GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Dec 2018, 22:24

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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST

December 14, 2018

December 14, 2018

09:00 AM PST

10:00 AM PST

10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.
• Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

December 14, 2018

December 14, 2018

10:00 PM PST

11:00 PM PST

Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.

Square ABCD has arc AC centred at B and arc BD centred at C

Author Message
TAGS:

Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

10 Sep 2014, 21:22
2
9
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:38) correct 38% (02:53) wrong based on 107 sessions

HideShow timer Statistics

As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is

A: $$\frac{16\pi}{3} - 4\sqrt{3}$$

B: $$4\sqrt{3} - \frac{4\pi}{3}$$

C: $$4 + 4\sqrt{3} + \frac{4\pi}{3}$$

D: $$16 + 2\sqrt{3} - \frac{8\pi}{3}$$

E: $$16 - 4\sqrt{3} - \frac{8\pi}{3}$$

Attachment:

square.png [ 9.08 KiB | Viewed 2583 times ]

_________________

Kindly press "+1 Kudos" to appreciate

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8676
Location: Pune, India
Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

10 Sep 2014, 22:40
2
PareshGmat wrote:
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is

A: $$\frac{16\pi}{3} - 4\sqrt{3}$$

B: $$4\sqrt{3} - \frac{4\pi}{3}$$

C: $$4 + 4\sqrt{3} + \frac{4\pi}{3}$$

D: $$16 + 2\sqrt{3} - \frac{8\pi}{3}$$

E: $$16 - 4\sqrt{3} - \frac{8\pi}{3}$$

Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb

Anxiously awaiting for breakthrough....

Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this.

Area of this shaded region = Area of square - Area of a quarter of a circle - (Area of Quarter of circle - Area of the other shaded region)

Area of this shaded region = Area of square$$- (1/4)*\pi*4^2 - ((1/4)*\pi*4^2 - (16/3)*\pi + 4*\sqrt{3})$$ - Area obtained in that question if radius is 4

Area of this shaded region $$= 16 - 4*\pi - 4*\pi + (16/3)*\pi - 4*\sqrt{3}$$

Area of this shaded region $$= 16 - (8/3)*\pi - 4\sqrt{3}$$
_________________

Karishma
Veritas Prep GMAT Instructor

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

10 Sep 2014, 23:58
VeritasPrepKarishma wrote:
PareshGmat wrote:
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is

Attachment:
square.png

A: $$\frac{16\pi}{3} - 4\sqrt{3}$$

B: $$4\sqrt{3} - \frac{4\pi}{3}$$

C: $$4 + 4\sqrt{3} + \frac{4\pi}{3}$$

D: $$16 + 2\sqrt{3} - \frac{8\pi}{3}$$

E: $$16 - 4\sqrt{3} - \frac{8\pi}{3}$$

Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb

Anxiously awaiting for breakthrough....

Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this.

Area of this shaded region = Area of square - Area of a quarter of a circle - (Area of Quarter of circle - Area of the other shaded region)

Area of this shaded region = Area of square$$- (1/4)*\pi*4^2 - ((1/4)*\pi*4^2 - (16/3)*\pi + 4*\sqrt{3})$$ - Area obtained in that question if radius is 4

Area of this shaded region $$= 16 - 4*\pi - 4*\pi + (16/3)*\pi - 4*\sqrt{3}$$

Area of this shaded region $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

Yes Karishma, thank you so much for the calculus.

That question placed was in line for this one. Can you kindly tell as to what level this question would be...?
_________________

Kindly press "+1 Kudos" to appreciate

Intern
Joined: 17 Apr 2012
Posts: 15
Location: United States
WE: Information Technology (Computer Software)
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

11 Sep 2014, 02:50
2
2
Took sometime, but learned some new concepts. There is always alternate ways of solution one problem.

Step 1: lets name the point of intersection of arc AC and arc BD as O. Connecting BO, we will get sector AOB with AB,OB as radius and arc AO. Similiarly connecting OC, we will get sector sector COD with CD ,OC as radius and arc DO.

Step 2. OB = OC = BC = radius of the both the circle = side of square ABCD. BOC will form a equilateral triangle.

Step 3: Area of the shaded area = Area of Square - (Area of sector AOB + Area of triangle BOC + Area of sector COD)

Step 4: Calculate area of sector AOB. angle AOB = 30° (Since BOC is equilateral triangle, all angles will be 60°).
Area of sector will be (30/360) of area of circle. = (30/360)Πr² = (1/12)Π×4² = 4Π/3

Step 5: Area of equilateral triangle BOC with side r = (1/2)*(r√3/2)*r = 4√3

Step 6. Area of sector COD will be equal to area of Sector AOB (same radius, same angle).

Step 7. Area of shaded area = 16 - (4Π/3 + 4√3 + 4Π/3 ) = 16 - 8Π/3 - 4√3
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8676
Location: Pune, India
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

11 Sep 2014, 03:21
1
PareshGmat wrote:
VeritasPrepKarishma wrote:
PareshGmat wrote:
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is

Attachment:
square.png

A: $$\frac{16\pi}{3} - 4\sqrt{3}$$

B: $$4\sqrt{3} - \frac{4\pi}{3}$$

C: $$4 + 4\sqrt{3} + \frac{4\pi}{3}$$

D: $$16 + 2\sqrt{3} - \frac{8\pi}{3}$$

E: $$16 - 4\sqrt{3} - \frac{8\pi}{3}$$

Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb

Anxiously awaiting for breakthrough....

Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this.

