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If X, Y and Z are positive integers, is X greater than Z Y?

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If X, Y and Z are positive integers, is X greater than Z Y? [#permalink] New post 05 Apr 2006, 00:44
00:00
A
B
C
D
E

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If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2
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 [#permalink] New post 05 Apr 2006, 01:06
B....

We need to know whether x>z-y

1) statement.... x>z+y... insufficient
2) statement.... x=z-y... sufficient.

Answer is No...
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Re: DS: integers [#permalink] New post 05 Apr 2006, 02:28
Dilshod wrote:
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2


(1) --> X> Z+Y > Z-Y (because Z and Y are positive integers ---> Z+Y> Z-Y) ---> answer to the question: YES ---> suff

(2) Z^2 = X^2 + Y^2 --> X^2 = Z^2 - Y^2 = (Z-Y) ( Z+Y) > (Z-Y) (Z-Y)
---> X^2 > (Z-Y)^2 ---> (X- (Z-Y)) * (X+ (Z-Y)) > 0
---> 2 cases:

X> Z-Y AND X > Y-Z
OR X< Z-Y AND X < Y-Z ( This can't happen because either Z-Y or Y-Z must be smaller than 0 ---> X <0 ---> against the assumption in the problem)
---> The only case is X> Z-Y ---> suff

Go for D.
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Re: DS: integers [#permalink] New post 05 Apr 2006, 03:17
laxieqv wrote:
Dilshod wrote:
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2


(1) --> X> Z+Y > Z-Y (because Z and Y are positive integers ---> Z+Y> Z-Y) ---> answer to the question: YES ---> suff

(2) Z^2 = X^2 + Y^2 --> X^2 = Z^2 - Y^2 = (Z-Y) ( Z+Y) > (Z-Y) (Z-Y)
---> X^2 > (Z-Y)^2 ---> (X- (Z-Y)) * (X+ (Z-Y)) > 0
---> 2 cases:

X> Z-Y AND X > Y-Z
OR X< Z-Y AND X < Y-Z ( This can't happen because either Z-Y or Y-Z must be smaller than 0 ---> X <0 ---> against the assumption in the problem)
---> The only case is X> Z-Y ---> suff

Go for D.


agree with you, but the oa is not d
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Re: DS: integers [#permalink] New post 05 Apr 2006, 03:40
laxieqv wrote:
Dilshod wrote:
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2


(1) --> X> Z+Y > Z-Y (because Z and Y are positive integers ---> Z+Y> Z-Y) ---> answer to the question: YES ---> suff

(2) Z^2 = X^2 + Y^2 --> X^2 = Z^2 - Y^2 = (Z-Y) ( Z+Y) > (Z-Y) (Z-Y)---> X^2 > (Z-Y)^2 ---> (X- (Z-Y)) * (X+ (Z-Y)) > 0
---> 2 cases:

X> Z-Y AND X > Y-Z


OR X< Z-Y AND X < Y-Z ( This can't happen because either Z-Y or Y-Z must be smaller than 0 ---> X <0 ---> against the assumption in the problem)
---> The only case is X> Z-Y ---> suff

Go for D.


Well, i had some flaw in the red part.

Correct it :
Z^2 = X^2+ Y^2 --> X^2= Z^2 - Y^2 --> X^2= (Z-Y) * (Z+Y)
---> X= [(Z-Y) * (Z+Y)]/ X = (Z-Y) * ( ( Z+Y) / X)
+If Z-Y < 0 ---> Z-Y < 0 < X --> Z-Y< X
+If Z-Y > 0
If the bold part < 1 ( that is Z+Y< X) --> X< Z-Y ---> Z+Y < X< Z-Y --> Z+Y < Z-Y ---> unreasonable --> this case can't happen
If the bold part >1 ( that is Z+Y > X) ---> X > Z-Y
--> finally, we still conclude that X> Z-Y

I'm still stuck to D.

