November 14, 2018 November 14, 2018 07:00 PM PST 08:00 PM PST Join the webinar and learn timemanagement tactics that will guarantee you answer all questions, in all sections, on time. Save your spot today! Nov. 14th at 7 PM PST November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299)
Author 
Message 
Math Expert
Joined: 02 Sep 2009
Posts: 50572

In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Jul 2015, 02:25
Question Stats:
68% (01:43) correct 32% (02:40) wrong based on 218 sessions
HideShow timer Statistics



Manager
Joined: 13 Sep 2014
Posts: 88
WE: Engineering (Consulting)

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Jul 2015, 02:47
Let Area of larger square be x
Therefore side of smaller square is x/_/2
Given:
(X^2 / 2)* (1/4) = 5/2
X^2 = 20
Which is area of larger Square



Manager
Joined: 26 Dec 2011
Posts: 114

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Jul 2015, 03:59
Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution.Solution  Lets say, area of the shaded square A. A/4 = 5/2 A = 4 × (5/2) A = 10 So, the area of the shaded square is equal to 10, and each side is equal to √10. Diagonal of shaded square is side of a largest square = √(side^2+side^2) = √((√10)^2+(√10)^2) = √20 Area of largest square = √20 * √20 = 20. ANS A
_________________
Thanks, Kudos Please



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2698
Location: India
GMAT: INSIGHT
WE: Education (Education)

In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Jul 2015, 04:51
Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution.Area of Smallest Square = 2^2 = 4 area of the shaded square / the area of the smallest square = 5/2 i.e. area of the shaded square = (5/2)*4 = 10 Since the Shaded Square is bisecting the sides of Largest square therefore the Area of Largest Square will be twice the area of Shaded Square Area of Shaded Square = (1/2) Area of Biggest Squarei.e. Area of Biggest Square = 2 * (10) = 20
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Intern
Status: My heart can feel, my brain can grasp, I'm indomitable.
Affiliations: Educator
Joined: 16 Oct 2012
Posts: 39
Location: Bangladesh
WE: Social Work (Education)

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Jul 2015, 06:22
let side length of larger square be x so, side length of shaded square is (x/2)*√2 or x/√2 now, (x/√2)^2 : 2^2 = 5 : 2 or, x^2 = 4*5 = 20 area of larger square is 20 ans: A
_________________
please press "+1 Kudos" if useful



Intern
Joined: 11 Sep 2013
Posts: 7

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Jul 2015, 21:44
Area of shaded region = 5/2 *4 = 10 So side =√10
Therefore diagonal of shaded square =√20 which is the side of larger square so the area of larger square = (√20)^2 = 20



Manager
Joined: 20 Jul 2011
Posts: 80

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
17 Jul 2015, 15:58
Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution.Area of the Smallest Sq = 4 Ratio of Shaded to Smallest sq = 5/2 = x/4 =>Area of shaded sq. x = 10. If Area = 10, Side =\(\sqrt{10}\) Draw a diagonal to the Shaded square, x. x^2 = 10 + 10 = 20 => Area of the big square (Side ^ 2) or in other way, Since the shaded sq. bisects the vertices of big sq., the area of the big square will be twice the shaded sq. = 2 * 10 = 20. Option A



Math Expert
Joined: 02 Sep 2009
Posts: 50572

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
19 Jul 2015, 12:09
Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution. 800score Official Solution:The area of the smallest square must be equal to 2^2 = 4. Since the ratio of the area of the shaded square (it consists of the shaded region and the smallest square) to the area of the smallest square is 5/2, we can solve for the area of the shaded square by solving for A in the following equation: A/4 = 5/2 A = 4 × (5/2) A = 10 So, the area of the shaded square is equal to 10, and each side is equal to √10. Now, all we have to do is recognize that the triangles made at each corner of the large square where, two sides are equal to half the side length and the hypotenuse is equal to the edge length of the shaded square, are right triangles and to use the Pythagorean Theorem to solve for half the edge length. Let each side of the triangle be equal to x, then 2x^2 = 10, so x^2 = 5, and x = √5. This means that the entire side length of the larger square is twice the length of a side of that triangle, or 2√5. Finally, the area of a square is equal to a side squared, so the answer must be (2√5 )^2 = 20, or answer choice (A).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 14 Apr 2015
Posts: 14

