Is (-y)/2 > y^2 ?

Y squ =1/25
y=1/5 or -1/5.
If Y =1/5 then -y/2>ysquare

-1/10<1/25 not sufficient.

y=-1/5 then 1/10>1/25.
Option 1 is insufficient.


2 Mod y = -y => y=-ve

so put values y=-1 then 1/2<1
y=-1/2 then 1/4=1/4

y=-1/3 then 1/6>1/9

y=-1/4 then 1/8>1/16.

from y>=-1/3.

Combine 1 and 2 y=-1/5. so Option is C sufficient.

Hi All,

This DS question involves a bit of math and a bit of Number Properties; by TESTing VALUES, you can prove what the correct answer is.

We're ask if (-Y)/2 is greater than Y^2. This is a YES/NO question.

Fact 1: Y^2 = 1/25

Here, we're dealing with a straight-forward equation with two solutions: 1/5 and -1/5. We still have to make sure that we're answering the question that's ASKED though:

IF....
Y = 1/5
(-1/5)/2 = -1/10
-1/10 is NOT greater than 1/25 so the answer to the question is NO

IF....
Y = -1/5
(-(-1/5))/2 = 1/10
1/10 IS greater than 1/25 so the answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: |Y| = -Y

This Fact tell us that Y CANNOT be positive. It can be 0 or ANY negative number though.

IF....
Y = -1/5
(-(-1/5))/2 = 1/10
1/10 IS greater than 1/25 so the answer to the question is YES

IF...
Y = 0
0/2 is NOT greater than 0 so the answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know....
Y = 1/5 or -1/5
Y is 0 or ANY negative

There's only 1 value that fits both Facts: -1/5....so the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Missed the case that y could be zero. Good explanation
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Hi spetznaz,

As a general rule, you should keep "0" in mind when working through every DS question that you face. Beyond the fact that it's a remarkably easy number to TEST, it's often a 'pattern-breaker' and helps to prove that a particular Fact is insufficient. The difference between the correct answer and any of the incorrect answers often comes down to the 'thoroughness' of your thinking; just a little more (focused) work will get you the correct answer.

GMAT assassins aren't born, they're made,
Rich

y^2 will always be +ve; hence for LHS to be greater than y^2, it should be +ve.
Also, it should fulfill two conditions:-

1) y should be -ve
2) y should be >-1/2 (because if y= -1/2 then LHS=RHS and if it is <-1/2 then LHS < RHS)

1st statement-
y^2= 1/25
y= 1/5 or -1/5

If its +1/5 then the does not hold good, but if it is -1/5 then yes. Insufficient.

2nd statement-
|y| = -y
That means y is -ve, but doesn't tell u sit y is >-1/2 Not sufficient

Combining statement 1 and 2, we get the answer that y=-1/5
Hence C is the answer
_________________

If we modify the original condition and the question, we get 0>y^2+(y/2)?, 0>y(y+1/2)?. This leads to -1/2<y<0?. However, using 1) and 2), we get y=-1/5. The answer is yes and the conditions are sufficient. Hence, the correct answer is C.

did algebraic

is -y/2>y^2? means y(2y+1)<0 means -1/2<y<0? So, that is the question

St1. y=+-1/5. Not Suff

St.2 y<0. Not Suff

St1+St2 means y=-1/5. Suff

C

forgot to test for 0 as well :S

C

(1) y = +1/5; y = -1/5
-1/10 < 1/25; 1/10 > 1/25
Insufficient

(2) thus y is negative
Use the above: y = -1/5
1/10 > 1/25
Test: y = -1
1/2 < 1
Insufficient

Taking (1) and (2) thus y=-1/5

Given is -y/2>y^2
or
y^2+y/2<0
or if -1/2<y<0 (We have to find is this is correct or not)
Statement 1
y^2 = 1/25
or y =1/5 or -1/5
For y=1/5 it does not satisfy our condition.
but for y=-1/5 it does. Hence statement 1 is clearly insufficient.

Statement 2

y+|y| =0
only possible if y is negative. But we don't know if it is greater than -1/2. Hence it is insufficient

Combining both statements we have y=-1/5
hence answer is C

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