MYP 3 · Maths
SETS AND VENN DIAGRAMS
65 questions across 5 sub-topics
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Sets
Complement of a set
Intersection and union
Venn diagrams
Problem solving with Venn diagrams
Sets 11 questions
QUESTION 1
3 marks
Criterion A
Medium
Let $A = \{2, 4, 6, 8, 10, 12\}$ and $B = \{3, 6, 9, 12, 15\}$.
a.
List the elements that belong to both $A$ and $B$.
[1] b.
State $n(A)$ and $n(B)$ (the number of elements in each set).
[1] c.
Is the statement $9 \in A$ true or false? Justify your answer.
[1] Show complete worked solution
(a)
$$\{6, 12\}$$
(b)
$n(A)=6$, $n(B)=5$
(c)
False — 9 is not one of the listed elements of $A$.
QUESTION 2
4 marks
Criterion A
Medium
Consider the set $P = \{x : x \text{ is a factor of } 24\}$.
a.
List all the elements of set $P$.
[2] b.
State $n(P)$.
[1] c.
Is $P$ a finite or infinite set? Explain.
[1] Show complete worked solution
(a)
$$P = \{1, 2, 3, 4, 6, 8, 12, 24\}$$
(b)
$n(P) = 8$
(c)
Finite — it has a fixed, countable number of elements (8), unlike a set such as 'all even numbers' which continues forever.
QUESTION 3
6 marks
Criterion B
Medium
Investigate the relationship between the number of elements in a set and the number of possible subsets it has.
a.
List all the subsets of $\{a\}$ (a set with 1 element), then all the subsets of $\{a,b\}$ (2 elements).
[2] b.
List all the subsets of $\{a,b,c\}$ (3 elements). How many are there?
[2] c.
Describe the pattern (2, 4, 8, ...) and predict the number of subsets for a set with 5 elements.
[2] Show complete worked solution
(a)
$\{a\}$: subsets are $\emptyset, \{a\}$ — 2 subsets. $\{a,b\}$: subsets are $\emptyset, \{a\}, \{b\}, \{a,b\}$ — 4 subsets.
(b)
$\emptyset, \{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}$ — 8 subsets.
(c)
Each additional element doubles the number of subsets, giving powers of 2: $2^1=2$, $2^2=4$, $2^3=8$. For 5 elements: $2^5=32$ subsets.
QUESTION 4
4 marks
Criterion C
Medium
A student writes: "The set $\{1, 2, 2, 3\}$ has 4 elements."
a.
Explain, using the definition of a set, why the student's statement is incorrect.
[2] b.
State the correct value of $n(\{1,2,2,3\})$ and explain your reasoning using correct set notation.
[2] Show complete worked solution
(a)
A set cannot contain duplicate elements — each element is either in the set or not, regardless of how many times it might be 'listed'. So $\{1,2,2,3\}$ is really just $\{1,2,3\}$, which has 3 elements, not 4.
(b)
$$\{1,2,2,3\} = \{1,2,3\}, \text{ so } n(\{1,2,2,3\}) = 3$$
QUESTION 5
5 marks
Criterion D
Medium
A school offers 3 elective subjects. Let $M=\{\text{students taking Music}\}$ with $n(M)=45$, and $A=\{\text{students taking Art}\}$ with $n(A)=38$. There are 210 students in total, and every student takes at least one elective.
a.
If 12 students take both Music and Art, find the number of students taking Music OR Art (but possibly not both).
[3] b.
Find how many students take neither Music nor Art (i.e. they take only other electives).
[2] Show complete worked solution
(a)
Using $n(M \cup A) = n(M)+n(A)-n(M\cap A) = 45+38-12=71$.
(b)
$$210 - 71 = 139 \text{ students}$$
QUESTION 6
6 marks
Criterion A
Hard
Let $U=\{1,2,\ldots,20\}$, $A=\{x\in U : x \text{ is a multiple of 3}\}$, $B=\{x\in U : x \text{ is a multiple of 4}\}$.
a.
List the elements of $A$ and $B$.
[2] b.
Find $A\cap B$, and explain what these elements represent in terms of multiples.
[2] c.
Find $n(A\cup B)$, using the inclusion-exclusion formula.
[2] Show complete worked solution
(a)
$A=\{3,6,9,12,15,18\}$. $B=\{4,8,12,16,20\}$.
(b)
$A\cap B=\{12\}$ — this is the multiple of BOTH 3 and 4, i.e. the multiple of LCM$(3,4)=12$ within the range.
(c)
$n(A\cup B)=n(A)+n(B)-n(A\cap B)=6+5-1=10$.
QUESTION 7
5 marks
Criterion A
Hard
A set $S=\{2,4,6,8\}$ has 4 elements.
a.
List ALL subsets of $S$ (there should be $2^4=16$ of them, including the empty set and $S$ itself).
[4] b.
How many PROPER subsets does $S$ have (a proper subset excludes the set itself, but includes the empty set)?
[1] Show complete worked solution
(a)
$\emptyset$; $\{2\},\{4\},\{6\},\{8\}$; $\{2,4\},\{2,6\},\{2,8\},\{4,6\},\{4,8\},\{6,8\}$; $\{2,4,6\},\{2,4,8\},\{2,6,8\},\{4,6,8\}$; $\{2,4,6,8\}$ — 16 subsets total.
(b)
$16-1=15$ proper subsets (excluding $S$ itself, but the empty set still counts).
QUESTION 8
8 marks
Criterion B
Hard
Investigate the pattern connecting the number of elements in a set to the number of subsets containing EXACTLY 2 elements.
a.
For $S=\{a,b,c\}$ (3 elements), list all subsets with exactly 2 elements.
[2] b.
For $S=\{a,b,c,d\}$ (4 elements), list all subsets with exactly 2 elements.
[3] c.
Based on 3 elements give 3 subsets and 4 elements give 6 subsets, predict the number of 2-element subsets for 5 elements, and verify by listing them.
[3] Show complete worked solution
(a)
$\{a,b\},\{a,c\},\{b,c\}$ — 3 subsets.
(b)
$\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\},\{c,d\}$ — 6 subsets.
