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Re: GMat club test M11 doubt [#permalink]
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
Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]
16 Sep 2012, 18:21
1
This post received KUDOS
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.
Re: GMat club test M11 doubt [#permalink]
17 Sep 2012, 04:00
1
This post received KUDOS
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: If x, y and z are positive integers, is x% of y bigger than [#permalink]
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. _________________
I will rather do nothing than be busy doing nothing - Zen saying
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 \(x>z\).
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 ?
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 ?
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 .
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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