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.

Region IV comprises the rational numbers that are not integers. A number is rational if and only if it can be expressed as the quotient of integers. 

From Statement 1 alone, it can be inferred that  is rational, and that it is not an integer. Since , it follows that . However, this is not sufficient to narrow it down completely.

For example:

If , then , a natural number, putting it in Region I.

If , then , a rational number but not an integer, putting it in Region IV.

From Statement 2 alone, it can be inferred that  is rational, and that it is not an integer. From , it follows that . The nonzero rational numbers are closed under division, so  must be a rational number. However, since  is not an integer,  cannot be an integer, since the integers are closed under multiplication. Therefore, Statement 2 alone proves that  belongs in Region IV.

Region I comprises the natural numbers - 

From Statement 1 alone,  is a natural number; since , it follows that  is the difference of a natural number and 7 - that is, 

 could be in any of three regions - I, II, or III.

Conversely, from Statement 2 alone,   is the sum of a natural number and 7 - that is,

 must be a natural number and it must be in Region I.

Assume Statement 1 alone. It cannot be determined what region  is in. 

For example, suppose , which is in Region I (the set of natural numbers, or positive integers). It is possible that , putting it in Region I, or , putting it in Region III (the set of integers that are not whole numbers - that is, the set of negative integers).

Assume Statement 2 alone. It cannot be determined what region  is in. 

For example, suppose , which is in Region III; then , which is also in Region III. But suppose ; then , which, as an irrational number, is in Region V.

Now assume both statements. Then  has an integer as a square and an integer as a cube.  must either be an integer or an irrational number. But 

, making it the quotient of integers, which is rational. Therefore,  is an integer. Furthermore, its cube is negative, so  is negative. The two statements together prove that  is a negative integer, which belongs in Region III.

A set with  elements has exactly  subsets in all, and  proper subsets (every subset except one - the set itself). 

From Statement 1, since  has  subsets, it follows that it has 6 elements. From Statement 2, since  has 63 proper subsets, it has 64 subsets total, and, again, 6 elements. Either statement alone is sufficient.

Statement 1 alone gives insufficient information. For example, if , then:

 or 

Since , it is unclear which of  and  is greater, if either.

Statement 2 gives insufficient information; if  is positive,  is negative, and vice versa.

Assume both to be true. The two statements form a system of equations that can be solved using substitution:

Case 1: 

Case 2: 

This equation has no solution.

Therefore, the only possible solution is . Therefore, it can be concluded that .