(c)
Prediction: following the pattern (3, 6, ...), for 5 elements there should be 10 such subsets. Verifying with $\{a,b,c,d,e\}$: $\{a,b\},\{a,c\},\{a,d\},\{a,e\},\{b,c\},\{b,d\},\{b,e\},\{c,d\},\{c,e\},\{d,e\}$ — exactly 10, confirming the prediction.
QUESTION 9
5 marks
Criterion B
Hard
Investigate whether $n(A)+n(B)$ always overcounts by exactly $n(A\cap B)$, testing with sets that have NO overlap at all.
a.
Let $A=\{1,2,3\}$ and $B=\{7,8\}$ (no elements in common). Find $n(A)$, $n(B)$, $n(A\cap B)$, and $n(A\cup B)$.
[2] b.
Verify the formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$ still works even when the overlap is 0, and explain what this special case (called 'disjoint sets') tells you about when simple addition ($n(A)+n(B)$) IS valid.
[3] Show complete worked solution
(a)
$n(A)=3$, $n(B)=2$, $A\cap B=\emptyset$ so $n(A\cap B)=0$, $A\cup B=\{1,2,3,7,8\}$ so $n(A\cup B)=5$.
(b)
$3+2-0=5$, matching $n(A\cup B)$. This confirms that when two sets are DISJOINT (no shared elements, $n(A\cap B)=0$), the formula simplifies to plain addition $n(A)+n(B)$ — simple addition is only exactly correct in this special no-overlap case; whenever sets DO overlap, the intersection must be subtracted to avoid double-counting.
QUESTION 10
5 marks
Criterion C
Hard
A student writes $\{1,2,3\} = \{3,2,1\}$ and $\{1,1,2,3\}=\{1,2,3\}$, then asks: 'if order and repetition don't matter for sets, does that mean $(1,2)$ and $(2,1)$ are also the same thing?'
a.
Confirm the student's two set equalities are correct, briefly explaining why.
[2] b.
Answer the student's question: is $(1,2)$ the same as $(2,1)$? Explain the key difference between SET notation $\{\}$ and ORDERED PAIR notation $()$.
[3] Show complete worked solution
(a)
Both are correct: sets are defined purely by WHICH elements they contain, not the order they're listed in, or how many times an element is mentioned — so $\{1,2,3\}=\{3,2,1\}$ (same elements) and $\{1,1,2,3\}=\{1,2,3\}$ (duplicates don't create 'extra' elements).
(b)
No — $(1,2)$ and $(2,1)$ are generally considered DIFFERENT ordered pairs, because ordered pairs (used e.g. for coordinates) specifically DO care about order — $(1,2)$ might represent 'x=1, y=2' while $(2,1)$ represents 'x=2, y=1', which are different points. This is the key distinction: curly-brace set notation $\{\}$ ignores order and repetition, while round-bracket ordered-pair notation $()$ preserves order as meaningful information.
QUESTION 11
7 marks
Criterion D
Hard
A survey of 200 people found $M=\{\text{people who own a car}\}$ with $n(M)=142$, and asked about bicycle ownership too.
a.
If $n(M')=58$ people do NOT own a car, verify this is consistent with the total of 200 people surveyed.
[2] b.
Additional data shows 95 people own a bicycle, and 40 people own BOTH a car and a bicycle. A city planner wants to estimate how many people own NEITHER a car NOR a bicycle, to plan public transport needs. Find this number, and explain one assumption the planner must make about the survey data for this estimate to be reliable.
[5] Show complete worked solution
(a)
$n(M)+n(M')=142+58=200$, matching the total surveyed — consistent.
(b)
$n(\text{car}\cup\text{bike})=142+95-40=197$. Neither: $200-197=3$ people. Key assumption: this calculation assumes the survey accurately captured EVERY person's true ownership status (no measurement error, no people misreporting) — if the survey had a non-representative sample or reporting errors, this estimate of 'people needing public transport' could be inaccurate for real planning purposes.
Complement of a set 11 questions
QUESTION 1
4 marks
Criterion A
Medium
The universal set is $U = \{1,2,3,\ldots,15\}$. Let $A = \{2,4,6,8,10,12,14\}$ (the even numbers in $U$).
a.
List the elements of $A'$ (the complement of $A$).
[2] b.
State $n(A')$, and verify that $n(A) + n(A') = n(U)$.
[2] Show complete worked solution
(a)
$$A' = \{1,3,5,7,9,11,13,15\}$$
(b)
$n(A')=8$. Check: $n(A)+n(A')=7+8=15=n(U)$
QUESTION 2
4 marks
Criterion A
Medium
$U=\{x : x \text{ is a whole number}, 1\le x \le 20\}$. Let $B=\{x \in U : x \text{ is a multiple of 3}\}$.
a.
List the elements of $B$.
[1] b.
List the elements of $B'$.
[2] c.
Find $n(B')$.
[1] Show complete worked solution
(a)
$$B = \{3,6,9,12,15,18\}$$
(b)
$$B' = \{1,2,4,5,7,8,10,11,13,14,16,17,19,20\}$$
(c)
$n(B')=14$
QUESTION 3
6 marks
Criterion B
Medium
Investigate what happens when you find the complement of a complement, i.e. $(A')'$.
a.
With $U=\{1,2,3,4,5,6\}$ and $A=\{1,3,5\}$, find $A'$, then find $(A')'$ (the complement of $A'$).
[3] b.
Compare $(A')'$ to the original set $A$. What do you notice?
[1] c.
Explain why this makes sense conceptually (what does 'the complement of the complement' actually mean?).
[2] Show complete worked solution
(a)
$A' = \{2,4,6\}$. $(A')' = U \setminus A' = \{1,3,5\}$.
(b)
$(A')' = A$ — they are identical.
(c)
The complement of $A$ is everything NOT in $A$. Taking the complement again means everything NOT in 'everything not in $A$' — which brings you back to exactly the elements that ARE in $A$. So $(A')'=A$ always.
QUESTION 4
4 marks
Criterion C
Medium
A student claims that if $n(U)=30$ and $n(A)=18$, then $n(A')$ must also be calculable, and states $n(A')=12$.
a.
Verify whether the student's calculation is correct, showing the rule used.
[2] b.
Explain, in words, why this rule ($n(A')=n(U)-n(A)$) is always true for any set $A$ within a universal set $U$.
[2] Show complete worked solution
(a)
$n(A') = n(U)-n(A) = 30-18=12$. The student's answer is correct.