Last edited by laxieqv on 05 Apr 2006, 03:49, edited 1 time in total.
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 [#permalink] New post 05 Apr 2006, 03:41
Missinterpreted 2x with x^2 here....

ignore my answer....
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Re: DS: integers [#permalink] New post 05 Apr 2006, 08:07
laxieqv wrote:
laxieqv wrote:
Dilshod wrote:
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2


(1) --> X> Z+Y > Z-Y (because Z and Y are positive integers ---> Z+Y> Z-Y) ---> answer to the question: YES ---> suff

(2) Z^2 = X^2 + Y^2 --> X^2 = Z^2 - Y^2 = (Z-Y) ( Z+Y) > (Z-Y) (Z-Y)---> X^2 > (Z-Y)^2 ---> (X- (Z-Y)) * (X+ (Z-Y)) > 0
---> 2 cases:

X> Z-Y AND X > Y-Z


OR X< Z-Y AND X < Y-Z ( This can't happen because either Z-Y or Y-Z must be smaller than 0 ---> X <0 ---> against the assumption in the problem)
---> The only case is X> Z-Y ---> suff

Go for D.


Well, i had some flaw in the red part.

Correct it :
Z^2 = X^2+ Y^2 --> X^2= Z^2 - Y^2 --> X^2= (Z-Y) * (Z+Y)
---> X= [(Z-Y) * (Z+Y)]/ X = (Z-Y) * ( ( Z+Y) / X)
+If Z-Y < 0 ---> Z-Y < 0 < X --> Z-Y< X
+If Z-Y > 0
If the bold part < 1 ( that is Z+Y< X) --> X< Z-Y ---> Z+Y < X< Z-Y --> Z+Y < Z-Y ---> unreasonable --> this case can't happen
If the bold part >1 ( that is Z+Y > X) ---> X > Z-Y
--> finally, we still conclude that X> Z-Y

I'm still stuck to D.


Honestly, I didn't get your "bold part" approach. One thing for sure is that you shouldn't take gmat questions that serious. Remember that you have less than 2 minutes for each question.

So can anybody explain, why st.2 is insufficient?
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Re: DS: integers [#permalink] New post 05 Apr 2006, 09:03
Z^2 = X^2+ Y^2 --> X^2= Z^2 - Y^2 --> X^2= (Z-Y) * (Z+Y)
Because we want to know the relation btw X and ( Z-Y) , I divide both sides by X

---> X= [(Z-Y) * (Z+Y)]/ X = (Z-Y) * ( ( Z+Y) / X)
At this point, we have to consider whether (Z-Y) is (-) or (+) as we examine the value of the both part
+If Z-Y < 0 ---> Z-Y < 0 < X --> Z-Y< X
+If Z-Y > 0:
If the bold part < 1 ( that is Z+Y< X) --> X< Z-Y ---> Z+Y < X< Z-Y --> Z+Y < Z-Y ---> unreasonable --> this case can't happen
If the bold part >1 ( that is Z+Y > X) ---> X > Z-Y
--> finally, we still conclude that X> Z-Y

I'm still stuck to D.

It looks lengthy but as soon as you jot down your reasoning, it's quick!
If not, plugging number may be the least time-consuming but you never know if you take ALL cases into consideration.
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 [#permalink] New post 05 Apr 2006, 11:30
A..
X > Z+Y.. is sufficient to know X > Z-Y.. try out any values..
X^2 = Z^2-Y^2.. Depends on whether Z > Y or Y > Z.
i.e X = Z -Y
or X = Y-Z.
Now we dont know which one is true because we dont know which one is greater to make sure X > 0.

Hence A.

OA?
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Re: DS: integers [#permalink] New post 05 Apr 2006, 11:53
laxieqv wrote:
I'm still stuck to D.

It looks lengthy but as soon as you jot down your reasoning, it's quick!
If not, plugging number may be the least time-consuming but you never know if you take ALL cases into consideration.


I agree with Laxie, although I had a hard time following the reasoning :shock:
Here is my explanation for D.

1. SUFF as explained above
2. Z^2 = X^2+Y^2,
From Pythagoras theorem, we can conclude that X, Y & Z are sides of a right angled triangle. And in any traingle, the third side is always greater than difference of sides and sum of the other 2 sides.
Hence Z-Y < X < Y+Z or Y-Z < X < Y+Z

In either case X > Y-Z..., Hence SUFF!
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 [#permalink] New post 05 Apr 2006, 12:05
willget800 wrote:
A..
X > Z+Y.. is sufficient to know X > Z-Y.. try out any values..
X^2 = Z^2-Y^2.. Depends on whether Z > Y or Y > Z.
i.e X = Z -Y
or X = Y-Z.
Now we dont know which one is true because we dont know which one is greater to make sure X > 0.

