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If x, y and z are positive integers, is x% of y bigger than
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20 Jul 2009, 10:24
Question Stats:
67% (01:34) correct 33% (01:39) wrong based on 231 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 What not B is also ans.
Because statement B says x<y<z, so the statement is enough to ans the statement.
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Re: GMat club test M11 doubt
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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
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16 Sep 2012, 18:30
How is it possible that ZY=YX 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
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16 Sep 2012, 19:21
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 xyyz >0 or y(xz) >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 (xz) [Sign or Value either should do] 1. x=z Putting in TQS we can see y(xz) = y* 0 = 0. so we know the relation x% y = y% z. Therefore SUFFICIENT 2. zy = yx 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 xz. Therefore we do not know about TQS. Therefore , INSUFFICIENT. So Ans is "A"
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Re: GMat club test M11 doubt
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17 Sep 2012, 05:00
ctiger100 wrote: How is it possible that ZY=YX 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: If x, y and z are positive integers, is x% of y bigger than
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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
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26 Sep 2012, 07:25
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 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) zy=yx > \(x+z=2y\). 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
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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 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) zy=yx > \(x+z=2y\). 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
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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 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) zy=yx > \(x+z=2y\). 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
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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
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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 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) zy=yx > \(x+z=2y\). 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 zy = yx 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 zy = yx 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
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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 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) zy=yx > \(x+z=2y\). 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 zy = yx 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 zy = yx 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 zy=yx.
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Re: If x, y and z are positive integers, is x% of y bigger than
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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
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