In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
22 Jul 2015, 10:26
Bunuel wrote: Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution. 800score Official Solution:The area of the smallest square must be equal to 2^2 = 4. Since the ratio of the area of the shaded square (it consists of the shaded region and the smallest square) to the area of the smallest square is 5/2, we can solve for the area of the shaded square by solving for A in the following equation: A/4 = 5/2 A = 4 × (5/2) A = 10 So, the area of the shaded square is equal to 10, and each side is equal to √10. Now, all we have to do is recognize that the triangles made at each corner of the large square where, two sides are equal to half the side length and the hypotenuse is equal to the edge length of the shaded square, are right triangles and to use the Pythagorean Theorem to solve for half the edge length. Let each side of the triangle be equal to x, then 2x^2 = 10, so x^2 = 5, and x = √5. This means that the entire side length of the larger square is twice the length of a side of that triangle, or 2√5. Finally, the area of a square is equal to a side squared, so the answer must be (2√5 )^2 = 20, or answer choice (A). Hello Bunuel , Can i not take as area of shaded region + area of small square = 10+4 = 14 so area of square (shaded region + small square)=14 so its side is root(14) since this particular square bisects the larger square So area of larger square =2(area of this square)= 2*14=28 Is this not correct ? problem is i could not understand how area of shaded region can be interpreted to get the side Thanks



Intern
Joined: 04 Sep 2015
Posts: 7

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
30 Oct 2015, 06:50
Hi I have the Same Problem with this one. Since the area of These shaded Square is 10 you have to add the small Square to get the whole area. Thats 14 then and one side is the suare Root of 14. Why is that wrong?



Veritas Prep GMAT Instructor
Joined: 15 Jul 2015
Posts: 110
GPA: 3.62
WE: Corporate Finance (Consulting)

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
30 Oct 2015, 07:09
Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution.The area of the small square is 4. If the area of the shaded square has a a ratio of 5/2 that of the small square, then its area is 10. We can therefore determine each side of the shaded square has a length of \sqrt{10} If each side of the shaded squre bisects the side of the large square, the triangle created by side of the shaded square and 1/2 each side of the large square is a 45:45:90 suqare with the shaded square's side as the hypotenuse. The proportion of the sides of a 45:45:90 triangle are 1:1:\sqrt{2} and, in this case the hypotenuse is \sqrt{10}, so the sides of the large square are each 2\sqrt{5}. The area, therefore, is 2\sqrt{5} squared, or 20. Answer A
_________________
Dennis Veritas Prep  GMAT Instructor
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



SVP
Joined: 12 Dec 2016
Posts: 1674
Location: United States
GPA: 3.64

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
15 Sep 2017, 07:58
Bunuel wrote: In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices? A. 20 B. 30 C. 40 D. 50 E. 60 Kudos for a correct solution.this is gonna be tested in gmat?



Manager
Joined: 21 Jun 2017
Posts: 131
Concentration: Finance, Economics
WE: Corporate Finance (Commercial Banking)

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
02 Nov 2018, 20:43
Area of shaded region =x^2 ,,,,, then (x^2  4) / 4 = 5/2 with this i am getting x =sqrt14. Please point out my mistake chetan2uRegards
_________________
Even if it takes me 30 attempts, I am determined enough to score 740+ in my 31st attempt. This is it, this is what I have been waiting for, now is the time to get up and fight, for my life is 100% my responsibility.



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2698
Location: India
GMAT: INSIGHT
WE: Education (Education)

In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
03 Nov 2018, 20:10
ShankSouljaBoi wrote: Area of shaded region =x^2 ,,,,, then (x^2  4) / 4 = 5/2 with this i am getting x =sqrt14. Please point out my mistake [/url]
Regards ShankSouljaBoi \((x^2  4) / 4 = 5/2\) is incorrect and should be written as \((x^2) / 4 = 5/2\) for the simple reason that the language of question reads the ratio of the area of the shaded square to the area of the smallest square is 5/2,It says area of shaded square and doesn't say Area of shaded REGION I hope this helps!!!
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Math Expert
Joined: 02 Aug 2009
Posts: 7025

Re: In the figure above, each square is contained within the area of the l
[#permalink]
Show Tags
04 Nov 2018, 05:34
ShankSouljaBoi wrote: Area of shaded region =x^2 ,,,,, then (x^2  4) / 4 = 5/2 with this i am getting x =sqrt14. Please point out my mistake chetan2uRegards The question is flawed. It should have named the squares say ABCD for smaller one and EFGH for middle one and then said ratio of areas of EFGH and ABCD is 5/2. There is NO shaded square here, it is a shaded region. Bunuel, May I request you to change the wordings. It is ambiguous.
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor




Re: In the figure above, each square is contained within the area of the l &nbs
[#permalink]
04 Nov 2018, 05:34






