Sets - GMAT Quantitative

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 .

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.

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

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.

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.

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 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.

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.

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 .

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

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.

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.

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.

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 .

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.

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

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.