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Re: If x, y and z are positive integers, is x% of y bigger than [#permalink]
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: If x, y and z are positive integers, is x% of y bigger than [#permalink]
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]
Expert Reply
If \(x\), \(y\), and \(z\) are positive integers, is \(x\%\) of \(y\) greater than \(y\%\) of \(z\)?

The question essentially asks: is \(\frac{x}{100} * y > \frac{y}{100} * z\)? Given that \(y\) is a positive integer, we can safely divide both sides by \(\frac{y}{100}\). Thus, the question simplifies to: is \(x > z\)? It's important to note that the answer does not depend on the value of \(y\).

(1) \(x=z\).

The answer to the question is NO. This is sufficient.

(2) \(z-y=y-x\).

If we rearrange, we get \(x+z=2y\). This doesn't provide enough information to determine whether \(x > z\). Thus, it's insufficient.


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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