135
Solving problems with sets
Textbook: Part 4, lesson 135, pp. 78–80
GoalSolving problems about sets with Euler–Venn diagrams; “only” and “at least one”.
New words
only · faqatat least one · kamida bittasiboth · ikkalasi hamdifference · farq
Explanation
If a problem has two groups, we draw a diagram and write the number of children who belong to both in the overlap. The number belonging “only to A” = n(A) − n(A ∩ B). If everyone takes part in at least one, total = n(A) + n(B) − n(A ∩ B). Those taking part in neither = whole class − those taking part. The elements of A that are not in B also form a separate set.
Worked examples
A = {11, 22, 33, 44, 55}, B = {22, 44, 66}. A ∩ B = {22, 44}; A ∪ B = {11, 22, 33, 44, 55, 66}; the elements of A that are not in B: {11, 33, 55}.
17 children go to football, 12 to swimming and 7 to both. Only football: 17 − 7 = 10, only swimming: 12 − 7 = 5. At least one: 10 + 5 + 7 = 22 children.
Class activity
“Class survey”: the class is asked two questions (for example “Do you drink tea?”, “Do you drink milk?”). Children raise hands, a diagram is drawn and we count those who chose only one, both or neither.
Practice
1
18 children go to football, 15 to swimming and 9 to both. How many children go to at least one club?
24
2
In the same problem, how many children go only to football?
9
3
A class has 30 pupils. 20 learn English, 14 learn Russian and 6 learn both. How many pupils learn neither?
2
4
A class has 20 pupils and each goes to the chess or drawing club. 12 go to chess and 10 to drawing. How many pupils go to both?
2
5
A = {2, 4, 6, 8, 10}, B = {4, 8, 12}. Write the set of elements of A that are not in B.
{2, 6, 10}