ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Venn Diagrams And Set Notation
17 questions · exam conditions
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Venn Diagrams And Set NotationQuestion 1 of 17

Let A and B be two non-empty finite sets. If n(A∪B)=n(A)+n(B)n(A \cup B) = n(A) + n(B), which of the following must be true?

Sets A and B are disjoint (A∩B=∅A \cap B = \emptyset).
Set A is a subset of set B.
Set A and set B are equal.
The universal set is the union of A and B.
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Venn Diagrams And Set Notation

Practice Venn Diagrams And Set Notation in ACCUPLACER Quantitative Reasoning, Algebra & Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Venn Diagrams And Set Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let A and B be two non-empty finite sets. If n(A∪B)=n(A)+n(B)n(A \cup B) = n(A) + n(B), which of the following must be true?

  1. Sets A and B are disjoint (A∩B=∅A \cap B = \emptyset). (correct answer)
  2. Set A is a subset of set B.
  3. Set A and set B are equal.
  4. The universal set is the union of A and B.
Explanation: The general formula for the union of two sets is n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B). If n(A∪B)=n(A)+n(B)n(A \cup B) = n(A) + n(B), it implies that n(A∩B)=0n(A \cap B) = 0. A set with 0 elements is the empty set (∅\emptyset). Sets with an empty intersection are called disjoint sets.

Question 2

In a group of 60 pet owners, 35 own a dog and 28 own a cat. If 12 owners have both a dog and a cat, how many owners have a dog but not a cat?

  1. 9
  2. 16
  3. 23 (correct answer)
  4. 51
Explanation: Let D be the set of dog owners and C be the set of cat owners. We are given n(D)=35n(D) = 35 and n(D∩C)=12n(D \cap C) = 12. The question asks for the number of owners who have a dog but not a cat, which corresponds to the elements in D but not in C. This can be calculated as n(D)−n(D∩C)n(D) - n(D \cap C). So, the number of owners with only a dog is 35−12=2335 - 12 = 23. The total number of pet owners (60) is extra information not needed for this specific question.

Question 3

A box contains 100 tiles. 40 are red (R), 30 are square (S), and 12 are both red and square. If a tile is chosen at random, what is the probability that it is red or square?

  1. 0.12
  2. 0.42
  3. 0.58 (correct answer)
  4. 0.70
Explanation: The probability of an event is the number of favorable outcomes divided by the total number of outcomes. We want the probability of a tile being red or square, P(R ∪\cup S). First, we find the number of tiles that are red or square using the formula n(R∪S)=n(R)+n(S)−n(R∩S)n(R \cup S) = n(R) + n(S) - n(R \cap S). This gives 40+30−12=5840 + 30 - 12 = 58. The total number of tiles is 100. Therefore, the probability is 58100=0.58\frac{58}{100} = 0.58.

Question 4

Let S be the set of all students at a college. Let M be the set of students taking a music class and A be the set of students on an athletic team. Which of the following describes the set M∩A′M \cap A'? Note: A' represents the complement of set A.

  1. Students taking a music class or who are not on an athletic team.
  2. Students taking a music class and who are also on an athletic team.
  3. Students taking neither a music class nor an athletic team.
  4. Students taking a music class but who are not on an athletic team. (correct answer)
Explanation: The notation M∩A′M \cap A' represents the intersection of set M and the complement of set A. The intersection symbol (∩\cap) corresponds to "and". The complement A' represents everything not in set A. Therefore, M∩A′M \cap A' is the set of students who are in the music class (M) AND are not on an athletic team (A'). This is best described as 'students taking a music class but who are not on an athletic team'.

Question 5

At a conference, every attendee is either an engineer or a scientist. There are 120 engineers and 90 scientists. If 35 attendees are both engineers and scientists, what is the total number of attendees at the conference?

  1. 55
  2. 85
  3. 175 (correct answer)
  4. 210
Explanation: The phrase 'every attendee is either an engineer or a scientist' means the total number of attendees is the number of people in the union of the two sets. Let E be the set of engineers and S be the set of scientists. We want to find n(E∪S)n(E \cup S). Using the formula n(E∪S)=n(E)+n(S)−n(E∩S)n(E \cup S) = n(E) + n(S) - n(E \cap S), we calculate the total as 120+90−35=210−35=175120 + 90 - 35 = 210 - 35 = 175.

Question 6

A high school has 300 seniors. Of these, 110 are in the band and 150 are in the athletics program. There are 40 seniors who are in both band and athletics. How many seniors are in neither band nor athletics?

