Is y < 2x?

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Is y < 2x?

by melguy » Sat Jun 22, 2013 1:06 am
Please help with the question. Thanks

I chose A as the answer considering we do not know if y and x are negative or positive for statement 2. But OA does not matches. I did it as per below to match my ans with OA

1. multiply both sides by 3
2. y-2x < 0
3. i.e. y<2x

Am I correct in doing the problem this way?

OA is D
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by GMATGuruNY » Sat Jun 22, 2013 3:00 am
melguy wrote:Please help with the question. Thanks

I chose A as the answer considering we do not know if y and x are negative or positive for statement 2. But OA does not matches. I did it as per below to match my ans with OA

1. multiply both sides by 3
2. y-2x < 0
3. i.e. y<2x

Am I correct in doing the problem this way?

OA is D
Your approach to statement 2 is perfect.

Is y < 2x?

Statement 1: y/4 < x/2.
Multiplying each side by 4, we get:
4 * (y/4) < 4 * (x/2)
y < 2x.
SUFFICIENT.

Statement 2: (y-2x)/3 < 0
Multiplying each side by 3, we get:
3 * (y-2x)/3 < 3 * 0
y-2x < 0
y < 2x.
SUFFICIENT.

The correct answer is D.
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by melguy » Sat Jun 22, 2013 4:26 am
GMATGuruNY wrote:
Statement 2: (y-2x)/3 < 0
Multiplying each side by 3, we get:
3 * (y-2x)/3 < 3 * 0
y-2x < 0
y < 2x.
SUFFICIENT.
Please correct me if I am wrong. I remember studying that we need to be careful of signs in inequalities. So how can we consider the possibility of variables being negative.

y < 2x or y > -2x in this case. As per OA we just assumed both variables are positive. If not in this case, plz let me know what are the scenarios in which we need to be careful about negatives and signs.
Thanks for your help.

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by GMATGuruNY » Sat Jun 22, 2013 5:58 am
melguy wrote: Please correct me if I am wrong. I remember studying that we need to be careful of signs in inequalities. So how can we consider the possibility of variables being negative.

y < 2x or y > -2x in this case. As per OA we just assumed both variables are positive. If not in this case, plz let me know what are the scenarios in which we need to be careful about negatives and signs.
Thanks for your help.
If the process of simplifying an inequality requires that we multiply or divide each side by a VARIABLE WHOSE SIGN IS UNKNOWN, then we have to consider what will happen if the unknown variable is positive, 0, or negative.
To illustrate:

Is y > 2x?

Statement 1: (y-x)/x > 1
Case 1: x>0
Here, x is positive, so the direction of the inequality doesn't change when we simplify.
x * (y-x)/x > x * 1
y-x > x
y > 2x.

Case 2: x=0
Not possible, since (y-x)/x is undefined if x=0, and statement 2 indicates that (y-x)/x > 1.

Case 3: x<0
Here, x negative, so we must flip the direction of the inequality when we multiply each side by x:
x * (y-x)/x < x * 1
y-x < x
y < 2x.

Since y>2x in Case 1 but y<2x in Case 3, INSUFFICIENT.

Statement 2: x>0
No information about y.
INSUFFICIENT.

Statements combined:
Since x>0, only Case 1 above is viable, implying that y>2x.
SUFFICIENT.

The correct answer is C.

In my solution to the problem that you posted, at no point do I multiply or divide each side of an inequality by a variable whose sign is unknown.
Thus, different cases -- x>0, x=0, x<0 -- do not have to be considered.
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by melguy » Sat Jun 22, 2013 6:07 am
GMATGuruNY wrote:
In my solution to the problem that you posted, at no point do I multiply or divide each side of an inequality by a variable whose sign is unknown.
Thus, different cases -- x>0, x=0, x<0 -- do not have to be considered.
Thanks a lot GMATGuru :-) I was not considering the above key point.