Mar 20 09:00 PM EDT  10:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Wednesday, March 20th at 9 PM EDT Mar 20 07:00 AM PDT  09:00 AM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. Mar 23 07:00 AM PDT  09:00 AM PDT Christina scored 760 by having clear (ability) milestones and a trackable plan to achieve the same. Attend this webinar to learn how to build trackable milestones that leverage your strengths to help you get to your target GMAT score. Mar 27 03:00 PM PDT  04:00 PM PDT Join a free live webinar and learn the winning strategy for a 700+ score on GMAT & the perfect application. Save your spot today! Wednesday, March 27th at 3 pm PST
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 06 Jan 2015
Posts: 605
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)

The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
Updated on: 30 Jul 2016, 07:16
Question Stats:
79% (02:23) correct 21% (02:49) wrong based on 360 sessions
HideShow timer Statistics
The table above is a 3 by 3 grid of 9 numbers, 5 of which are given. The 4 numbers that are not given are denoted by A, B, C, and D. The product of the numbers in each row and in each column is the same for all rows and columns. What is the value of the product ABCD ? A) 36 B) 72 C) \(36\sqrt{2}\) D) \(36\sqrt{3}\) E) \(72\sqrt{6}\) Attachment:
Table.jpg [ 6.52 KiB  Viewed 8777 times ]
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च
Resource: GMATPrep RCs With Solution
Originally posted by NandishSS on 30 Jul 2016, 06:05.
Last edited by Bunuel on 30 Jul 2016, 07:16, edited 1 time in total.
Edited the question.




SC Moderator
Joined: 13 Apr 2015
Posts: 1686
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
30 Jul 2016, 08:42
A * B * sqrt(3) = 6*sqrt(6) A * B = 6 * sqrt(2)
C * D * 2sqrt(3) = 6*sqrt(6) C * D = 3 * sqrt(2)
A * B * C * D = 6 * 3 * 2 = 36
Answer: A




Manager
Joined: 18 Feb 2015
Posts: 84

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
14 Aug 2016, 10:50
Hi all,
Does anyone know where 6*sqrt(6) is coming from?
Thank you.



SC Moderator
Joined: 13 Apr 2015
Posts: 1686
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
14 Aug 2016, 20:48
HarveyKlaus wrote: Hi all,
Does anyone know where 6*sqrt(6) is coming from?
Thank you. Refer to the center column that has numbers 1, 6 and sqrt(6). Multiply and you will get 6*sqrt(6).



Manager
Joined: 26 Mar 2017
Posts: 115

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
22 Jun 2017, 08:40
just square each number in the table and then the solution becomes obvious A 1 12 B 6 6 3 6 D so we need 6*6=36 cheers
_________________
I hate long and complicated explanations!



Current Student
Joined: 30 May 2017
Posts: 66
Concentration: Finance, General Management
GPA: 3.23

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
22 Jun 2017, 12:32
daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate?
_________________
Veritas Prep 6/18/17 600 Q:38 V:35 IR:5 Veritas Prep 6/29/17 620 Q:43 V:33 IR:4 Manhattan 7/12/17 640 Q:42 V:35 IR:2.4 Veritas Prep 7/27/17 640 Q:41 V:37 IR:4 Manhattan 8/9/17 670 Q:44 V:37 IR:3 Veritas Prep 8/21/17 660 Q:45 V:36 IR:7 GMAT Prep 8/23/17 700 Q:47 V:38 IR:8 GMAT Prep 8/27/17 730 Q:49 V:40 IR:8 Veritas Prep 8/30/17 690 Q:47 V:37 IR:8



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3361
Location: India
GPA: 3.12

The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
22 Jun 2017, 13:24
Smokeybear00 wrote: daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate? If you look at the question, the middle column reads 1,6,\(\sqrt{6}\) not 1,6,6 Hence squaring it will give you 1,36 and 6. Btw, why are you squaring the numbers? The solution to this problem is as follows :
Since the product of each of the rows and columns must be equal, only the center row has all three values, the product of which is 6*\(\sqrt{6}\) We need to find the product of A,B,C,D. From the first row : A * B * sqrt(3) = 6*sqrt(6) ==> A * B = 6 * sqrt(2) From the third row :C * D * 2sqrt(3) = 6*sqrt(6) ==> C * D = 3 * sqrt(2) Therefore product of A,B,C and D as asked in the question is (6*sqrt(2))*(3*sqrt(2)) =6*3*2 = 36 (Option A)Hope it helps!
_________________
You've got what it takes, but it will take everything you've got



Manager
Joined: 26 Mar 2017
Posts: 115

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
23 Jun 2017, 01:40
Smokeybear00 wrote: daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate? well, Im sorry I made a careless mistake, but you should get the point A 1 12 B 6 6 3 36 D
_________________
I hate long and complicated explanations!