Area of this shaded region = Area of square - Area of a quarter of a circle - (Area of Quarter of circle - Area of the other shaded region)

Area of this shaded region = Area of square$$- (1/4)*\pi*4^2 - ((1/4)*\pi*4^2 - (16/3)*\pi + 4*\sqrt{3})$$ - Area obtained in that question if radius is 4

Area of this shaded region $$= 16 - 4*\pi - 4*\pi + (16/3)*\pi - 4*\sqrt{3}$$

Area of this shaded region $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

Yes Karishma, thank you so much for the calculus.

That question placed was in line for this one. Can you kindly tell as to what level this question would be...?

This will not be a GMAT question - too convoluted and very cumbersome calculation. The first one, perhaps, is workable since it needs only one step but even that one would be 700+.
_________________

Karishma
Veritas Prep GMAT Instructor

Intern
Joined: 12 May 2014
Posts: 14
Location: United States
Concentration: Strategy, Operations
Schools: IIMC'17
GMAT Date: 10-22-2014
GPA: 1.9
WE: Engineering (Energy and Utilities)
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

11 Sep 2014, 04:04
Solved with the same method as Prasanna.
I believe that most questions are solvable within time if the mind strikes in the right direction (Think that can comes with practice only)

TIP: Most geometry questions that seem unsolvable with the figure provided can be solved with some constructions. (Again comes with practice):P
Manager
Joined: 04 May 2015
Posts: 71
Concentration: Strategy, Operations
WE: Operations (Military & Defense)
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

05 Sep 2015, 11:40
1
VeritasPrepKarishma wrote:
PareshGmat wrote:
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is

Attachment:
square.png

A: $$\frac{16\pi}{3} - 4\sqrt{3}$$

B: $$4\sqrt{3} - \frac{4\pi}{3}$$

C: $$4 + 4\sqrt{3} + \frac{4\pi}{3}$$

D: $$16 + 2\sqrt{3} - \frac{8\pi}{3}$$

E: $$16 - 4\sqrt{3} - \frac{8\pi}{3}$$

Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb

Anxiously awaiting for breakthrough....

Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this.

Area of this shaded region = Area of square - Area of a quarter of a circle - (Area of Quarter of circle - Area of the other shaded region)

Area of this shaded region = Area of square$$- (1/4)*\pi*4^2 - ((1/4)*\pi*4^2 - (16/3)*\pi + 4*\sqrt{3})$$ - Area obtained in that question if radius is 4

Area of this shaded region $$= 16 - 4*\pi - 4*\pi + (16/3)*\pi - 4*\sqrt{3}$$

Area of this shaded region $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

Hi VeritasPrepKarishma

I'm very confused

I understand, calculating the square and then subtracting the area of a quarter of a circle with radius four, but I am lost after that as to how we calculate the leftover for the second arc? are you able to elaborate for me please?

I had a read of the other post you linked to which asks you to calculate the opposite area. Once again, I can understand how one section is calculated, just not how this works in a simultaneous sense?

Anyone else feel free to jump in too.

Edit: I worked on this for a while and finally got to the bottom of it. It's about finding the difference between the arc and an equilateral triangle... My first mistake was that I did not know the formula for the area of an equilateral triangle and had to google it. Second mistake... Not being clever enough to recognize the process. Working on it though

_________________

If you found my post useful, please consider throwing me a Kudos... Every bit helps

Director
Joined: 23 Jan 2013
Posts: 568
Schools: Cambridge'16
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

Updated on: 09 Sep 2015, 20:38
Option E is most favourable format in which we subtract unneeded area from 16. Answer in 1.5 min

E

Originally posted by Temurkhon on 09 Sep 2015, 20:32.
Last edited by Temurkhon on 09 Sep 2015, 20:38, edited 1 time in total.
Intern
Joined: 29 Oct 2014
Posts: 38
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

10 Sep 2015, 03:00
VeritasPrepKarishma wrote:
PareshGmat wrote:
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is

Attachment:
square.png

A: $$\frac{16\pi}{3} - 4\sqrt{3}$$

B: $$4\sqrt{3} - \frac{4\pi}{3}$$

C: $$4 + 4\sqrt{3} + \frac{4\pi}{3}$$

D: $$16 + 2\sqrt{3} - \frac{8\pi}{3}$$

E: $$16 - 4\sqrt{3} - \frac{8\pi}{3}$$

Experts: Please jump in. This question went top of my head. Either its too hard or too easy to make me dumb

Anxiously awaiting for breakthrough....

Now that you know how to calculate the area of the other shaded region (the question you have put in the Quantitative forum), you should be able to solve this.

Area of this shaded region = Area of square - Area of a quarter of a circle - (Area of Quarter of circle - Area of the other shaded region)

Area of this shaded region = Area of square$$- (1/4)*\pi*4^2 - ((1/4)*\pi*4^2 - (16/3)*\pi + 4*\sqrt{3})$$ - Area obtained in that question if radius is 4

Area of this shaded region $$= 16 - 4*\pi - 4*\pi + (16/3)*\pi - 4*\sqrt{3}$$

Area of this shaded region $$= 16 - (8/3)*\pi - 4\sqrt{3}$$

Nice explanationVeritasPrepKarishma
Non-Human User
Joined: 09 Sep 2013
Posts: 9157
Re: Square ABCD has arc AC centred at B and arc BD centred at C  [#permalink]

Show Tags

12 Aug 2018, 18:44
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Square ABCD has arc AC centred at B and arc BD centred at C &nbs [#permalink] 12 Aug 2018, 18:44
Display posts from previous: Sort by