posted the OA yet. Whoever responded made an error.
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by GMATGuruNY » Wed Mar 15, 2017 10:35 am

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Mo2men wrote:If x and y are integers and x + y = 21, is x > 19?

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

Statement 1:
2y > 19 + y
y > 19.
Since y>19, it is not possible that x>19, given that x+y = 21.
Thus, the answer to the question stem is NO.
SUFFICIENT.

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.

The correct answer is D.
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by Jeff@TargetTestPrep » Wed Dec 13, 2017 7:03 am

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Mo2men wrote:If x and y are integers and x + y = 21, is x > 19?

(1) 2y > 19 + y
(2) 3x/y > 51x
We are given that x and y are integers and that x + y = 21. We must determine whether x > 19.

Statement One Alone:

2y > 19 + y

Simplifying the inequality in statement one, we have:

2y > 19 + y

y > 19

Since y is greater than 19, and since x and y add up to 21, x must be less than 19 (in fact, x must be less than 2). Statement one alone is sufficient to answer the question.

Statement Two Alone:

3x/y > 51x

In order for x > 19, x has to be at least positive. If x is positive, then we can divide both sides of the inequality by 3x and obtain:

1/y > 17

0 < y < 1/17

However, y is an integer and there is no integer between 0 and 1/17. Thus, x can't be positive. If x is not positive, then it must be either 0 or negative. It turns out x can't be 0, either, since both sides of the inequality would then be 0, but 0 is not greater than 0. Therefore, x must be negative (for example, if x = -1, then y = 22, and 3(-1)/22 is indeed greater than 51(-1). Since we've determined that x is negative, x is not greater than 19. Statement two alone is sufficient to answer the question.

Answer: D

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