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Is y^3 < |y|

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Source: — Data Sufficiency |
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by GMATGuruNY » Tue Jun 27, 2017 3:07 am
gmattoend wrote:Is y^3 < |y|

(1) y < 1

(2) y < 0

OA: B
Here, testing cases seems easier and more efficient.

Statement 1:
If y=0, then y³ = |y|, with the result that the answer to the question stem is NO.
If y=-1, then y³ < |y|, with the result that the answer to the question stem is YES.
INSUFFICIENT.

Statement 2:
Plugging a negative value into the question stem, we get:
(negative)³ < | negative | ?
negative < positive ?
The answer to the question in blue is YES.
SUFFICIENT.

The correct answer is B.
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by Jay@ManhattanReview » Tue Jun 27, 2017 3:40 am
gmattoend wrote:Is y^3 < |y|

(1) y < 1

(2) y < 0

OA: B

Dear Sirs,

How can I solve this problem with algebra?
This question can be solved logically. There is no need to go the Algebraic way.

Question: Is y^3 < |y|?

We see that RHS, |y| is either a positive quantity or zero, while y^3 can be anything positive, negative, or zero.

So, if y = 0, the answer is No. y^3 = |y|; however, if y < 0, the answer is yes, y^3 < |y|.

Statement 1: y < 1

y < 1 has three ranges.

1. 0 < y < 1: It implies that y is a positive fraction. Cube of a positive fraction that is less than 1 is always less than the number itself. Ex.: Say y = 1/2, then y^3 = 1/8 and |y| = 1/2. Here. 1/8 < 1/2. The answer is Yes.

2. y = 0: We have already seen that @ y = 0, y^3 = |y|. The answer is No. No unique answer. Insufficient.

Though we concluded that Statement 1 is not sufficient, let's take the third range too for the sake of understanding.

3. y < 0: It implies that y is negative. We already saw that if y is negative, y^3 < |y|. The answer is Yes.

Statement 2: y < 0

We have discussed this above. Thus, Statement 2 is sufficient. The answer is Yes.

The correct answer: B

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by gmattoend » Tue Jun 27, 2017 7:25 am
GMATGuruNY wrote:
gmattoend wrote:Is y^3 < |y|

(1) y < 1

(2) y < 0

OA: B
Here, testing cases seems easier and more efficient.

Statement 1:
If y=0, then y³ = |y|, with the result that the answer to the question stem is NO.
If y=-1, then y³ < |y|, with the result that the answer to the question stem is YES.
INSUFFICIENT.

Statement 2:
Plugging a negative value into the question stem, we get:
(negative)³ < | negative | ?
negative < positive ?
The answer to the question in blue is YES.
SUFFICIENT.

The correct answer is B.
Mitch Sir,

Because LHS is positive, I can divide both sides by |y| and I can rephrase the question to be :

Is y^2 < 1?

However, Statement 2 will be insufficient.

What is my mistake here?
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by GMATGuruNY » Tue Jun 27, 2017 7:57 am
gmattoend wrote:Mitch Sir,

Because LHS is positive, I can divide both sides by |y| and I can rephrase the question to be :

Is y^2 < 1?

However, Statement 2 will be insufficient.

What is my mistake here?
If we divide both sides of the question stem by |y|, we get:
y³/|y| < 1 ?

Case 1: y>0
Here, y³/|y| = y².
Substituting y³/|y| = y² into the blue expression above, we get:
y² < 1?
Since y here is constrained to POSITIVE VALUES, the answer will be YES if y is a positive fraction between 0 and 1.

Case 2: y<0
Here, y³/|y| = -(y²).
Substituting y³/|y| = -(y²) into the blue expression above, we get:
-(y²) < 1?
y² > -1?
In this case, the answer will be YES regardless of the value of y, since the square of any value will be greater than -1.

Since statement 2 indicates that y<0, the answer to the question stem is YES. as discussed under Case 2 above.
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by [email protected] » Tue Jun 27, 2017 10:15 am
Hi gmattoend,

The Quant section of the GMAT is NOT a 'math test' - it's a 'critical thinking' test that will involve lots of little calculations as you work through it. As such, you have to be careful about taking a "just do math" approach to this section of the GMAT (since you will likely end up making the section harder than it actually is and limiting how high you score). Most GMAT questions can be approached in more than one way, so one of your goals during your studies should be to learn the various approaches - so that you can use the most efficient approach for each question that you face. With this prompt, TESTing VALUES or using Number Properties would almost certainly be faster and easier than doing Algebra.

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by gmattoend » Tue Jun 27, 2017 10:27 am
GMATGuruNY wrote: If we divide both sides of the question stem by |y|, we get:

y³/|y| < 1 ?
Why did not you reduce the fraction to y^2 <1??

y^3 / |y| = y * y^2 / |y|. Because y^2 is positive and |y| is positive , we can divide them without concern.

y^3 / |y| = y * y^2 / |y| = y * y = y^2

So the question could be y³/|y| < 1 or y^2 < 1

Is there any problem with above?
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by GMATGuruNY » Tue Jun 27, 2017 12:00 pm
gmattoend wrote:y^3 / |y| = y * y^2 / |y| = y * y = y^2
The step in red is valid only if y>0.
y²/|y| = y/|y| * y.
If y>0, then y/|y| = 1, with the result that y²/|y| = y/|y| * y = 1 * y = y.
If y<0, then y/|y| = -1, with the result that y²/|y| = y/|y| * y = -1 * y = -y.
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by Kaustubhk » Wed Jun 28, 2017 4:56 am
Hi Experts,

I solved this in a different way. Kindly correct if I'm wrong.

Y^3 < |y| ..
|y| > y^3 .. Reverse

Y > y^3 .. For positive

-y > y^3
Y < -y^3.. For Negative



y^3<y < -Y^3., combining the two


A) y< 1

For y= 0,

0< 0 < -0 .. Not satisfied

B) y< 0

For y= -1


-1< -1< 1

For Y= -2

-8 < -2 < 8

The answer is B
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