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# In the xy-plane shown, the shaded region consists of all points that l

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Re: In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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17 Dec 2018, 00:11
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louhit wrote:
Hi Guys,

I have spent too many minutes on this questions which is also on OG quant review 2019 DS01613, I am still not able to get through the questions ask and the explanation given here. I think i have a thick brain it seems, Can Bunuel or VeritasKarishma help me in understanding the question stem and theory/concept behind it. Thanks

First go through these three posts:
https://www.veritasprep.com/blog/2010/1 ... he-graphs/
http://www.veritasprep.com/blog/2010/12 ... s-part-ii/
https://www.veritasprep.com/blog/2011/0 ... -part-iii/

They explain the concept of how the xy co-ordinate plane is divided into two parts by a line. The parabola does the same. It divides the plane into two regions:
y > x^2 - 4x
y < x^2 - 4x

Stmnt 2 gives us that (a, b) lies in the region y > x^2 - 4x which is the region inside the parabola (if you want to verify, check (2, 1). It satisfies y > x^2 - 4x.
Since the question stem tells us that b < 0, so we are looking at a point below the x axis. The region inside the parabola below x axis is the shaded region. So the point must lie on it.
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Re: In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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20 Jul 2019, 23:25
1
If you are getting confused with the normal way of solving but you are great at visualizing and drawing coordinate geometry, here's the video solution for you

https://gmatquantum.com/official-guides ... nt-review/
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In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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Updated on: 18 Nov 2019, 00:18
Statement 1- tells us nothing much. Point can lie inside or outside the parabola in range 0<x<4.
Insufficient

Statement 2- Locus of point (a,b) is $$y>x^2-4x$$ , where y<0. In other words locus of the (a, b) lies below the y-axis and inside the parabola $$y=x^2-4x$$.

Sufficient.

dgboy765 wrote:

In the xy-plane shown, the shaded region consists of all points that lie above the graph of y=x^2 - 4x and below the the x-axis. Does the point (a,b) (not shown) lie in the shaded region if b<0?

(1) 0 < a < 4
(2) a^2 - 4a < b

Source: Official GMAT Quantitative Review 2016
P. 162 DS #124

Can someone explain the process to solving this problem in the simplest way possible? (but please don't be overly brief. I'm not as intuitive as you.)

Attachment:
2016-01-24_1416.png

Originally posted by nick1816 on 15 Sep 2019, 15:10.
Last edited by nick1816 on 18 Nov 2019, 00:18, edited 1 time in total.
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Re: In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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16 Nov 2019, 20:35

Posted from my mobile device
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Re: In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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18 Nov 2019, 00:13
LeenaSai wrote:

Posted from my mobile device

I have talked about it here: https://gmatclub.com/forum/in-the-xy-pl ... l#p2191933
Let me know if you still have doubts.
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In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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Updated on: 15 Mar 2020, 14:28
Hi there!

In the image attached, we realize point (a,b) is in the given shaded region if, and only if,

$$\left\{ \begin{gathered} \,0 < a < 4\,\,\,\left( {\text{I}} \right) \hfill \\ \,{a^2} - 4a < b < 0\,\,\,\left( {{\text{II}}} \right) \hfill \\ \end{gathered} \right.$$

$$\left( {\text{I}} \right)\,\,0 < a < 4\,\,\,$$ Although $$\,b < 0\,$$ is given, we still can have a "yes" and a "no" possibilities:

$$\left. \begin{gathered} \left( {a,b} \right) = \left( {1, - 2} \right)\,\,\,\,;\,\,\,{a^2} - 4a\mathop < \limits^? b\,\,\,:{\text{yes}}\,\, \hfill \\ \left( {a,b} \right) = \left( {1, - 3} \right)\,\,\,\,;\,\,\,{a^2} - 4a\mathop < \limits^? b\,\,\,:{\text{no}}\,\, \hfill \\ \end{gathered} \right\}\,\,\,\,{\text{INS}}{\text{.}}$$

$$\left( {{\text{II}}} \right)\,\,{a^2} - 4a < b < 0\,\,\,$$

$$\left. {{a^2} - 4a < b < 0\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered} \,\left( {{\text{II}}} \right)\,\,\,{\text{immediately}} \hfill \\ \,a\left( {a - 4} \right) = {a^2} - 4a < 0\,\,\,\, \Rightarrow \,\,\,\,0 < a < 4\,\,\, \Rightarrow \,\,\,\,\left( {\text{I}} \right)\,\,\,\,\, \hfill \\ \end{gathered} \right.} \right\}\,\,\,\,{\text{SUF}}{\text{.}}$$

Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

Originally posted by fskilnik on 04 Mar 2020, 06:51.
Last edited by fskilnik on 15 Mar 2020, 14:28, edited 1 time in total.
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In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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13 Mar 2020, 23:03
louhit wrote:
Hi Guys,

