The number of subsets of any set can be calculated by raising 2 to the power of the number of elements in the set. The first statement gives you that information immediately. The second gives you enough information to find the number of elements, as there are eight primes between 1 and 20: 2, 3, 5, 7, 11, 13, 17, and 19. From either statement alone, you can deduce the answer to be .

The multiples of 3 between 29 and 50 are 30, 33, 36, 39, 42, 45, and 48 - therefore,  has seven elements total. 

The number of subsets in a set can be calculated by raising 2 to the power of the number of elements. Therefore, the answer to our question is .

If we only know Martindale got 240 votes, since 240 is not a majority, we cannot determine the winner with certainty. For example, the following two results are possible:

Martindale: 240, Nance: 0, Osgood 373 - Osgood wins

Martindale 240, Nance 373, Osgood 0 - Nance wins.

A similar argument shows the second statement insufficient as well, since 244 is not a majority. 

But the two statements together allow us a complete picture;

Martindale 240, Nance 244, Osgood 129 - Nance wins

From Statement 1 alone, it can be immediately deduced that Benson won, since 251 of the 457 votes consititute a majority:

Benson got 54.9% of the vote, so there is no way that Anderson or Carter could have bested Benson.

Statement 2 tells only that Carter did not win, since either Anderson or Benson had to have won at least half, or ; we cannot, however, tell which one it was without further information.

, which is a set of 12 elements. A set this size has  subsets.

 and  are the sets of positive multiples of 4 and 5, respectively. For a number to be in both sets, the number must be divisible by both 4 and 5. This happens if and only if it is also divisible by .

The elements of   are precisely the mulitples of 20:

All of the numbers end in 0 and have 2, 4, 6, 8,or 0 as their second-to-last digit.

Statement 1 does not, by itself, prove or disprove that , since there are numbers like 10 and 30 that do not fall in this set. But none of the elements of  have 5 as their second-to-last digit, so Statement 2 proves  to be false.

Suppose you know both statements. From the first statement, you can calculate that  seniors are taking calculus, physics, or both. However, as illustrated by these two examples, you cannot tell which course has more seniors enrolled.

Example 1: 50 seniors are enrolled in both courses.

Then  seniors are enrolled in calculus;  seniors are enrolled in physics but not calculus;  seniors total are enrolled in physics. This means that seniors in physics outnumber seniors in calculus. 

Example 2: 100 seniors are enrolled both courses.

Then  seniors are enrolled in calculus;  seniors are enrolled in physics but not calculus;  seniors total are enrolled in physics. This means that seniors in calculus outnumber seniors in physics. 

  is the complement of  - that is, the set of all elements in the universal set  that are not in . To find  given , we need to know the elements in . Statement 1 gives us this information; Statement 2 does not.

Statement (1) allows us to find B:

B = S + 3 = {4 , 6 , 10} Therefore:

S + B = {1+4 , 1+6 , 1+10 , 3+4 , 3+6 , 3+10 , 7+4 , 7+6 , 7+10}

S + B = {5 , 7 , 9 , 11 , 13 , 17}. SO statement (1) is sufficient to find S+B

Statement (2) does not give us enough information to find B. It could be any set between  {4 , 6 , 10} and {1 , 3 , 4 , 6 , 7 , 10} if it includes the same numbers as S. Therefore Statement 2 is not sufficient.

Statement 1 alone only eliminates Regions I and II (since whole numbers are nonnegative integers); negative numbers can be found in Regions III, IV, and V.

Statement 2 alone states that  is an integer that as not a whole number - that is,  is a negative integer. Since , as a consequence, . , the product of a positive integer and a negative integer, is a negative integer, and it would be placed in Region III, which comprises exactly the negative integers.