(b)
Every element of $U$ is either in $A$ or not in $A$ (there's no third option), so the elements of $U$ split exactly into two non-overlapping groups: $A$ and $A'$. This means their counts must add up to the total: $n(A)+n(A')=n(U)$, which rearranges to give the rule.
QUESTION 5
4 marks
Criterion D
Medium
At a company with 150 employees, the universal set $U$ is all employees. Let $R=\{\text{employees who drive to work}\}$, with $n(R)=97$.
a.
Find $n(R')$, the number of employees who do NOT drive to work.
[2] b.
The company is installing bike racks and wants to provide enough for 60% of the non-drivers, assuming they cycle. How many bike racks should be installed?
[2] Show complete worked solution
(a)
$$n(R') = 150-97=53$$
(b)
$$60\% \times 53 = 31.8 \approx 32 \text{ bike racks (rounding up to ensure enough capacity)}$$
QUESTION 6
5 marks
Criterion A
Hard
$U=\{x : x \text{ is a whole number}, 1\le x\le 30\}$. $P=\{x\in U : x \text{ is prime}\}$.
a.
List the elements of $P$.
[2] b.
List the elements of $P'$ (numbers from 1 to 30 that are NOT prime), and state $n(P')$.
[3] Show complete worked solution
(a)
$P=\{2,3,5,7,11,13,17,19,23,29\}$ — 10 primes under 30.
(b)
$P'=\{1,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30\}$ — 20 elements. ($n(P')=30-10=20$.)
QUESTION 7
6 marks
Criterion A
Hard
$U=\{1,2,\ldots,50\}$. $A=\{x\in U : x \text{ is a multiple of 6}\}$, $B=\{x \in U : x \text{ is a multiple of 8}\}$.
a.
Find $n(A)$ and $n(B)$.
[2] b.
Find $n(A\cap B)$ (multiples of LCM$(6,8)=24$), then find $n((A\cup B)')$ — the number of elements in NEITHER set.
[4] Show complete worked solution
(a)
Multiples of 6 up to 50: $6,12,\ldots,48$ — $n(A)=8$. Multiples of 8 up to 50: $8,16,\ldots,48$ — $n(B)=6$.
(b)
Multiples of 24 up to 50: $24,48$ — $n(A\cap B)=2$. $n(A\cup B)=8+6-2=12$. $n((A\cup B)')=50-12=38$.
QUESTION 8
7 marks
Criterion B
Hard
Investigate the relationship between $(A\cap B)'$ and $A'\cup B'$ (this is one of De Morgan's Laws).
a.
With $U=\{1,2,\ldots,10\}$, $A=\{1,2,3,4,5\}$, $B=\{3,4,5,6,7\}$: find $A\cap B$, then find $(A\cap B)'$.
[2] b.
Now find $A'$ and $B'$ separately, then find $A'\cup B'$.
[3] c.
Compare $(A\cap B)'$ from part (a) to $A'\cup B'$ from part (b). State the general rule (De Morgan's Law) this demonstrates, in words.
[2] Show complete worked solution
(a)
$A\cap B=\{3,4,5\}$. $(A\cap B)'=\{1,2,6,7,8,9,10\}$.
(b)
$A'=\{6,7,8,9,10\}$. $B'=\{1,2,8,9,10\}$. $A'\cup B' = \{1,2,6,7,8,9,10\}$.
(c)
Both equal $\{1,2,6,7,8,9,10\}$ — identical! This demonstrates De Morgan's Law: $(A\cap B)'=A'\cup B'$, i.e. 'the complement of an intersection equals the union of the complements' — being OUTSIDE both sets' overlap means you're either outside $A$, or outside $B$ (or both).
QUESTION 9
5 marks
Criterion B
Hard
Investigate whether $(A\cup B)'=A'\cap B'$ (the SECOND De Morgan's Law), using the SAME sets as before: $U=\{1,\ldots,10\}$, $A=\{1,2,3,4,5\}$, $B=\{3,4,5,6,7\}$.
a.
Find $A\cup B$, then $(A\cup B)'$.
[2] b.
Using $A'=\{6,7,8,9,10\}$ and $B'=\{1,2,8,9,10\}$ (found earlier), calculate $A'\cap B'$, and confirm whether it matches $(A\cup B)'$.
[3] Show complete worked solution
(a)
$A\cup B=\{1,2,3,4,5,6,7\}$. $(A\cup B)'=\{8,9,10\}$.
(b)
$A'\cap B' = \{8,9,10\}$ — matches $(A\cup B)'$ exactly, confirming the second De Morgan's Law: $(A\cup B)'=A'\cap B'$.
QUESTION 10
5 marks
Criterion C
Hard
A student says: '$(A\cap B)'$ and $A'\cap B'$ look almost identical (just moved the dash), so they must be equal too.'
a.
Using $U=\{1,\ldots,10\}$, $A=\{1,2,3,4,5\}$, $B=\{3,4,5,6,7\}$ (as before), calculate $A'\cap B'$ directly, and compare it to $(A\cap B)'=\{1,2,6,7,8,9,10\}$ found earlier.
[3] b.
Explain why it's important to keep track of exactly which De Morgan's Law applies to which combination (intersection-then-complement vs complement-then-intersection), rather than assuming similar-looking expressions are automatically equal.
[2] Show complete worked solution
(a)
$A'=\{6,7,8,9,10\}$, $B'=\{1,2,8,9,10\}$. $A'\cap B'=\{8,9,10\}$. This is clearly DIFFERENT from $(A\cap B)'=\{1,2,6,7,8,9,10\}$ — the student's claim is false.
(b)
The two De Morgan's Laws are specifically $(A\cap B)'=A'\cup B'$ and $(A\cup B)'=A'\cap B'$ — notice the operation SWAPS ($\cap$ becomes $\cup$ and vice versa) when the complement moves inside/outside. Assuming expressions with similar symbols are equal without checking can lead to genuinely wrong conclusions, as shown here — precise attention to which specific rule applies matters in mathematics.
QUESTION 11
7 marks
Criterion D
Hard
A hospital's patient database uses set $U$ = all 500 patients this month. $D=\{\text{patients with diabetes}\}$, $n(D)=85$.
a.
