Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
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.
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.
Attachments
04Mar20_1.gif [ 16.15 KiB | Viewed 12896 times ]
14 Mar 2020, 00:03
Kudos
Bookmarks
VeritasKarishma
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)
15 Mar 2020, 01:03
Kudos
Bookmarks
Kritisood
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).