  1. 0
  2. 80 (correct answer)
  3. 220
  4. 260
Explanation: First, find the number of seniors who are in at least one of the two activities (the union). Let B be band and A be athletics. n(B∪A)=n(B)+n(A)−n(B∩A)=110+150−40=220n(B \cup A) = n(B) + n(A) - n(B \cap A) = 110 + 150 - 40 = 220. This means 220 seniors are in band, athletics, or both. To find the number of seniors in neither activity, subtract this from the total number of seniors: 300−220=80300 - 220 = 80.

Question 7

Of the 200 cars at a dealership, 80 are sedans and 70 have a sunroof. There are 30 sedans that have a sunroof. If a car is selected at random from the sedans, what is the probability that it has a sunroof?

  1. 38\frac{3}{8} (correct answer)
  2. 720\frac{7}{20}
  3. 320\frac{3}{20}
  4. 37\frac{3}{7}
Explanation: The question asks for a conditional probability: the probability that a car has a sunroof given that it is a sedan. The sample space is restricted to the 80 sedans. Within this group of 80 sedans, 30 have a sunroof. Therefore, the probability is the number of sedans with a sunroof divided by the total number of sedans, which is 3080=38\frac{30}{80} = \frac{3}{8}.

Question 8

Given sets A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5}, what is A ∩ B'?

  1. A ∩ B' = {6, 8, 10} (correct answer)
  2. A ∩ B' = {2, 4}
  3. A ∩ B' = {1, 3, 5}
  4. A ∩ B' = {1, 2, 3, 4, 5, 6, 8, 10}
Explanation: B' (complement of B) relative to the universal set A ∪ B contains elements not in B. A ∩ B' means elements in A but not in B. Since A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5}, the elements in A that are not in B are {6, 8, 10}. Choice B gives A ∩ B instead. Choice C gives elements in B but not in A. Choice D incorrectly attempts A ∪ B.

Question 9

In a group of 80 people, 45 own cars (C), 35 own motorcycles (M), and 15 own both cars and motorcycles. If someone is randomly selected from this group, what is the probability they own a car or a motorcycle?

  1. The probability is 15/80 = 3/16
  2. The probability is 80/80 = 1
  3. The probability is 50/80 = 5/8
  4. The probability is 65/80 = 13/16 (correct answer)
Explanation: When you encounter probability questions involving two overlapping groups, you need to use the principle of inclusion-exclusion to avoid double-counting people who belong to both categories. To find the probability that someone owns a car OR a motorcycle, you can't simply add the car owners (45) and motorcycle owners (35) because this counts the 15 people who own both vehicles twice. Instead, use the formula: |C ∪ M| = |C| + |M| - |C ∩ M|. So the number of people who own at least one vehicle is: 45 + 35 - 15 = 65 people. Therefore, the probability is 6580=1316\frac{65}{80} = \frac{13}{16}, making answer D correct. Answer A (15/80) represents only those who own both vehicles, not those who own at least one. This misinterprets "or" as "and." Answer B (80/80 = 1) would mean everyone owns a vehicle, but we can calculate that 15 people own neither cars nor motorcycles (80 - 65 = 15). Answer C (50/80) incorrectly subtracts the overlap from just one category (45 - 15 + 35 = 65, not 50), showing a misunderstanding of the inclusion-exclusion principle. Remember: when dealing with "or" probability questions involving overlapping sets, always subtract the intersection to avoid double-counting. Draw a Venn diagram if it helps visualize the overlap—this is a common pattern on standardized tests.

Question 10

In a survey about streaming services, 40% subscribe to Netflix (N), 30% subscribe to Hulu (H), and 15% subscribe to both. If a person is selected at random, what is the probability they subscribe to Netflix given that they subscribe to Hulu?

  1. P(N|H) = 0.15
  2. P(N|H) = 0.5 (correct answer)
  3. P(N|H) = 0.375
  4. P(N|H) = 0.55
Explanation: When you see conditional probability questions, you're looking for the probability of one event given that another event has already occurred. The key formula is P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}, which reads as "the probability of A given B equals the probability of both A and B divided by the probability of B." Here, you need P(N|H) - the probability someone subscribes to Netflix given they subscribe to Hulu. You know that 15% subscribe to both services and 30% subscribe to Hulu. Using the formula: P(N∣H)=P(N∩H)P(H)=0.150.30=0.5P(N|H) = \frac{P(N \cap H)}{P(H)} = \frac{0.15}{0.30} = 0.5 This makes intuitive sense: among the 30% who have Hulu, exactly half of them (15% out of 30%) also have Netflix. Choice A gives 0.15, which is simply P(N ∩ H) - the probability of having both services. This ignores the conditioning on Hulu subscribers. Choice C shows 0.375, which you'd get by incorrectly dividing 0.15 by 0.40 (using Netflix subscribers as the denominator instead of Hulu subscribers). Choice D gives 0.55, which might result from adding probabilities incorrectly or confusing this with P(N ∪ H). The correct answer is B: P(N|H) = 0.5. Remember: conditional probability questions always restrict your sample space to those who satisfy the "given" condition. Don't use the total population - only consider the subset that meets the condition after "given that."