Find $n(D')$, the number of patients without diabetes.
[2] b.
The hospital wants to allocate nursing staff proportionally: diabetes-care nurses should be assigned in proportion to $n(D)$, and general-care nurses in proportion to $n(D')$. If there are 30 nursing staff total to allocate this way, find how many should be assigned to each group (rounding sensibly to whole numbers, and explaining your rounding choice), and comment on one limitation of using this SINGLE condition (diabetes) to allocate ALL nursing staff, when patients may have multiple different needs.
[5] Show complete worked solution
(a)
$$500-85=415$$
(b)
Diabetes-care: $30\times\frac{85}{500}=5.1\to5$ nurses (rounding to nearest whole number). General-care: $30\times\frac{415}{500}=24.9\to25$ nurses (check: $5+25=30$). Limitation: allocating staff based on ONLY the diabetes condition ignores that patients may have OTHER overlapping conditions requiring specialized care too (e.g. a patient could have diabetes AND a separate condition needing different attention) — a single-criterion allocation like this oversimplifies real patient care needs, which likely require a more nuanced, multi-factor staffing model in practice.
Intersection and union 12 questions
QUESTION 1
3 marks
Criterion A
Medium
Let $C = \{1,2,3,4,5,6\}$ and $D = \{4,5,6,7,8,9\}$.
a.
Find $C \cap D$ (the intersection).
[1] b.
Find $C \cup D$ (the union).
[1] c.
State $n(C \cup D)$.
[1] Show complete worked solution
(a)
$$C \cap D = \{4,5,6\}$$
(b)
$$C \cup D = \{1,2,3,4,5,6,7,8,9\}$$
(c)
$n(C\cup D) = 9$
QUESTION 2
4 marks
Criterion A
Medium
$E = \{x : x \text{ is a multiple of 4}, 1\le x\le30\}$ and $F=\{x : x \text{ is a multiple of 6}, 1\le x\le30\}$.
a.
List the elements of $E$ and $F$.
[2] b.
Find $E \cap F$, and explain what these numbers represent in terms of multiples.
[2] Show complete worked solution
(a)
$E=\{4,8,12,16,20,24,28\}$. $F=\{6,12,18,24,30\}$.
(b)
$E\cap F = \{12,24\}$ — these are the multiples of BOTH 4 and 6, i.e. multiples of $\text{LCM}(4,6)=12$.
QUESTION 3
7 marks
Criterion B
Medium
Investigate the relationship between $n(A)$, $n(B)$, $n(A\cap B)$, and $n(A\cup B)$.
a.
For $A=\{1,2,3,4,5\}$ and $B=\{3,4,5,6,7\}$, find $n(A)$, $n(B)$, $n(A\cap B)$, and $n(A\cup B)$.
[3] b.
Calculate $n(A)+n(B)-n(A\cap B)$ and compare it to $n(A\cup B)$.
[2] c.
Explain why adding $n(A)+n(B)$ directly (without subtracting the intersection) would give the wrong count for $n(A\cup B)$.
[2] Show complete worked solution
(a)
$n(A)=5$, $n(B)=5$, $A\cap B=\{3,4,5\}$ so $n(A\cap B)=3$, $A\cup B=\{1,2,3,4,5,6,7\}$ so $n(A\cup B)=7$.
(b)
$5+5-3=7$, which matches $n(A\cup B)=7$ exactly.
(c)
Simply adding $n(A)+n(B)$ counts every element that's in BOTH sets twice (once from each set), so the overlap (intersection) must be subtracted once to correct for this double-counting.
QUESTION 4
4 marks
Criterion C
Medium
A classmate says $A \cap B$ and $A \cup B$ are 'basically the same thing, just written differently.'
a.
Using $A=\{1,2,3\}$ and $B=\{3,4,5\}$, calculate both $A\cap B$ and $A\cup B$ to show they are different.
[2] b.
Explain, in your own words, the conceptual difference between intersection and union.
[2] Show complete worked solution
(a)
$A\cap B=\{3\}$ (only the shared element). $A\cup B=\{1,2,3,4,5\}$ (all elements from either set). These are clearly different sets.
(b)
Intersection ($\cap$) contains only elements that appear in BOTH sets — it's the overlap. Union ($\cup$) contains every element that appears in EITHER set (or both) — it's everything combined, with no duplicates.
QUESTION 5
5 marks
Criterion D
Medium
At a sports club, $S=\{\text{members who play soccer}\}$ with $n(S)=64$, and $T=\{\text{members who play tennis}\}$ with $n(T)=41$. There are 89 members who play at least one of the two sports.
a.
Find the number of members who play BOTH soccer and tennis.
[3] b.
The club has 120 members in total. How many play NEITHER soccer nor tennis?
[2] Show complete worked solution
(a)
$n(S\cup T)=n(S)+n(T)-n(S\cap T) \Rightarrow 89 = 64+41-n(S\cap T) \Rightarrow n(S\cap T)=105-89=16$.
(b)
$$120-89=31 \text{ members}$$
QUESTION 6
6 marks
Criterion A
Hard
$A=\{x : x \text{ is a factor of 36}\}$, $B=\{x : x \text{ is a factor of 48}\}$.
a.
List the elements of $A$ and $B$.
[3] b.
Find $A\cap B$, and explain the connection between this set and HCF$(36,48)$.
[3] Show complete worked solution
(a)
$A=\{1,2,3,4,6,9,12,18,36\}$. $B=\{1,2,3,4,6,8,12,16,24,48\}$.
(b)
$A\cap B=\{1,2,3,4,6,12\}$. This set contains exactly the COMMON FACTORS of 36 and 48 — its LARGEST element (12) is precisely HCF$(36,48)$.
QUESTION 7
5 marks
Criterion A
Hard
In a class of 40 students, $R=\{\text{students who play rugby}\}$ with $n(R)=22$, and $S=\{\text{students who play soccer}\}$ with $n(S)=19$. It is known that 8 students play NEITHER sport.
a.
Find $n(R\cup S)$.
[2] b.
Find $n(R\cap S)$, the number of students who play BOTH sports.
[3] Show complete worked solution
(a)
$$40-8=32$$
(b)
$n(R\cup S)=n(R)+n(S)-n(R\cap S) \Rightarrow 32=22+19-n(R\cap S) \Rightarrow n(R\cap S)=41-32=9$.
