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

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

Hi,
In very simple terms to solve this Q..

The type of parabola etc is amplified by mike in the above post...
A parabola of Quadratic equation will have a minimum or a maximum value depending on the coeff of $$x^2$$..
here it is positive, so the parabola will be open upwards and will have a mininmum value at$$x=\frac{-b}{2a}$$ or 4/2=2.. and the value is $$2^2-4*2=-4$$...

lets see the statements now..

(1) $$0 < a < 4$$..
so point (a,b) can be anywhere depending on value of b
at a=3.99, b can be -4, so will be outside the graph or at some point inside..
and at a=2, b can be -3.99, it will be inside the graph..
so insuff..

(2) $$a^2 - 4a < b$$
the moment you see this equation, its similarity with the original equation y=x^2 - 4x should strike you..
we substitute a and b as x and y in the eq we get b=a^2-4a...
since the equation $$b=a^2-4a$$ is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it...
so suff..
you can test this with, say at the x axis..
at a=4, b=0..
$$a^2-4a=b... 4^2-4*4=b=0$$..
so if $$a^2-4a<0, 0<a<4$$ satisfies the condition for within the shaded portion and so suff..
the moment $$a^2-4a>0$$, a>5 or a<0 on x axis, and this point will be outside the parabola..
hope it helped you in some way
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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Hi Mike
Actually, I can not get the concept behind your explanation.
Please explain more taking into account the basic concepts since I did not master the Coordinate concepts.
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hatemnag wrote:
Hi Mike
Actually, I can not get the concept behind your explanation.
Please explain more taking into account the basic concepts since I did not master the Coordinate concepts.

Dear hatemnag,
I'm happy to respond.

My friend, before I answer your question, I am going to challenge you. I am going to challenge you to ask a better, more thorough question. See this blog article:

You see, the question you asked was very vague and general. Vague questions are poor questions. It's perfectly fine that you don't understand some basic concepts about Coordinate Geometry and that you want to learn more--in fact, it's wonderful that you are asking for help! The trouble is, I have absolutely no idea what you already know and what you need. Your vague question completely leaves me in the dark.

An excellent question would involve making explicitly clear, exactly and specifically, what parts you understand and what parts you don't understand. For example, you could go through each point I make in my Jan. 27, 2016 post above, and tell me exactly what you understand and exactly what confuses you about each item.

Of course, a more specific, more detailed question, will help me respond to you more effectively, but what students often fail to understand: I recommend crafting an excellent question, not for me, but for you! The process of putting all the effort into writing an excellent question will be tremendously helpful to you: by explaining all this, you will force yourself to make connections and have realizations, and all this effort will prime your mind so that my more detailed response will be that much more helpful for you. It's much harder to produce an excellent question, and, in fact, all the effort it takes to craft an excellent question is actually an investment in your own understanding and learning. When a student poses an excellent question to the teacher, it's a total win-win scenario. This is precisely why asking excellent questions is one of the habits of excellence.

Therefore, my friend, I am going to challenge you to write an excellent question. Craft the highest quality question you can, making explicitly and specifically clear precisely which parts you understand and don't understand about coordinate geometry. It's fine to have a ton of questions, that's great, but it's important for both of us to appreciate the basic ideas that you do understand, because all learning is based on what you already grasp.

Does all this make sense?

Mike
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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hi there, mikemcgarry and chetan2u can someone help me out? I am taking the GMAT in 8 days (my practice CATs have been in the 700 range Q49, V40), but this question came as a bit of a shock to me. Note, I have not read either of your explanations yet because this concept looks new to me. What is "a" and "b" referring to? Is it the a and b within the quadratic equation (ax^2+bx+c=0)? For the most part, the only questions I have seen related to graphs in my practice is lines (y=mx+b). I will read your explanations shortly, but first, is there a gap in my understanding of the basics? I think I remember from grade school this U shape being a "Parabola." Is that correct? If so, what about Parabola's do we need to know for the GMAT?
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
gzimmer wrote:
hi there, mikemcgarry and chetan2u can someone help me out? I am taking the GMAT in 8 days (my practice CATs have been in the 700 range Q49, V40), but this question came as a bit of a shock to me. Note, I have not read either of your explanations yet because this concept looks new to me. What is "a" and "b" referring to? Is it the a and b within the quadratic equation (ax^2+bx+c=0)? For the most part, the only questions I have seen related to graphs in my practice is lines (y=mx+b). I will read your explanations shortly, but first, is there a gap in my understanding of the basics? I think I remember from grade school this U shape being a "Parabola." Is that correct? If so, what about Parabola's do we need to know for the GMAT?

