Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 19 Jun 2013, 17:51

m14 q28

Author Message
Manager
Joined: 29 Jul 2009
Posts: 226
Followers: 1

Kudos [?]: 22 [0], given: 6

m14 q28 [#permalink]  05 Jan 2010, 22:39
00:00

Question Stats:

50% (01:52) correct 50% (01:27) wrong based on 32 sessions
If curves x^2 + y^2 = 4 and y = |x| enclose a sector on the top part of XY-plane, what is the area of this sector?

(A) \frac{\pi}{4}
(B) \frac{\pi}{2}
(C) \pi
(D) 2\pi
(E) 3\pi

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The area of the circle is \pi 2^2 = 4\pi . The area of the sector = \text{(the area of the circle)}*\frac{90}{360} = \pi .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.
[Reveal] Spoiler: OA
 Kaplan Promo Code Knewton GMAT Discount Codes Manhattan GMAT Discount Codes
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879

Kudos [?]: 10130 [2] , given: 965

Re: m14 q28 [#permalink]  06 Jan 2010, 05:11
2
KUDOS
11MBA wrote:
If curves x^2 + y^2 = 4 and y = |x| enclose a sector on the top part of XY-plane, what is the area of this sector?

(C) 2008 GMAT Club - m14#28

* \frac{\pi}{4}
* \frac{\pi}{2}
* \pi
* 2\pi
* 3\pi

The area of the circle is \pi 2^2 = 4\pi . The area of the sector = \text{(the area of the circle)}*\frac{90}{360} = \pi .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

x^2 + y^2 = 4 is an equation of a circle centered at the origin and the radius \sqrt{4}=2.

Graph of the function y = |x| is given below:
Attachment:

graph_modulus.png [ 7.61 KiB | Viewed 4210 times ]

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle ={\pi}{r^2}=4\pi, 1/4 of this value = \pi.
_________________
Manager
Joined: 29 Jul 2009
Posts: 226
Followers: 1

Kudos [?]: 22 [0], given: 6

Re: m14 q28 [#permalink]  06 Jan 2010, 08:07
My bad. I graphed abso(y)=x, and that's why i only had half of the graph be above the xy plane. Sorry.
Intern
Joined: 29 Dec 2009
Posts: 42
Location: India
Concentration: Finance, Real Estate
Schools: Duke (Fuqua) - Class of 2014
GMAT 1: 770 Q50 V44
GPA: 3.5
WE: General Management (Real Estate)
Followers: 3

Kudos [?]: 8 [0], given: 1

Re: m14 q28 [#permalink]  14 Jun 2010, 07:59
I know these are supposed to be the toughest challenges, but is this something which is included in the scope of the GMAT? I thought coordinate plane on GMAT was restricted to straight line equations only....

Posted from my mobile device
Director
Joined: 21 Dec 2009
Posts: 590
Concentration: Entrepreneurship, Finance
Followers: 14

Kudos [?]: 142 [0], given: 20

Re: m14 q28 [#permalink]  14 Jun 2010, 22:14
From x^2 + y^2 = r^2 => radius = 2
Area of the circle = [img]\pi[/img]r^2
The line |x| has a reflection about the origin at [img]90^{\circ}[/img]
Area of sector is 1/4 x area of circle = [img]\pi[/img]4/4 = [img]\pi[/img]
Therefore ans is C
_________________

KUDOS me if you feel my contribution has helped you.

Manager
Joined: 14 Apr 2010
Posts: 241
Followers: 2

Kudos [?]: 8 [0], given: 1

Re: m14 q28 [#permalink]  02 Aug 2010, 02:22
The top most sector makes an angle of 90 degrees. But, here we are to find the enclosed angle as shown in the figure. How can that be 90 degree? Am i missing something here???
Manager
Joined: 14 Apr 2010
Posts: 241
Followers: 2

Kudos [?]: 8 [0], given: 1

Re: m14 q28 [#permalink]  02 Aug 2010, 20:44
Oh!!!! i got it....hehehe
Manager
Joined: 12 Jul 2010
Posts: 68
Followers: 1

Kudos [?]: 2 [0], given: 3

Re: m14 q28 [#permalink]  13 Aug 2010, 01:48
Its C.