QUESTION 8
7 marks
Criterion B
Hard
Investigate the relationship between $n(A)$, $n(A\cap B)$, and $n(A \text{ only})$ (elements in $A$ but NOT in $B$).
a.
For $A=\{1,2,3,4,5,6\}$ and $B=\{4,5,6,7,8\}$, find $A\cap B$ and the elements that are in $A$ ONLY (not in $B$).
[2] b.
Compare $n(A\text{ only})$ to $n(A)$ and $n(A\cap B)$. Write a general formula connecting these three quantities.
[2] c.
Use your formula to find $n(B\text{ only})$ WITHOUT listing elements, given $n(B)=5$ and $n(A\cap B)=3$, then verify by direct counting.
[3] Show complete worked solution
(a)
$A\cap B=\{4,5,6\}$. $A$ only (elements in $A$ but not $B$): $\{1,2,3\}$.
(b)
$n(A\text{ only})=3$, $n(A)=6$, $n(A\cap B)=3$. Since $6-3=3$, the formula is: $$n(A\text{ only}) = n(A) - n(A\cap B)$$
(c)
$n(B\text{ only})=n(B)-n(A\cap B)=5-3=2$. Verify: $B$ only $=\{7,8\}$, which indeed has 2 elements.
QUESTION 9
7 marks
Criterion C
Hard
Investigate whether $A\cap(B\cup C)$ is always the SAME as $(A\cap B)\cup(A\cap C)$ (this is the distributive law for sets).
a.
With $A=\{1,2,3,4\}$, $B=\{3,4,5\}$, $C=\{4,5,6\}$: find $B\cup C$, then find $A\cap(B\cup C)$.
[2] b.
Now find $A\cap B$ and $A\cap C$ separately, then find $(A\cap B)\cup(A\cap C)$.
[3] c.
Compare your two results. Do they match, supporting the distributive law $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$?
[2] Show complete worked solution
(a)
$B\cup C=\{3,4,5,6\}$. $A\cap(B\cup C)=\{3,4\}$.
(b)
$A\cap B=\{3,4\}$. $A\cap C=\{4\}$. $(A\cap B)\cup(A\cap C)=\{3,4\}$.
(c)
Both give $\{3,4\}$ — matching exactly, supporting that $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$ for these sets (consistent with the general distributive law for sets).
QUESTION 10
4 marks
Criterion C
Hard
A student says: 'Since $A\cup B$ always has MORE elements than $A\cap B$, the union is always the bigger operation.'
a.
Test this claim using $A=B=\{1,2,3\}$ (two IDENTICAL sets). Find $A\cup B$ and $A\cap B$.
[2] b.
Explain precisely under what condition $n(A\cup B)=n(A\cap B)$ can happen, correcting the student's overgeneralization.
[2] Show complete worked solution
(a)
$A\cup B=\{1,2,3\}$ and $A\cap B=\{1,2,3\}$ — they are EXACTLY THE SAME, not one bigger than the other.
(b)
$n(A\cup B)=n(A\cap B)$ occurs precisely when $A=B$ (the two sets are identical) — in this special case, the union and intersection are literally the same set. The student's claim ('union is ALWAYS bigger') is only true when $A$ and $B$ are NOT identical; equal sets are the exception.
QUESTION 11
6 marks
Criterion D
Hard
A company's project requires skills in both Coding ($C$) and Design ($D$). Among 60 job applicants, $n(C)=38$, $n(D)=27$, and $n(C\cap D)=15$ (applicants with BOTH skills).
a.
Find how many applicants have AT LEAST one of the two skills, and how many have NEITHER.
[3] b.
The project needs exactly 4 people, ideally all with BOTH skills, for a specialized sub-team. Given only 15 applicants have both, the manager wants to know: if 4 people are selected at random from JUST the 15 dual-skilled applicants, discuss (without calculating exact probability) why this selection process would likely lead to a MORE effective sub-team than randomly selecting 4 from all 60 applicants, referencing the relevant set.
[3] Show complete worked solution
(a)
$n(C\cup D)=38+27-15=50$. Neither: $60-50=10$ applicants.
(b)
Selecting from the 15 dual-skilled applicants ($C\cap D$) guarantees every team member has BOTH required skills, directly matching the project's specific needs. Selecting randomly from all 60 could easily result in team members who have only ONE of the two skills (or neither), likely making the team less effective at a task requiring both — deliberately drawing from the intersection set targets exactly the qualified sub-group needed.
QUESTION 12
7 marks
Criterion D
Hard
At a school carnival, 250 tickets were sold. $F=\{\text{ticket holders who visited the Food stalls}\}$, $n(F)=180$. $G=\{\text{ticket holders who visited the Games area}\}$, $n(G)=165$. Every ticket holder visited AT LEAST one of the two areas.
a.
Find $n(F\cap G)$, the number who visited BOTH areas.
[3] b.
Find the number who visited only the Food stalls. Then decide whether this number or the total number who visited the Food stalls is more useful for planning food-stall queue capacity, and justify your choice.
[4] Show complete worked solution
(a)
$n(F\cup G)=250$ (since everyone visited at least one). $n(F\cap G)=n(F)+n(G)-n(F\cup G)=180+165-250=95$.
(b)
$$n(F\cap G)=180+165-250=95.$$ Therefore $$n(F\text{ only})=180-95=\boxed{85}.$$ For planning the total demand on the food-stall queues, $\boxed{n(F)=180}$ is the more useful figure because visitors who also went to the Games area still used the food stalls. The food-only value would be useful only for a question specifically about visitors who did not use the Games area.
Venn diagrams 13 questions
QUESTION 1
4 marks
Criterion A
Medium
The Venn diagram shows the number of students in a class who play Chess (C) and Checkers (K).
a.
How many students play only Chess (not Checkers)?
[1] b.
How many students play both games?
[1] c.
How many students are in the class in total?
[2] Show complete worked solution
(a)
9 students (the region inside Chess only).
(b)
5 students (the overlapping region).
(c)
$$9+5+6+4 = 24 \text{ students}$$
QUESTION 2
4 marks
Criterion A
Medium
Draw and interpret a Venn diagram for the following: $U=\{1,2,\ldots,10\}$, $X=\{2,4,6,8,10\}$, $Y=\{3,6,9\}$.
a.