Question 11

Let U be the set of integers from 1 to 20, inclusive. Let A be the set of multiples of 3 in U, and let B be the set of multiples of 4 in U. How many elements are in the set A∪BA \cup B?

  1. 9
  2. 10 (correct answer)
  3. 11
  4. 12
Explanation: First, identify the elements in each set. Set A (multiples of 3) is {3, 6, 9, 12, 15, 18}, so it has 6 elements. Set B (multiples of 4) is {4, 8, 12, 16, 20}, so it has 5 elements. The intersection of A and B, A∩BA \cap B, contains elements in both sets, which is {12}, so it has 1 element. To find the number of elements in the union, A∪BA \cup B, we use the formula: n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B). Substituting the values, we get 6+5−1=106 + 5 - 1 = 10.

Question 12

A survey of 85 students found that 50 students take a math class, 45 students take a science class, and 15 students take neither. How many students take both a math class and a science class?

  1. 10
  2. 15
  3. 25 (correct answer)
  4. 40
Explanation: Let M be the set of students taking math and S be the set taking science. The total number of students is 85. Since 15 students take neither, the number of students taking at least one of the subjects is 85−15=7085 - 15 = 70. This is the size of the union, n(M∪S)=70n(M \cup S) = 70. Using the Principle of Inclusion-Exclusion, n(M∪S)=n(M)+n(S)−n(M∩S)n(M \cup S) = n(M) + n(S) - n(M \cap S). We have 70=50+45−n(M∩S)70 = 50 + 45 - n(M \cap S), which simplifies to 70=95−n(M∩S)70 = 95 - n(M \cap S). Solving for the intersection, n(M∩S)=95−70=25n(M \cap S) = 95 - 70 = 25.

Question 13

At a community picnic with 120 people, 70 people ate a hamburger and 60 people ate a hot dog. If 25 people ate neither a hamburger nor a hot dog, how many people ate only a hamburger?

  1. 25
  2. 35 (correct answer)
  3. 45
  4. 60
Explanation: This is a multi-step problem. First, determine the number of people who ate at least one item. Total people = 120, Neither = 25. So, n(H∪D)=120−25=95n(H \cup D) = 120 - 25 = 95. Next, find the number who ate both using the inclusion-exclusion principle: 95=n(H)+n(D)−n(H∩D)95 = n(H) + n(D) - n(H \cap D) which is 95=70+60−n(H∩D)95 = 70 + 60 - n(H \cap D). This gives 95=130−n(H∩D)95 = 130 - n(H \cap D), so n(H∩D)=35n(H \cap D) = 35. Finally, the number who ate only a hamburger is the total who ate a hamburger minus those who ate both: n(H)−n(H∩D)=70−35=35n(H) - n(H \cap D) = 70 - 35 = 35.

Question 14

If P(A) = 0.6, P(B) = 0.4, and P(A ∩ B) = 0.2, what is P(A' ∩ B')?

  1. P(A' ∩ B') = 0.6
  2. P(A' ∩ B') = 0.8
  3. P(A' ∩ B') = 0.2 (correct answer)
  4. P(A' ∩ B') = 0.4
Explanation: When you encounter probability questions involving complements and intersections, think about how events relate to each other and use De Morgan's laws to find efficient solution paths. To find P(A′∩B′)P(A' \cap B'), you can use De Morgan's law, which states that the complement of a union equals the intersection of complements: P(A′∩B′)=P((A∪B)′)P(A' \cap B') = P((A \cup B)'). This means you need to find the probability that neither A nor B occurs, which is 1 minus the probability that at least one occurs. First, calculate P(A∪B)P(A \cup B) using the addition rule: P(A∪B)=P(A)+P(B)−P(A∩B)=0.6+0.4−0.2=0.8P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 - 0.2 = 0.8 Therefore: P(A′∩B′)=1−P(A∪B)=1−0.8=0.2P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.8 = 0.2 Looking at the wrong answers: Choice A (0.6) incorrectly uses just P(A)P(A), missing the relationship between events. Choice B (0.8) mistakenly gives you P(A∪B)P(A \cup B) instead of its complement - this is the probability that at least one event occurs, not that neither occurs. Choice D (0.4) incorrectly uses just P(B)P(B), ignoring the intersection relationship entirely. Remember this pattern: when you see questions asking for P(A′∩B′)P(A' \cap B'), immediately think "neither A nor B occurs" and use the complement of the union. The formula P(A′∩B′)=1−P(A∪B)P(A' \cap B') = 1 - P(A \cup B) will save you time compared to calculating complements separately.

