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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Manbehindthecurtain wrote:
Are x and y both positive?

(1) $$2x - 2y = 1$$

(2) $$\frac{x}{y} > 1$$

(1) $$2x - 2y = 1$$
$$2(x - y) = 1$$
$$x - y = \frac{1}{2}$$
$$x = y + \frac{1}{2}$$

INSUFFICIENT

(2) $$\frac{x}{y} > 1$$
Only thing we know is that x and y are either both positive or both negative.

INSUFFICIENT

Together

$$\frac{x}{y} > 1$$
$$1 + \frac{y}{2} > 1$$
$$\frac{y}{2} > 0$$

Thus, y is greater than 0 and as x is greater than y, we know that both are positive.

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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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1
1
Manbehindthecurtain wrote:
Are x and y both positive?

(1) $$2x - 2y = 1$$

(2) $$\frac{x}{y} > 1$$

Quote:
If we are given a statement x/y>1, and we are asked is x and y both positive: is my below mentioned approach correct or is there some gap in my understanding, please help!.

x/y >1
Multiplying above equation by by y^2 both side-------------------------->we can write it as xy>y^2
Therefore,
y(x-y)>0
y(y-x)<0 (sign flips on converting x-y to y-x)
so we have 2 points on number line now 0 and x. hence, y lies between 0 and x --->(0<y<x)
this gives us both x and y are more than zero.
hence, x n y are both positive.

Interesting manipulations! But here is the problem: y lies between 0 and x --->(0<y<x)

y lies between 0 and x does not translate to 0 < y < x.
What if x is negative? Then x < y < 0
Say if x is -3, y will be between -3 and 0.

SO both could be positive BUT both could be negative too.
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Top Contributor
Manbehindthecurtain wrote:
Are x and y both positive?

(1) $$2x - 2y = 1$$

(2) $$\frac{x}{y} > 1$$

Target question: Are x and y both positive?

Statement 1: 2x - 2y = 1
There are several pairs of numbers that satisfy this condition. Here are two:
Case a: x = 1 and y = 0.5, in which case x and y are both positive
Case b: x = -0.5 and y = -1, in which case x and y are not both positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x/y > 1
This tells us that x/y is positive. This means that either x and y are both positive or x and y are both negative. Here are two possible cases:
Case a: x = 4 and y = 2, in which case x and y are both positive
Case b: x = -4 and y = -2, in which case x and y are not both positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2
Statement 1 tells us that 2x - 2y = 1.
Divide both sides by 2 to get: x - y = 1/2
Solve for x to get x = y + 1/2

Now take the statement 2 inequality (x/y > 1) and replace x with y + 1/2 to get:
(y + 1/2)/y > 1
Rewrite as: y/y + (1/2)/y > 1
Simplify: 1 + 1/(2y) > 1
Subtract 1 from both sides: 1/(2y) > 0
If 1/(2y) is positive, then y must be positive.

Statement 2 tells us that either x and y are both positive or x and y are both negative.
Now that we know that y is positive, it must be the case that x and y are both positive
Since we can now answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Manbehindthecurtain wrote:
Are x and y both positive?

(1) $$2x - 2y = 1$$

(2) $$\frac{x}{y} > 1$$

(1) $$2x - 2y = 1$$

(x - y) = 1/2

So I imagine it on the number line. x - y = 1/2 tells me that x is to the right of y, a distance of 1/2 away. They could be placed on the number line in any way.

----------------------------- 0 -------------------y --(1/2) -- x ------------

------y --(1/2) -- x ----------- 0 -------------------------

Not sufficient

(2) $$\frac{x}{y} > 1$$

x and y could be both positive or both negative.

Not sufficient.

Using both statements,
If x and y are to the right of 0 i.e. both positive, x > y and hence x/y > 1
e.g. 3/2.5 > 1
If x and y are to the left of 0 i.e. both negative, x> y but x/y < 1 (since y is negative, the inequality sign flips)
e.g. -2.5/-3 < 1

Hence, to satisfy both statements, x and y both must be positive.

