Is x > y ?

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Is x > y ?

by Mo2men » Tue Jul 16, 2019 3:44 am

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Is x > y ?

1) (2/5) x > (3/8) y

2) (3/7) x > (15/29) y

OA: C
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My approach

1) (2/5) x > (3/8) y
Simplifying

16x > 15y

Let x =y=1.................16>15.........answer is No

Let x =2 &y=1.................32>15.........answer is Yes

Insufficient

2) (3/7) x > (15/29) y
Simplifying

29x > 35y

Let x =y=-1.................-29>-35.........answer is No

Let x =2 &y=1.................32>15.........answer is Yes

Insufficient

My problem starts here
combine 1 & 2

As the sign is same direction we can add both in inequalities

45x > 50 y......9x > 10y

Let x =y=-1.................-9>-10.........answer is No

Let x =2 &y=1.................18>10.........answer is Yes

Answer is E.

However, the OA is C

1- Why combing both inequalities did not help here? While it helps in other situation when combing both statements like the following question:
Is xy >0
1) 3x -y >0
2) 3x - 2y < 0
After combing subtracting 1 from 2........we can safely say that y > 0 and then from 1, we can deduce e that x >0.

2- What is the best when combining such inequalities for question at hand?

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by Ian Stewart » Tue Jul 16, 2019 6:54 am

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It's possible to answer a question like this by adding inequalities, but I think it's a difficult way to solve this type of problem. Neither Statement is sufficient alone (though you need to consider a negative value for x to see that Statement 2 is not sufficient). Using both Statements, we know:

16x > 15y
29x > 35y

If we just add these inequalities immediately, we find 45x > 50y, or 9x > 10y. That is true here, but the problem is it's just not very useful - it doesn't answer our question. But if we multiply the first inequality by 6 on both sides first, and then add the inequalities:

96x > 90y
29x > 35y

we find 125x > 125y, and x > y, which is what we wanted to know. I chose to multiply by 6 because the numbers 29 and 35 differ by 6, but the numbers 16 and 15 differ by only 1. We need to make that difference 6 so we'll get the same numbers in front of x and y after adding the inequalities.

That's not a method most test takers will find intuitive, I don't think (I just wanted to show that adding inequalities is still correct in this situation). Instead, you might notice that, using both Statements, we have

x > (15/16)y
x > (35/29)y

If y > 0, then it is always true that

(15/16)y < y < (35/29)y

so if y > 0, then because x > (35/29)y, it must be true that x > y. And if y < 0, it is always true that

(35/29)y < y < (15/16)y.

so if y < 0, then because x > (15/16)y, it must be true that x > y.

So in either case, x > y is true (and x > y is also true if y = 0), so the answer is C.
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x > y

by GMATGuruNY » Thu Jul 18, 2019 3:46 am

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Mo2men wrote:Is x > y ?

1) (2/5) x > (3/8) y

2) (3/7) x > (15/29) y
I received a PM requesting that I post a response.
The issue at hand is how to combine the two inequalities.
Visually-oriented test-takers might find it helpful to plot the values on a number line.

Statement 1:
(2/5)x > (3/8)y
x > (5/2)(3/8)y
x > (15/16)y

Statement 2:
(3/7)x > (15/29)y
x > (7/3)(15/29)y
x > (35/29)y

If y > 0, then plotting y and the blue values above yields the following number line:
0..........(15/16)y..........y..........(35/29)y..........
Statement 2 indicates that x > (35/29)y, with the result that x is to the right of y on the number line.
Thus, x > y.

If y < 0, then plotting y and the blue values above yields the following number line:
...........(35/29)y..........y..........(15/16)y..........0
Statement 1 indicates that x > (15/16)y, with the result that x is to the right of y on the number line.
Thus, x > y.

If y = 0, then each statement implies that x > 0, with the result that x > y.

Since x > y in all 3 cases, the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.
Last edited by GMATGuruNY on Thu Jul 18, 2019 4:03 am, edited 1 time in total.
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by Mo2men » Thu Jul 18, 2019 3:59 am

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Thanks GMATGuru and Ian,

I still do not understand why combing both inequalities did not produce one conclusive solution.

Should do not any pair verifying the combine inequality 9x > 10y also apply directly to each inequality separately ? Take for example x=y=-1 it makes 9x> 10x valid but it does not apply for the statement 1...why? The combined inequality must work for both as i understand.

Where did I go wrong in my understanding?

Thanks in advance

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x > y

by GMATGuruNY » Thu Jul 18, 2019 4:51 am

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If A, then B does not imply If B, then A.
True statement:
If John is in Times Square, then John is in New York.
The following statement is NOT necessarily true.
If John is in New York, then John is in Times Square.
Mo2men wrote:Should do not any pair verifying the combine inequality 9x > 10y also apply directly to each inequality separately ? Take for example x=y=-1 it makes 9x> 10x valid but it does not apply for the statement 1...why? The combined inequality must work for both as i understand.
Is x > y?

Statement 1: 5x > 3y
Statement 2: 4x > 7y

Each statement alone is clearly INSUFFICIENT.

Adding the two inequalities, we get the following true statement:
If 5x > 3y and 4x > 7y, then 9x > 10y.
However, the following statement is NOT necessarily true.
If 9x > 10y, then 5x > 3y and 4x > 7y.
As a result, some values that satisfy 9x > 10y will not satisfy the individual statements in red.
While x=y=-1 satisfies 9x > 10y, this case is not valid because it does not satisfy the condition in Statement 1 that 5x > 3y.
For this reason, it is not helpful to add the two inequalities as written.

We can prove that x > y as follows:

5x > 3y --> the distance between the coefficients = 5-3 = 2
4x > 7y --> the distance between the coefficients = 7-4 = 3
If we multiply the first inequality by 3 and the second inequality by 2, the distance between the coefficients will be the same in each case:
5x > 3y --> 15x > 9y
4x > 7y --> 8x > 14y
In each of the resulting inequalities, the distance between the coefficients is 6:
15-9 = 6
14-8 = 6

Why is this helpful?
Because adding together the resulting inequalities will yield the same coefficient for x and y:
15x + 8x > 9y + 14y
23x > 23y
x > y
Thus, the answer to the question stem is YES.
SUFFICIENT.

The correct answer is C.
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