Solving problems with Venn diagrams – Sets (Senior 1 Mathematics) | Ganzaa

Sets

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Senior 1 · Mathematics

Sets

5 lessons

Lesson 4 of 5

Solving problems with Venn diagrams

25 min

When adding counts twice

A sports teacher at a school in Kicukiro, Kigali, reads two lists: 27 learners of S1 play football and 19 play basketball. Does she have $27 + 19 = 46$ players? Not necessarily. A learner who plays both games is on both lists, so adding the lists counts that learner twice. A Venn diagram shows exactly how many learners sit in each region, and one formula corrects the double count.

Important

Rule (two sets): for any two finite sets $A$ and $B$ ,

n(A \cup B) = n(A) + n(B) - n(A \cap B).

Why: write $p$ , $q$ , $r$ and $s$ for the numbers of elements in the four regions of the diagram below. Then $n (A) = p + q$ and $n (B) = q + r$ , so $n (A) + n (B) = p + 2 q + r$ , which counts the overlap twice. Subtracting $n(A \cap B) = q$ once leaves $p + q + r = n(A \cup B)$ .

$n (A) = p + q$ , $n (B) = q + r$ , $n(A \cup B) = p + q + r$ and $n (U) = p + q + r + s$ .

Word problems describe regions in words. Learn to translate the phrases:

Phrase in the problem	Set	Number of elements
in both $A$ and $B$	$A \cap B$	$q$
in $A$ only	$A - B$	$p = n(A) - n(A \cap B)$
in at least one of the sets	$A \cup B$	$p + q + r$
in exactly one of the sets	$A \,\Delta\, B$	$p + r$
in neither set	$(A \cup B)'$	$s = n(U) - n(A \cup B)$

Worked example 1: football and basketball

In a class of 44 learners at a school in Kicukiro, 27 play football ( $F$ ), 19 play basketball ( $B$ ) and 8 play both. How many play (a) football only, (b) basketball only, (c) at least one of the games, (d) neither game?

Solution Start with the overlap, then work outwards.

Football only: $27 - 8 = 19$ .
Basketball only: $19 - 8 = 11$ .
At least one: $n(F \cup B) = 27 + 19 - 8 = 38$ .
Neither: $44 - 38 = 6$ .

$n (U) = 44$ . Check: $19 + 8 + 11 + 6 = 44$ . ✓

Worked example 2: an unknown overlap

In a class of 45 learners, 28 like mathematics ( $M$ ), 22 like English ( $E$ ) and 5 like neither subject. How many like both?

Solution Let $x$ be the number who like both. Fill the diagram in terms of $x$ :

$n (U) = 45$

The four regions make up the whole class:

(28 - x) + x + (22 - x) + 5 = 45 \;\Rightarrow\; 55 - x = 45 \;\Rightarrow\; x = 10.

So 10 learners like both subjects. Check with the formula: $n(M \cup E) = 45 - 5 = 40$ , and $28 + 22 - 10 = 40$ . ✓

Three sets: eight regions

Three overlapping circles split the rectangle into 8 regions. Each region is described by saying, for each set, whether it is in or out.

The eight regions of a three-set Venn diagram.

Region	Description	Region	Description
1	$A$ only	5	$A$ and $C$ , not $B$
2	$B$ only	6	$B$ and $C$ , not $A$
3	$C$ only	7	all three: $A \cap B \cap C$
4	$A$ and $B$ , not $C$	8	none: $(A \cup B \cup C)'$

Important

Rule (three sets)

\begin{aligned} n(A \cup B \cup C) = {} & n(A) + n(B) + n(C) \\ & - n(A \cap B) - n(A \cap C) - n(B \cap C) \\ & + n(A \cap B \cap C) \end{aligned}

Why: adding $n (A) + n (B) + n (C)$ counts an element of region 4, 5 or 6 twice and an element of region 7 three times. Subtracting the three pairwise intersections removes one count from regions 4, 5 and 6 (now counted once), but three counts from region 7 (now counted zero times). Adding $n(A \cap B \cap C)$ puts region 7 back once.

Tip

Fill a three-set diagram from the centre outwards. Write $n(A \cap B \cap C)$ in region 7 first. Then fill regions 4, 5 and 6 by subtracting region 7 from each pairwise intersection. Fill regions 1, 2 and 3 last, and finally region 8.

Worked example 3: languages in a school in Musanze

100 learners were asked which languages they speak besides Kinyarwanda. 60 speak English ( $E$ ), 35 speak French ( $F$ ) and 40 speak Kiswahili ( $K$ ); 15 speak English and French, 20 speak English and Kiswahili, 10 speak French and Kiswahili, and 5 speak all three. How many speak (a) none of the three, (b) exactly one of them?

