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If x and y are integers and x > 0, is y > 0?

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If x and y are integers and x > 0, is y > 0? [#permalink]

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If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x


[Reveal] Spoiler:
From 1, y can have +ve and -ve values. For example, x = 2, y = 1 or x = 2, y = -1. Thus, not sufficient to answer whether y >0

From 2, -y < x or x +y > 0, again not sufficient. x = 0.9, y = -0.8 or x = 0.9, y = 0.8

Combining 1 & 2
7x - 2y > 0
x + y > 0

----------------

7x - 2y > 0
7x + 7y > 0 (multiplying earlier equation 2 by 7)

summing the above two equations, can't we derive 5y > 0 and thus y > 0?
[Reveal] Spoiler: OA

Last edited by mvictor on 27 Jul 2017, 05:07, edited 1 time in total.
included author's explanations in a spoiler

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 11 Sep 2010, 14:17
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Orange08 wrote:
If x and y are integers and x > 0, is y > 0?
(1) 7x – 2y > 0
(2) -y < x


From 1, y can have +ve and -ve values. For example, x = 2, y = 1 or x = 2, y = -1. Thus, not sufficient to answer whether y >0

From 2, -y < x or x +y > 0, again not sufficient. x = 0.9, y = -0.8 or x = 0.9, y = 0.8

Combining 1 & 2
7x - 2y > 0
x + y > 0

----------------

7x - 2y > 0
7x + 7y > 0 (multiplying earlier equation 2 by 7)

summing the above two equations, can't we derive 5y > 0 and thus y > 0?


If \(y=0\) and \(x\) is any positive integer both statements hold true and the answer to the question is NO.
If \(y=1\) and \(x=2\) again both statements hold true and the answer to the question is YES.
Two different answers. Not sufficient.

Answer: E.

As for your question: when we sum \(7x - 2y > 0\) and \(7x + 7y > 0\) we'll get \(14x+5y>0\) and not \(5y > 0\).

Hope it helps.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 11 Sep 2010, 14:26
Bunuel wrote:
Orange08 wrote:
x + y > 0

----------------

As for your question: when we sum \(7x - 2y > 0\) and \(7x + 7y > 0\) we'll get \(14x+5y>0\) and not \(5y > 0\).

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to
9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 11 Sep 2010, 14:34
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Orange08 wrote:
Bunuel wrote:
Orange08 wrote:
x + y > 0

----------------

As for your question: when we sum \(7x - 2y > 0\) and \(7x + 7y > 0\) we'll get \(14x+5y>0\) and not \(5y > 0\).

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to
9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?


I don't think simultaneous equations are solved on inequalities.

If A > 0 and B> 0; it doesn't mean A-B>0. A=2, B 5 will prove this.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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Orange08 wrote:
Bunuel wrote:
Orange08 wrote:
x + y > 0

----------------

As for your question: when we sum \(7x - 2y > 0\) and \(7x + 7y > 0\) we'll get \(14x+5y>0\) and not \(5y > 0\).

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to
9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?


You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So we can only add \(7x - 2y > 0\) and \(7x + 7y > 0\) as their signs are in the same direction (>).

Hope it helps.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 12 Sep 2010, 00:37
Bunuel wrote:

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So we can only add \(7x - 2y > 0\) and \(7x + 7y > 0\) as their signs are in the same direction (>).

Hope it helps.


Absolutely fantastic. Thanks Bunuel once again. This has indeed cleared one of the flaws in my understanding.

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 03 May 2014, 20:22
If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x


Bunuel, please have a look at this.

statement1. 7x - 2y > 0
=> x > 2y/7---------(1)

y can be negative or positive. Not sufficient

statement2. -y < x
=> x + y > 0---------(2)

Again, y can be negative or positive. Not Sufficient.

statement 1 & 2 Together.

from eq. (1) & (2),
[a value > 2y/7] + y > 0
lets consider least value of x (actually this would be lesser than the least).
2y/7 + y > 0
=> 9y/7 > 0 or (9/7)y>0
therefore, y must be > than 0.

Answer C.

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 04 May 2014, 02:09
honey86 wrote:
If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x


Bunuel, please have a look at this.

statement1. 7x - 2y > 0
=> x > 2y/7---------(1)

y can be negative or positive. Not sufficient

statement2. -y < x
=> x + y > 0---------(2)

Again, y can be negative or positive. Not Sufficient.

statement 1 & 2 Together.

from eq. (1) & (2),
[a value > 2y/7] + y > 0
lets consider least value of x (actually this would be lesser than the least).
2y/7 + y > 0
=> 9y/7 > 0 or (9/7)y>0
therefore, y must be > than 0.

Answer C.


Why are you considering the least value of x??? Nothing prevents x to be greater than this value and in this case your logic will not hold.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 11 Feb 2017, 00:17
1) 7x-2y>0

7(8)-2(3) = positive
7(8)-2(-3) = positive

y can be neg or pos. NOT sufficient.

2) -y<x

x = 8 , y can equal 3
x = 8, -3<8, so y can be positive

x = 8, y can equal -3
x=8, -(-3)<8, 6<8, y can be negative too. Not sufficient.

c) combined: Look at the breakdown from statement (1) with numbers that also satisfy statement (2)'s criteria. y can be neg OR positive.

Answer is Choice E.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 27 Jul 2017, 05:11
from 1 alone
y can be negative and y can be positive
say y=-1 and x=1.
7+2=9. 9>0

say y=1 and x=1
7-2=5. 5>0

1 alone is insufficient. A and D are out.

statement 2 says -y<x.
rewrite 0<x+y
y can be negative and y can be positive.
say y=-1 and x=2
2-1 = 1. 1>0. works

say y=1 and x=1
1+1=2. 2>0. works.

B is out, so we are left with either C or E - 50/50 chance we'd get it right, not bad
but combining, we see that both statements say that y can be > or < 0.
C is out, and E is the answer.

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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New post 19 Oct 2017, 11:10
Bunuel , could you illustrate the graphical solution for this problem too?

The graph of the first statement passes through the origin and lies to the right of the line in the 1st quadrant and to its left in the third quadrant. If x is positive, only values in the first quadrant hold. Then y>0, is it not? OA a?

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Re: If x and y are integers and x > 0, is y > 0?   [#permalink] 19 Oct 2017, 11:10
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