Set Problems

Solve Venn diagram problems: find a set from clues, find a missing number or an unknown x from the total, then use the regions to find a probability.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024 · 6 real past-paper questions below

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What a Set Problem Is

A set problem gives you a Venn diagram with something missing. It might be a set you must build, a number in one region, or an unknown $x$ .

Most counting problems use one fact.

Rule

The numbers in all four regions add up to the total in the box. Use that to write an equation.

Region	Holds
Left circle only	in A, not in B
Overlap	in both
Right circle only	in B, not in A
Outside the circles	in neither

Build a Set From Clues

The most common short question, 1 or 2 marks: find a set C that fits some clues, or explain why two sets share nothing.

Say $E = {1, 2, \dots, 15}$ , A is the multiples of 3 and B is the odd numbers. C is a subset of A, every member of C is even, and $n (C) = 2$ .

Worked example

1
$A = {3, 6, 9, 12, 15}$ — Write A out in full first. You can only test the clues once the numbers are listed.
2
$A \cap B = {3, 9, 15}$ — Exams often ask for this first. These members of A are also odd, so they are on both lists.
3
$C \subset A$ — Clue 1: C is a subset of A, so C may only use the five numbers in A.
4
$even in A : 6, 12$ — Clue 2: every member of C is even. Cross out the odd ones, 3, 9 and 15. That leaves 6 and 12.
5
$n (C) = 2$ — Clue 3: C has exactly 2 members. Exactly two numbers passed, so C uses both.
6
$C = {6, 12}$ — Write the answer as a set, in curly brackets.

Key idea

"Explain why $A \cap D = \emptyset$ " when D is 2, 4, 8: say A and D have no members in common, because none of 2, 4, 8 is a multiple of 3.

Common mistake

Putting 9 or 15 in C. They are in A, but they are odd, so they break a clue.

Find a Set From Set Facts

The 2-mark question gives facts written in symbols and asks you to list a set C. Read each fact in words, then place the numbers one fact at a time.

Say $E = {1, 2, \dots, 10}$ , $A = {1, 4, 9}$ and $B = {2, 3, 5, 7}$ . Also $A \cup B \cup C = E$ , $A \cap C = {4}$ and $B \cap C^{'} = {2, 5}$ . List C.

Worked example

1
$not in A or B : 6, 8, 10$ — First find the numbers in the box that are in neither A nor B.
2
$6, 8, 10 \in C$ — $A \cup B \cup C = E$ says every number is in at least one of the three sets. 6, 8 and 10 are not in A or B, so they must be in C.
3
$4 \in C, 1, 9 \in / C$ — $A \cap C = {4}$ says 4 is the only member of A that is also in C. So 1 and 9 stay out.
4
$2, 5 \in / C$ — $C^{'}$ means not in C. So $B \cap C^{'} = {2, 5}$ says 2 and 5 are the members of B that are not in C.
5
$3, 7 \in C$ — The other members of B, 3 and 7, are not on that list, so they are in C.
6
$C = {3, 4, 6, 7, 8, 10}$ — Collect every number you put in C and write them as a set.

Key idea

A close cousin: D has 3 members and $D \cap (A \cup B) = \emptyset$ . D shares nothing with A or B, so it can only use 6, 8 and 10: $D = {6, 8, 10}$ .

Common mistake

Forgetting 6, 8 and 10. The fact $A \cup B \cup C = E$ is what puts them in C.

Find the Number in Both

The 3-mark question: you get each total and the number in neither, but not the number in both. Call the overlap $x$ and add up the regions.

Worked example

1
$45 students$ — The question: 26 play football, 21 play tennis and 7 play neither.
2
$both = x$ — We do not know how many play both, so call it x. It goes in the overlap.
3
$F only = 26 - x$ — The 26 footballers include the x who also play tennis. Take them away to leave football only.
4
$T only = 21 - x$ — Same for tennis: the 21 include the x in the overlap.
5
$(26 - x) + x + (21 - x) + 7 = 45$ — Every student is in exactly one region, so the four regions add up to 45.
6
$54 - x = 45$ — Collect: 26 + 21 + 7 = 54, and $x - x - x = - x$ .
7
$54 = 45 + x$ — Add x to both sides, so x is no longer negative.
8
$x = 9$ — Take 45 from both sides: 54 − 45 = 9. So 9 play both.
9
$F only = 26 - 9 = 17$ — Put x = 9 back in to find football only.
10
$T only = 21 - 9 = 12$ — Same for tennis only.
11
$17 + 9 + 12 + 7 = 45$ — Check: the four regions add up to the total, so the numbers are right.

Key idea

Quick check: 45 take away 7 leaves 38 inside the circles. The totals 26 and 21 add to 47. The extra 9 is the overlap, counted twice.

Common mistake

Writing 26 in football only. The 26 includes the 9 who play both.

The Formula for n(A ∪ B)

Many textbooks print the last card's idea as a formula. It finds any one of the numbers when you know the others.

Formula

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

$n (A \cup B)$ is how many are in A or B or both: everyone inside the circles. $n (A)$ and $n (B)$ are the two circle totals. $n (A \cap B)$ is the number in both. You take it away because adding the two totals counts the overlap twice.

Say 50 people: 31 have a cat (A), 24 have a dog (B) and 9 have neither. How many have both?

