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# Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0

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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
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1) x>5/6y
clearly x>y

2)xy<0

we don't know whether x is + or -

A for me.
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
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1. not sufficient
If x is 4 and y is 3, then we get 24 > 15. X is greater than Y. However, if we use x=3 and y= 3.1, we get 18>15.5. We still don't know if x is greater than y or not. We cannot assume x or y are integers.

2. not sufficient
This tells us that between x and y, exactly 1 is even and exactly 1 is odd. We do not know which one is negative and which is positive so this is not sufficient.

We are now down to answers C and E.

Both statements combined will give us the answer.
Since statement 1 is multiplying each x and y by positive numbers and since statement 2 tells us x or y is negative, x must be greater than y since no negative value of x would work in statement 1.

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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
2013gmat wrote:
1) x>5/6y
clearly x>y

2)xy<0

we don't know whether x is + or -

A for me.

statement 1 allows possibilities for x to be bigger than y and possibilities of y being bigger than x.
If x=4 and y=3, then x is larger since 6x>5y = 24>15 - x is larger
However, if x=3 and y=3.1, 6x>5y = 18>15.5 - y is larger

Since statement 2 lets us know one of the variables is negative, we can use statement 1 to know that one of the variables is positive and one is negative. Therefore, plugging in values to statement 1 we can find that x always has to be the positive variable and y always has to be negative.
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
Kudos
Question: X>Y ?

Statement 1:
6x>5y
6x-5y >0
x-5/6y >0
we can deduce from above inequality that
If y is positive then x would be greater than or equal to y.
If y is negative then x would be greater than y

not sufficient

Statement 2
xy<0 i.e. x and y are of opposite sign
So x<y or x>y
Not Sufficient

Combine Statement 1 & 2
from statement 2, x and y have to be of opposite sign.
In order to make x-5/6y>0 valid , x has to be +ve and y has to be -ve.
Hence X > Y
Sufficient

Regards,
Ammu
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
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Plugged in a few negative and positive numbers, and that sometimes gives us x<y (but st 1 holds true), and sometimes x>y (st 1 still holds true). so not sure

Down to BCE:
St 2 just tells us that x and y have opposite signs. But which is which?
Not sufficient!

C: we know they have opposite signs, then y must be negative (since a negative x would make statement 1 invalid, since both of the constants are positive)

C!

I would like to know a shortcut to getting somewhere with statement 1. All I do is try numbers, and that takes really long!
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
Kudos
Statement 1. 6x > 5y
If x =1, y=1 . The answer is no
If x =3 , y =2. The answer is yes
Insufficient

Statement 2. xy < 0

Either of x or y is negative. We don't know which of 2 numbers is negative. Insufficient

Combining 1 and 2.

Since either of the 2 numbers is negative and since 6x > 5y, we can say that x is positive and y is negative
Therefore x> y . The answer should be C
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
Kudos
2013gmat wrote:
1) x>5/6y
clearly x>y

2)xy<0

we don't know whether x is + or -

A for me.

It has to be C...
1. 6x>5y
=>x>5/6y
Not suffcient.

2. xy<0
so either x<0 or y<0
Not Sufficient

x>5/6y and xy<0
so x has to be positive. because 6x>5y
so we can conclude that x>y.

Hope this solution helps
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
1
Kudos
Bunuel wrote:

Tough and Tricky questions: Inequalities.

Is $$x \gt y$$?

(1) $$6x \gt 5y$$

(2) $$xy \lt 0$$

Kudos for a correct solution.

We need to know whether x>y

St 1 says 6x>5y...Dividing both side by 6 we get x>[$$fraction]5/6[/fraction]$$y

So x>0.833y so we can have x=y or x>y

2 says xy<0 now so both are of opposite sign but we don't know which one is greater

Combining we get x>5y/6...now if x+ and y- clearly x>y

if x- and y+ is impossible because x>0.833 y is not met consider y=100 x>83.33 but then x is positive which cannot be the case

ans c
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x>y ?

(1) 6x>5y
(2) xy<0

In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 equations. for 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), in xy<0, if x>0, y<0, which is yes. If x<0, y>0. In this case, 6x>5y is impossible. Thus, it is always yes and sufficient when x>0, y<0.

 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
ammuseeru wrote:
Question: X>Y ?

