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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

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. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

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. _________________

After so many confusing thoughts, I decided to write this post. I am yet to finalize the list of B-Schools I am applying to. For a background check, I...