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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] New post 20 Jul 2009, 09:24
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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] New post 16 Sep 2012, 18:21
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:lol:
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] New post 17 Sep 2012, 04:00
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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] New post 20 Jul 2009, 09:30
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


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Re: GMat club test M11 doubt [#permalink] New post 16 Sep 2012, 17:30
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] New post 17 Sep 2012, 04:53
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] New post 26 Sep 2012, 06:25
Expert's post
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] New post 13 Sep 2013, 03:47
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] New post 13 Sep 2013, 03:49
Expert's post
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] New post 13 Sep 2013, 04:18
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] New post 29 May 2014, 04:13
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] New post 29 May 2014, 06:00
Expert's post
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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x, y and z are positive integers, is x% of y bigger than   [#permalink] 29 May 2014, 06:00
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