Manager
Joined: 26 Mar 2017
Posts: 115

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
23 Jun 2017, 01:41
pushpitkc wrote: Smokeybear00 wrote: daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate? If you look at the question, the middle column reads 1,6,\(\sqrt{6}\) not 1,6,6 Hence squaring it will give you 1,36 and 6. Btw, why are you squaring the numbers? The solution to this problem is as follows :
Since the product of each of the rows and columns must be equal, only the center row has all three values, the product of which is 6*\(\sqrt{6}\) We need to find the product of A,B,C,D. From the first row : A * B * sqrt(3) = 6*sqrt(6) ==> A * B = 6 * sqrt(2) From the third row :C * D * 2sqrt(3) = 6*sqrt(6) ==> C * D = 3 * sqrt(2) Therefore product of A,B,C and D as asked in the question is (6*sqrt(2))*(3*sqrt(2)) =6*3*2 = 36 (Option A)Hope it helps! yes thanks, I made a careless mistake I squared the numbers, because for me the problem becomes much easier
_________________
I hate long and complicated explanations!



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2825

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
26 Jun 2017, 18:01
NandishSS wrote: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given. The 4 numbers that are not given are denoted by A, B, C, and D. The product of the numbers in each row and in each column is the same for all rows and columns. What is the value of the product ABCD ? A) 36 B) 72 C) \(36\sqrt{2}\) D) \(36\sqrt{3}\) E) \(72\sqrt{6}\) We can start with the middle column: 1 x √6 x 6 = 6√6. Thus, we know that the entries in each row and in each column multiply to yield 6√6. For Row 1: A x 1 x 2√3 = 6√6 A = (6√6)/(2√3) A = 3√2 For Column 1: 3√2 x B x √3 = 6√6 B x 3√6 = 6√6 B = (6√6)/(3√6) B = 2 For Row 2: 2 x √6 x C = 6√6 C = (6√6)/(2√6) C = 3 For Row 3: √3 x 6 x D = 6√6 D = (6√6)/(6√3) D = √2 Thus, ABCD = 3√2 x 2 x 3 x √2 = 18√2 x √2 = 36 Answer: A
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews



Retired Moderator
Joined: 19 Mar 2014
Posts: 933
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
27 Jun 2017, 14:49
The table above is a 3 by 3 grid of 9 numbers, 5 of which are given. The 4 numbers that are not given are denoted by A, B, C, and D. The product of the numbers in each row and in each column is the same for all rows and columns. What is the value of the product ABCD ? From Second column we see that the multiplication for each row/columns should be \(= 6 * \sqrt{6}\) \(= 2 * 3 * \sqrt{3} * \sqrt{2}\) \(AB * \sqrt{3}\) \(= 2 * 3 * \sqrt{3} * \sqrt{2}\) \(AB = 6 * \sqrt{2}\) \(CD * 2\sqrt{3}\) \(= 2 * 3 * \sqrt{3} * \sqrt{2}\) \(CD = 3 * \sqrt{2}\) \(AB * CD = 6 * \sqrt{2} * 3 * \sqrt{2}\) \(AB * CD = 18 * 2\) \(AB * CD = 36\) Hence, Answer is A
_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."
Best AWA Template: https://gmatclub.com/forum/howtoget60awamyguide64327.html#p470475



Manager
Joined: 28 Jan 2018
Posts: 157
Location: Taiwan
GPA: 3.34

Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
Show Tags
26 Nov 2018, 00:27
A faster way to solve this
All product = 6√6
A x B x √3 = 6√6, AB = (6√6 / √3) C x D x 2√3 = 6√6, CD = (6√6 / 2√3) A x B x C x D = (6√6 / 2√3) x (6√6 / √3) = (6 x 6 x 6) / ( 2 x 3) = 36
Don't solve for individual value, waste too much time




Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
[#permalink]
26 Nov 2018, 00:27






