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# If x, y, and z are numbers, is z = 18 ?

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If x, y, and z are numbers, is z = 18 ? [#permalink]  04 Feb 2014, 00:52
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81% (01:37) correct 19% (00:46) wrong based on 215 sessions
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x, y, and z are numbers, is z = 18 ?

(1) The average (arithmetic mean) of x, y, and z is 6.
(2) x = -y

Data Sufficiency
Question: 74
Category: Arithmetic Statistics
Page: 158
Difficulty: 600

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Math Expert
Joined: 02 Sep 2009
Posts: 30392
Followers: 5093

Kudos [?]: 57402 [0], given: 8811

Re: If x, y, and z are numbers, is z = 18 ? [#permalink]  04 Feb 2014, 00:52
Expert's post
SOLUTION

If x, y, and z are numbers, is z = 18 ?

(1) The average (arithmetic mean) of x, y, and z is 6 --> x + y + z = 6*3 = 18. We cannot say whether z = 18. Not sufficient.

(2) x = -y --> x + y = 0. Clearly insufficient.

(1)+(2) x + y + z = 18 --> 0 + z = 18 --> z = 18. Sufficient.

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Kudos [?]: 65 [0], given: 29

Re: If x, y, and z are numbers, is z = 18 ? [#permalink]  04 Feb 2014, 03:38
Ans. C

From S1: x+y+z=18.Not sufficient

From S2: x+y=0.Not sufficent to determine z

Combining the two we get z=18 definitely.Both are together sufficient.
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Joined: 29 Oct 2013
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Kudos [?]: 9 [1] , given: 37

Re: If x, y, and z are numbers, is z = 18 ? [#permalink]  04 Feb 2014, 04:37
1
KUDOS
My answer is also C (explanation is already provided, so not doing honours)
Math Expert
Joined: 02 Sep 2009
Posts: 30392
Followers: 5093

Kudos [?]: 57402 [0], given: 8811

Re: If x, y, and z are numbers, is z = 18 ? [#permalink]  08 Feb 2014, 04:38
Expert's post
SOLUTION

If x, y, and z are numbers, is z = 18 ?

(1) The average (arithmetic mean) of x, y, and z is 6 --> x + y + z = 6*3 = 18. We cannot say whether z = 18. Not sufficient.

(2) x = -y --> x + y = 0. Clearly insufficient.

(1)+(2) x + y + z = 18 --> 0 + z = 18 --> z = 18. Sufficient.

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Kudos [?]: 36 [0], given: 216

If x, y, and z are numbers, is z = 18 ? [#permalink]  30 Sep 2015, 12:10
Bunuel wrote:
SOLUTION

If x, y, and z are numbers, is z = 18 ?

(1) The average (arithmetic mean) of x, y, and z is 6 --> x + y + z = 6*3 = 18. We cannot say whether z = 18. Not sufficient.

(2) x = -y --> x + y = 0. Clearly insufficient.

(1)+(2) x + y + z = 18 --> 0 + z = 18 --> z = 18. Sufficient.

Dear Bunnel,
Here,Is it possible to guess that number could be in a fraction format too?like $$\frac{1}{5}$$+$$\frac{-1}{5}$$=0 for x+y=0
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Math Expert
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Kudos [?]: 57402 [1] , given: 8811

Re: If x, y, and z are numbers, is z = 18 ? [#permalink]  30 Sep 2015, 22:16
1
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Expert's post
AbdurRakib wrote:
Bunuel wrote:
SOLUTION

If x, y, and z are numbers, is z = 18 ?

(1) The average (arithmetic mean) of x, y, and z is 6 --> x + y + z = 6*3 = 18. We cannot say whether z = 18. Not sufficient.

(2) x = -y --> x + y = 0. Clearly insufficient.

(1)+(2) x + y + z = 18 --> 0 + z = 18 --> z = 18. Sufficient.

Dear Bunnel,
Here,Is it possible to guess that number could be in a fraction format too?like $$\frac{1}{5}$$+$$\frac{-1}{5}$$=0 for x+y=0

Yes, x and y could be fractions. In fact they could be any real numbers satisfying x + y = 0.
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Re: If x, y, and z are numbers, is z = 18 ?   [#permalink] 30 Sep 2015, 22:16
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