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.
Jun 0808:00 PM EDT
-10:00 PM EDT
A powerful GMAT course taught live online + 6 months of access to TTP OnDemand GMAT Masterclass included! Mon/Wed June 8, 2026 →August 12, 2026 8:00pm-10:00pm EST
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
Jun 1111:00 AM EDT
-01:00 PM EDT
TTP GMAT OnDemand gives serious students 400+ hours of expert video instruction, the full TTP course, AI support, weekly office hours, and a 715+ score guarantee—all built for elite GMAT score improvement.
15 Jul 2021, 05:41
Kudos
Bookmarks
The key to solving the problem is finding the coordinates where the equation meets the axes with the extremems we can get the value of the figure
|x/2| + |Y/2| = 5
so when x = 0, y= |10|
when y=0 , x=|10|
we get the extrremums as (0,10),(10,0),(0,-10) and (-10,0).
it's a square whose one side has a value can be traced to as 10*2^1/2
area of the figure being 100*2 = 200
Hence IMO D
|x/2| + |Y/2| = 5
so when x = 0, y= |10|
when y=0 , x=|10|
we get the extrremums as (0,10),(10,0),(0,-10) and (-10,0).
it's a square whose one side has a value can be traced to as 10*2^1/2
area of the figure being 100*2 = 200
Hence IMO D
rajshreeasati
Joined: 11 Jul 2018
Last visit: 16 Mar 2025
Posts: 76
Own Kudos:
Given Kudos: 43
Schools: ISB '27 (A)
29 Apr 2022, 04:45
Kudos
Bookmarks
The best approach to such Coordinate Geometry Questions is always diagrams-
we have |x/2| + |y/2| = 5,
we will get 4 coordinates. x = 0, y = 10, y = 0, x = 10, x = 0, y = -10 and y = 0 and x = -10.
by drawing a diagram we get a figure where we have 4 triangles with 10 as the height and 10 as the base.
so area of region will be = 4 * 1/2 * 10*10 => 200.
we have |x/2| + |y/2| = 5,
we will get 4 coordinates. x = 0, y = 10, y = 0, x = 10, x = 0, y = -10 and y = 0 and x = -10.
by drawing a diagram we get a figure where we have 4 triangles with 10 as the height and 10 as the base.
so area of region will be = 4 * 1/2 * 10*10 => 200.
17 Nov 2022, 01:28
Kudos
Bookmarks
Equation of a line whose intercept on X-axis is (a,0) and Y-axis is (0,b) is
\(\frac{x}{a}+\frac{y}{b}=1\)
We get intercept points as (10,0) and (0,10).
Area of a triangle formed by these points and origin = \(\frac{(10*10)}{2}=50\)
Area of the region = 50*4 = 200
\(\frac{x}{a}+\frac{y}{b}=1\)
We get intercept points as (10,0) and (0,10).
Area of a triangle formed by these points and origin = \(\frac{(10*10)}{2}=50\)
Area of the region = 50*4 = 200
