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# If x, y and z are positive integers, is x% of y bigger than

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If x, y and z are positive integers, is x% of y bigger than [#permalink]

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20 Jul 2009, 10:24
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35% (medium)

Question Stats:

63% (00:54) correct 37% (00:49) wrong based on 200 sessions

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If x, y and z are positive integers, is x% of y bigger than y% of Z?

(1) x= z
(2) z –y = y - x

[Reveal] Spoiler:
What not B is also ans.

Because statement B says x<y<z, so the statement is enough to ans the statement.
[Reveal] Spoiler: OA

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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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16 Sep 2012, 19:21
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Probably this approach of solving the question should help you understand:

x% of y is bigger than y% of z

essentially question is asking whether
xy >yz
or xy-yz >0
or y(x-z) >0 ? ->Lets call it 'The question statement' or TQS

We know that x, y and z are positive integers. Therefore y >0 and hence we would be able to answer if we know about (x-z) [Sign or Value either should do]

1. x=z

Putting in TQS we can see y(x-z) = y* 0 = 0. so we know the relation x% y = y% z. Therefore SUFFICIENT

2. z-y = y-x
or (z+x) =2Y

So we know z+x is positive (But we already knew it since X and Z both are positive) However we still do not know about x-z. Therefore we do not know about TQS. Therefore , INSUFFICIENT.

So Ans is "A"
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Re: GMat club test M11 doubt [#permalink]

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17 Sep 2012, 05:00
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Expert's post
ctiger100 wrote:
How is it possible that Z-Y=Y-X if X=Z according to (1)? I read somewhere 1 and 2 can't conflict on DS questions

The statements are not contradictory here; they can both be true if x=y=z. For example, if x = y = z = 6, you'll find that both statements are true, and these values agree with the restrictions in the question itself.
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Re: GMat club test M11 doubt [#permalink]

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20 Jul 2009, 10:30
No where is it mentioned,nor can you deduce that x < y < z,

Stmt B says Z –y = y - x

which means y is equidistant from x and z

could be

z.................y..............x
or
x............y................z

So InSUFF

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Re: GMat club test M11 doubt [#permalink]

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16 Sep 2012, 18:30
How is it possible that Z-Y=Y-X if X=Z according to (1)? I read somewhere 1 and 2 can't conflict on DS questions

If z and x are equal, you'd get a negative on one side of the equation and a positive on the other side... Just saying (1) x=z conflicts with
(2) z - y = y- x unless they are all zero, but that isn't possible since the question says they're positive.

skpMatcha wrote:
No where is it mentioned,nor can you deduce that x < y < z,

Stmt B says Z –y = y - x

which means y is equidistant from x and z

could be

z.................y..............x
or
x............y................z

So InSUFF

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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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17 Sep 2012, 05:53
Answer is A.
Question is (X/100)Y > (Y/100)Z --> Is X>Z ?
Statement 1 Directly states that X=Z, so X is not greater than Z.
Statement 2 Y=(X+Z)/2 ---> We just get to know that Y is the midpoint of X & Z. Magnitude of either X or Z is not known.
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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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26 Sep 2012, 07:25
Expert's post
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vaivish1723 wrote:
If x, y and z are positive integers, is x% of y bigger than y% of Z?

(1) x= z
(2) z –y = y - x

[Reveal] Spoiler:
What not B is also ans.

Because statement B says x<y<z, so the statement is enough to ans the statement.

If x, y, and z are positive integers, is x% of y greater than y% of z?

The question asks: is $$\frac{x}{100}*y>\frac{y}{100}*z$$? --> Since $$y$$ is a positive integer we can safely reduce by $$\frac{y}{100}$$ and the question becomes: is $$x>z$$? Notice that the answer to the question does not depend on the value of $$y$$.

(1) x=z. Answer to the question is NO. Sufficient.
(2) z-y=y-x --> $$x+z=2y$$. Clearly insufficient to say whether $$x>z$$.

Answer: A.
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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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13 Sep 2013, 04:47
Bunuel wrote:
vaivish1723 wrote:
If x, y and z are positive integers, is x% of y bigger than y% of Z?

(1) x= z
(2) z –y = y - x

[Reveal] Spoiler:
What not B is also ans.

Because statement B says x<y<z, so the statement is enough to ans the statement.

If x, y, and z are positive integers, is x% of y greater than y% of z?

The question asks: is $$\frac{x}{100}*y>\frac{y}{100}*z$$? --> Since $$y$$ is a positive integer we can safely reduce by $$\frac{y}{100}$$ and the question becomes: is $$x>z$$? Notice that the answer to the question does not depend on the value of $$y$$.

(1) x=z. Answer to the question is NO. Sufficient.
(2) z-y=y-x --> $$x+z=2y$$. Clearly insufficient to say whether $$x>z$$.

Answer: A.

Bunuel, can you please explain how you reduce by $$\frac{y}{100}$$ ? If we divide the inequality by $$\frac{y}{100}$$ then we get 100 on one side and 1 on the other, no ?

