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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. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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]
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. _________________

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...