Worked example

1
$n (A \cup B) = 50 - 9 = 41$ — Everyone except the 9 with neither is inside the circles.
2
$41 = 31 + 24 - n (A \cap B)$ — Put the three numbers you know into the formula.
3
$41 = 55 - n (A \cap B)$ — Add the two circle totals: 31 + 24 = 55.
4
$n (A \cap B) = 55 - 41 = 14$ — Add $n (A \cap B)$ to both sides and take 41 from both sides. So 14 have both.
5
$cat only = 31 - 14 = 17$ — Take the overlap from the cat total.
6
$dog only = 24 - 14 = 10$ — Take the overlap from the dog total.
7
$17 + 14 + 10 + 9 = 50$ — Check: the four regions add up to the total.

Common mistake

Putting 50 in for $n (A \cup B)$ . The 9 with neither are outside the circles, so take them off first.

Greatest and Least Number in Both

Some questions give only the totals and ask for the greatest or least possible number in both. Picture how far the circles can slide.

Say $n (E) = 30$ , $n (A) = 18$ and $n (B) = 15$ .

Worked example

1
$n (A \cap B) \leq n (B) = 15$ — The overlap is part of B, so it can never be bigger than B's 15.
2
$greatest n (A \cap B) = 15$ — This happens when all of B sits inside A. B is the smaller set, so it fits.
3
$n (A \cup B) \leq 30$ — Everyone is inside the box, so the two circles together hold at most 30.
4
$18 + 15 - n (A \cap B) \leq 30$ — Write $n (A \cup B)$ using the formula from the last card.
5
$33 - n (A \cap B) \leq 30$ — Add the totals: 18 + 15 = 33.
6
$n (A \cap B) \geq 3$ — 33 is 3 more than 30, so at least 3 people must be counted in both.
7
$least n (A \cap B) = 3$ — This happens when nobody is outside the circles.

Key idea

If the two totals add up to $n (E)$ or less, the least number in both is 0: the circles can sit apart.

Common mistake

Giving 18 as the greatest. The overlap must fit inside both circles, so it can be no bigger than the smaller total.

Find x From the Total

The 3 or 4-mark question: each region is written with $x$ . Add every region, set it equal to the total, and solve.

Say a group has 42 students. Art only is $3 x$ , both is $x$ , Music only is $2 x + 1$ and 5 take neither.

Worked example

1
$3 x + x + (2 x + 1) + 5 = 42$ — Every student is in exactly one region, so the four regions add up to 42.
2
$6 x + 6 = 42$ — Collect the x terms: 3x + x + 2x = 6x. Then the numbers: 1 + 5 = 6.
3
$6 x = 36$ — Take 6 from both sides, so only the x term is left.
4
$x = 6$ — Divide both sides by 6.
5
$Art only = 3 \times 6 = 18$ — Put x = 6 back into Art only.
6
$Music only = 2 \times 6 + 1 = 13$ — Put x = 6 into Music only: 12 + 1 = 13.
7
$18 + 6 + 13 + 5 = 42$ — Check: the four regions add up to the total.
8
$n (Art) = 18 + 6 = 24$ — The Art circle holds Art only and both, so add those two regions.

Common mistake

Stopping at $x = 6$ . If the question asks how many take Art, you must add Art only and both: 24.

Key idea

Write the equation down first. It usually earns a mark on its own, even if you slip later. And x counts people, so it must be a whole number, not negative. If it is not, recheck your adding.

Probability From a Venn Diagram

The follow up question, 1 or 2 marks: one person is chosen at random. Count the people who fit, and put them over the people you choose from.

Formula

P = \frac{number who fit}{number you choose from}

The top is the regions that fit. The bottom is everyone you pick from. Use the Art and Music diagram: 18, 6, 13 and 5.

Worked example

1
$P (both) = \frac{6}{42}$ — 6 of the 42 students take both, and you choose from all 42.
2
$= \frac{1}{7}$ — Simplify: divide the top and the bottom by 6.
3
$P (Music) = \frac{6 + 13}{42}$ — Music students fill the whole Music circle: both, and Music only.
4
$= \frac{19}{42}$ — 6 + 13 = 19. Nothing divides 19 and 42, so it does not simplify.
5
$choose from Music: 19$ — Now the question says one Music student is chosen. That is a smaller group: only the 19 in the Music circle.
6
$P (also Art) = \frac{6}{19}$ — Of those 19, the 6 in the overlap also take Art. So the bottom is 19, not 42.

Common mistake

"One of the students who take Music is chosen." The bottom is 19, the Music circle, not 42.

Which Method? Read the Question

Look at what is missing and what the question asks for.

You see	You do
Find a set C with clues	write each set out, pick numbers that pass every clue
Explain why the overlap is empty	say they have no members in common
Facts like $A \cap C = {4}$	numbers outside A and B first, then one fact at a time
Totals and neither, no both	call both x, add the regions
Greatest or least in both	smaller total, or the two totals take away the box
Regions written with x	add them, equal to the total, solve
Chosen at random	fit over total
Chosen from people who...	bottom is that circle only
Only lists to sort	Venn Diagrams
What does a symbol mean?	Set Notation

Key idea

Always finish with a check: the four regions must add up to the total in the box.

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