Hence A.

OA?


Doesnt matter which one is Greater! U know X=Z-Y so X CANNOT be > Z-Y!

U have to either prove X>Z-Y or prove that X is not > Z - Y
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Re: DS: integers [#permalink] New post 05 Apr 2006, 12:28
Dilshod wrote:
laxieqv wrote:
laxieqv wrote:
Dilshod wrote:
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2


(1) --> X> Z+Y > Z-Y (because Z and Y are positive integers ---> Z+Y> Z-Y) ---> answer to the question: YES ---> suff

(2) Z^2 = X^2 + Y^2 --> X^2 = Z^2 - Y^2 = (Z-Y) ( Z+Y) > (Z-Y) (Z-Y)---> X^2 > (Z-Y)^2 ---> (X- (Z-Y)) * (X+ (Z-Y)) > 0
---> 2 cases:

X> Z-Y AND X > Y-Z


OR X< Z-Y AND X < Y-Z ( This can't happen because either Z-Y or Y-Z must be smaller than 0 ---> X <0 ---> against the assumption in the problem)
---> The only case is X> Z-Y ---> suff

Go for D.


Well, i had some flaw in the red part.

Correct it :
Z^2 = X^2+ Y^2 --> X^2= Z^2 - Y^2 --> X^2= (Z-Y) * (Z+Y)
---> X= [(Z-Y) * (Z+Y)]/ X = (Z-Y) * ( ( Z+Y) / X)
+If Z-Y < 0 ---> Z-Y < 0 < X --> Z-Y< X
+If Z-Y > 0
If the bold part < 1 ( that is Z+Y< X) --> X< Z-Y ---> Z+Y < X< Z-Y --> Z+Y < Z-Y ---> unreasonable --> this case can't happen
If the bold part >1 ( that is Z+Y > X) ---> X > Z-Y
--> finally, we still conclude that X> Z-Y

I'm still stuck to D.


Honestly, I didn't get your "bold part" approach. One thing for sure is that you shouldn't take gmat questions that serious. Remember that you have less than 2 minutes for each question.

So can anybody explain, why st.2 is insufficient?


Here is my two cents

From st 2

X= [(Z-Y) * (Z+Y)]/ X

---> X = (Z-Y) * (Z+Y)/X --------(A)
Now the value of X dependeds on value of (Z+Y)/X.
Here (Z+Y)/X can be greater than 1 or less than depends on the values of X, Y, Z.

if Z+ Y > X then A will be greater than Xand if Z+ Y < X then A will be less than X.

So from St 2 we can't say any thing.

So answer is A
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Re: DS: integers [#permalink] New post 05 Apr 2006, 18:31
gmat_crack wrote:

Here is my two cents

From st 2

X= [(Z-Y) * (Z+Y)]/ X

---> X = (Z-Y) * (Z+Y)/X --------(A)
Now the value of X dependeds on value of (Z+Y)/X.
Here (Z+Y)/X can be greater than 1 or less than depends on the values of X, Y, Z.

if Z+ Y > X then A will be greater than Xand if Z+ Y < X then A will be less than X.

So from St 2 we can't say any thing.

So answer is A


buddy,
If Z+Y < X --> (Z+Y)/ X <1 ---> X= (Z-Y) * sth smaller than 1 --->
X< Z-Y. At this point, it's tempting to conclude but when we look back:
we have Z+Y< X then X< Z-Y --> Z+Y< Z-Y ---> Y< 0 ---> unreasonable.
Thus, this case can't be correct.
The only case is Z+Y >X --> X> Z-Y
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Re: DS: integers [#permalink] New post 05 Apr 2006, 18:36
giddi77 wrote:
laxieqv wrote:
I'm still stuck to D.

It looks lengthy but as soon as you jot down your reasoning, it's quick!
If not, plugging number may be the least time-consuming but you never know if you take ALL cases into consideration.


I agree with Laxie, although I had a hard time following the reasoning :shock:
Here is my explanation for D.

1. SUFF as explained above
2. Z^2 = X^2+Y^2,
From Pythagoras theorem, we can conclude that X, Y & Z are sides of a right angled triangle. And in any traingle, the third side is always greater than difference of sides and sum of the other 2 sides.
Hence Z-Y < X < Y+Z or Y-Z < X < Y+Z

In either case X > Y-Z..., Hence SUFF!


