Problem Solving

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

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:
y≠0, since 0<0² is not valid.
Thus, y² must be positive, implying that if we divide each side of y < y² by y², the direction of the inequality doesn't change:
y/y² < y²/y²
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<y².
Look for values that work in some answer choices but not in others.

Let y=2, since 2<2².
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)².
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.

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....

[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..

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?

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