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# r|s|t u|v|w x|y|z Each of the letters in the table above

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SVP
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r|s|t u|v|w x|y|z Each of the letters in the table above [#permalink]

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11 Nov 2005, 05:31
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r|s|t
u|v|w
x|y|z

Each of the letters in the table above represents one of the numbers 1, 2, or 3, and each of these numbers occurs exactly once in each row and exactly once in each column. What is the value of r?
(1) v + z = 6
(2) s + t + u + x = 6
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hey ya......

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Manager
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11 Nov 2005, 05:55
(1)

tells us nothing. Sum of the diagona can range between 3 and 9...
INSUFF

(2)
s+t+u+x=6
since all four share a row/column, we have to tistrubute the 6 among s+t and u+x, whereas each letter pair must receive at least 3. note, you cannot get a value of 2 out of two dissimilar numbers larter than 0. And with the same reasoning, you cannot give s+t more than 3, because then u+x would only get 2...
then you have s+t=3 and u+x=3
since no letter may be zero, s+t and u+x must be 2 and 1 or vice versa.
Hence r =3
SUFF

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Manager
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11 Nov 2005, 07:11
D for me
1=>v=z=3
you can fill the grid 3 1 2
2 3 1
1 2 3
r=3 suff
2=> fill the grid r= 3

suff

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Manager
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11 Nov 2005, 07:27
gotoknow3 wrote:
D for me
1=>v=z=3
you can fill the grid 3 1 2
2 3 1
1 2 3
r=3 suff
2=> fill the grid r= 3

suff

True! I change to D
Didnt realize taht V=Z=3 must be true.

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Director
Joined: 27 Jun 2005
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11 Nov 2005, 10:35
A,
r = 3
v+z=6 the only way its possible is if both v and z have value 3.
now the row 2,3 already has value 3. First row needs a 3, can't have 3 on the position of s,t cause those column already has 3.

s,t will have value (1,2) or (2,1)
same is the logic with u,v can have only valuse (1,2) or (2,1)

B just suggests that s+t+u+x = 6 = 1+1+1+3=1+1+2+2 so any variable can have value from 1 2 3 ....

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Director
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11 Nov 2005, 10:43

B is also sufficient ........s+t+u+x= 6 can use 3 as one . hence all of these 4 variable will have value 1 or 2 so r will have value 3.

it should be D

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SVP
Joined: 28 May 2005
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11 Nov 2005, 13:47
OA is indeed D.

good job.
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hey ya......

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11 Nov 2005, 13:47
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