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Re: Tuesday Q2 - Value of Z [#permalink]
13 Oct 2009, 06:07

hogann wrote:

If 2xy + z = 9, what is the value of the positive integer z?

(1) xyz – z2 = 0

(2) x + y – 3z = -5

not entirely sure of this but A seems sufficient

1. xyz -z2 = 0 then z(xy-2) =0 since z is +ve so xy -2 =0 so xy = 2 and we get value of z. so A suff 2. using the equation given in question and option we are left with 6xy +x+y =22. So B is insuff

Re: Tuesday Q2 - Value of Z [#permalink]
13 Oct 2009, 08:39

asterixmatrix wrote:

not entirely sure of this but A seems sufficient

1. xyz -z2 = 0 then z(xy-2) =0 since z is +ve so xy -2 =0 so xy = 2 and we get value of z. so A suff 2. using the equation given in question and option we are left with 6xy +x+y =22. So B is insuff

I think with 1 you have xy - 2 = 0 => xy = 2 => 2z - 2z = 0

This does not give you the value of Z i'm sorry ! Z can be 1 or 4324236543253465,4 !

Re: Tuesday Q2 - Value of Z [#permalink]
14 Oct 2009, 02:27

hi,,, pl dont substitute the value of xy in the eq from which u have derived the value xy.. it will ofcourse be 0.. substitute it in main eq 2xy + z = 9....ie..2*2+z=9....z=5... sufficient

Re: Tuesday Q2 - Value of Z [#permalink]
15 Oct 2013, 06:31

yangsta8 wrote:

hogann wrote:

If 2xy + z = 9, what is the value of the positive integer z?

(1) xyz – z^2 = 0

(2) x + y – 3z = -5

Had a problem with exponents this morning!

Reorganising question stem: 2xy+z=9 => xy=(9-z)/2

Using Equation 1) xyz-z^2 =0 Substitute in the question stem we get \frac{z(9-z)}{2} - z^2=0 z(9-z) - 2z^2=0 9z-z^2-2z^2=0 9z-3z^2=0 3z(3-z)=0

Z=0 or Z=3 but we know from quesiton z is positive. Hence A is suff.

Using Equation 2) We cannot solve for Z. We can only get it down to two variables.

Hence ANS = A

Good job buddy, I didn't see that one coming either. Guys, always remember that when you are given something like in this case z is a positive integer, then they usually if not always say it for some reason Just to keep in mind

Re: If 2xy + z = 9, what is the value of the positive integer z? [#permalink]
15 Oct 2013, 06:50

3

This post received KUDOS

Expert's post

If 2xy + z = 9, what is the value of the positive integer z?

(1) xyz - z^2 = 0 --> \(z(xy-z)=0\) --> \(z=0\) or \(z=xy\). Since given that z is a positive integer, then discard \(z=0\) and we are left with \(z=xy\). From \(2xy + z = 9\) we get that \(2z + z = 9\) --> \(z=3\). Sufficient.

(2) x + y - 3z = -5. We cannot solve for z. Not sufficient.

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