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MGMAT Question

by voodoo_child » Fri Feb 24, 2012 3:23 am
If X!= 0 ; Is ((x^2 + 1)/ x ) > y?

a)x=y
b)y>0

OA = C
Obviously, b) is not sufficient.
However, I believe that the answer must be A.
If x=y and x>0 => x^2 + 1 > x^2 => 1>0 => TRUE
However, if x=y and x<0 => x^2 + 1 < x^2 => 1 < 0 => NOT POSSIBLE.

Therefore, x=y and x<0 is not possible. Hence, the answer is A.

Any thoughts?

Thanks
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Source: — Data Sufficiency |
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by rijul007 » Fri Feb 24, 2012 6:35 am
voodoo_child wrote:If X!= 0 ; Is ((x^2 + 1)/ x ) > y?

a)x=y
b)y>0

OA = C
Obviously, b) is not sufficient.
However, I believe that the answer must be A.
If x=y and x>0 => x^2 + 1 > x^2 => 1>0 => TRUE
However, if x=y and x<0 => x^2 + 1 < x^2 => 1 < 0 => NOT POSSIBLE.

Therefore, x=y and x<0 is not possible. Hence, the answer is A.

Any thoughts?

Thanks
Check out the highlighted part
You cross multiplied x, which is supposed to be a negative number, at both sides without changing the inequality sign.

1 < 2
If we multiply both sides by -1, we will have to reverse the inequality sign
-1 > -2



Solution to the problem

Statement 1
x = y

(x^2 +1)/x > x
x + 1/x > x

Case 1: x>0
x=1/2 --> 1/2+2 > 1/2
x=2 --> 2+1/2 > 2
TRUE

Case 2: 0<x
x= -1 --> -1-1 > -1 (Not true)
x= -2 --> -2-(1/2) > -2 (Not true)


Insufficient


Statement 2
y>0

Clearly not Sufficient


Combining the two statements
x=y>0

x+1/x > x
True


Option C
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by voodoo_child » Fri Feb 24, 2012 7:10 am
rijul007 wrote: Case 2: 0<x
x= -1 --> -1-1 > -1 (Not true)
x= -2 --> -2-(1/2) > -2 (Not true)


Insufficient
Nope. The fact that the equation doesn't hold good for X<0 implies that X<0 is not possible. You could draw a graph and check the values of X for which the equation will hold good.

I think that I changed the sign correctly. See this :

((x^2 + 1)/ x ) > y becomes x^2 + 1 < x^2. I did change the sign.

I am still not convinced.
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by rijul007 » Fri Feb 24, 2012 8:13 am
voodoo_child wrote:
rijul007 wrote: Case 2: 0<x
x= -1 --> -1-1 > -1 (Not true)
x= -2 --> -2-(1/2) > -2 (Not true)


Insufficient
Nope. The fact that the equation doesn't hold good for X<0 implies that X<0 is not possible. You could draw a graph and check the values of X for which the equation will hold good.

I think that I changed the sign correctly. See this :

((x^2 + 1)/ x ) > y becomes x^2 + 1 < x^2. I did change the sign.

I am still not convinced.
Oh yeah, you did change the sign..Sorry about that.

I see where you are going wrong
In statement 1, is it anywhere explicitely mentioned that x<0 is not possible?

It just says that x=y
the value of x could be anything other than 0.
You just need to check whether for all those possible values of x, the inequality stands correct or not.
If both the cases, x<0 and x>0, yielded that the statement is false or that the statement is true,
then the statement would have been sufficient
But in this ques, x>0 suggests that the inequality is true and x<0 suggests that the inequality is not true. Hence the statement is not sufficient.
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by GMATGuruNY » Fri Feb 24, 2012 8:28 am
voodoo_child wrote:If X!= 0 ; Is ((x^2 + 1)/ x ) > y?

a)x=y
b)y>0

OA = C
Obviously, b) is not sufficient.
However, I believe that the answer must be A.
If x=y and x>0 => x^2 + 1 > x^2 => 1>0 => TRUE
However, if x=y and x<0 => x^2 + 1 < x^2 => 1 < 0 => NOT POSSIBLE.

