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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

05 Jan 2008, 13:28

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 --> v=3, z=3 --> s<3, t<3 --> r=3 2. s + t + u + x = 6. if any term were 3 other terms would be 1. But it is impossible, because there are 2 terms in one raw among any 3 terms. r=3
_________________

1. v+z = 6 so v and z must both be 3. this places a 3 in two columns AND in two rows. r MUST be 3 as well. SUFFICIENT

2. (r+s+t)+(r+u+x) = 12 since each row and column must have (1+2+3). 2r+(s+t+u+x) = 12 2r+6 = 12 2r = 6 r = 3 SUFFICIENT

Answer D

for statement two you could have seen that it's impossible for s, t, u or x to be equal to 3 because the 3 remaining numbers would have to total 3. This is impossible unless all 3 numbers are 1, which is impossible because we can only have one 1 per column and row.

EDIT: aaahhh!!! walker again! you're everywhere and always one step ahead!

Answer is D, ie either one can give the answer. Look at the rows and columns clearly before reading further. From S1, v+z=6, so both v=z= 3 ( all are 1, 2 or 3) only one 3 can be present in any row or column, eliminate rows and columns containing 3, then we see that only r can take the value of 3, so, r=3 and s1 is sufficient

From S2, (s+t) + (u+x) = 6, s+t belongs to a row and u+x belongs to a column, the minimum sum of two numbers in a row/column is 3 (sum 1 +2) (since nos are 1,2 & 3 only) so, from above given equation we see that s+t=3 and u+x=3 so, the only number in that row/column is r and it should be 3. since we have already taken 1 & 2. so, r=3 and S2 is sufficient