The sector makes an angle of 90 degrees at the center of the circle. So area of the sector = area of the circle/4 = pi
Intern
Joined: 15 May 2011
Posts: 40
WE 1: IT Consulting - 5 Years
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: m14 q28 [#permalink]  16 Jun 2011, 09:59
I did not know the equation for a circle, so had to spend extra seconds figuring out that x^2 + Y^2 = 4 was indeed the equation of a circle with radius 2. I ended up with C, but probably took 30-60s more than I should have.

Question - do these type of problems represent the problems at the 700-800 level on the GMAT? I thought the equation for a circle was pretty much out of scope for the GMAT. I hear that the GMAT quant section is getting tougher but is this (problems like this) the level where its headed?
Manager
Joined: 03 Jun 2010
Posts: 191
Location: United States (MI)
Concentration: Marketing, General Management
Followers: 4

Kudos [?]: 18 [0], given: 40

Re: m14 q28 [#permalink]  16 Jun 2011, 10:04
I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.
Intern
Joined: 15 May 2011
Posts: 40
WE 1: IT Consulting - 5 Years
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: m14 q28 [#permalink]  16 Jun 2011, 10:33
ulm wrote:
I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.

Well, then I'm glad I came across this problem Could you tell me in which paper and digital tests you saw questions with circle formulas? I've pretty much based my Math theory on the Manhattan guides and don't remember seeing the circle formula there (or for that matter OG11).
Intern
Status: Don't worry about finding inspiration. It eventually come >>>>>>>
Joined: 31 May 2011
Posts: 23
Location: Î Ñ D Ï Â
Followers: 0

Kudos [?]: 4 [0], given: 2

Re: m14 q28 [#permalink]  16 Jun 2011, 10:42
Area of sector = Θ/360°*πr²
Now we know that r= 2 and Θ= 90°
So, area comes to be π.
Attachments

Blank.png [ 196.1 KiB | Viewed 2825 times ]

Manager
Joined: 03 Jun 2010
Posts: 191
Location: United States (MI)
Concentration: Marketing, General Management
Followers: 4

Kudos [?]: 18 [0], given: 40

Re: m14 q28 [#permalink]  16 Jun 2011, 11:15
AriBenCanaan wrote:
ulm wrote:
I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.

Well, then I'm glad I came across this problem Could you tell me in which paper and digital tests you saw questions with circle formulas? I've pretty much based my Math theory on the Manhattan guides and don't remember seeing the circle formula there (or for that matter OG11).

If we're talking about must-have tests, it definitely was in GMAT Prep.
Senior Manager
Joined: 20 Dec 2010
Posts: 260
Schools: UNC Duke Kellogg
Followers: 3

Kudos [?]: 19 [0], given: 4

Re: m14 q28 [#permalink]  17 Jun 2011, 07:43
y = |x| ---- the slope of this line is (+1) if X is positive and (-1) if X is negative...which means the lines make a 45 degree angle with the x axis...and both the lines sweep a total of 90 degrees...

The circle is 360 degrees -- so the area swept is 1/4 of the total area.

(x^2) + (y^2) = (2)^2

Area of the circle = pi*(2^2)

Area of the sector = (1/4)*Area of the circle

Ans: pi (C)
Manager
Joined: 28 Feb 2011
Posts: 93
Followers: 0

Kudos [?]: 29 [0], given: 2

Re: m14 q28 [#permalink]  17 Jun 2011, 11:04
C

the sector has a 90 degree angle..
area of a sector theta(90 degree)/360 x (pi x sq. radius)
_________________

Fight till you succeed like a gladiator..or doom your life which can be invincible

Senior Manager
Joined: 15 Sep 2009
Posts: 272
GMAT 1: 750 Q V
Followers: 5

Kudos [?]: 36 [0], given: 6

Re: m14 q28 [#permalink]  20 Jun 2012, 06:04
Straight C.

