IMO D

Stmt I

Lets pick numbers that can satisfy the statement

x = 3,9,15,21,27...
all these numbers when divided by 2 have remainder 1, and when divided by 3 remainder is 0

Any of these numbers when divided by 6 will always have a remainder 3

Sufficient

Stmt II

Again lets pick numbers to satisfy the statement

x = 3,15,27,39..
all these numbers when divided by 12 have remainder 3

Any of these numbers when divided by 6 will have a remainder 3

[spoiler]Sufficient

D[/spoiler]

Correct me if I am but 3/12 does not produce a remainder of 3, nor does 3/6 produce a remainder of 3. 3 is a smaller number and there is nothing remaining.

Therefore Statement I is insufficient because we can't rule out 3 as an option.

Statement II however tells us that x must be larger than 12 to produce a remainder of 3. So by my reasoning the answer should be B.

Correct me if I am but 3/12 does not produce a remainder of 3, nor does 3/6 produce a remainder of 3. 3 is a smaller number and there is nothing remaining.

Therefore Statement I is insufficient because we can't rule out 3 as an option.

Statement II however tells us that x must be larger than 12 to produce a remainder of 3. So by my reasoning the answer should be B.

IMO it is B.

lets consider stmt 1:
The foll numbers satisfy stmt1.
3, 6, 9, 12, 15.

Dividing by 6 will give 0 and 3 as remainders. So insufficient.

Stmt2:
The foll # satisfy stmt2.
3, 15, 27
In each case the remainder is 3, when divided by 6.
hence Sufficient.

One could prove this algebraically but picking numbers is faster here.

HT helps

rossmj wrote:Correct me if I am but 3/12 does not produce a remainder of 3, nor does 3/6 produce a remainder of 3. 3 is a smaller number and there is nothing remaining.

.

3 divided by 12 AND 3 divided by 6 DOES produce 3 as remainder.
When the number is smaller than the divisor, the number itself is the remainder... Very important concept.

Also search this forum for a post by Ian Stewart /Lunarpower(Ron purewal) for a good discussion on remainders when negative numbers are involved.

vittalgmat wrote: lets consider stmt 1:
The foll numbers satisfy stmt1.
3, 6, 9, 12, 15.

Dividing by 6 will give 0 and 3 as remainders. So insufficient.

HT helps

6 and 12 do not satisfy the statement, when divided by 2 we do not get remainder 1

sacx wrote:

vittalgmat wrote: lets consider stmt 1:
The foll numbers satisfy stmt1.
3, 6, 9, 12, 15.

Dividing by 6 will give 0 and 3 as remainders. So insufficient.

HT helps

6 and 12 do not satisfy the statement, when divided by 2 we do not get remainder 1

Ooops my bad.. thanks sacx for pointing out.
Stmt1 includes all Odd multiples of 3.
3, 9, 15...
remainder is 3 in each case when divided by 6.
So stmt 1 is sufficient.!!!.

So D is the answer.

Hope I dint miss anything.