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Re: DS: integers [#permalink] New post 05 Apr 2006, 18:47
laxieqv wrote:
gmat_crack wrote:

Here is my two cents

From st 2

X= [(Z-Y) * (Z+Y)]/ X

---> X = (Z-Y) * (Z+Y)/X --------(A)
Now the value of X dependeds on value of (Z+Y)/X.
Here (Z+Y)/X can be greater than 1 or less than depends on the values of X, Y, Z.

if Z+ Y > X then A will be greater than Xand if Z+ Y < X then A will be less than X.

So from St 2 we can't say any thing.

So answer is A


buddy,
If Z+Y < X --> (Z+Y)/ X <1 ---> X= (Z-Y) * sth smaller than 1 --->
X< Z-Y. At this point, it's tempting to conclude but when we look back:
we have Z+Y< X then X< Z-Y --> Z+Y< Z-Y ---> Y< 0 ---> unreasonable.
Thus, this case can't be correct.
The only case is Z+Y >X --> X> Z-Y


Hey buddy, I got your point. You are genius :good
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 [#permalink] New post 06 Apr 2006, 00:03
Hi laxie,

Why you think that number picking will take more time?

You just need to know the number tendency:
pick 5, 4 and 3
x will be greater than z-y by 2 or 1
pick 10, 8 and 6
x will be greater than z-y by 2 or 4
Thus the greater number you give for x,y,z. the greater difference it will be between x and z-y, where x will always be greate than z-y
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 [#permalink] New post 06 Apr 2006, 01:50
Dilshod wrote:
Hi laxie,

Why you think that number picking will take more time?



I did say it's the least time-consuming (refer to my above post) ...just that it may not be generic, coz you may ignore some ilicit cases.

For GMAT questions, there's no point that we shouldn't take it seriously. Especially when the GMAT conducted by new test provider contains more difficult quantitive questions than the old one.

As questions posted here to discuss, I just want to try to explain quantitively how and why a certain outcome is. That is to say, we should try to fathom the precision as much as possible. That is my point of view.
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 [#permalink] New post 06 Apr 2006, 05:19
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0.

(2) Z2 = X2 + Y2

as X,Y,Z are positive integers and X-Z-Y>0 => X>Z+Y
so, X is greatest in the set => X also has to be greater than Z-Y and so A will give the answer

in B, Z 2=X2+Y2 I was forced to think in terms of pythagoras theorem and so Z is greatest length.
and X might or might not be greater than Z-Y so I we cannot get answer from B

Answer is A
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 [#permalink] New post 06 Apr 2006, 05:27
Friends, please help me.

Sometime i realy get dejected looking at the way i arrive at the solution. My answer might be right but the way i think (sometime) is diffrent from the way this whole forum look at the problem.

PLEASE LOOK AT MY REPLY ABOVE. HERE I THOUGHT IN TERMS OF PYTHAGORAS THEOREM!!

I am worried some of my habits might cost me extra bits .... Any inputs will be helpful.
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 [#permalink] New post 06 Apr 2006, 05:47
laxieqv wrote:
Dilshod wrote:
Hi laxie,

Why you think that number picking will take more time?



I did say it's the least time-consuming (refer to my above post) ...just that it may not be generic, coz you may ignore some ilicit cases.

For GMAT questions, there's no point that we shouldn't take it seriously. Especially when the GMAT conducted by new test provider contains more difficult quantitive questions than the old one.

As questions posted here to discuss, I just want to try to explain quantitively how and why a certain outcome is. That is to say, we should try to fathom the precision as much as possible. That is my point of view.

yes, I agree with you that it is useful to try to solve the question "the right" way here on forum. But not on real GMAT. On real test you should learn to analize and find the easiest and fastest way while not harming your score. That is why, I think we should at least try to think outside of the box. Hope, I wasn't harsh.

Quote:
in B, Z 2=X2+Y2 I was forced to think in terms of pythagoras theorem and so Z is greatest length.
and X might or might not be greater than Z-Y so I we cannot get answer from B

Can you give me ANY example where Z-Y>X, using positive integers?
  [#permalink] 06 Apr 2006, 05:47
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