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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Bunuel wrote:
sam2010 wrote:
Here is my confusion.
Here is how I approached the question
1. 2x-2y=1
so x-y=.5

now x=1, y=.5
or x=1/4, y=-1/4
so can't tell

2. x/y>1
x>y so again can't tell.

Now if we combine both
still the options x=1, x=.5 is true
and so is the option x=1/4, y=-1/4 true

So can't tell hence E. I know this is not the correct answer but what am I missing?

Problem with your solution is that the red part is not correct.

$$\frac{x}{y}>1$$ does not mean that $$x>y$$. If both x and y are positive, then $$x>y$$, BUT if both are negative, then $$x<y$$.

From (2) $$\frac{x}{y}>1$$, we can only deduce that x and y have the same sigh (either both positive or both negative).

Posted from my mobile device
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Bunuel @chetan4u

Please tell me if my approach is correct, 1st and 2nd are insufficient on its own.
Importantly statement II tells us that both are same sign hence from, x/y > 1, we can infer that |x| > |y| (hope this is correct inference).

Also, when combining both "x - y = 1/2"; if both X and Y are positive than we can obtain 1/2 as answer. However if both are negative i.e [-x -(-y) ]
X being the greater, we cannot get a positive answer =1/2.

I hope this is correct. Thanks
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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Bunuel wrote:
Are x and y both positive?

GRAPHIC APPROACH.

Notice that the question is basically asks whether the point (x, y) is in the first quadrant.

(1) $$2x - 2y = 1$$. Draw line $$y=x-\frac{1}{2}$$: Not sufficient.

(2) $$\frac{x}{y} > 1$$. Draw line $$\frac{x}{y}=1$$. The solutions is the green region: Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Attachment:
graph.png

Attachment:
graph %282%29.png

Can you please tell me where is the intersection point? Thanks Math Expert V
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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D4kshGargas wrote:
Bunuel wrote:
Are x and y both positive?

GRAPHIC APPROACH.

Notice that the question is basically asks whether the point (x, y) is in the first quadrant.

(1) $$2x - 2y = 1$$. Draw line $$y=x-\frac{1}{2}$$: Not sufficient.

(2) $$\frac{x}{y} > 1$$. Draw line $$\frac{x}{y}=1$$. The solutions is the green region: Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Attachment:
The attachment graph.png is no longer available

Attachment:
The attachment graph %282%29.png is no longer available

Can you please tell me where is the intersection point? Thanks Common points of a blue line from (1) and the green areas from (2) is the portion of the blue line which lies in the first quadrant.
Attachments Untitled.png [ 16.06 KiB | Viewed 198 times ]

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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Bunuel wrote:
D4kshGargas wrote:
Bunuel wrote:
Are x and y both positive?

GRAPHIC APPROACH.

Notice that the question is basically asks whether the point (x, y) is in the first quadrant.

(1) $$2x - 2y = 1$$. Draw line $$y=x-\frac{1}{2}$$: Not sufficient.

(2) $$\frac{x}{y} > 1$$. Draw line $$\frac{x}{y}=1$$. The solutions is the green region: Not sufficient.

(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Attachment:
graph.png

Attachment:
graph %282%29.png

Can you please tell me where is the intersection point? Thanks Common points of a blue line from (1) and the green areas from (2) is the portion of the blue line which lies in the first quadrant.

Oh okay, thanks.
I thought my browser wasn't rendering the Blue Line...
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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Bunuel: in the graph, could you please explain how did you define the green region from X>Y?? thanks!
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1  [#permalink]

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NikC wrote:
Bunuel: in the graph, could you please explain how did you define the green region from X>Y?? thanks!

NikC

The line at 45º from x-axis is the link where y = x

x is greater on the right side than on left side

so area to the right of line y=x, will represent the area where x > y

I hope that help!
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