Question 15

Three sets are defined as follows: A = {x | x is even and 1 ≤ x ≤ 10}, B = {x | x is prime and 1 ≤ x ≤ 10}, and C = {x | x is a multiple of 3 and 1 ≤ x ≤ 10}. What is |A ∩ (B ∪ C)|?

  1. |A ∩ (B ∪ C)| = 3
  2. |A ∩ (B ∪ C)| = 2 (correct answer)
  3. |A ∩ (B ∪ C)| = 4
  4. |A ∩ (B ∪ C)| = 1
Explanation: When you encounter set theory problems involving intersections and unions, start by identifying each set clearly, then work systematically through the operations. First, let's list each set:
  • Set A (even numbers from 1 to 10): A = {2, 4, 6, 8, 10}
  • Set B (prime numbers from 1 to 10): B = {2, 3, 5, 7}
  • Set C (multiples of 3 from 1 to 10): C = {3, 6, 9}
Next, find B ∪ C (the union of B and C, containing all elements in either set): B ∪ C = {2, 3, 5, 6, 7, 9} Now find A ∩ (B ∪ C) (elements that are in both A and the union): Looking at A = {2, 4, 6, 8, 10} and B ∪ C = {2, 3, 5, 6, 7, 9}, the common elements are 2 and 6. So A ∩ (B ∪ C) = {2, 6}, which has 2 elements. Answer B is correct: |A ∩ (B ∪ C)| = 2. Answer A (3 elements) likely comes from miscounting or including an extra element like 3, which isn't even. Answer C (4 elements) might result from confusing intersection with union or making computational errors. Answer D (1 element) could come from only finding one of the two correct elements, possibly missing that 6 is both even and a multiple of 3. Always write out your sets explicitly when working with set operations. This prevents errors and makes it easier to spot elements that satisfy multiple conditions, like 2 being both even and prime.

Question 16

In a survey of 120 students, 75 students like pizza (P), 60 students like burgers (B), and 25 students like neither pizza nor burgers. How many students like both pizza and burgers?

  1. 40 students like both pizza and burgers (correct answer)
  2. 35 students like both pizza and burgers
  3. 20 students like both pizza and burgers
  4. 15 students like both pizza and burgers
Explanation: Using the principle of inclusion-exclusion: |P ∪ B| = |P| + |B| - |P ∩ B|. Since 25 students like neither, 120 - 25 = 95 students like at least one. So 95 = 75 + 60 - |P ∩ B|, which gives |P ∩ B| = 40. Choice B incorrectly subtracts the 'neither' group from the calculation. Choice C assumes the intersection is the difference between total pizza lovers and the union. Choice D incorrectly uses 25 as part of the intersection calculation.

Question 17

Consider the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with subsets A = {2, 4, 6, 8} and B = {3, 6, 9}. What is (A ∪ B)'?

  1. (A ∪ B)' = {6}
  2. (A ∪ B)' = {2, 3, 4, 6, 8, 9}
  3. (A ∪ B)' = {1, 5, 7, 10} (correct answer)
  4. (A ∪ B)' = {1, 2, 3, 4, 5, 7, 8, 9, 10}
Explanation: When you encounter set operations with complements, you need to work step-by-step: first find the union, then find its complement within the universal set. Let's find A∪BA \cup B first. The union contains all elements that appear in either set A or set B (or both). From A={2,4,6,8}A = \{2, 4, 6, 8\} and B={3,6,9}B = \{3, 6, 9\}, we get A∪B={2,3,4,6,8,9}A \cup B = \{2, 3, 4, 6, 8, 9\}. Notice that 6 appears in both sets but is only listed once in the union. The complement (A∪B)′(A \cup B)' contains all elements in the universal set U that are NOT in A∪BA \cup B. Since U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} and A∪B={2,3,4,6,8,9}A \cup B = \{2, 3, 4, 6, 8, 9\}, the complement is {1,5,7,10}\{1, 5, 7, 10\}. Choice A gives {6}, which is actually an element that's IN the union, not outside it. Choice B lists the union itself, not its complement—this represents the common error of forgetting to take the complement. Choice D includes almost all elements from U, missing only 6, which suggests confusion about what complement means. The correct answer is C: (A∪B)′={1,5,7,10}(A \cup B)' = \{1, 5, 7, 10\}. Study tip: For complement problems, always work in two clear steps: find the original set operation first, then identify which elements from U are missing from that result. Double-check by ensuring your complement and original set together make up the entire universal set with no overlap.