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If x > 0, y > 0, and z > 0, then 2/x + (y + 1/z)/2 ?

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Math Expert
Joined: 02 Sep 2009
Posts: 58452
If x > 0, y > 0, and z > 0, then 2/x + (y + 1/z)/2 ?  [#permalink]

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19 Feb 2019, 22:36
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Difficulty:

5% (low)

Question Stats:

96% (01:48) correct 4% (03:24) wrong based on 24 sessions

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If x > 0, y > 0, and z > 0, then $$\frac{2}{x} + \frac{y + \frac{1}{z}}{2}$$?

A. 2x/(2x + 2y)

B. (4 + xy + x)/(2x)

C. (y + z)/(2x)

D. (xyz + 4)/(x + y + z)

E. (4z + xyz + x)/(2xz)

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Re: If x > 0, y > 0, and z > 0, then 2/x + (y + 1/z)/2 ?  [#permalink]

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19 Feb 2019, 22:49
Option E.

The question stem translates to E when

w.r.t. (y + 1/z)/2 where Z is the LCM for y+1/z. It results in ((yz+1)/z)/2. it can be simplified to (yz+1)/2z.

The LCM between X and 2z is 2xz and the overall term of 2/x and the above is simplified to option E
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Re: If x > 0, y > 0, and z > 0, then 2/x + (y + 1/z)/2 ?  [#permalink]

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20 Feb 2019, 01:30
Bunuel wrote:
If x > 0, y > 0, and z > 0, then $$\frac{2}{x} + \frac{y + \frac{1}{z}}{2}$$?

A. 2x/(2x + 2y)

B. (4 + xy + x)/(2x)

C. (y + z)/(2x)

D. (xyz + 4)/(x + y + z)

E. (4z + xyz + x)/(2xz)

solve expression
$$\frac{2}{x} + \frac{y + \frac{1}{z}}{2}$$

we get
(4z + xyz + x)/(2xz)
IMO E
Re: If x > 0, y > 0, and z > 0, then 2/x + (y + 1/z)/2 ?   [#permalink] 20 Feb 2019, 01:30
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If x > 0, y > 0, and z > 0, then 2/x + (y + 1/z)/2 ?

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