Sets - GMAT Quantitative

Let and be the set of all Toastmasters and Elks, respectively, and let be the set of all people. and , so the set to which John belongs is the shaded set in this Venn diagram:

the logical opposite of this is that John belongs to the shaded set in the diagram:

A way of saying this is or , or, equivalently, if , then .

In plain English, if John is not an Elk, then John is not a Toastmaster.

We need to flip a coin until we get two heads in a row. The smallest number of possible flips is 2, which would occur if our first two flips are both heads. This eliminates three of our answer choices, because we know the sample space must start at 2.

This leaves us with {x : x = 2, 3, 4 . . .} and {x : x = 2, 3, 4, 5, 6}. Let's think about {x : x = 2, 3, 4, 5, 6}. What if I flip a coin 6 times and get 6 tails? Then I have to keep flipping beyond 6 flips until I get two heads in a row; therefore the answer must be {x : x = 2, 3, 4 . . .}, because we don't have an upper limit on the number of flips it will take to produce two successive heads.

Let and be the sets of employees that speak Spanish and French, respectively. Then we know that

and

Since 517 employees speak neither, employees speak either Spanish, French, or both.

We are looking for , which can be found by using the formula:

, making 1,728 a multiple of 3, and thus, an element of .

1,728 is not a perfect square; . Thus, 1,728 is not an element of .

1,728 is a perfect cube: . Thus, 1,728 is an element of .

, which is represented by the region inside circles and and outside . This is region .

is the set of elements that fall either in or the complement of , or both - that is, either in , outside of , or both. This union is intersected with the complement of , meaning that only the elements of the union that also fall outside of are considered.

"Color" in all of and everything outside of - but then, uncolor everything inside . That makes the correct choice:

In order to find the median, the set needs to be written in numerical order:

Since and are both the middle numbers, taking their average will give the median of the set.

525 is a multiple of all three of the integers 3, 5, and 7:

Therefore, 525 is an element of each of sets , and, subsequently, falls into region , which represents .

Let be the number of students who don't take any of the three classes.

Prime numbers are numbers with no other factors than themselves and one. Two is the first prime number and the only even prime number. Other examples are 5, 7, 11, etc.

Even numbers are numbers divisible by 2. Set C includes all numbers ending in 0, 2, 4, 6, or 8.

Thus, there is one number common to both sets: 2.

The sets and do not intersect, so no senior is enrolled in both French IV and physics; the sets and do not intersect, so no senior is enrolled in both French IV and calculus.

, so every senior enrolled in calculus is also enrolled in physics; contrapositively, every senior not enrolled in physics is also not enrolled in calculus.

The correct choice is the remaining statement - every senior enrolled in physics is also enrolled in calculus - since is not a subset of .

Let be the sets of Moogs, Noogs, Poogs, and Roogs, respectively, From the premise of the problem, we have that

.

It follows by transitivity that and .

Assume Statement 1 alone. If Stumpy is a Moog, then

.

, makng Stumpy a Poog. But since Stumpy is NOT a Poog, then it follows, contrapositively, that Stumpy is not a Moog.

Assume Statement 2 alone. If Stmpy is a Moog, then

,

making Stumpy a Roog. Since Stumpy is NOT a Roog, then it follows, contrapositively, that Stumpy is not a Moog.

In set theory, "Wendy likes neither catfish nor flounder" can be stated as

.

The opposite of this is that is in the complement of this set:

.

By DeMorgan's law, this is equivalent to saying

, or

.

If , then falls in the intersection of two sets. One is , the union of , the set of people who like The Rolling Stones, and the complement of , the set of people not in - the set of all people who do not like The Who. The other set is the complement of , the set of all people not in - the set of people who don't like Lynyrd Skynyrd.

Therefore, if , both of the following must happen: Nancy must like The Rolling Stones or not like The Who, or both, and Nancy must not like Lynyrd Skynyrd.

From Statement 1 alone, Nancy does not like The Who, but it is not clear who else she likes. She is in , but it is not clear whether she is in the set of people who don't like Lynyrd Skynyrd, so the question of whether she falls in is open.

From Statement alone, since Nancy likes Lynyrd Skynyrd, she cannot be in . This makes false.

The opposite of the statement is the statement

,

which, by DeMorgan's Law, is equivalent to

or

This means that Kim is in the intersection of and , or, equivalently, she is in both , the set of people who can swim, and , the complement of the set of people who can ride a bicycle. Therefore, Kim can swim, but she cannot ride a bicycle.

The opposite of the statement is the statement

,

which, by DeMorgan's Law, is equivalent to

, or

Victor belongs to either or both of , the set of all people who like cola, and , the set of all people who like milk. Therefore, Victor likes cola, milk, or both.

To fall in the set , a restaurant would have to be in the union of three sets - that is, any one or more. The three sets are:

, the set of all restaurants in Bancroft that accept the America's Best credit card;

, the set of all restaurants in Bancroft that accept the Bank Zero credit card; and

, the complement of the set of all restaurants in Bancroft that accept the Chargitall credit card, or, equivalently, set of all restaurants in Bancroft that do not accept the Chargitall credit card.

The complement of this set, , is the set of all restaurants that belong to none of , so the correct choice must be a restaurant that belongs to and . The restaurant must accept Chargitall, and must not accept the other two cards.

The only one of the five restaurants fitting this description is Benny's.

We can best answer this question using a Venn diagram. We want to examine the union of two sets:

, the intersection of and , which is the region the two circles have in common, and

, the complement , which is the region outside the circle representing this set.

Since we are examing the union, we want to shade in everything in one or both of these two sets. This is shown below:

If Tommy likes Frank Sinatra, The Beatles, and Elvis Presley, falls in all three circles, placing him in Region (i).

If Tommy does not like Frank Sinatra, The Beatles, or Elvis Presley, falls in none of the three circles, placing him in Region (ii).

If Tommy likes Frank Sinatra, but neither The Beatles nor Elvis Presley, falls in circle A, but neither B nor C, placing him in Region (iii).

If Tommy likes Frank Sinatra and The Beatles, but not Elvis Presley, falls in circles A and B, but not C, placing him in Region (iv).

If Tommy likes Frank Sinatra and Elvis Presley, but not The Beatles, falls in circles A and C, but not B, placing him in Region (v).

In all of the first four cases, is a true statement, since the region is part of the grayed area. In the fifth case, is false. Therefore, the only impossible statement among the choices is "Tommy likes Frank Sinatra and Elvis Presley, but not The Beatles."

, which is the intersection of and . It follows that and .

, so it follows that Julianna likes Band A. The three choices that state that she does not can be eliminated.

. This is the union of , the complement of - that is, the set of people not in - and . It follows that either Julianna does not like Band B, does like Band C, or both. Therefore, it is not true that she likes Band B and does not like Band C. This can be eliminated.

The only possible choice is that Julianna likes Band A and Band C, but not Band B.

In set notation, "Jeremy is not a dancer, but he is a singer" can be stated as

That is, Jeremy falls in the intersection of the complement of the set of dancers and the set of singers .

The opposite of this is that is in the complement of this set:

By DeMorgan's law, this is equivalent to saying

, or

.