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In the xy-plane, region A consists of all the points (x,y)

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In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

(1) a > 2b
(2) b = 1

I am having a hard time conceptualizing how to solve this problem. I read the explanation in Total GMAT Math(which is where problem came from) and understand how to do the algebra. However I do NOT understand how the inequality translates as a graphical representation on the coordinate plane.

Originally posted by josemnz83 on 23 Jun 2013, 14:00.
Last edited by chetan2u on 10 Dec 2015, 02:26, edited 3 times in total.
Renamed the topic and edited the question.
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 23 Jun 2013, 14:26
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In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

The area defined by \(1 - x < 2y\) or \(y>\frac{1}{2}-\frac{x}{2}\) is the area OVER the black line.
We are asked if point (a,b) is in that region

1) a >2b
The area defined by this is x>2y or \(y<\frac{x}{2}\): BELOW the red line. Is this sufficient to say that the point is also in the area specified above?
No, not sufficient.

2) b=1
So it stays on the green line.
Clearly not sufficient.

1+2)Now we know that the point is in the area below the red line, and is on the line y=1. This is sufficient to answer the question, as the area defined by both statements is always over the black line. Sufficient
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 23 Jun 2013, 15:32
Mil gracias amigo! The graphic helped tons! Very good job!

Zarrolou wrote:
In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

The area defined by \(1 - x < 2y\) or \(y>\frac{1}{2}-\frac{x}{2}\) is the area OVER the black line.
We are asked if point (a,b) is in that region

1) a >2b
The area defined by this is x>2y or \(y<\frac{x}{2}\): BELOW the red line. Is this sufficient to say that the point is also in the area specified above?
No, not sufficient.

2) b=1
So it stays on the green line.
Clearly not sufficient.

1+2)Now we know that the point is in the area below the red line, and is on the line y=1. This is sufficient to answer the question, as the area defined by both statements is always over the black line. Sufficient
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 23 Jun 2013, 18:22
Zarrolou wrote:
In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

The area defined by \(1 - x < 2y\) or \(y>\frac{1}{2}-\frac{x}{2}\) is the area OVER the black line.
We are asked if point (a,b) is in that region

1) a >2b
The area defined by this is x>2y or \(y<\frac{x}{2}\): BELOW the red line. Is this sufficient to say that the point is also in the area specified above?
No, not sufficient.

2) b=1
So it stays on the green line.
Clearly not sufficient.

1+2)Now we know that the point is in the area below the red line, and is on the line y=1. This is sufficient to answer the question, as the area defined by both statements is always over the black line. Sufficient


Hi,
How u got 1 not sufficient?
why 1-x>2y changed to x>2y?
Can u plz explain me.
Thankz in advance,
rrsnathan.
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 23 Jun 2013, 19:38
1
1
You are confusing the two inequalities. The inequality 1 - x <2y is the inequality that defines region A.

The inequality x >2y is the inequality that defines statement 1.

I'm certainly not an expert on coordinate geometry but will try to explain the problem to you. I will refer to the graph that Zorralou kindly provided. The graphical representation of the inequalities helped me understand the problem and I hope that it helps you as well

The first thing that you want to do is rewrite the given inequality (1-x <2y) in slope intercept form. The inequality will then look like this y >-1/2x + 1/2. Recognize that the y intercept is positive 1/2 and that the inequality has a negative slope that is half of 1, or 1/2. Use this information to help you sketch the inequality on the coordinate plane. See Zorrolou's graph; he has represented this inequality with the black line.

Any point above the "black line" is part of region A so the question is essentially asking whether point (a,b) is above the black line.

Statement 1 says that a>2b
You first want to rewrite the variables in terms of x and y. a >2b becomes x >2y. Again I found it helpful to rewrite the inequality in slope intercept form. Divide both sides by 2 and you get x/2 > y or y< x/2.

Graph this inequality on the coordinate plane. See Zorrolou's diagram again. The inequality y <x/2 is represented by the red line. Notice that the inequality states that y is LESS than 1/2x which means that the region of this inequality is BELOW the red line.

Back to our original question. Is point (a,b) part of region A? We now know that region A is above the black line and below the red line. This is helpful but is not sufficient. As it is now point (a,b) could be above the black line or below the black line. For instance point (5,1) is above the black line (part of region A) and below the red line while point (-1,2) is also below the red line but is NOT part of region A (above the black line).

I hope that my explanation helps you. I'm still trying to learn all this myself! I took the GMAT 3 times and this is the first time that I'm actually trying to get some of these math concepts down. I have not taken any math since taking calculus in high school over a decade ago! Best of luck!

rrsnathan wrote:
Zarrolou wrote:
In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

The area defined by \(1 - x < 2y\) or \(y>\frac{1}{2}-\frac{x}{2}\) is the area OVER the black line.
We are asked if point (a,b) is in that region

1) a >2b
The area defined by this is x>2y or \(y<\frac{x}{2}\): BELOW the red line. Is this sufficient to say that the point is also in the area specified above?
No, not sufficient.

2) b=1
So it stays on the green line.
Clearly not sufficient.

