If equation |x|+|y|= 5 encloses a certain region on the : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 16:19

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If equation |x|+|y|= 5 encloses a certain region on the

Author Message
Intern
Joined: 20 Jun 2007
Posts: 13
Followers: 0

Kudos [?]: 3 [0], given: 0

If equation |x|+|y|= 5 encloses a certain region on the [#permalink]

### Show Tags

05 Jun 2008, 16:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If equation |x|+|y|= 5 encloses a certain region on the graph, what is the area of that region?

a) 5
b) 10
c) 25
d) 50
e) 100

Can someone help explain this little bit more into detail? Would greatly be appreciated

The end point of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals to $$5*\sqrt{2}$$ and the area of the figure is that expression squared or 50.
Director
Joined: 10 Sep 2007
Posts: 947
Followers: 8

Kudos [?]: 287 [0], given: 0

### Show Tags

05 Jun 2008, 16:52
|x| and |y| are always >= 0, so their minimum value can be 0.
As We have to satisfy the inequality |X| + |Y| = 5, so if one of them is 0 then other one has to be 5.
Taking Y as 0 and X as 5 or -5 give 2 coordinates (5,0), (-5,0)
Similarly if X is 0 then Y can be 5 or -5 give 2 coordinates (0,5), (0,-5)

You have 4 coordinates of this graph. Due to symmetry you can say that it is square and not rohombus.
Length of any side = Sqrt(25 + 25) = 5Sqrt(2)

So Area of Square = Side * Side = 5Sqrt(2) * 5Sqrt(2) = 50

Intern
Joined: 20 Jun 2007
Posts: 13
Followers: 0

Kudos [?]: 3 [0], given: 0

### Show Tags

05 Jun 2008, 17:18
can't it x & y also be 3 & 2 or 4 & 1. it would still satisfy the equation no?
Manager
Joined: 21 Mar 2008
Posts: 244
Followers: 3

Kudos [?]: 30 [0], given: 0

### Show Tags

05 Jun 2008, 21:04
dzelkas wrote:
can't it x & y also be 3 & 2 or 4 & 1. it would still satisfy the equation no?

3 and 2 and 4 and 1 are points on this line. But we are trying to see what kind of a figure it is and hence looking for intersection of the line with x (y=0) and y(x=) axis to give us reference coordinates.Thats how we come up with the coordinates as (0, 5) and (5, 0).
Similarly , for the other 3 equations( one each for a variation of the sign of x and y), we will have the other 2 corodinates(we will have common corodinates as well) and finally get the figure we intend to find the area of.
You actually dont need to go all the way. By the symmetry of the equation, you can figure out that it will be a square.
Director
Joined: 14 Aug 2007
Posts: 733
Followers: 10

Kudos [?]: 180 [0], given: 0

### Show Tags

05 Jun 2008, 21:36
vertex at (-5,0) (0,5) (5,0) (0,-5)
length = sqrt(50)
area = 50
D
Re: graph equation question   [#permalink] 05 Jun 2008, 21:36
Display posts from previous: Sort by