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Difficulty:
25%
(medium)
Question Stats:
81% (02:26) correct
19%
(03:00)
wrong
based on 1835
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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 59090 times ]
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
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
JeffTargetTestPrep
Expert
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
NandishSS
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
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General Discussion
Posts: 69
Hi all,
Does anyone know where 6*sqrt(6) is coming from?
Thank you.
Does anyone know where 6*sqrt(6) is coming from?
Thank you.
HarveyKlaus
Refer to the center column that has numbers 1, 6 and sqrt(6). Multiply and you will get 6*sqrt(6).
Posts: 57
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
A 1 12
B 6 6
3 6 D
so we need 6*6=36
cheers
Smokeybear00
Joined: 30 May 2017
Last visit: 05 Apr 2018
Posts: 55
Concentration: Finance, General Management
GMAT 1: 690 Q47 V38
GPA: 3.23
GMAT 1: 690 Q47 V38
Posts: 55
daviddaviddavid
6^2 does not equal 6, (see column 2, row 3). Can you elaborate?
Smokeybear00
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!
Posts: 57
Smokeybear00
well, Im sorry I made a careless mistake, but you should get the point
A 1 12
B 6 6
3 36 D
ydmuley
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Last visit: 01 Dec 2019
Posts: 807
Location: India
Concentration: Finance, Entrepreneurship
Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20
GPA: 3.5
Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20
Posts: 807
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
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
lary301254M7
Joined: 28 Jan 2018
Last visit: 17 Apr 2023
Posts: 107
Location: Taiwan
GMAT 1: 690 Q47 V38
GPA: 3.34
GMAT 1: 690 Q47 V38
Posts: 107
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
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
Posts: 5,200
Hello, people. Let's get into this!
The third column lets us know that the product of a single row or column is \(6\sqrt{6}\)
The product of two columns is therefore \(6\sqrt{6}\) x \(6\sqrt{6}\)
If we multiply the two columns that contain AB and CD, we can create the following expression:
ABCD x \(\sqrt{3}\) x \(2\sqrt{3}\) = \(6\sqrt{6}\) x \(6\sqrt{6}\)
ABCD = (\(6\sqrt{6}\) x \(6\sqrt{6}\)) / (\(\sqrt{3}\) x \(2\sqrt{3}\))
ABCD = (36 x 6) / [2 (3)]
ABCD = 36
A is your answer.
The third column lets us know that the product of a single row or column is \(6\sqrt{6}\)
The product of two columns is therefore \(6\sqrt{6}\) x \(6\sqrt{6}\)
If we multiply the two columns that contain AB and CD, we can create the following expression:
ABCD x \(\sqrt{3}\) x \(2\sqrt{3}\) = \(6\sqrt{6}\) x \(6\sqrt{6}\)
ABCD = (\(6\sqrt{6}\) x \(6\sqrt{6}\)) / (\(\sqrt{3}\) x \(2\sqrt{3}\))
ABCD = (36 x 6) / [2 (3)]
ABCD = 36
A is your answer.
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