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by kevincanspain » Thu Apr 14, 2011 11:45 pm
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
Last edited by kevincanspain on Fri Apr 15, 2011 6:50 pm, edited 1 time in total.
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by HSPA » Fri Apr 15, 2011 12:46 am
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
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by manpsingh87 » Fri Apr 15, 2011 1:04 am
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)..!!!
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by HSPA » Fri Apr 15, 2011 1:26 am
Very nice assumption Manap... E is independent of X and Z..

E > D though it is controversial.. agreed
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by jaymw » Fri Apr 15, 2011 3:14 am
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.
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by force5 » Fri Apr 15, 2011 4:29 am
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.
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by Stuart@KaplanGMAT » Fri Apr 15, 2011 10:02 am
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.
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by kevincanspain » Fri Apr 15, 2011 6:52 pm
Sorry about the typo in E- it should read y > y^2. Stuart, what other possible interpretation of the question did you think of?
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by pemdas » Sat Apr 16, 2011 2:33 am
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
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by force5 » Sat Apr 16, 2011 2:56 am
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??
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by Ian Stewart » Sat Apr 16, 2011 1:06 pm
force5 wrote:
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??
You can certainly square both sides of an equation; if a = b, then it is certainly true that a^2 = b^2. That's not what you're doing. If you square the right side of the equation:

sqrt[(y-1)^2] = y-1

you don't get |y-1|; you get (y-1)^2. I think you're trying to use the identity sqrt(a^2) = |a|, which is true, and which we can use on the left side (only) of the equation. Then we learn

|y-1| = y-1

which is only true when y > 1, since the right side will be negative when y < 1, and the left side is never negative.
kevincanspain wrote:Sorry about the typo in E- it should read y > y^2. Stuart, what other possible interpretation of the question did you think of?
each of the following indicates that y is positive

The GMAT would not use the word 'indicates' (which simply means 'points to', rather than 'guarantees') in this kind of question, which might be the reason Stuart thought the question sounded somewhat informal. One could rephrase the question in several ways; the word 'indicates' could be replaced with 'ensures', for example, or one could reverse the language and ask 'y could be negative if which of the following is true?' among other possibilities.
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by force5 » Sun Apr 17, 2011 6:56 am
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.
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by Ian Stewart » Sun Apr 17, 2011 10:23 am
force5 wrote: 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.
No, there's a logical problem with what you're doing. I'll explain first with a simple example. If you know that x=1, you can certainly conclude that x > 0. But you cannot then conclude that x can be any number greater than 0; you lost information when you rewrote 'x = 1' as 'x > 0'. Similarly, if you know that x=y, you can certainly conclude that x^2 = y^2, but it is not then a possibility that x=1 and y=-1, since we know x and y are equal; when you squared both sides, you 'erased' negative signs, and you lost information about the relationship between x and y.

In this question, if sqrt[ (y-1)^2 ] = y-1, you can certainly square both sides to learn that (y-1)^2 = (y-1)^2. But you lose information about y when you do that (because squaring both sides 'erases' negative signs); in fact you arrive at something which is true for any value of y. That does not mean that the original equation is true for any value of y, however; just as in the two simpler examples I gave in the first paragraph, your new equation gives you less information than the original.
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by force5 » Sun Apr 17, 2011 10:36 am
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.
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