If a and y are integers and 4a+y is odd, is a^2/y 0?

Question

If a and y are integers and 4a+y is odd, is a^2/y > 0?

(1) y^3+1>0
(2) ay<0

Fig · Answer

(C) for me

4a+y is odd implies that y is odd. From 1 y^3+1>0 <=> y^3 > -1 => y > 0 as y is odd. As a^2 >= 0 Thus, a^2/y >=0 : We cannot conclude. INSUFF. From 2 ay<0 implies that a and y must have opposite signs. Hence, a^2/y > 0 if a < 0 and y > 0 or a^2/y < 0 if a > 0 and y < 0 INSUFF. The 2 statments together As from 1 : y > 0, the second statment say us that a > 0 (the case of a =0 is out). Thus, a^2/y > 0 SUFF.

yezz · Answer

If a and y are integers and 4a+y is odd, is a^2/y > 0?

(1) y^3+1>0
(2) ay<0

THE QUESTION ASKS IF Y IS POSITIVE

4A+Y IS ODD THUS Y IS ODD

Y^3>-1 THUS Y IS INTEGER , NOT ZERO AND IS +VE AT LEAST = 1 ......SUFF

FROM TWO

AY<0......NOT SUFF EITHER A OR Y IS -VE

MY ANSWER IS A

Fig · Answer

Sorry Yezz.... But in stat 1, a could be equal to 0. So we cannot state if a^2/y > 0.

yezz · Answer

I VE SEEN THE TRICK , SEEM I WILL NEVER GET IT :cry:

THANKS FIG

Fig · Answer

U will have it... u will be ok

(I'm the first to fall in trap

)

trivikram · Answer

E for me

From 1) y>-1

So y could be +ve or -ve

NOT SUFF

From 2) ay<0 NOT SUFF which is positive or -ve

Cobining (1) and (2)

Since y could be positive or -ve from (1) we cannot deduce anything for 2 whether a is positive or -ve.

So E should be it

Fig · Answer

Sorry Trivikram.... but, as 4*a + y is odd, we know that y must be odd. From (1), y > -1. That means that the first odd number possible is y = 1 (>0) and y cannot be negative in stat 1

trivikram · Answer

I realized it when I was at work today...such freaking errors cost dearly...

ywilfred · Answer

Since 4a is always even, then y must be odd.

Also, since a^2 is always positive, the question is really asking if y is a positive odd integer or a negative odd integer.

St1:
y^3 > -1. Since the next biggest odd value of y^3 is positive, then y must be an odd integer. (** y^3 is odd since y is odd)

St2:
ay < 0 --> not useful since we can't tell if y is positive or negative as the sign of 'a' determines what the sign of 'y' would take

Ans A

netcaesar · Answer

A too, because of the same reasons that ywilfred said.

From the statements, we know that y is ODD:

4a+y = ODD (4a is always even,so y must be ODD).

We know that a^2 is positive too, so we only need to know if y is POSITIVE or NEGATIVE.

Then:

1) y^3+1>0

Then y^3>-1, then y>-1.

We know that y is odd so y=1, 3, ......SUFFICIENT

2) a x y < 0

Then a<0 and y>0 OR a>0 or y<0. NOT SUF.

typicalgmatter · Answer

A

I agree because remember a^2 is the numerator. So if y is definitely a positive integer and a^2 must be a positive integer. a^2/y must be greater than 0. 

Remember 0^2 is 1...

yogeshsheth · Answer

C for me 2
A nice one..
trick here is that a could be zero.

typicalgmatter · Answer

I'm so dumb... When I was considering 0 for some reason I reversed things and had 2^0 power. 

C is correct

If a and y are integers and 4a+y is odd, is a^2/y > 0?

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