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Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3

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Bunuel
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Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3 [#permalink] New post   20 Jul 2022, 01:30
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Question Stats:

35% (02:07) correct 65% (02:30) wrong based on 52 sessions
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Is \(x = 1, y = 2,\) and \(z = 3\)?

(1) \(5x + 2z + 3 = 3x + 4y = y + 2z + 3\)

(2) \(5x + z + 3 = 3x + 4y\) and \(3x + 4y = y + 2z + 3\)


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AUG1523
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Re: Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3 [#permalink] New post   20 Jul 2022, 07:17
Is the answer B? I simply substituted and found that 1 is not true, 14=11=11. Statement 2 behaves perfectly (all 11). Is it B? I am suspicious since this is marked 700 but seemed too easy
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Re: Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3 [#permalink] New post   12 Aug 2022, 05:08
AUG1523 wrote:
Is the answer B? I simply substituted and found that 1 is not true, 14=11=11. Statement 2 behaves perfectly (all 11). Is it B? I am suspicious since this is marked 700 but seemed too easy



I did the same. I don't understand how the answer is A and yes this seems too easy.
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Re: Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3 [#permalink] New post   12 Aug 2022, 09:26
The answer should be E

1) put the value of xyz into the formula, then 14=11=11, insufficient

2) from the formula we know that:
2x-4y+z+3=0
3x+3y-2z-3=0
Then 5x-y-z=0
So y=5x-z

Put this result into the formula “5x+z+3=3x+4y”
Then 5x+z+3=3x+4*(5x-z)=3x+20x-4z
Then 18x=5z+3
Then z=(18x-3)/5
If z is integer, then when 18x is divided by 5, the reminder must be 3.
When x=1, then y=2 and z=3
When x=6, then y=9 and z=21
Insufficient

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Re: Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3 [#permalink] New post   13 Aug 2022, 00:55
AUG1523 wrote:
Is the answer B? I simply substituted and found that 1 is not true, 14=11=11. Statement 2 behaves perfectly (all 11). Is it B? I am suspicious since this is marked 700 but seemed too easy


Answer IMO is A because by using Statement 1, we know for sure that for the tri-equation to be true, values given for x,y and z cannot hold true. I looked at the statement as true and then evaluated whether the given values can hold or not.

In Statement 2, the values CAN hold true and cannot hold true too. Thus, it's insufficient.

Hope this helps.
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Re: Is x = 1, y = 2, and z = 3? (1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3 [#permalink]
13 Aug 2022, 00:55
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