Hi gzimmer,

a and b refer to a point in the plane where a is on x-axis and B on y-axis..
You are correct that such equation refers to parabola.
Also parabola if asked would be basic, as for instance one can see the similarities in two equations- one in Q and other in statement II.
And that would also mean you are likely to be doing well in Quant till then.

Now with just 8 days left, just do the basic and do not spend too much time on topics rarely seen ..
All the best.
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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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

Hi,
I solved it this way:

remember that stem says b<0.

st1. clearly insufficient because say y= -5 then it is not in shade region but for y greater that -4 it is in region.

st2. you can rewrite it as a^2-4a+4<b+4 (add 4 to both sides) then LHS is (a-2)^2 and RHS is b+4 ---> (a-2)^2<a+4 as LHS is something always positive then it can not be smaller than something negative thus b+4 must be greater than 0, therefore we have -4<b<0 hence 0<b+4<4. ---> (a-2)^2<4 ---> |x-2|<2 --> -2<x-2<2 --> 0<x<4. it is in the region. suff.
B
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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chetan2u wrote:
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

Hi,
In very simple terms to solve this Q..

The type of parabola etc is amplified by mike in the above post...
A parabola of Quadratic equation will have a minimum or a maximum value depending on the coeff of x^2..
here it is positive, so the parabola will be open upwards and will have a mininmum value at x=-b/2a or 4/2=2.. and teh value is 2^2-4*2=-4...

lets see the statements now..

(1) 0 < a < 4..
at a=3.99, b can be -4, so will be outside the graph ..
and at a=2, b can be -3.99, it will be inside the graph..
so insuff..

(2) a^2 - 4a < b
the moment you see this equation, its similarity with the original equation y=x^2 - 4x should strike you..
we substitute a and b as x and y in the eq we get b=a^2-4a...
since the equation b=a^2-4a is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it...
so suff..
you can test this with, say at the x axis..
at a=4, b=0..
a^2-4a=b... 4^2-4*4=b=0..
so if a^2-4a<0, 0<a<4 satisfies the conditi0n for within the shaded portion and so suff..
the moment a^2-4a>0, a>5 or a<0 on x axis, and this point will be outside the parabola..
hope it helped you in some way

In Statement 1 couldn't Point B be anything? For instance -100000 ? little bit confused why you picked -3.99 and and -4 for b ? or am I missing something ?

Im sure Im missing something or the questions is just very basic

I mean from statement 1: no information about b at all --> so clearly insufficient.

Statement 2: it explicitly tells us that a,b lies above the graph and it is given that b<0
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
Hi VeritasPrepKarishma mikemcgarry shashankism Bunuel

Can you suggest flaw in below approach:

Given: Question has asked me whether a point (a,b) lies on shaded curve.
I hope there is no discrepancy between whether the shaded curve includes the parabola.
As per mine understanding it does not.

Essentially the parabola is in quadrant IV and hence (x,-y) should be co-ordinates of points in this quadrant.

St 1: 0<a<4 -> I simply took in random values between 0 to 3.9 and found that since x is positive st 1 is sufficient.
After reading the solutions and above explanation found that for two different values of a/x, say 1 or 8 I am not alway
getting positive values of y hence st 1 is insufficient.

St 2: at first glance the equation translates to x^2 - 4x < y since (a,b) have to in shaded region but for these how did we
intuitively assume that (a,b) is on curve? this query is especially pertaining to above explanation by mikemcgarry

As per my understanding it seems we have put an upper gap on parabola by saying b<0 in question stem. Any view on this?
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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daviddaviddavid wrote:
chetan2u wrote:
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

Hi,
In very simple terms to solve this Q..

The type of parabola etc is amplified by mike in the above post...
A parabola of Quadratic equation will have a minimum or a maximum value depending on the coeff of x^2..
here it is positive, so the parabola will be open upwards and will have a mininmum value at x=-b/2a or 4/2=2.. and teh value is 2^2-4*2=-4...

lets see the statements now..

(1) 0 < a < 4..
at a=3.99, b can be -4, so will be outside the graph ..
and at a=2, b can be -3.99, it will be inside the graph..
so insuff..

(2) a^2 - 4a < b
the moment you see this equation, its similarity with the original equation y=x^2 - 4x should strike you..
we substitute a and b as x and y in the eq we get b=a^2-4a...
since the equation b=a^2-4a is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it...
so suff..
you can test this with, say at the x axis..
at a=4, b=0..
a^2-4a=b... 4^2-4*4=b=0..
so if a^2-4a<0, 0<a<4 satisfies the conditi0n for within the shaded portion and so suff..
the moment a^2-4a>0, a>5 or a<0 on x axis, and this point will be outside the parabola..
hope it helped you in some way

In Statement 1 couldn't Point B be anything? For instance -100000 ? little bit confused why you picked -3.99 and and -4 for b ? or am I missing something ?

