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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
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