Problems on Sets

Lessons

Senior 3 · Mathematics

Problems on Sets

4 lessons

Lesson 4 of 4

Choosing a method for set problems

25 min

Before you trust a survey

A report on 100 learners says that 60 speak English, 50 French and 40 Kiswahili; 20 speak English and French, 15 English and Kiswahili, 10 French and Kiswahili; 5 speak all three languages and 20 speak none. It looks reasonable. It is impossible, and a Venn diagram shows why in two minutes. This lesson brings together everything in the unit: choosing regions from the words of a question, using De Morgan's laws to count, drawing subsets inside a three-set diagram, reading data from a table, and testing whether data can be true.

From the words to the regions

Use the region numbers of the three-set diagram (1, 2, 3 in one set only; 4, 5, 6 in two; 7 in all three; 8 in none).

Words in the question	Set	Regions
$A$ only	$A \cap B' \cap C'$	1
$A$ and $B$	$A \cap B$	4, 7
$A$ and $B$ but not $C$	$A \cap B \cap C'$	4
$A$ or $B$	$A \cup B$	1, 2, 4, 5, 6, 7
not $A$	$A^{'}$	2, 3, 6, 8
at least one	$A \cup B \cup C$	1 to 7
exactly one	—	1, 2, 3
at least two	$(A \cap B) \cup (A \cap C) \cup (B \cap C)$	4, 5, 6, 7
none	$(A \cup B \cup C)' = A' \cap B' \cap C'$	8

Definition

Two-way table: a table that sorts the members of one group by two yes-or-no questions at once, with one question for the rows and one for the columns.

Notation: the four inner cells of the table are the four regions of a two-set Venn diagram.

Example: the cell "electricity: yes, phone: no" is region 1, $E \cap P'$ , of the Venn diagram for $E$ and $P$ .

Worked example 1: counting with De Morgan's law

$n (U) = 50$ , $n (A^{'}) = 22$ , $n (B^{'}) = 26$ and $n(A' \cap B') = 8$ . Find $n(A \cap B)$ .

Solution $n (A) = 50 - 22 = 28$ and $n (B) = 50 - 26 = 24$ . By De Morgan's law, $A' \cap B' = (A \cup B)'$ , so $n(A \cup B) = 50 - 8 = 42$ . Then

n(A \cap B) = 28 + 24 - 42 = 10.

Check: $18 + 10 + 14 + 8 = 50$ . ✓

Worked example 2: testing a report

Show that the language report at the start of this lesson cannot be correct.

Solution Method 1: the rule. The number who speak at least one of the languages would be

60 + 50 + 40 - 20 - 15 - 10 + 5 = 110.

But only $100 - 20 = 80$ learners speak at least one of them. The figures contradict each other.

Method 2: the regions. From the centre outwards: pairs only $15, 10, 5$ ; English only $60 - 15 - 10 - 5 = 30$ , French only $50 - 15 - 5 - 5 = 25$ , Kiswahili only $40 - 10 - 5 - 5 = 20$ . The eight regions add up to $30 + 25 + 20 + 15 + 10 + 5 + 5 + 20 = 130$ , not 100. At least one of the figures in the report is wrong.

Worked example 3: a subset inside a three-set problem

In an S3 class of 45 learners, 25 sing in the choir ( $R$ ), 20 are in the drama club ( $D$ ) and 12 are in the traditional dance troupe ( $T$ ). Every member of the dance troupe also sings in the choir. 8 learners are in both the choir and the drama club, and 3 of these are also in the dance troupe. How many learners (a) are in none of the three groups (b) sing in the choir only?

Solution Because $T \subseteq R$ , draw circle $T$ entirely inside circle $R$ . It still overlaps $D$ , since 3 dancers are in drama.

$T \cap D = 3$ ; these learners are also in $R$ .
Choir and drama but not dance: $8 - 3 = 5$ .
Dance but not drama: $12 - 3 = 9$ .
Choir only: $25 - 5 - 3 - 9 = 8$ . Drama only: $20 - 5 - 3 = 12$ .

$T$ sits inside $R$ , so there is no region for $T$ outside $R$ .

(a) In at least one group: $8 + 5 + 9 + 3 + 12 = 37$ , so $45 - 37 = 8$ are in none. (b) 8 sing in the choir only.

Check: $n (R) = 8 + 5 + 9 + 3 = 25$ ✓, $n (D) = 5 + 3 + 12 = 20$ ✓, $n (T) = 9 + 3 = 12$ ✓.

