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What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7

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What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7  [#permalink]

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New post 02 Nov 2018, 02:54
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A
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C
D
E

Difficulty:

  15% (low)

Question Stats:

75% (00:44) correct 25% (00:56) wrong based on 44 sessions

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Re: What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7  [#permalink]

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New post 02 Nov 2018, 03:52
2
carcass wrote:
What is the average (arithmetic mean) of 3x and 6y?

(1) x + 2y = 7
(2) x + y = 5



Question: (3x+6y)/2 = ?

Statement 1: x + 2y = 7

i.e. (3x+6y)/2 = 3*(x+2y)/2 = 3*7/2 = 10.5
SUFFICIENT

Statement 2: x + y = 5
i.e. (3x+6y)/2 = 3*(x+2y)/2
so we need x+2y but we have x+y which can not give us the value of x+2y hence
NOT SUFFICIENT

Answer: Option A
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Re: What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7  [#permalink]

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New post 02 Nov 2018, 05:30
Can we solve it by using a linear equation system?

(1) x + 2y = 7
(2) x + y = 5

Substitute:

= > y = 2;
3x + 6y/2 -> 3x+12/2

x=-4 ?

Avarage = 3*(-4) + 12 /2 = 0
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Re: What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7  [#permalink]

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New post 02 Nov 2018, 05:47
Mich18 wrote:
Can we solve it by using a linear equation system?

(1) x + 2y = 7
(2) x + y = 5

Substitute:

= > y = 2;
3x + 6y/2 -> 3x+12/2

x=-4 ?

Avarage = 3*(-4) + 12 /2 = 0


You can NOT substitute values like this...
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What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7  [#permalink]

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New post 04 Nov 2018, 02:14
GMATinsight wrote:
Mich18 wrote:
Can we solve it by using a linear equation system?

(1) x + 2y = 7
(2) x + y = 5

Substitute:

= > y = 2;
3x + 6y/2 -> 3x+12/2

x=-4 ?

Avarage = 3*(-4) + 12 /2 = 0


You can NOT substitute values like this...





I have confuse, it can be solved as well by Liner equesion.

(1) x + 2y = 7
(2) x + y = 5

Substitute:

= > y = 2;
(1) x + 2y = 7 => x=3
or
(2) x + y = 5 => x=3

Avarage = 3*(3) + 6*2/2 = 10.5

Does it mean, that C is also O.k ??
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What is the average (arithmetic mean) of 3x and 6y? (1) x + 2y = 7 &nbs [#permalink] 04 Nov 2018, 02:14
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