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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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13 Sep 2013, 04:49
Skag55 wrote:
Bunuel wrote:
vaivish1723 wrote:
If x, y and z are positive integers, is x% of y bigger than y% of Z?

(1) x= z
(2) z –y = y - x

[Reveal] Spoiler:
What not B is also ans.

Because statement B says x<y<z, so the statement is enough to ans the statement.

If x, y, and z are positive integers, is x% of y greater than y% of z?

The question asks: is $$\frac{x}{100}*y>\frac{y}{100}*z$$? --> Since $$y$$ is a positive integer we can safely reduce by $$\frac{y}{100}$$ and the question becomes: is $$x>z$$? Notice that the answer to the question does not depend on the value of $$y$$.

(1) x=z. Answer to the question is NO. Sufficient.
(2) z-y=y-x --> $$x+z=2y$$. Clearly insufficient to say whether $$x>z$$.

Answer: A.

Bunuel, can you please explain how you reduce by $$\frac{y}{100}$$ ? If we divide the inequality by $$\frac{y}{100}$$ then we get 100 on one side and 1 on the other, no ?

No.

$$\frac{x}{100}*y>\frac{y}{100}*z$$ --> $$\frac{y}{100}*x>\frac{y}{100}*z$$ --> $$x>z$$.
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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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13 Sep 2013, 05:18
Ah, my eyes got confused, for some reason I didn't think of xy/100, I somehow thought of x/100 + y or something.
Thanks anyway!

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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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29 May 2014, 05:13
Bunuel wrote:
vaivish1723 wrote:
If x, y and z are positive integers, is x% of y bigger than y% of Z?

(1) x= z
(2) z –y = y - x

[Reveal] Spoiler:
What not B is also ans.

Because statement B says x<y<z, so the statement is enough to ans the statement.

If x, y, and z are positive integers, is x% of y greater than y% of z?

The question asks: is $$\frac{x}{100}*y>\frac{y}{100}*z$$? --> Since $$y$$ is a positive integer we can safely reduce by $$\frac{y}{100}$$ and the question becomes: is $$x>z$$? Notice that the answer to the question does not depend on the value of $$y$$.

(1) x=z. Answer to the question is NO. Sufficient.
(2) z-y=y-x --> $$x+z=2y$$. Clearly insufficient to say whether $$x>z$$.

Answer: A.

Hi Bunnel,

I have a query.

in this question st1 i.e. ans A is quite clear. but I stuck with st2

here this is mentioned that z-y = y-x and x,y, And z are positive integers.

so now question asks.

(x/100) *y > (y/100)*z

in st2 i can take following values

z-y = y-x

z y y x

4 3 3 2

or
10 7 7 4

so we put this xyz values and we can get ans from st2 also. i choose D .

Please clarify this

Thanks

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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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29 May 2014, 07:00
PathFinder007 wrote:
Bunuel wrote:
vaivish1723 wrote:
If x, y and z are positive integers, is x% of y bigger than y% of Z?

(1) x= z
(2) z –y = y - x

[Reveal] Spoiler:
What not B is also ans.

Because statement B says x<y<z, so the statement is enough to ans the statement.

If x, y, and z are positive integers, is x% of y greater than y% of z?

The question asks: is $$\frac{x}{100}*y>\frac{y}{100}*z$$? --> Since $$y$$ is a positive integer we can safely reduce by $$\frac{y}{100}$$ and the question becomes: is $$x>z$$? Notice that the answer to the question does not depend on the value of $$y$$.

(1) x=z. Answer to the question is NO. Sufficient.
(2) z-y=y-x --> $$x+z=2y$$. Clearly insufficient to say whether $$x>z$$.

Answer: A.

Hi Bunnel,

I have a query.

in this question st1 i.e. ans A is quite clear. but I stuck with st2

here this is mentioned that z-y = y-x and x,y, And z are positive integers.

so now question asks.

(x/100) *y > (y/100)*z

in st2 i can take following values

z-y = y-x

z y y x

4 3 3 2

or
10 7 7 4

so we put this xyz values and we can get ans from st2 also. i choose D .

Please clarify this

Thanks

You cannot say whether a statement is sufficient based only on couple of examples.

Try to choose x and z, so that x>z and see what you get.

Another thing is that it seems that you don't understand the solution provided in my post:
The question asks: is $$\frac{x}{100}*y>\frac{y}{100}*z$$? --> Since $$y$$ is a positive integer we can safely reduce by $$\frac{y}{100}$$ and the question becomes: is $$x>z$$? Notice that the answer to the question does not depend on the value of $$y$$.

How can you get whether $$x>z$$ from $$x+z=2y$$? Also, it's almost always better to combine like terms in equations, so it's better to write $$x+z=2y$$ from z-y=y-x.
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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]

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12 Jul 2017, 20:20
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Re: If x, y and z are positive integers, is x% of y bigger than   [#permalink] 12 Jul 2017, 20:20
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