Find $X \cap Y$ and $X \cup Y$.
[2] b.
State how many elements lie outside both circles (in neither $X$ nor $Y$), and list them.
[2] Show complete worked solution
(a)
$X\cap Y=\{6\}$. $X\cup Y=\{2,3,4,6,8,9,10\}$.
(b)
Elements of $U$ not in $X\cup Y$: $\{1,5,7\}$ — 3 elements.
QUESTION 3
6 marks
Criterion B
Medium
Investigate how the overlap region size in a 2-circle Venn diagram affects the total count $n(A \cup B)$, keeping $n(A)$ and $n(B)$ fixed at 10 each.
a.
If the overlap ($A\cap B$) contains 0 elements (the sets don't overlap at all), find $n(A\cup B)$.
[2] b.
If the overlap contains 10 elements (one set is a subset of the other, or they're identical), find $n(A\cup B)$.
[2] c.
Describe the relationship between overlap size and total union size.
[2] Show complete worked solution
(a)
$n(A\cup B) = n(A)+n(B)-n(A\cap B) = 10+10-0=20$.
(b)
$n(A\cup B)=10+10-10=10$.
(c)
As the overlap size increases, the union size decreases (for fixed $n(A)$ and $n(B)$) — more shared elements means fewer NEW elements are added by the second set, right down to a minimum union size equal to the larger individual set when full overlap occurs.
QUESTION 4
4 marks
Criterion C
Medium
A Venn diagram shows 3 in the 'Dogs only' region, 7 in the overlap ('Dogs and Cats'), and 5 in the 'Cats only' region, for pet ownership among 20 surveyed families.
a.
Explain how to find the number of families who own dogs (in total, including those who also own cats), using the diagram.
[2] b.
Explain how to find the number of families who own neither a dog nor a cat, given 20 families were surveyed in total.
[2] Show complete worked solution
(a)
Add the 'Dogs only' region and the overlap region together: $3+7=10$ families own dogs (some of whom also own cats).
(b)
Add all three visible regions ($3+7+5=15$) and subtract from the total surveyed: $20-15=5$ families own neither.
QUESTION 5
5 marks
Criterion D
Medium
A survey of 80 gym members found: 52 use the weights room, 38 use the cardio room, and 15 use neither.
a.
Find the number of members who use AT LEAST one of the two facilities.
[2] b.
Find the number of members who use BOTH the weights room and the cardio room, and hence complete a Venn diagram description (state the count in each of the 3 regions: weights only, cardio only, both).
[3] Show complete worked solution
(a)
$$80-15=65 \text{ members}$$
(b)
$n(W\cup C)=n(W)+n(C)-n(W\cap C) \Rightarrow 65=52+38-n(W\cap C) \Rightarrow n(W\cap C)=90-65=25$. So: weights only $=52-25=27$, cardio only $=38-25=13$, both $=25$.
QUESTION 6
5 marks
Criterion A
Hard
The Venn diagram shows sports played by 50 students.
a.
How many students play ONLY football?
[1] b.
How many students play AT LEAST one of the two sports?
[2] c.
Find the total number of students represented in the diagram, and verify it matches the given total of 50.
[2] Show complete worked solution
(a)
15 students.
(b)
$$15+22+8=45 \text{ students}$$
(c)
$$15+22+8+5=50$$ — matches.
QUESTION 7
4 marks
Criterion A
Hard
The Venn diagram shows subject choices among 100 students.
a.
How many students study History (in total, including those who also study Geography)?
[2] b.
What PERCENTAGE of the 100 students study NEITHER subject?
[2] Show complete worked solution
(a)
$$45+30=75 \text{ students}$$
(b)
$$\frac{7}{100}\times100=7\%$$
QUESTION 8
8 marks
Criterion B
Hard
Investigate how the shape of a Venn diagram's regions changes as the OVERLAP between two sets increases, while keeping $n(A)=30$ and $n(B)=25$ fixed.
a.
If the overlap ($n(A\cap B)$) is 0, find $n(A\text{ only})$, $n(B\text{ only})$, and $n(A\cup B)$.
[2] b.
If the overlap increases to 15, recalculate $n(A\text{ only})$, $n(B\text{ only})$, and $n(A\cup B)$.
[3] c.
Describe the pattern: as the overlap increases (with $n(A)$ and $n(B)$ fixed), what happens to the 'only' regions and to the total union size? Explain why this makes sense visually on a Venn diagram.
[3] Show complete worked solution
(a)
$A$ only $=30$, $B$ only $=25$, $A\cup B=30+25-0=55$.
(b)
$A$ only $=30-15=15$. $B$ only $=25-15=10$. $A\cup B=30+25-15=40$.
(c)
As overlap increases, BOTH 'only' regions shrink (since more of each set's elements are being 'shared' rather than exclusive), and the total union shrinks too. Visually, this makes sense because a bigger overlapping middle region means the two circles are covering MORE of the same space, so their combined (union) area covers LESS total ground than if they barely overlapped.
QUESTION 9
4 marks
Criterion B
Hard
Investigate whether a Venn diagram for 2 sets can EVER have a NEGATIVE number in any region, and what this would mean if it appeared in a calculation.
a.
Given $n(A)=20$, $n(B)=15$, and someone claims $n(A\cap B)=25$, calculate $n(A\text{ only})=n(A)-n(A\cap B)$. What do you get?
[2] b.
Explain what this negative result reveals about the ORIGINAL claim ($n(A\cap B)=25$) — is it actually possible, given $n(A)=20$?
[2] Show complete worked solution
(a)
$n(A\text{ only})=20-25=-5$ — a NEGATIVE value, which is impossible for a real region of a Venn diagram (you can't have $-5$ elements).
(b)
This reveals the original claim is IMPOSSIBLE: the intersection $A\cap B$ is a SUBSET of $A$ itself, so it can never contain MORE elements than $A$ does. Since $n(A)=20$, we must have $n(A\cap B)\le20$ — the claimed value of 25 violates this basic requirement, which is exactly why the calculation produced a nonsensical negative result.
QUESTION 10
5 marks
Criterion C
Hard
A survey Venn diagram (partially completed) shows 12 people who drink only Coffee, and 20 who drink both Tea and Coffee, out of 60 people surveyed in total, with 8 drinking neither.
a.
