hi phoenixhazard
it is very strange but i got E can you verify text or oa, below my reasoning
if (x+y)/z<0 then for that to be true two cases are possible

x+y>0 and z<0. or also x+y<0 and z>0

to the statements
(1) x<y, no other solving than picking simple answer.
x=1,y=2 x<y, 1+2=3>0 ( here z must be -ve) and x>0 answer no
x=-1,y=2, x<y.-1+2=1>0 ( again z>0) x<0 the ans is yes
insuff
(2) z<0 it means that x+y>0,
again x=1, y=2 1+2=3>0 x=1>0 the answer no
x=-1, y=2 -1+2=1>0 x=-1<0 the answer is yes insuff
both z<0, x+y>0, x=1,y=2-no , x=-1,y=2,yes to me insuff, the only i can derive from here is that y>0 but the value of x depends on...
E to me,
guys help me to find my mistake

e also..

because if z is a negative.... (x+y) > 0 but x could be a negative or a positive depends on the value of x which is unknown

5abi wrote:e also..

because if z is a negative.... (x+y) > 0 but x could be a negative or a positive depends on the value of x which is unknown

Its C in the book.

They explain that if z<0 then it means x+y must be less than 0 to make a number that is positive. I just dont understand how x<y would give you x+y has to be less than 0 since it doesnt!

please check again the whole problem
if z<0 then it means x+y must be less than 0 to make a number that is positive
in the problem we have number that is -ve
(x+y)/z<0

the same can be said for (x+y)/z <0 if z<0 then (x+y) > 0 but there is nothing indicating this as a certainity.

clock60 wrote:hi phoenixhazard
it is very strange but i got E can you verify text or oa, below my reasoning
if (x+y)/z<0 then for that to be true two cases are possible

x+y>0 and z<0. or also x+y<0 and z>0

to the statements
(1) x<y, no other solving than picking simple answer.
x=1,y=2 x<y, 1+2=3>0 ( here z must be -ve) and x>0 answer no
x=-1,y=2, x<y.-1+2=1>0 ( again z>0) x<0 the ans is yes
insuff
(2) z<0 it means that x+y>0,
again x=1, y=2 1+2=3>0 x=1>0 the answer no
x=-1, y=2 -1+2=1>0 x=-1<0 the answer is yes insuff
both z<0, x+y>0, x=1,y=2-no , x=-1,y=2,yes to me insuff, the only i can derive from here is that y>0 but the value of x depends on...
E to me,
guys help me to find my mistake

Even i am surprised How the answer is C, It should be E
Can you post the official explanation..........
I am sure that will Help a lot.........

clock60 wrote:hi phoenixhazard
it is very strange but i got E can you verify text or oa, below my reasoning
if (x+y)/z<0 then for that to be true two cases are possible

x+y>0 and z<0. or also x+y<0 and z>0

to the statements
(1) x<y, no other solving than picking simple answer.
x=1,y=2 x<y, 1+2=3>0 ( here z must be -ve) and x>0 answer no
x=-1,y=2, x<y.-1+2=1>0 ( again z>0) x<0 the ans is yes
insuff
(2) z<0 it means that x+y>0,
again x=1, y=2 1+2=3>0 x=1>0 the answer no
x=-1, y=2 -1+2=1>0 x=-1<0 the answer is yes insuff
both z<0, x+y>0, x=1,y=2-no , x=-1,y=2,yes to me insuff, the only i can derive from here is that y>0 but the value of x depends on...
E to me,
guys help me to find my mistake

Hi

Basically you're given a situation where

x+y/z>0 has to be true because that's the situation for the problem

Now if this is true and y<0

then for it to be applicable to the above given situation

then x + y must be negative

Now you're also told that x<y then the x=-1,y=2 which you mentioned cannot be taken

because it cannot apply to this situation ie

where x + y/z has to be greater than zero (which is given)

Remember the question is asking you for the answer when x + y/z is greater than 0

and when z is negative you cannot take x=-1 and y=2 because then x+y/z is not equal to 0

because that would give you a positive numerator while it has to be negative to
meet the above equation

because z is also negative.

Both cannot be positive because numerator is positive which it cannot be

Both can be negative then x is less than 0
One can be positive and one can be negative (the negative number has to be x because x is the smaller number by statement 1)

So both ways x is less than 0 and so answer is C

a+b

for z<0 meaning x+y > 0
thus x = -1 y = 2 or x=1 and y = 2.

E it is.

x and y cannot be both positive because it's given that x plus y by z is less than 0 and z is equal to less than 0

so denominator is negative and you're given that the equation is positive and therefore the numerator must also be negative to make it positive as a whole

if you take x=1 and x=2 it can't be positive hence you cannot take those 2 numbers

Can only take 2 negative numbers or 1 positive and 1 negative

with 2 negatives x is less than zero and with the other we know that x<y so x is negative

either ways x is negative

C is the answer and you can check that on the OG

I Think OA should be E

phoenixhazard wrote:This is from official GMAC book.

If (x+y)/z < 0, is x <0?
(1) x<y
(2)z<0


I have the explanation from the book but it doesn't make sense to me. The OA is C. Can you please explain to me in your own words, perhaps that will clarify it.

We know that x+y and z have opposite signs and we are asked whether x is negative. (2) tells us that z <0, so x + y > 0 . Since x < y, x could be positive, negative or 0.

For example (-1 + 2)/-1 < 0 and (1+2)/-1 < 0

The answer for the question above is E. However, if we had been told that (x+y)/z > 0, (1) and (2) would allow us to conclude that x + y <0 and thus x < 0.

The question is wrong. Please refer D41 question of the OG12 guide.the actual questionis as below.

If (x+y)/z > 0, is x <0?
(1) x<y
(2)z<0

Could an expert reply to the corrected question please?

singalong wrote:The question is wrong. Please refer D41 question of the OG12 guide.the actual questionis as below.

If (x+y)/z > 0, is x <0?
(1) x<y
(2)z<0

(x+y)/z>0 implies that the quantities (x+y) and z are either both positive or both negative.

Statement 1: x<y. This is clearly insufficient, as it tells us nothing about z, so we can't use the information given in the stem. It could be that x>0 (x=1, y=2, z=1) or x<0 (x=-1, y=2, z=1). INSUFFICIENT.

Statement 2: z<0. If z<0, then x+y<0. This could be true if x<0 (x=-3, y=1) or x>0 (y=-3, x=1). INSUFFICIENT.

Statements 1&2: We know that x+y<0 and x<y or x-y<0. When we have two inequalities with the inequality symbol pointing in the same direction, we can add the left and right sides to produce a new, true inequality: Adding x+y<0 and x-y<0 gives 2x<0, which implies that x<0. SUFFICIENT.

Ans: C