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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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27 Jun 2017, 00:34
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5
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Difficulty:

55% (hard)

Question Stats:

62% (01:45) correct 38% (02:11) wrong based on 65 sessions

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Is y > 0?

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

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

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

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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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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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27 Jun 2017, 03:45
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.......

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

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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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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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