Therefore, x=y and x<0 is not possible. Hence, the answer is A.

Any thoughts?

Thanks
You've reversed the process.
We don't use the question stem to determine whether the statement is possible.
We use the statement to determine whether the QUESTION STEM is possible.

Here's what you've proved algebraically:
If x=y and x<0, then the relationship in the question stem is not possible.
This simply means that, if x=y and x<0, then the answer to the question stem -- Is ((x²+ 1)/ x ) > y? -- is NO.
In short:
If x=y and x<0, is ((x^2 + 1)/ x ) > y?
NO, it is NOT POSSIBLE.

The situation is easier to see if we plug in numbers.

If x=y=-1, is ((x² + 1)/ x ) > y?
((-1)² + 1)/(-1) > -1
-2 > -1.
NO. (In other words, NOT POSSIBLE.)

If x=y=1, is ((x² + 1)/ x ) > y?
(1²+1)/1 > 1
2 > 1.
YES.

Since in the first case the answer is NO, and in the second case the answer is YES, INSUFFICIENT.
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by voodoo_child » Sat Feb 25, 2012 4:38 am
Thanks Mitch and rijul007. It seems to me that I need to correct my basics then. I need your help with similar but confusing MGMAT example.

Is X>0?

a) |x+3| = 4x-3
b) |x+1| = 2x-1

OA = D

I will only solve a) because b) can be easily solved similar to a)

a) =>

x+3=4x-3 (x+3 >=0) .........Eq(i)

AND

-(x+3)=4x-3 (x+3 < 0) .........Eq(ii)

Solving Eq (i), x=2 for x>= -3 (FINE)

However, solving Eq(ii), x=0 ; for x<-3 (NOT POSSIBLE)

If we go by the logic that you mentioned that if any of the solution is "NOT POSSIBLE", the answer is "Not Sufficient."

The answer here is sufficient. Can you please explain the gap?

Appreciate your help.

Thanks
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by rijul007 » Sat Feb 25, 2012 6:29 am
voodoo_child wrote: If we go by the logic that you mentioned that if any of the solution is "NOT POSSIBLE", the answer is "Not Sufficient."
Thats wrong.
It seems you have a doubt with Data suficiency questions in general.
I suugest you take a look at the Data sufficiency videos here
https://www.gmatprepnow.com/module/gmat-data-sufficiency#
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by rijul007 » Sat Feb 25, 2012 6:36 am
voodoo_child wrote: Is X>0?

a) |x+3| = 4x-3
b) |x+1| = 2x-1
Statement 1:
a) |x+3| = 4x-3

case 1: x>-3
x+3 = 4x-3
3x = 6
x = 2

Case 2: x<-3
-x-3 = 4x-3
5x = 0
x = 0
0 is not less than -3
Hence, x is not equal to 0

x = 2

Is x>0?
Yes

Sufficient


Statement 2
Case 1: x>-1
x+1 = 2x-1
x = 2

Case 2: x<-1
-x-1 = 2x-1
3x = 0
x=0 is not less than -1
x is not equal to 0

x=2
Is x>0?
Yes

Sufficient


Option D
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by voodoo_child » Sun Feb 26, 2012 7:51 pm
Rijul,
Thanks for your reply. I did go through the video set. I still don't understand the reason why the absolute value problem is sufficient. We were able to find values of x for which the equation |x+3| = 4x-3 doesn't hold good. Hence, by the logic used in the first example, the answer must "not sufficient".

Why is it that the similar concept is "Not Sufficient" in the original problem?

Any thoughts?
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by MBAD » Thu Mar 01, 2012 6:52 pm
Going back to this question, I still think the answer would be A. Can someone please explain the error in my thinking.

