GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 01 Jun 2020, 04:25 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the figure above, SAND and SURF are squares, and O is the center of

Author Message
TAGS:

### Hide Tags

Senior SC Moderator V
Joined: 14 Nov 2016
Posts: 1343
Location: Malaysia
In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

11 00:00

Difficulty:   75% (hard)

Question Stats: 56% (02:44) correct 44% (03:07) wrong based on 108 sessions

### HideShow timer Statistics In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1

Attachment: Untitled.png [ 36.59 KiB | Viewed 2106 times ]

_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Originally posted by hazelnut on 26 May 2017, 04:02.
Last edited by Bunuel on 26 May 2017, 05:02, edited 2 times in total.
Edited the question.
Senior Manager  B
Joined: 13 Oct 2016
Posts: 352
GPA: 3.98
Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

2
hazelnut wrote: In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1

Attachment:
Untitled.png

Side of SAND – x

Side of SURF – y

Radius of the circle – r

Now let’s take a look at the right triangle SUA

x^2 + y^2 = 4r^2

r^2 = (x^2 + y^2)/4

Taking the required fraction:

C/Q = π*r^2 / (x^2 + y^2) = π/4 = 3.14/4 is slightly more than 3/4.

Current Student P
Joined: 18 Aug 2016
Posts: 588
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

1
hazelnut wrote: In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1

Attachment:
Untitled.png

Let side of SAND be "a" and SURF be "b"

AREA of SAND & SURF will be $$a^2+b^2$$

Radius of circle = \sqrt{a^2 + b^2}
On solving
Area of circle/Area of Squares = 3.14/4 which lies between 3/4 & 7/8 hence option C
_________________
We must try to achieve the best within us

Thanks
Luckisnoexcuse
Intern  Joined: 13 May 2017
Posts: 8
Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

Then Q= 4(x )square
C= 2*22/7*x*x
So C/Q= 11/7

Sent from my iPhone using GMAT Club Forum mobile app
Intern  B
Joined: 28 May 2015
Posts: 36
Location: India
GMAT 1: 710 Q51 V34
GPA: 4
Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

1
Since Triangle USA is a right angled triangle with UA as diameter of the circle.

If we assume sides of sqaure as 6 and 8 respectively
Then 6^2 + 8^2 = 10^2 = UA^2

Which is basically sum of the areas of two sqaures I.e. P

Area of circle C =( π•10^2 )/4

Ratio C /P = [ ( π•10^2 )/4 ] ÷ 10^2
= π / 4
= 3.14 / 4 , greater than 3/4

Option C

Sent from my ONE A2003 using GMAT Club Forum mobile app
VP  V
Joined: 18 Dec 2017
Posts: 1373
Location: United States (KS)
GMAT 1: 600 Q46 V27 Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

hazelnut wrote: In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1

Attachment:
Untitled.png

I have a slightly different approach.
You can see the right triangle. It is right there. So why not to just assume the sides? Let's say it is 6,8,10 Triplet.
Area of squares is 64+36 which is 100.
Area of Circle is Pi*5*5 which works out to be approximately 75. Since you know actual value is higher the fraction will be higher than 3/4. Hence C.
_________________
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long

Why You Don’t Deserve A 700 On Your GMAT

Learn from the Legend himself: All GMAT Ninja LIVE YouTube videos by topic
You are missing on great learning if you don't know what this is: Project SC Butler
CEO  V
Joined: 03 Jun 2019
Posts: 2919
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

### Show Tags

hazelnut wrote: In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1

Attachment:
Untitled.png

Given: In the figure above, SAND and SURF are squares, and O is the center of the circle.

Asked: If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

$$Q = a^b + b^2 = D^2$$ where D is the diameter of the circle

$$C = \pi r^2 = \frac{\pi D^2}{4} = \frac{\pi}{4} Q$$

$$\frac{C}{Q} = \frac{\pi}{4} =\frac{7}{8} > \frac{3.14}{4} > \frac{3}{4}$$

IMO C
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com In the figure above, SAND and SURF are squares, and O is the center of   [#permalink] 25 Aug 2019, 01:59

# In the figure above, SAND and SURF are squares, and O is the center of  