As seen in Microstrip's diagram, the area is a quarter of the total area of the circle.
_________________

+1 Kudos me - I'm half Irish, half Prussian.

Intern
Status: Preparing for GMAT
Joined: 11 Dec 2012
Posts: 15
Location: India
Concentration: Healthcare, Entrepreneurship
Schools: LBS '14
GMAT 1: 640 Q49 V27
GPA: 3.7
WE: Consulting (Non-Profit and Government)
Followers: 0

Kudos [?]: 26 [1] , given: 104

Re: m14 q28 [#permalink]  22 Jan 2013, 00:28
1
KUDOS
Why don't we extend both the lines and account for the 45 degree angle formed in the quadrants III & IV ? I am sorry if its a silly question.
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879

Kudos [?]: 10130 [0], given: 965

Re: m14 q28 [#permalink]  22 Jan 2013, 06:40
jainpiyushjain wrote:
Why don't we extend both the lines and account for the 45 degree angle formed in the quadrants III & IV ? I am sorry if its a silly question.

Meanwhile go through this post for a solution: m14-q28-88810.html#p670247

Hope it helps.
_________________
Manager
Joined: 28 Jul 2011
Posts: 212
Followers: 0

Kudos [?]: 15 [0], given: 12

Re: m14 q28 [#permalink]  28 Jan 2013, 10:53
Bunuel wrote:
11MBA wrote:
If curves x^2 + y^2 = 4 and y = |x| enclose a sector on the top part of XY-plane, what is the area of this sector?

(C) 2008 GMAT Club - m14#28

* \frac{\pi}{4}
* \frac{\pi}{2}
* \pi
* 2\pi
* 3\pi

The area of the circle is \pi 2^2 = 4\pi . The area of the sector = \text{(the area of the circle)}*\frac{90}{360} = \pi .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

x^2 + y^2 = 4 is an equation of a circle centered at the origin and the radius \sqrt{4}=2.

Graph of the function y = |x| is given below:
Attachment:
graph_modulus.png

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle ={\pi}{r^2}=4\pi, 1/4 of this value = \pi.

How did you find the radius "x^2 + y^2 = 4 is an equation of a circle centered at the origin and the radius \sqrt{4}=2." can you please explain?
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12116
Followers: 1879

Kudos [?]: 10130 [0], given: 965

Re: m14 q28 [#permalink]  28 Jan 2013, 11:00
kuttingchai wrote:
Bunuel wrote:
11MBA wrote:
If curves x^2 + y^2 = 4 and y = |x| enclose a sector on the top part of XY-plane, what is the area of this sector?

(C) 2008 GMAT Club - m14#28

* \frac{\pi}{4}
* \frac{\pi}{2}
* \pi
* 2\pi
* 3\pi

The area of the circle is \pi 2^2 = 4\pi . The area of the sector = \text{(the area of the circle)}*\frac{90}{360} = \pi .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

x^2 + y^2 = 4 is an equation of a circle centered at the origin and the radius \sqrt{4}=2.

Graph of the function y = |x| is given below:
Attachment:
graph_modulus.png

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle ={\pi}{r^2}=4\pi, 1/4 of this value = \pi.

How did you find the radius "x^2 + y^2 = 4 is an equation of a circle centered at the origin and the radius \sqrt{4}=2." can you please explain?

Check here: math-coordinate-geometry-87652.html (Circle on a plane chapter).

Hope it helps.
_________________
Re: m14 q28   [#permalink] 28 Jan 2013, 11:00
Similar topics Replies Last post
Similar
Topics:
10 M14 #13 23 11 Nov 2008, 21:40
5 M14 #27 19 13 Nov 2008, 17:55
2 M14 #18 17 02 Feb 2009, 21:55
14 M14 #19 18 02 Feb 2009, 22:06
11 M14#10 18 18 Mar 2009, 13:41
Display posts from previous: Sort by

m14 q28

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.