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If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x

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If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x  [#permalink]

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New post 09 Oct 2018, 02:56
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (01:55) correct 34% (01:35) wrong based on 62 sessions

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Re: If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x  [#permalink]

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New post 09 Oct 2018, 03:16
The answer is C.

1) let's use numbers to see the logic: say y=2. let's see if a "yes" answer is possible - that is, if y is larger than x: say x=1. This means 2/1=4/2 (z=4). definitely possible! now, is a "no" answer possible - can y-x<0? let's try: x=4. This means 2/4=1/2 (z=1). also possible!
we have found two contradicting answers, thus this data is insufficient - . eliminate answers A and D.
2) this tells us nothing about y, and thus nothing about the relationship between x and y, clearly insufficient - eliminate B

Combined: if z > x, then the second scenario we checked before (where y-x<o) is impossible, since it had z=1 and x=2. In fact, we can now see that any "no" answer will be impossible: if z>x, this means that z/x > 1, which means that the left side of the equation - y/x, must also be larger than 1: if y/x>1 ==> y - x >0 - sufficient! answer C.
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Re: If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x  [#permalink]

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New post 09 Oct 2018, 03:18
1
Bunuel wrote:
If x, y and z are positive integers, is y − x > 0?


(1) y/x = z/y

(2) z > x



we are looking for the positive difference between y and x.

Statement 1: As we don't the order of these integers it is not possible to determine the difference. NOT sufficient.

Statement : z>x. but information about y. NOT sufficient.


Combining 1 and 2:


y/x = z/y

\(z= y^2 / x.\)

we already know,

z>x

\(y^2/x >x\)
\(y^2>x^2.\)

we already know all these are integers.

thus, y -x >0.


The best answer is C.
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Re: If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x  [#permalink]

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New post 09 Oct 2018, 21:48
Bunuel wrote:
If x, y and z are positive integers, is y − x > 0?


(1) y/x = z/y

(2) z > x



chetan2u - your approach
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If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x  [#permalink]

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New post 09 Oct 2018, 22:09
1
saurabh9gupta wrote:
Bunuel wrote:
If x, y and z are positive integers, is y − x > 0?


(1) y/x = z/y

(2) z > x



chetan2u - your approach


Hi..

Info given
a) x, y and z are positive integers. SO you can cross multiply
b) is y-x>0..... or Is y>x?

(1) y/x = z/y..
\(\frac{y}{x} =\frac{z}{y}.........y^2=xz\)..
    case I : if \(x<z\), so \(z=x+a\), where a is a positive number....
    \(y^2=x(x+a)=x^2+ax\).... now ax is positive, so \(y^2>x^2.......y>x\) as y and x are positive
    case II : if \(x>z\), so \(z=x-a\), where a is a positive number....
    \(y^2=x(x-a)=x^2-ax\).... now ax is positive, so \(y^2<x^2.......y<x\) as y and x are positive
insuff

(2) z > x
nothing about y
insuff

combined..
case I of statement 1 is left
case I : if x<z, so z=x+a, where a is a positive number....\(y^2=x(x+a)=x^2+ax\).... now ax is positive, so \(y^2>x^2.......y>x\) as y and x are positive
so y>x....y-x>0
sufficient

C
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Re: If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x  [#permalink]

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New post 26 Nov 2019, 06:13
1) y/x = z/y => y^2 = xz. that means y is the middle no, however we dont know if it is in order of x,y,z or z,y,x. NOT SUFFICIENT.

2) z>x. Dont know about y. INSUFFICIENT.

1+2 , we now see z>x, so the the integers are in order of x<y<z. Thus y-x =0 .SUFFICIENT.

Correct Answer is C
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Re: If x, y and z are positive integers, is y−x>0? (1) y/x = z/y (2) z > x   [#permalink] 26 Nov 2019, 06:13
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