y<y^2 ??

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neerajkumar1_1: Master | Next Rank: 500 Posts; Posts: 270; Joined: Wed Apr 07, 2010 9:00 amFollowed by: 2 members; Thanked: 24 times

by neerajkumar1_1 » Fri Feb 17, 2012 3:09 am

which of the following describes all the values of y for which y<y^2.

1 < y
-1 < y < 0
y < -1
1/y < 1
0 < y < 1

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Source: Beat The GMAT — Problem Solving | Fri Feb 17, 2012 3:09 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Feb 17, 2012 4:21 am

neerajkumar1_1 wrote:which of the following describes all the values of y for which y<y^2.

1 < y
-1 < y < 0
y < -1
1/y < 1
0 < y < 1

Approach 1:
yâ‰ 0, since 0<0Â² is not valid.
Thus, yÂ² must be positive, implying that if we divide each side of y < yÂ² by yÂ², the direction of the inequality doesn't change:
y/yÂ² < yÂ²/yÂ²
1/y < 1.

The correct answer is D.

Approach 2:
Use POE (process of elimination).
Each answer choice represents a RANGE of values.
Plug in values that satisfy y<yÂ².
Look for values that work in some answer choices but not in others.

Let y=2, since 2<2Â².
Eliminate any answer whose range does not include y=2.
A: 1<y
1<2.
This works. Hold onto A.

B: -1 < y < 0
-1 < 2 < 0.
Doesn't work. Eliminate B.

C: y < -1
2 < -1.
Doesn't work. Eliminate C.

D: 1/y < 1
1/2 < 1.
This works. Hold onto D.

E: 0 < y < 1
0 < 2 < 1.
Doesn't work. Eliminate E.

Let y=-2, since -2<(-2)Â².
Eliminate any remaining answer whose range does not include y=-2.
A: 1<y
1<-2.
Doesn't work. Eliminate A.

The correct answer is D.

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[email protected]: Legendary Member; Posts: 934; Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members; Location: AAMCHI MUMBAI LOCAL; Thanked: 63 times

by [email protected] » Fri Feb 17, 2012 7:31 am

which of the following describes all the values of y for which y<y^2.

1 < y
-1 < y < 0
y < -1
1/y < 1
0 < y < 1

Solution:

Basically the thing required for satisfying the above equation is that y should be > 1 or y should be < -1 that is extreme cases...

A and C are good options but they only consider only one case at hand...

Like option A only considers y being > 1

and

Like option B only considers y being < -1

So option D is the only option that considers both the options.

Try putting all the values for the option D and you will find that the answer turns out to be correct

Option A is completely out of scope as it is a positive fraction and in positive fraction's case the equation is always y > y^2...

But the interesting thing about option D is that it also considers the negative fraction that also hold in the above condition i.e y < y^2...

Hope this explanation really helped....

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neerajkumar1_1: Master | Next Rank: 500 Posts; Posts: 270; Joined: Wed Apr 07, 2010 9:00 amFollowed by: 2 members; Thanked: 24 times

by neerajkumar1_1 » Fri Feb 17, 2012 10:53 pm

[email protected] wrote:which of the following describes all the values of y for which y<y^2.

1 < y
-1 < y < 0
y < -1
1/y < 1
0 < y < 1

Solution:

Basically the thing required for satisfying the above equation is that y should be > 1 or y should be < -1 that is extreme cases...

A and C are good options but they only consider only one case at hand...

Like option A only considers y being > 1

and

Like option B only considers y being < -1

So option D is the only option that considers both the options.

Try putting all the values for the option D and you will find that the answer turns out to be correct

Option A is completely out of scope as it is a positive fraction and in positive fraction's case the equation is always y > y^2...

But the interesting thing about option D is that it also considers the negative fraction that also hold in the above condition i.e y < y^2...

Hope this explanation really helped....

yeah it did make sense.. in fact i was looking at the question in the wrong way..
of course y^2 > y in all cases except 0<y<1 ...
Instead of looking for the entire range i was looking at the possible ranges..
anyways..
thanks for the clarification..

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krishna239455: Master | Next Rank: 500 Posts; Posts: 114; Joined: Mon Jun 13, 2011 9:10 am; Thanked: 1 times

by krishna239455 » Thu May 10, 2012 5:58 am

Dear Aneesh

Since I love the "Critical Point Method" explained by you in previous forums, i tried to solve this problem with the same method.
y<y^2 is nothing but 0<y(y-1)
I got 0,1 critical points and by the graph i got y>1 and y<0
This satisfies only 4th answer choice 1/y<1

Am I correct with this approach?

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aneesh.kg: Master | Next Rank: 500 Posts; Posts: 385; Joined: Mon Apr 16, 2012 8:40 amFollowed by: 29 members; Location: Pune, India; Thanked: 186 times

by aneesh.kg » Thu May 10, 2012 10:00 am

Absolutely correct.

I hope you understand that after getting
y > 1
you divided both sides by y
and got
1 > 1/y
because y is positive when y > 1.

For the others who're wondering what the Critical Points method is and how it makes our life really really easy in problems like these, read this:
https://www.beatthegmat.com/critical-poi ... tml#470867

Aneesh Bangia
GMAT Math Coach
[email protected]

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