Data Sufficiency

Does anyone know how to solve this?

If P is a set of integers and 3 is in P, is evervy positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P.

AD
BCE

Jason

I believe that the answer is E. Here's how I got the answer.

A. INSUFFICIENT: 3+1=4 which is not a multiple of 3, however, 3+3 is multiple of 3.
B. INSUFFICIENT: 6-3= 3 which is a multiple of 3, however, 2-3 is not a multiple of 3.

D. CANNOT BE: Because if we combined the two statements, it does not say anything about the answer. It always comes down to having two or more different answers.

I dont know if I am right but that's how I analyzed the problem.

jtsgmat wrote:Does anyone know how to solve this?

If P is a set of integers and 3 is in P, is evervy positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P.

AD
BCE

Jason

I see it as A

if you have 3 in the set, and for any integer in P you add 3, then you would have 3, (3+3), (6 +3), etc...

but that's just my take.

The answer could be A depending upon the interpretation of the problem.

In another interpretation of the problem, the answer could be (C).

I just took the set like {-3, 0 3}

I looked at statement (1)
-3 + 3 = 0 is in the set
0 + 3 is in the set
3 + 3 should be there as well, so set should be {-3, 0 , 3 , 6}
So you could go ad infinitum to show that the set is really {-3, 0, 3, 6, ...+infinity}
Sufficient

Then I looked at statement 2, Again started with {-3,0,3} backwards
3 - 3 = 0 is in the set
0 - 3 = -3 is in the set
-3 - 3 = -6 should also be there, so set should be {-6, -3, 0, 3}
So you could go ad infinitum to show that the set is really {-infinity, ...-3, 0, 3}

OK, so Statement 2 tells me that all positive multiples of 3 are not in P, UNLESS P already has ALL positive multiples of 3. That is, a situation like

{-infinity, ..., -3, 0, 3, 6, ....+infinity}

If that were not the case, I could argue that Statement 2 is ALSO SUFFICIENT TO TELL ME THAT THE SET WOULD NOT CONTAIN ALL +VE MULTIPLES OF 3. In this interpretation, Statement (2) is also SUFFICIENT.

But given that "infinity" itself is an ethereal concept, take your pick!

Rehana

My Pick would be A

As per statement 1 if 3 is there then all postive multiples of 3 will be there so it is sufficient.
As per statement 2 you can definately say that all negative multiples of 3 are present but nothing concrete can be said about positive multiples of 3. So statement 2 is insufficient.

So my pick is A

jtsgmat wrote:Does anyone know how to solve this?

If P is a set of integers and 3 is in P, is evervy positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P.

AD
BCE

Jason

My pick is D, since it is asking for a positive multiple .. I guess, the answer is D.. the set might have numbers like 0, 0+3, 3+3, 6+3, etc...

or 3, 3-3, 6-3, 9-3 etc...

jtsgmat wrote:Does anyone know how to solve this?

If P is a set of integers and 3 is in P, is evervy positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P.

AD
BCE

Jason

1. Since 3 is given to be in P then 3+3 is also in P, then 3+3+3 is also in P. So all multiples of 3 are in P. sufficient

2. 3 is given to be in P. No other integer inside P is given. all we can say is 3-3 is in P. 3-3-3 is also in P. So negative multiples are there in P. But nothing more known about the set P so cannot definitely say that P doesnt contain all the positive multiples of 3. so insufficient

I think A. What is OA?

Any replies, I think it is E because the questions doesn't say that the set contains only multiples of 3, it is asking that.

So the set can have 4 also, so 3+4 =7 and 4-3 = 1. Both non multiples of 3.

Any OA for this one?

vineetbatra wrote:Any replies, I think it is E because the questions doesn't say that the set contains only multiples of 3, it is asking that.

So the set can have 4 also, so 3+4 =7 and 4-3 = 1. Both non multiples of 3.

Any OA for this one?

I think it is E too. IMO "any integer" can be used loosely compared to "every integer"

1) set can be {1,3,4,6} or {3,6,9}
2) set can be {0,3,4,6,7} or {0,3,6}

IMO - A

jtsgmat wrote:Does anyone know how to solve this?

If P is a set of integers and 3 is in P, is evervy positive multipe of 3 in P?

1) For any integer in P, the sum of 3 and that integer is aslo in P.

2) For any integer in P, that integer minus 3 is also in P.

AD
BCE

Jason

IMO D
1) if there is atleast one multiple of 3 int he set, then its true.
since 3 is there in the set, 3=3=6,3=6=9... they all will be there in the set.

SUFF
2) lets say only 3 is there in P, the set will be {3,0, and negative multiples of 3}.
so whatever be the integer, the set can at max only contain multiple of 3 less than the greatest number in the set.
SUFF