Problem Solving

If z > x, each of the following indicates that y is positive EXCEPT

(A) The positive square root of the square of y-1 is equal to y-1.
(B) yz > xy
(C) 7y/8 < y
(D) x(y^3) > 0
(E) y > y^2

A) says y=2
B) we know z>x if Y is negitve when multiplying on both sides the sign changes
c) Is y an integer?? let y = 1/2 so 0.5 > 7/8 * 0.5
E) y is a positive integer or a -ve integer
d) x(y^3)>0
we have two cases 1) X and Y^3 are both positve 2) both negitive
if y^3 is negitive the Y is negitive... We dont know what is X???

Here we cannot come to a conclusion.. Kevin I am stuck with D and E .. did I miss something basic

kevincanspain wrote:If z > x, each of the following indicates that y is positive EXCEPT

(A) The positive square root of the square of y-1 is equal to y-1.
(B) yz > xy
(C) 7y/8 < y
(D) x(y^3) > 0
(E) y^2 > y

lets consider every option..!!!

sqrt(x^2)=|x|= x if x>=0 and -x if x<0;
A) sqrt(y-1)^2=y-1; therefore y-1>0; y>1; i.e. y is positive.
B)yz>xy; y(z-x)>0 from initial condition we know that z-x>0; i.e. z-x is positive, therefore y must also be positive.
C)7y/8<y; y-7y/8>0; y/8>0; therefore y is positive.
D)x(y^3)>0;this expression will be greater than zero if both x and y^3 must have same sign, if x>0, then y must be greater than zero, if x<0, then y must also be negative.
E)y^2>y; y(y-1)>0; this inequality holds true if y<0, or y>1..!!!

i think some more additional information is required to answer the question...!!!

if this question appears in my exam then i would pick option E(i'll assume that x and z are positive)..!!!

Very nice assumption Manap... E is independent of X and Z..

E > D though it is controversial.. agreed

What's the source of this question?

I doubt that something like this would ever come up on the GMAT because, as discussed, answer choice D and E both qualify as the right answer. Nowhere in the prompt does it say that Z and X must be positive, therefore it can't just be assumed.

What's also really strange about this question is that information about Z and X is given but Z and X TOGETHER only show up in one of the answer choices, rendering this information completely useless for all other answer choices.

A) sqrt (y-1)^2= y-1
squaring both sides
|y-1|= |y-1|
y can be both positive or Negative

In B and C y has to be +ve

D) Y can be negative if x is -ve.
E) y^2>y or
y(y-1)>0
again y can be +ve or -ve

Hence A, D, E indicate that y can be both +ve or -ve.

kevincanspain wrote:If z > x, each of the following indicates that y is positive EXCEPT

(A) The positive square root of the square of y-1 is equal to y-1.
(B) yz > xy
(C) 7y/8 < y
(D) x(y^3) > 0
(E) y^2 > y

The question could definitely be worded more clearly (I'm not sure of the source, but it sounds "homemade"). Let's assume, for the sake of the question, that it reads "MUST BE positive EXCEPT" instead of "IS positive EXCEPT". Even with that change, (D) and (E) both satisfy the condition.

When that happens, it's not an indication that you should choose the "better" answer; rather, it's an indication that the question is flawed.

Let's deal with the simpler choices first:

(B) yz > yx

Well, if y were negative, then when we divide both sides by y we'd get:

z < x

which we know is untrue.

If y=0, then we get:

0>0

which we know is untrue.

Therefore, y must be positive - eliminate (B).

(C) 7y/8 < y

Exact same reasoning as (B). If y<0, then when we divide both sides by y we get:

7/8 > 1

which is clearly untrue. If y=0, then we again get:

0 < 0

which is also untrue. Accordingly, y must be greater than 0; eliminate (C).

(D) x(y^3)>0

We know that x and y must have the same sign; since we don't know whether x is positive or negative, y could be EITHER positive OR negative. Accordingly, y does not HAVE TO be positive: choose (D)!

(E) y^2 > y

If y is negative, this statement will always be true, since a negative squared is positive. If y is 0 or a positive fraction or 1, this statement is false. If y is greater than 1, this statement will be true. Consequently, (E) indicates that y may or may not be positive. Ummm.. choose (E) too!

and finally:

(A) The positive square root of the square of y-1 is equal to y-1.

Let's think about this one logically rather than getting involved in algebra. By common sense, "the positive square root" of something is... wait for it... positive!

So, from (A) we can conclude that:

y - 1 > 0
y > 1

So of course (A) shows that y MUST be positive: eliminate (A).

To "fix" the question, change (E) to:

(E) y^2 < y

so that y must be a positive fraction.

Sorry about the typo in E- it should read y > y^2. Stuart, what other possible interpretation of the question did you think of?

given: z>x, y>0
a) |sqrt(y-1)|=y-1, y-1=y^2-2y+1, y^2-3y+2=0, (y-1)(y-2)=0, y>0
b) y(z-x)>0 two options y>0 and z-x>0 {z>x} OR y<0 and z-x<0 {z<x}; the second option is out, y>0
c) (7y-8y)/8 <0, 7y-8y<0, y>0
d) two options x>0 and y^3>0 {y>0} OR x<0 and y^3<0 {y<0}
e) y>y^2, y(y-1)<0, two options y<0 and y-1>0 {y>1} OR y>0 and y-1<0 {y<1} --> the only viable solution area is 0<y<1

IOM d

kevincanspain wrote:If z > x, each of the following indicates that y is positive EXCEPT

(A) The positive square root of the square of y-1 is equal to y-1.
(B) yz > xy
(C) 7y/8 < y
(D) x(y^3) > 0
(E) y > y^2

Hi Stuart i agree to your solution but there is never a definitive way to do such problems. May i know what actually is wrong in what i am doing...

sqrt (y-1)^2= y-1
squaring both sides
|y-1|= |y-1|
y can be both positive or Negative

are we not allowed to square both sides in such situations??

Ian thank a lot. Yes you are correct i was trying to use sqrt(a^2) = |a| but made a mistake. I'm just going to trouble you one more time.

so if i understand i can square both the sides. after squaring we get

(y-1)^2= (y-1)^2

now this means Y will have infinite solutions. Hence Y can be both positive and negative including zero.

great Ian. thanks a lot it really helped. I guess i was the only one with this confusion but i'm happy its clear now. All credit to you.
i can't see the virtual beer or else i would have given it to you.

thanks any ways.