Worked example 4: from a two-way table to a Venn diagram

60 learners were asked whether their home has electricity ( $E$ ) and whether anyone at home has a smartphone ( $P$ ). The illustrative results are in the table.

	smartphone: yes	smartphone: no	total
electricity: yes	28	7	35
electricity: no	15	10	25
total	43	17	60

Draw the Venn diagram and find $n(E \cup P)$ , $n (P^{'})$ and $n(E' \cap P')$ .

Solution The four inner cells are the four regions: $E$ only 7, both 28, $P$ only 15, neither 10.

The table and the diagram hold the same four numbers.

$n(E \cup P) = 7 + 28 + 15 = 50$ ; $n (P^{'}) = 7 + 10 = 17$ (the "smartphone: no" column); $n(E' \cap P') = 10$ .

Warning

A negative region means a mistake. If a region works out below zero, either the data are impossible or you subtracted the same learners twice (often by forgetting that " $A$ and $B$ " already includes the centre). Go back and check before you answer.

Tip

Three questions before you write anything. Which regions does the question ask for? Which regions do I already know? Where does the unknown go? Answer these, and the rest of the problem is arithmetic.

The Organic Chemistry Tutor: Venn Diagram Word Problems With 3 Categories (19 min). Watch how each problem is filled from the centre outwards and how the questions about "only", "exactly two" and "none" are answered from the finished diagram.

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Practice

The last three questions are harder.

$n (U) = 40$ , $n (A^{'}) = 15$ , $n (B^{'}) = 22$ and $n(A' \cap B') = 6$ . Find $n(A \cap B)$ .
$n (U) = 30$ , $n(A \cup B) = 26$ and $n(A' \cup B') = 21$ . Find $n(A \cap B)$ and $n(A' \cap B')$ .
A survey says: "Of 40 learners, 25 like Mathematics, 20 like Physics, 12 like both and 5 like neither." Are these data consistent?
In a class of 40, 22 learners play football and 9 play handball. Every handball player also plays football. How many play neither game, and how many play football but not handball?
50 traders were asked whether they sell beans ( $B$ ) and maize ( $M$ ): 12 sell both, 18 sell beans but not maize, 9 sell maize but not beans, and 11 sell neither. Draw the two-way table and the Venn diagram. Find $n(B \cup M)$ , $n (B)$ and $n (M^{'})$ .
In a group of 60 learners, 30 are in the science club ( $S$ ), 25 in the mathematics club ( $M$ ) and 10 in the robotics club ( $R$ ). Every robotics member is in the science club. 8 learners are in robotics and mathematics, and 12 in science and mathematics. How many are in none of the three clubs?
Someone claims that $n (A) = 20$ , $n(A \cap B) = 12$ , $n(A \cap C) = 10$ and $n(A \cap B \cap C) = 1$ . Show that these figures cannot all be right. What is the least possible value of $n(A \cap B \cap C)$ for the other three figures to be possible?

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$n (A) = 25$ , $n (B) = 18$ , $n(A \cup B) = 40 - 6 = 34$ , so $n(A \cap B) = 25 + 18 - 34 = 9$ .
$A' \cup B' = (A \cap B)'$ , so $n(A \cap B) = 30 - 21 = 9$ . $n(A' \cap B') = 30 - 26 = 4$ .
No. $n(M \cup P) = 25 + 20 - 12 = 33$ , and $33 + 5 = 38$ , not 40: two learners are missing.
Neither: $40 - 22 = 18$ . Football but not handball: $22 - 9 = 13$ .
$n(B \cup M) = 12 + 18 + 9 = 39$ ; $n (B) = 12 + 18 = 30$ ; $n (M^{'}) = 18 + 11 = 29$ .
$R \cap M = 8$ (all in $S$ ); $S$ and $M$ but not $R$ : $12 - 8 = 4$ ; $R$ but not $M$ : $10 - 8 = 2$ ; $S$ only: $30 - 4 - 8 - 2 = 16$ ; $M$ only: $25 - 4 - 8 = 13$ . In at least one club: $16 + 4 + 8 + 2 + 13 = 43$ , so 17 are in none.
$A$ only $= 20 - (12 - 1) - (10 - 1) - 1 = - 1$ , which is impossible. With centre $x$ : $A$ only $= 20 - (12 - x) - (10 - x) - x = x - 2 \ge 0$ , so $x \ge 2$ .

1Before you trust a survey

2From the words to the regions

Before you trust a survey

From the words to the regions