1+2)Now we know that the point is in the area below the red line, and is on the line y=1. This is sufficient to answer the question, as the area defined by both statements is always over the black line. Sufficient


Hi,
How u got 1 not sufficient?
why 1-x>2y changed to x>2y?
Can u plz explain me.
Thankz in advance,
rrsnathan.
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 23 Jun 2013, 20:16
hey josemnz83 tkankz. Really ur explanation was really helpful

+kudos for ur great explanation


josemnz83 wrote:
You are confusing the two inequalities. The inequality 1 - x <2y is the inequality that defines region A.

The inequality x >2y is the inequality that defines statement 1.

I'm certainly not an expert on coordinate geometry but will try to explain the problem to you. I will refer to the graph that Zorralou kindly provided. The graphical representation of the inequalities helped me understand the problem and I hope that it helps you as well

The first thing that you want to do is rewrite the given inequality (1-x <2y) in slope intercept form. The inequality will then look like this y >-1/2x + 1/2. Recognize that the y intercept is positive 1/2 and that the inequality has a negative slope that is half of 1, or 1/2. Use this information to help you sketch the inequality on the coordinate plane. See Zorrolou's graph; he has represented this inequality with the black line.

Any point above the "black line" is part of region A so the question is essentially asking whether point (a,b) is above the black line.

Statement 1 says that a>2b
You first want to rewrite the variables in terms of x and y. a >2b becomes x >2y. Again I found it helpful to rewrite the inequality in slope intercept form. Divide both sides by 2 and you get x/2 > y or y< x/2.

Graph this inequality on the coordinate plane. See Zorrolou's diagram again. The inequality y <x/2 is represented by the red line. Notice that the inequality states that y is LESS than 1/2x which means that the region of this inequality is BELOW the red line.

Back to our original question. Is point (a,b) part of region A? We now know that region A is above the black line and below the red line. This is helpful but is not sufficient. As it is now point (a,b) could be above the black line or below the black line. For instance point (5,1) is above the black line (part of region A) and below the red line while point (-1,2) is also below the red line but is NOT part of region A (above the black line).

I hope that my explanation helps you. I'm still trying to learn all this myself! I took the GMAT 3 times and this is the first time that I'm actually trying to get some of these math concepts down. I have not taken any math since taking calculus in high school over a decade ago! Best of luck!

rrsnathan wrote:
Zarrolou wrote:
In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

The area defined by \(1 - x < 2y\) or \(y>\frac{1}{2}-\frac{x}{2}\) is the area OVER the black line.
We are asked if point (a,b) is in that region

1) a >2b
The area defined by this is x>2y or \(y<\frac{x}{2}\): BELOW the red line. Is this sufficient to say that the point is also in the area specified above?
No, not sufficient.

2) b=1
So it stays on the green line.
Clearly not sufficient.

1+2)Now we know that the point is in the area below the red line, and is on the line y=1. This is sufficient to answer the question, as the area defined by both statements is always over the black line. Sufficient


Hi,
How u got 1 not sufficient?
why 1-x>2y changed to x>2y?
Can u plz explain me.
Thankz in advance,
rrsnathan.
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 23 Jun 2013, 23:59
josemnz83 wrote:
In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

(1) a > 2b
(2) b = 1

I am having a hard time conceptualizing how to solve this problem. I read the explanation in Total GMAT Math (which is where problem came from) and understand how to do the algebra. However I do NOT understand how the inequality translates as a graphical representation on the coordinate plane.


Similar questions to practices:
in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html (DS)
point-x-y-is-a-point-within-the-triangle-what-is-the-139419.html (PS)

Hope it helps.
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 11 Aug 2013, 03:46
2
josemnz83 wrote:
In the xy-plane, region A consists of all the points (x,y) such that 1 - x < 2y. Is the point (a,b) in region A?

(1) a > 2b
(2) b = 1

I am having a hard time conceptualizing how to solve this problem. I read the explanation in Total GMAT Math (which is where problem came from) and understand how to do the algebra. However I do NOT understand how the inequality translates as a graphical representation on the coordinate plane.

Full marks to zarrolou, however here is an alternate approach which I felt was simpler.

Given inequality 1 - x < 2y does point (a,b) satisfy this equality
so is 1-a<2b ?

(1) a>2b

a= 10 and 2b= 5 putting the values in 1-a<2b we get, is -9<5 ? answer is yes
a=-5 and 2b=-10 ( Note: a>2b) putting the values in 1-a<2b we get is 6 <-10 answer is no
hence insufficient

(2) b=1
is 1-a<2 , clearly insufficient if a =-3 answer is no if a= 3 answer is yes

1+2
then a>2,
if a>2 then 1-a<2 is always true ,(LHS is always negative so always less than 2)Sufficient.

Hope it helps.
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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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New post 28 Mar 2014, 15:32
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Well let me chip in with an alternative approach that I thought was even easier

So 2y>1-x
Is 2b>1-a?

(1) a>2b

If we add we have the question is 2a>1? Is a>1/2? We don't know hence insufficient

(2) b=1

Then we have is 2>1-a? Is a>-1?

This statement is insufficient on its own

Both together we have is a>-1, well we know that a>2 if we replace b=1 on the first statement hence sufficient

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Re: In the xy-plane, region A consists of all the points (x,y)  [#permalink]

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Re: In the xy-plane, region A consists of all the points (x,y)   [#permalink] 27 Nov 2019, 22:36
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