Explain, step by step, how to find the missing 'Tea only' region (labelled $x$ in the diagram), using the given total.
[3] b.
A classmate tried to find $x$ by only using $60-12-20=28$ (forgetting to also subtract the 'neither' region). Explain their error and why it's important to account for ALL four regions of a 2-set Venn diagram (both 'only' regions, the overlap, AND 'neither') when the total is given.
[2] Show complete worked solution
(a)
Total accounted for by known regions plus 'neither': $12+20+8=40$. Since the grand total is 60, the missing 'Tea only' region must be $60-40=20$.
(b)
The classmate's method missed subtracting the 8 people who drink neither — a Venn diagram for 2 sets within a universal set always has FOUR distinct regions (not three), and the total must account for every person, including those OUTSIDE both circles. Forgetting the 'neither' region leads to an answer (28) that's too large by exactly the size of that missed region (8), since $28-8=20$, the correct answer.
QUESTION 11
4 marks
Criterion C
Hard
The Venn diagram shows how many students in a music class play guitar and/or piano.
a.
Explain, using full sentences and correct set notation ($n(\cdot)$, $\cap$, $\cup$), how you would communicate to someone WITHOUT the diagram exactly what each of the three visible numbers (9, 14, 6) represents.
[4] Show complete worked solution
(a)
The number 9 represents $n(\text{Guitar only})$ — students who play guitar but NOT piano. The number 14 represents $n(\text{Guitar}\cap\text{Piano})$ — students who play BOTH instruments. The number 6 represents $n(\text{Piano only})$ — students who play piano but NOT guitar. Together, $n(\text{Guitar})=9+14=23$ and $n(\text{Piano})=14+6=20$.
QUESTION 12
6 marks
Criterion D
Hard
A community organization has 60 registered volunteers, shown by which projects they help with in the Venn diagram.
a.
Project Manager for Project A wants to send a group email ONLY to volunteers exclusively dedicated to Project A (not also helping Project B), to avoid overwhelming shared volunteers with duplicate messages. How many volunteers should receive this email?
[2] b.
The organization wants to recruit MORE volunteers for a joint A+B event, and decides the 12 people already doing BOTH projects are the best candidates to lead small teams (since they understand both projects). If each of these 12 leads a team requiring 3 additional NEW volunteers (from OUTSIDE the current 60), find the total number of people (leaders + new recruits) involved in this joint event, and identify which group in the diagram (if any) should NOT be asked to be a 'new' recruit, and why.
[4] Show complete worked solution
(a)
24 volunteers (the 'Project A only' region).
(b)
New recruits needed: $12\times3=36$. Total people involved: $12\text{ leaders}+36\text{ new}=48$. The 8 people who help with 'neither' project currently should NOT be considered for 'new recruit' status if the goal is genuinely NEW volunteers from outside the current 60 — although they aren't in A or B, they ARE still part of the existing 60 registered volunteers, so recruiting them wouldn't bring in truly NEW people to the organization.
QUESTION 13
6 marks
Criterion D
Hard
A city council's environmental survey of 120 households is shown in the Venn diagram, tracking recycling and composting habits.
a.
The council wants to launch a targeted campaign for households doing NEITHER recycling NOR composting, offering a free starter kit. Find how many households qualify, and calculate the total budget needed if each starter kit costs \$35.
[3] b.
The council's budget is only \$400. Explain the shortfall, and suggest ONE reasonable modification to the campaign (e.g. targeting a subset of the 15 households, reducing kit cost, or another approach) that would fit within the \$400 budget, showing the relevant calculation.
[3] Show complete worked solution
(a)
Households with neither: $120-38-45-22=15$. Budget needed: $15\times35=\$525$.
(b)
Shortfall: $525-400=\$125$. One option: reduce the number of kits — with \$400 available, $400\div35\approx11.4$, so the council could afford kits for only 11 of the 15 qualifying households (prioritizing perhaps by need or by first-come-first-served), leaving 4 households without a kit under the current budget.
Problem solving with Venn diagrams 18 questions
QUESTION 1
5 marks
Criterion A
Medium
In a class of 35 students, 22 study French, 18 study Spanish, and 7 study neither language.
a.
Find the number of students who study at least one of French or Spanish.
[2] b.
Find the number of students who study BOTH French and Spanish.
[2] c.
Find the number of students who study French only (not Spanish).
[1] Show complete worked solution
(a)
$$35-7=28 \text{ students}$$
(b)
$n(F\cup S)=n(F)+n(S)-n(F\cap S) \Rightarrow 28=22+18-n(F\cap S) \Rightarrow n(F\cap S)=40-28=12$
(c)
$$22-12=10 \text{ students}$$
QUESTION 2
5 marks
Criterion A
Medium
A cafe surveyed 60 customers about their drink preferences: 34 like coffee, 29 like tea, and 11 like neither.
a.
Find the number of customers who like at least one of coffee or tea.
[2] b.
Find the number who like both coffee and tea, then state how many like ONLY tea (not coffee).
[3] Show complete worked solution
(a)
$$60-11=49$$
(b)
$n(C\cap T)=34+29-49=14$. Tea only $=29-14=15$.
QUESTION 3
6 marks
Criterion B
Medium
Investigate a three-set Venn diagram problem. At a school, 50 students were surveyed about membership of Art ($A$), Drama ($D$) and Robotics ($R$). The stated pairwise overlaps include students who joined all three clubs.
a.
Given $n(A)=25$, $n(D)=20$, $n(R)=18$, $n(A\cap D)=10$, $n(A\cap R)=9$, $n(D\cap R)=8$, and $n(A\cap D\cap R)=8$, find $n(A\cup D\cup R)$ using inclusion-exclusion.
[3] b.
How many of the 50 surveyed students joined NONE of the three clubs?
[1] c.
Explain why the '+$n(A\cap D\cap R)$' term at the end of the formula is needed (i.e. why simply subtracting all three pairwise overlaps isn't enough).