Solution Work from the centre outwards.

All three: 5.
Exactly two: $E$ and $F$ only $= 15 - 5 = 10$ ; $E$ and $K$ only $= 20 - 5 = 15$ ; $F$ and $K$ only $= 10 - 5 = 5$ .
One only: $E$ only $= 60 - 10 - 15 - 5 = 30$ ; $F$ only $= 35 - 10 - 5 - 5 = 15$ ; $K$ only $= 40 - 15 - 5 - 5 = 15$ .

$n (U) = 100$

(a) The seven regions inside the circles hold $30 + 15 + 15 + 10 + 15 + 5 + 5 = 95$ learners, so $100 - 95 = 5$ speak none of the three.
Check with the rule: $60 + 35 + 40 - 15 - 20 - 10 + 5 = 95$ . ✓

(b) Exactly one: $30 + 15 + 15 = 60$ learners.

Worked example 4: shading a region described in symbols

Shade $A \cap (B \cup C)$ on a three-set Venn diagram.

Solution Work inside the brackets first. $B \cup C$ is everything in $B$ or $C$ : regions 2, 3, 4, 5, 6 and 7. The intersection with $A$ keeps only the part of that inside circle $A$ : regions 4, 5 and 7.

$A \cap (B \cup C)$ : the part of $A$ that also lies in $B$ or in $C$ .

Warning

"Football" is not "football only". The most common mistake is to write $n (F) = 27$ in the "football only" region. The 27 footballers are spread over two regions: those who play football only and those who play both. Always fill the overlap first, then subtract.

Khan Academy India: Word problems on union and intersection (7 min). Watch how each problem is first turned into sets, and how the rule for the number of elements in a union finds the missing number.

✦

Practice

Draw a Venn diagram for every question. The last three are harder.

$n (A) = 12$ , $n (B) = 9$ and $n(A \cap B) = 4$ . Find $n(A \cup B)$ .
Of 50 traders at Kimironko market, 32 sell vegetables, 26 sell fruit and 10 sell both. How many sell (a) fruit only, (b) neither?
$n (U) = 60$ , $n (A) = 35$ , $n (B) = 30$ and $n((A \cup B)') = 8$ . Find $n(A \cap B)$ .
In a cooperative of 36 farmers, 20 grow coffee and 24 grow bananas. Every farmer grows at least one of the two crops. How many grow both?
In a class of 40, 18 learners are in the drama club $D$ , 20 in the music club $M$ and 15 in the sports club $S$ . 7 are in $D$ and $M$ , 6 in $D$ and $S$ , 8 in $M$ and $S$ , and 3 in all three clubs. How many are in (a) none of the clubs, (b) exactly one club?
Using the region numbers 1 to 8 of the three-set diagram, list the regions that make up (a) $(A \cap B) \cup C$ (b) $A' \cap B'$ .
Use the regions $p$ , $q$ , $r$ , $s$ to show that $n(A - B) = n(A) - n(A \cap B)$ .

Show answersHide answers

$12 + 9 - 4 = 17$
With $V$ = vegetables and $F$ = fruit: (a) $26 - 10 = 16$ (b) $n(V \cup F) = 32 + 26 - 10 = 48$ , so $50 - 48 = 2$ .
$n(A \cup B) = 60 - 8 = 52$ , so $n(A \cap B) = 35 + 30 - 52 = 13$ .
With $C$ = coffee and $B$ = bananas: $n(C \cup B) = 36$ , so $n(C \cap B) = 20 + 24 - 36 = 8$ .
(a) $n(D \cup M \cup S) = 18 + 20 + 15 - 7 - 6 - 8 + 3 = 35$ , so $40 - 35 = 5$ are in none. (b) Centre 3; pairs only: 4, 3, 5; $D$ only $= 18 - 4 - 3 - 3 = 8$ , $M$ only $= 20 - 4 - 5 - 3 = 8$ , $S$ only $= 15 - 3 - 5 - 3 = 4$ . Exactly one: $8 + 8 + 4 = 20$ .
(a) Regions 3, 4, 5, 6, 7 (b) regions 3 and 8.
$A - B$ is region $p$ . And $n(A) - n(A \cap B) = (p + q) - q = p$ . ✓

Sources and credits

1.4: Set Operations with Two Sets, from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 26 September 2026). Adapted by Ganzaa.
1.5: Set Operations with Three Sets, from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 26 September 2026). Adapted by Ganzaa.
Video: Word problems on union and intersection, Khan Academy India, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1When adding counts twice

2Three sets: eight regions

When adding counts twice

Three sets: eight regions