I have spent too many minutes on this questions which is also on OG quant review 2019 DS01613, I am still not able to get through the questions ask and the explanation given here. I think i have a thick brain it seems, Can Bunuel or VeritasKarishma help me in understanding the question stem and theory/concept behind it. Thanks

First go through these three posts:
https://www.veritasprep.com/blog/2010/1 ... he-graphs/
http://www.veritasprep.com/blog/2010/12 ... s-part-ii/
https://www.veritasprep.com/blog/2011/0 ... -part-iii/

They explain the concept of how the xy co-ordinate plane is divided into two parts by a line. The parabola does the same. It divides the plane into two regions:
y > x^2 - 4x
y < x^2 - 4x

Stmnt 2 gives us that (a, b) lies in the region y > x^2 - 4x which is the region inside the parabola (if you want to verify, check (2, 1). It satisfies y > x^2 - 4x.
Since the question stem tells us that b < 0, so we are looking at a point below the x axis. The region inside the parabola below x axis is the shaded region. So the point must lie on it.

Hi VeritasKarishma!! I understand that point "a" lies inside the parabola. But how do we ensure that "b" does too? it just says b<0. it could lie anywhere below the x-axis. how do we assume that it lies inside the parabola only below the x-axis?

Also, we derive this from st 2 "0<a<4", but even statement one says this!! So how come statement 1 isn't sufficiency then (I know i must be missing something very silly)
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Re: In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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15 Mar 2020, 00:03
1
Kritisood wrote:
louhit wrote:
Hi Guys,

I have spent too many minutes on this questions which is also on OG quant review 2019 DS01613, I am still not able to get through the questions ask and the explanation given here. I think i have a thick brain it seems, Can Bunuel or VeritasKarishma help me in understanding the question stem and theory/concept behind it. Thanks

First go through these three posts:
https://www.veritasprep.com/blog/2010/1 ... he-graphs/
http://www.veritasprep.com/blog/2010/12 ... s-part-ii/
https://www.veritasprep.com/blog/2011/0 ... -part-iii/

They explain the concept of how the xy co-ordinate plane is divided into two parts by a line. The parabola does the same. It divides the plane into two regions:
y > x^2 - 4x
y < x^2 - 4x

Stmnt 2 gives us that (a, b) lies in the region y > x^2 - 4x which is the region inside the parabola (if you want to verify, check (2, 1). It satisfies y > x^2 - 4x.
Since the question stem tells us that b < 0, so we are looking at a point below the x axis. The region inside the parabola below x axis is the shaded region. So the point must lie on it.

Hi VeritasKarishma!! I understand that point "a" lies inside the parabola. But how do we ensure that "b" does too? it just says b<0. it could lie anywhere below the x-axis. how do we assume that it lies inside the parabola only below the x-axis?

Also, we derive this from st 2 "0<a<4", but even statement one says this!! So how come statement 1 isn't sufficiency then (I know i must be missing something very silly)

Note that (a, b) specifies a single point (a point whose x co-ordinate is 'a' and y co-ordinate is 'b'). So (2, 6) specifies a single point. Only a or only b can't specify a point.
Now the entire plane is divided into two parts - one lying inside the parabola and the other outside.
The one lying inside the parabola is x^2 - 4x < y and the one outside is x^2 - 4x > y. The actual parabola is x^2 - 4x = y.

The shaded region is inside the parabola so all points (a, b) inside it satisfy a^2 - 4a < b (the inequality given by statement 2)

When b is negative (the y coordinate of the point is negative), this will be the shaded region.

So (a, b) are points such as (2, -1).
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Re: In the xy-plane shown, the shaded region consists of all points that l  [#permalink]

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15 Mar 2020, 00:37
dgboy765 wrote:

In the xy-plane shown, the shaded region consists of all points that lie above the graph of y=x^2 - 4x and below the the x-axis. Does the point (a,b) (not shown) lie in the shaded region if b<0?

(1) 0 < a < 4
(2) a^2 - 4a < b

Source: Official GMAT Quantitative Review 2016
P. 162 DS #124

Can someone explain the process to solving this problem in the simplest way possible? (but please don't be overly brief. I'm not as intuitive as you.)

Attachment:
2016-01-24_1416.png

Given: In the xy-plane shown, the shaded region consists of all points that lie above the graph of y=x^2 - 4x and below the the x-axis.

Asked: Does the point (a,b) (not shown) lie in the shaded region if b<0?

(1) 0 < a < 4
Value of b is unknown
NOT SUFFICIENT

(2) a^2 - 4a < b
0 > b > a^2 - 4a
Since b is inside the bound of 0 and a^2 - 4a
SUFFICIENT

IMO B
Re: In the xy-plane shown, the shaded region consists of all points that l   [#permalink] 15 Mar 2020, 00:37

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