Statement 1:
6x>5y
6x-5y >0
x-5/6y >0
we can deduce from above inequality that
If y is positive then x would be greater than or equal to y.
If y is negative then x would be greater than y

not sufficient

Statement 2
xy<0 i.e. x and y are of opposite sign
So x<y or x>y
Not Sufficient

Combine Statement 1 & 2
from statement 2, x and y have to be of opposite sign.
In order to make x-5/6y>0 valid , x has to be +ve and y has to be -ve.
Hence X > Y
Sufficient

Regards,
Ammu

Name One condition when X and Y are equal as you put it ..
I think that is never possible..
Here is my approach
Statement one and 2 are insufficient as
Statement 1 => x>0.8Y => this does not imply x>y as we take any fraction between 0 and 1 to prove it.
next statement 2 tells us that they are opposite signs => insufficient.
Combining them=>
case first X>0 and Y<0 => sufficient (as a positive value is always greater than the negative value.
Next case 2 => Y>0 and X<0 => here as Y>0 and X>0.8Y => X must be positive too.
This is contradictory to our case .
Hence this case is never possible ..
So we conclude X>0 and Y<0 which confirms our result of X>Y always
Hence Sufficient
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
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Top Contributor
Bunuel wrote:

Tough and Tricky questions: Inequalities.

Is $$x \gt y$$?

(1) $$6x \gt 5y$$

(2) $$xy \lt 0$$

Kudos for a correct solution.

Target question: Is x > y?

Statement 1: 6x > 5y
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 1 and y = -1. Notice that 6x > 5y becomes 6 > -5, which is true. In this case, the answer to the target question is YES, x IS greater than y
Case b: x = 9 and y = 10. Notice that 6x > 5y becomes 54 > 50, which is true. In this case, the answer to the target question is NO, x is NOT greater than y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: xy < 0
Let' s TEST some more numbers.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = -1. In this case, the answer to the target question is YES, x IS greater than y
Case b: x = -1 and y = 1. In this case, the answer to the target question is NO, x is NOT greater than y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that xy < 0 (i.e., xy is NEGATIVE)
If the product xy is NEGATIVE, then there are two possible cases:

Case a: x is POSITIVE and y is NEGATIVE .
Notice that this case ALSO satisfies statement 1, which says 6x > 5y. When we replace x and y, we get: 6(POSITIVE) > 5(NEGATIVE), which is always true.
In this case, the answer to the target question is YES, x IS greater than y

Case b: x is NEGATIVE and y is POSITIVE.
Notice that this case DOES NOT satisfy statement 1, which says 6x > 5y. When we replace x and y, we get: 6(NEGATIVE) > 5(POSITIVE), which is NEVER true.
So, it cannot be the case that x is NEGATIVE and y is POSITIVE.
In other words, only case a can be true, which means the answer to the target question must be YES, x IS greater than y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
Bunuel wrote:

Tough and Tricky questions: Inequalities.

Is $$x \gt y$$?

(1) $$6x \gt 5y$$

(2) $$xy \lt 0$$

Kudos for a correct solution.

Statement 1:

$$6x > 5y$$

$$x = 2, y = 1$$ , we get $$6x > 5y$$, hence $$x > y$$
$$x = 1, y = 1$$ , we get $$6x > 5y$$, hence $$x = y$$
$$x = -1, y = -2$$ , we get $$6x > 5y$$, hence $$x > y$$
$$x = 2, y = -1$$ , we get $$6x > 5y$$, hence $$x > y$$

Statement 1 alone is Not Sufficient.

Statement 2:

$$xy \lt 0$$

we get $$x > 0$$ & $$y < 0$$, hence $$x > y$$
$$x < 0$$ & $$y > 0$$, hence $$x < y$$

Statement 2 alone is Not Sufficient.

Combing the 2 statements, we get $$x >0$$ & $$y < 0$$, to satisfy both the statements.

Hence $$x > y$$

Combining the 2 statement is Sufficient.

Thanks,
GyM
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
x>y?

1. X>0.8333Y not necessarily x>0.9y.. - inSuff
2. xy<0 possible x>0 & y<0 OR y>0 & x<0 - inSuff

1&2:
if(x>0) so sufficiant - x indeed >y
if(x<0) so sufficiant - x is not >y
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Re: Is x > y ? (1) 6x ＞ 5y (2) xy ＜ 0 [#permalink]
Bunuel wrote:
Is $$x \gt y$$?

(1) $$6x \gt 5y$$
(2) $$xy \lt 0$$

(1) $$6x \gt 5y$$ insufic

x…y…6x…5y:
1…1…6>5
2…1…12>5

(2) $$xy \lt 0$$ insufic

$$(x,y)=different.signs$$

(1 & 2) sufic

if x<0, y>0, then 6x<0 and 5y>0; but 6x>5y, so x>0 and y<0;

Ans (C)
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