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Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z

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Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z  [#permalink]

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New post 27 Jun 2017, 00:34
1
5
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:48) correct 44% (02:15) wrong based on 77 sessions

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Re: Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z  [#permalink]

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New post 27 Jun 2017, 00:49
1
Bunuel wrote:
Is y > 0?

(1) \(3x + 2y > z + 4\)

(2) \(6x - 7 < 5y + 2z\)



3x + 2y > z + 4
y> (z + 4 -3x)/2
y> (2z + 8 - 6x )/4
4y< (6x-2z-8)

6x - 7 < 5y + 2z
5y> (6x - 7 - 2z)
let 6x-2z-7 = k

4y<k-1
5y>k

k/5<y<(k-1)/4

Hence k cannot be negative
if k is not negative then y has to be positive

Hence C
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Re: Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z  [#permalink]

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New post 27 Jun 2017, 01:02
We need to find out if y > 0

1. \(3x + 2y > z + 4\)
Consider statement 1,
since we are not sure about value of z and x
we can't say anything for sure about the value of y. Insufficient.

(2) \(6x - 7 < 5y + 2z\)
This can be rewritten as \(6x - 5y < 2z +7\)
Adding one to both sides, \(6x - 5y +1 < 2z +8\)
Dividing by 2, we get \(\frac{(6x - 5y +1)}{2} < z +4\)
Hence, we can say that \(3x + \frac{(1 - 5y)}{2} < z +4\)
Again, since we have no idea about the value of z and x,
we can't say anything for sure about the value of y. Insufficient.

Now, on combining both the statements, we can get the following relation
\(3x + 2y > z + 4 > 3x + \frac{(1 - 5y)}{2}\)
Since 3x is common on both the sides, if y is negative.
\(3x + 2y < 3x + \frac{(1 - 5y)}{2}\)
Therefore, this relation can only work, when y is positive. (Option C)
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Re: Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z  [#permalink]

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New post 27 Jun 2017, 03:45
1
1
Bunuel wrote:
Is y > 0?

(1) \(3x + 2y > z + 4\)

(2) \(6x - 7 < 5y + 2z\)


DS : y > 0

Option 1 :

\(3x + 2y > z + 4\)
\(2y > z - 3x + 4\)
\(4y > 2z - 6x + 8\)


NOT SUFFICIENT....

Option 2 :
\(6x - 7 < 5y + 2z\)
\(6x - 2z < 5y + 7\)
\(2z - 6x > - (5y+7)\)

NOT SUFFICIENT....

Combined :
4y > - (5y+7) +8
9y > 1
y > 1/9

SUFFICIENT.......

Answer C.
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Re: Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z  [#permalink]

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New post 27 Jun 2017, 05:22
Bunuel wrote:
Is y > 0?

(1) \(3x + 2y > z + 4\)

(2) \(6x - 7 < 5y + 2z\)


IMO C

St1: 2y>z+4-3x

Nothing can be made out with out knowing x and z.

St2: 6x - 7 < 5y + 2z
rearranging 5y>6x - 7-2z

again no clue about x and z

combining both
Multiplying st1 by 2 on both side

St1= 4y>2z+8-6x
st2 = 5y>6x - 7-2z
+
-------------------------------------
9y>1
y>1/9

hence C
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Re: Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z  [#permalink]

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New post 26 Mar 2019, 10:47
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Re: Is y > 0? (1) 3x + 2y > z + 4 (2) 6x - 7 < 5y + 2z   [#permalink] 26 Mar 2019, 10:47
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