Sets and Venn diagrams Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Sets and Venn diagrams questions. See exactly how to solve problems on Venn diagram, set notation, two-set Venn diagram, intersection and union.

Venn diagramset notationtwo-set Venn diagramintersection and unionprobability from a Venn diagramlisting the elements of a set
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
There are 3030 students in a class. AA is the set of students who study French. BB is the set of students who study German. 1818 of the students study French. 1414 of the students study German. 77 of the students study both French and German. Work out how many of the students study French but not German.

Worked solution

  1. Write down what you know

    n(ξ)=30,n(A)=18,n(B)=14,n(A∩B)=7n(\xi) = 30, \quad n(A) = 18 , \quad n(B) = 14 , \quad n(A \cap B) = 7

    There are 3030 students altogether, and the question gives n(A)=18n(A) = 18, n(B)=14n(B) = 14 and n(A∩B)=7n(A \cap B) = 7.

  2. Work out n(A∩B′)n(A \cap B')

    n(A∩B′)=n(A)−n(A∩B)=18−7=11n(A \cap B') = n(A) - n(A \cap B) = 18 - 7 = 11

    Every one of the students who study both French and German is also one of the students who study French, so taking them away leaves exactly the students who study French but not German: 18−7=1118 - 7 = 11.

  3. State the answer

    n(A∩B′)=11n(A \cap B') = 11

    So there are 1111 students who study French but not German.

Answer
1111
Question 2
1 markeasy
There are 3030 students in Year 11. AA is the set of students who play football. BB is the set of students who play tennis. 1616 of the students play football. 1414 of the students play tennis. 66 of the students play both football and tennis. Work out how many of the students play tennis but not football.

Worked solution

  1. Write down what you know

    n(ξ)=30,n(A)=16,n(B)=14,n(A∩B)=6n(\xi) = 30, \quad n(A) = 16 , \quad n(B) = 14 , \quad n(A \cap B) = 6

    There are 3030 students altogether, and the question gives n(A)=16n(A) = 16, n(B)=14n(B) = 14 and n(A∩B)=6n(A \cap B) = 6.

  2. Work out n(A′∩B)n(A' \cap B)

    n(A′∩B)=n(B)−n(A∩B)=14−6=8n(A' \cap B) = n(B) - n(A \cap B) = 14 - 6 = 8

    Every one of the students who play both football and tennis is also one of the students who play tennis, so taking them away leaves exactly the students who play tennis but not football: 14−6=814 - 6 = 8.

  3. State the answer

    n(A′∩B)=8n(A' \cap B) = 8

    So there are 88 students who play tennis but not football.

Answer
88
Question 3
2 markseasy
There are 3030 people in a survey. AA is the set of people who own a dog. BB is the set of people who own a cat. 1717 of the people own a dog. 1414 of the people own a cat. 55 of the people own both a dog and a cat. Work out how many of the people own a dog but not a cat.

Worked solution

  1. Write down what you know

    n(ξ)=30,n(A)=17,n(B)=14,n(A∩B)=5n(\xi) = 30, \quad n(A) = 17 , \quad n(B) = 14 , \quad n(A \cap B) = 5

    There are 3030 people altogether, and the question gives n(A)=17n(A) = 17, n(B)=14n(B) = 14 and n(A∩B)=5n(A \cap B) = 5.

  2. Work out n(A∩B′)n(A \cap B')

    n(A∩B′)=n(A)−n(A∩B)=17−5=12n(A \cap B') = n(A) - n(A \cap B) = 17 - 5 = 12

    Every one of the people who own both a dog and a cat is also one of the people who own a dog, so taking them away leaves exactly the people who own a dog but not a cat: 17−5=1217 - 5 = 12.

  3. State the answer

    n(A∩B′)=12n(A \cap B') = 12

    So there are 1212 people who own a dog but not a cat.

Answer
1212
Question 4
1 markeasy
There are 3030 students in a music club. AA is the set of students who play the piano. BB is the set of students who play the guitar. n(A)=17n(A) = 17. n(B)=14n(B) = 14. n(A∩B)=8n(A \cap B) = 8. Work out n(A∩B′)n(A \cap B').

Worked solution

  1. Write down what you know

    n(ξ)=30,n(A)=17,n(B)=14,n(A∩B)=8n(\xi) = 30, \quad n(A) = 17 , \quad n(B) = 14 , \quad n(A \cap B) = 8

    There are 3030 students altogether, and the question gives n(A)=17n(A) = 17, n(B)=14n(B) = 14 and n(A∩B)=8n(A \cap B) = 8.

  2. Work out n(A∩B′)n(A \cap B')

    n(A∩B′)=n(A)−n(A∩B)=17−8=9n(A \cap B') = n(A) - n(A \cap B) = 17 - 8 = 9

    Every one of the students who play both the piano and the guitar is also one of the students who play the piano, so taking them away leaves exactly the students who play the piano but not the guitar: 17−8=917 - 8 = 9.

  3. State the answer

    n(A∩B′)=9n(A \cap B') = 9

    So there are 99 students who play the piano but not the guitar.

Answer
99
Question 5
2 markseasy
There are 2020 children at a party. AA is the set of children who like ice cream. BB is the set of children who like jelly. n(A)=12n(A) = 12. n(B)=9n(B) = 9. n(A∩B)=4n(A \cap B) = 4. Work out n(A′∩B)n(A' \cap B).

Worked solution

  1. Write down what you know

    n(ξ)=20,n(A)=12,n(B)=9,n(A∩B)=4n(\xi) = 20, \quad n(A) = 12 , \quad n(B) = 9 , \quad n(A \cap B) = 4

    There are 2020 children altogether, and the question gives n(A)=12n(A) = 12, n(B)=9n(B) = 9 and n(A∩B)=4n(A \cap B) = 4.

  2. Work out n(A′∩B)n(A' \cap B)

    n(A′∩B)=n(B)−n(A∩B)=9−4=5n(A' \cap B) = n(B) - n(A \cap B) = 9 - 4 = 5

    Every one of the children who like both ice cream and jelly is also one of the children who like jelly, so taking them away leaves exactly the children who like jelly but not ice cream: 9−4=59 - 4 = 5.

  3. State the answer

    n(A′∩B)=5n(A' \cap B) = 5

    So there are 55 children who like jelly but not ice cream.

Answer
55

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