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

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) \(0 < a < 4\)..
nothing about b....
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

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:
Asking Excellent Questions

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

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.

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

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/

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.

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.
Thank you in advance.

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

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.

I am not sure of the logic of this approach. The value of a is not independent in statement 2. It depends on the value of b. So say if b = -3, then the range of values of a will be different, not 0 to 4. Of course, when we try numbers, they will satisfy because the statement is enough to answer. Can we establish that the statement will also work for all numbers based on 2 cases? I would worry about it.

Here is how you can be sure of the logic. The Veritas links I gave above don't work anymore so I will brief my thought process here.

The parabola is y = x^2 - 4x

It divides the co-ordinate plane into two regions - one inside the parabola and one outside.
For one of these regions, y > x^2 - 4x and for the other y < x^2 - 4x. Which is which, let's find out.

The point (2, -1) obviously lies inside the parabola within our shaded region. What relation of x and y holds for it?
\(x^2 - 4x = 2^2 - 8 = -4\)
\(y = -1\)
\(y > x^2 - 4x\) here
So every point inside the parabola satisfies y > x^2 - 4x and every point outside it satisfies y < x^2 - 4x.

Now, if we have a point (a, b) such that b > a^2 - 4a, it must lie inside the parabola and within the shaded region (because b < 0).

Answer (B)

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

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).

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

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 x-axis. Does the point (a,b) (not shown) lie in the shaded region if b < 0 ?

(1) 0 < a < 4

We don't know what b is. We know b < 0 from the question stem, but we can can a yes and no answer.

if (1,-2) then
\(1^2 - 4(1) < -2\); YES

if (1,-4) then
\(1^2 - 4(1) > -2;\) NO

INSUFFICIENT.

(2) \(a^2 – 4a < b\)

Since \(b > a^2 - 4a\), b will be located above the graph. And since \(b < 0\), point (a,b) MUST be in the shaded region. SUFFICIENT.

Answer is B.

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..

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

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