Im sure Im missing something or the questions is just very basic

I mean from statement 1: no information about b at all --> so clearly insufficient.

Statement 2: it explicitly tells us that a,b lies above the graph and it is given that b<0

hi..

since b is not given and no relation exists between a and b, b can take any value..
you are correct in your understanding...two points given were just to illustrate that b could be inside or outside graph
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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I actually spent a lot of time on this simple question.

Here is the mistake I did for anyone who is still confused: I assumed that "a" is on x2-2x. Therefore, when I saw condition (1), 0<a<4, I started plugging in values for a in X2-2x . This would have worked if the assumption was stated; however, at no point did the question say that a is on function x2-2x. Therefore, we can not determine "b" from "a" because we are not given any info that ties the two point together.
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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chetan2u wrote:
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

Hi,
In very simple terms to solve this Q..

The type of parabola etc is amplified by mike in the above post...
A parabola of Quadratic equation will have a minimum or a maximum value depending on the coeff of $$x^2$$..
here it is positive, so the parabola will be open upwards and will have a mininmum value at$$x=\frac{-b}{2a}$$ or 4/2=2.. and the value is $$2^2-4*2=-4$$...

lets see the statements now..

(1) $$0 < a < 4$$..
so point (a,b) can be anywhere depending on value of b
at a=3.99, b can be -4, so will be outside the graph or at some point inside..
and at a=2, b can be -3.99, it will be inside the graph..
so insuff..

(2) $$a^2 - 4a < b$$
the moment you see this equation, its similarity with the original equation y=x^2 - 4x should strike you..
we substitute a and b as x and y in the eq we get b=a^2-4a...
since the equation $$b=a^2-4a$$ is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it...
so suff..
you can test this with, say at the x axis..
at a=4, b=0..
$$a^2-4a=b... 4^2-4*4=b=0$$..
so if $$a^2-4a<0, 0<a<4$$ satisfies the condition for within the shaded portion and so suff..
the moment a^2-4a>0, a>5 or a<0 on x axis, and this point will be outside the parabola..
hope it helped you in some way

Hi, thanks for your explanation. I have a doubt

In condition B - I understand that a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it.

However, the question asks us about the shaded region. Whereas a^2-4a< b will include all area inside the parabola - both above x axis and below x axis. this to ensure that the point lies under the x axis we will need Condition A (Which we can't prove/ derive from Condition B)

Therefore I feel Answer should be C. Please let me know if I am missing out on something
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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Tanvi94 wrote:
chetan2u wrote:
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

Hi,
In very simple terms to solve this Q..

The type of parabola etc is amplified by mike in the above post...
A parabola of Quadratic equation will have a minimum or a maximum value depending on the coeff of $$x^2$$..
here it is positive, so the parabola will be open upwards and will have a mininmum value at$$x=\frac{-b}{2a}$$ or 4/2=2.. and the value is $$2^2-4*2=-4$$...

lets see the statements now..

(1) $$0 < a < 4$$..
so point (a,b) can be anywhere depending on value of b
at a=3.99, b can be -4, so will be outside the graph or at some point inside..
and at a=2, b can be -3.99, it will be inside the graph..
so insuff..

(2) $$a^2 - 4a < b$$
the moment you see this equation, its similarity with the original equation y=x^2 - 4x should strike you..
we substitute a and b as x and y in the eq we get b=a^2-4a...
since the equation $$b=a^2-4a$$ is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it...
so suff..
you can test this with, say at the x axis..
at a=4, b=0..
$$a^2-4a=b... 4^2-4*4=b=0$$..
so if $$a^2-4a<0, 0<a<4$$ satisfies the condition for within the shaded portion and so suff..
the moment a^2-4a>0, a>5 or a<0 on x axis, and this point will be outside the parabola..
hope it helped you in some way

Hi, thanks for your explanation. I have a doubt

In condition B - I understand that a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it.

However, the question asks us about the shaded region. Whereas a^2-4a< b will include all area inside the parabola - both above x axis and below x axis. this to ensure that the point lies under the x axis we will need Condition A (Which we can't prove/ derive from Condition B)

Therefore I feel Answer should be C. Please let me know if I am missing out on something

Hello

The point you make is valid, that a^2 - 4a < b will include all area inside the parabola, both above x axis and below x axis. But its also mentioned in the question stem that b < 0. Since y coordinate is negative, the area that we have to consider is the one below x-axis only.
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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
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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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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]
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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, 16:10.
Last edited by nick1816 on 18 Nov 2019, 01:18, edited 1 time in total.
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Re: In the xy-plane shown, the shaded region consists of all points that [#permalink]

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