Question: If x is not equal to 0, is (x^2+1)/x>y
1. x=y
2. y>0

Taking x=y couldn't you substitute in the equation to get...
(y^2+1)/y>y
...now exploring both possibilities of y being positive or negative
If Y is neg... y^2+1<y^2 (multiplying by negative value y flipped sign)
If Y is pos... y^2+1>y^2
Looking at the results above, there is no instance in which y^2+1 could be less than y^2, so y must be positive

Therefore we could see y^2+1>y^2 or (y^2+1)/y>y or (x^2+1)/x>y

1. Sufficient....A (no?)

Thanks
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by krusta80 » Thu Mar 01, 2012 8:05 pm
MBAD wrote:Going back to this question, I still think the answer would be A. Can someone please explain the error in my thinking.

Question: If x is not equal to 0, is (x^2+1)/x>y
1. x=y
2. y>0

Taking x=y couldn't you substitute in the equation to get...
(y^2+1)/y>y
...now exploring both possibilities of y being positive or negative
If Y is neg... y^2+1<y^2 (multiplying by negative value y flipped sign)
If Y is pos... y^2+1>y^2
Looking at the results above, there is no instance in which y^2+1 could be less than y^2, so y must be positive

Therefore we could see y^2+1>y^2 or (y^2+1)/y>y or (x^2+1)/x>y

1. Sufficient....A (no?)

Thanks
The question in part (1) is whether the given formula holds true, given x = y. For all negative values of x, the equation is NEVER true. BUT, for all positive values of x, the equation is ALWAYS true. Therefore, both possibilities exist when given x = y, which makes it INSUFFICIENT.
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by krusta80 » Thu Mar 01, 2012 8:08 pm
voodoo_child wrote:Rijul,
Thanks for your reply. I did go through the video set. I still don't understand the reason why the absolute value problem is sufficient. We were able to find values of x for which the equation |x+3| = 4x-3 doesn't hold good. Hence, by the logic used in the first example, the answer must "not sufficient".

Why is it that the similar concept is "Not Sufficient" in the original problem?

Any thoughts?
The difference is that the equation that does not hold in the second example is one of the two parts given to us and is therefore considered FACT. Therefore, we can only consider values of x that work for the FACTUAL equation.

In the original question, it is our job to determine whether the QUESTION is always correct or not. Therefore we need to consider ALL possible values (whether numerical or true/false) generated by the allowable input parameters, which are established by parts (1) and/or (2).
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by GMATGuruNY » Fri Mar 02, 2012 6:27 am
MBAD wrote:Going back to this question, I still think the answer would be A. Can someone please explain the error in my thinking.

Question: If x is not equal to 0, is (x^2+1)/x>y
1. x=y
2. y>0

Taking x=y couldn't you substitute in the equation to get...
(y^2+1)/y>y
...now exploring both possibilities of y being positive or negative
If Y is neg... y^2+1<y^2 (multiplying by negative value y flipped sign)
If Y is pos... y^2+1>y^2
Looking at the results above, there is no instance in which y^2+1 could be less than y^2, so y must be positive
I received a PM asking me to comment.

The error lies in the red portion above: you're using the QUESTION STEM to determine whether the STATEMENT is possible.
If (x²+1)/x>y, is it possible in statement 1 that y<0?
The process is just the opposite.

We need to use the STATEMENT to determine the answer to the QUESTION STEM.
Remember: the question stem is just that -- A QUESTION.
When we substitute into the question stem, we must INCLUDE THE QUESTION MARK.

Statement 1: x=y.
If x=y, is (x²+1)/x>y?
(y²+1)/y > y?

Case 1: y>0, in which case we multiply each side by y without flipping the inequality.
y²+1 > y²?
1 > 0?
YES.

Case 2: y<0, in which case we multiply each side by y while flipping the inequality.
y²+1 < y²?
1 < 0?
NO.

Thus, the answer to the question stem can be YES or NO, indicating that statement 1 is INSUFFICIENT.
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by MBAD » Fri Mar 02, 2012 9:12 am
Key misunderstanding, thanks for the detailed clarification!
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