[2] Show complete worked solution
(a)
Using inclusion-exclusion,
$$n(A\cup D\cup R)=25+20+18-10-9-8+8=\boxed{44}.$$
(b)
The number who joined none is
$$50-n(A\cup D\cup R)=50-44=\boxed{6\text{ students}}.$$
(c)
Students in all three clubs get counted three times in $n(A)+n(D)+n(R)$, then subtracted three times (once in each pairwise overlap) — leaving them counted zero times. Adding back $n(A\cap D\cap R)$ once corrects this, so they end up counted exactly once, as they should be.
QUESTION 4
5 marks
Criterion C
Medium
A classmate solving a 2-set problem writes: 'total = French + Spanish + neither', ignoring any overlap.
a.
Using the earlier French/Spanish class example ($n(F)=22$, $n(S)=18$, neither$=7$, total$=35$), show what the classmate's formula would give, and explain why it's wrong.
[3] b.
Write the CORRECT formula connecting total, French, Spanish, neither, and the overlap (both).
[2] Show complete worked solution
(a)
Classmate's formula: $22+18+7=47$, which does not equal the actual total of 35. It's wrong because students who study BOTH languages get counted twice — once in the French count, once in the Spanish count — so the overlap must be subtracted to correct for this.
(b)
$$\text{total} = n(F) + n(S) - n(F\cap S) + \text{neither}$$
QUESTION 5
6 marks
Criterion D
Medium
A company surveys 200 employees about which of two training programs (Leadership $L$, Technical $T$) they've completed. It's known that twice as many employees completed Technical training as completed Leadership training, 30 completed both, and 20 completed neither.
a.
Let $n(L)=x$. Write an expression for $n(T)$, and write an equation for the total using the Venn diagram relationship.
[3] b.
Solve your equation to find $n(L)$ and $n(T)$.
[3] Show complete worked solution
(a)
$n(T)=2x$. Using $\text{total}=n(L)+n(T)-n(\text{both})+\text{neither}$: $$200 = x + 2x - 30 + 20$$
(b)
$200=3x-10 \Rightarrow 3x=210 \Rightarrow x=70$. So $n(L)=70$ and $n(T)=2\times70=140$. Check: $70+140-30+20=200$
QUESTION 6
5 marks
Criterion A
Hard
At a language school with 90 students, 25 study only French, 30 study only Spanish, and 8 study neither language.
a.
Find the number of students who study BOTH French and Spanish.
[3] b.
Find the total number who study French (including those who also study Spanish).
[2] Show complete worked solution
(a)
Both $=90-25-30-8=27$.
(b)
French total $=25+27=52$.
QUESTION 7
5 marks
Criterion A
Hard
A survey of 150 gym members found 95 use the treadmill, 70 use the weights room, and 20 use neither facility.
a.
Find the number who use BOTH facilities.
[3] b.
Find the number who use EXACTLY ONE of the two facilities (not both, not neither).
[2] Show complete worked solution
(a)
$n(T\cup W)=150-20=130$. $n(T\cap W)=95+70-130=35$.
(b)
Exactly one $=130-35=95$ (i.e. the union minus the overlap).
QUESTION 8
7 marks
Criterion B
Hard
Investigate a 3-set problem: at a school carnival, 200 attendees were surveyed on which of 3 activities (Face-painting $F$, Games $G$, Food $D$) they visited, with all 200 visiting at least one activity.
a.
Given $n(F)=110$, $n(G)=95$, $n(D)=120$, $n(F\cap G)=45$, $n(F\cap D)=50$, $n(G\cap D)=40$, and $n(F\cap G\cap D)=20$, apply the 3-set inclusion-exclusion formula to verify the total matches 200: $$n(F\cup G\cup D)=n(F)+n(G)+n(D)-n(F\cap G)-n(F\cap D)-n(G\cap D)+n(F\cap G\cap D)$$
[4] b.
Given this discrepancy (210 calculated vs 200 actual), suggest ONE possible real-world explanation for why survey data like this might not perfectly reconcile, and explain why checking totals like this is an important step before trusting survey-based conclusions.
[3] Show complete worked solution
(a)
$$110+95+120-45-50-40+20=210$$ This does NOT match the stated total of 200 — suggesting an inconsistency in the given data (the numbers as stated are not mutually consistent with 200 total attendees).
(b)
Possible explanations: measurement/counting errors in the original tallying, some attendees being double-counted across categories, or the sub-totals being independently (and imperfectly) estimated rather than precisely cross-tabulated. This demonstrates why VERIFYING that detailed breakdown numbers actually reconcile with a known total is an essential check — inconsistent data can lead to flawed conclusions if used without this kind of validation.
QUESTION 9
5 marks
Criterion B
Hard
Investigate how changing one piece of information in a two-set problem can make the data inconsistent and produce an impossible result.
a.
A problem states: 60 people surveyed, 40 like tea, 35 like coffee, and 10 like neither. Find how many like both, and check whether this is a sensible, achievable answer.
[2] b.
Now change only the number who like neither from 10 to 45, while keeping the total and the two set counts unchanged. Recalculate the intersection and explain why the new data are impossible.
[3] Show complete worked solution
(a)
$n(T\cup C)=60-10=50$. $n(T\cap C)=40+35-50=25$ — sensible, since 25 is less than both 40 and 35 (the both-group can't exceed either individual group).
(b)
With 45 people in neither set,
$$n(T\cup C)=60-45=15.$$
Then $$n(T\cap C)=40+35-15=60.$$ This is impossible because an intersection cannot contain more elements than either set: $60>40$ and $60>35$. The changed value makes the data inconsistent.
QUESTION 10
6 marks
Criterion C
Hard
A student solving a 2-set Venn diagram problem finds a NEGATIVE value for the 'both' region, and simply writes '$n(A\cap B)=-3$' as their final answer without comment.
a.
Explain why a negative answer for ANY region of a Venn diagram is a signal that something has gone wrong — either in the given data, or in the student's calculation — rather than a valid final answer.
[3] b.
Describe TWO specific checks a student should perform if they get a negative or otherwise 'impossible' result (e.g. an intersection bigger than one of the original sets), to identify where the problem lies.
[3] Show complete worked solution
(a)
Every region of a Venn diagram represents a COUNT of real people/objects, which can never be negative (the smallest possible count for any group is 0, representing an empty group). A negative result like $-3$ is mathematically impossible for a real-world count, so it must indicate either an arithmetic error in the working, or that the ORIGINAL problem's given numbers are inherently inconsistent (e.g. the stated totals don't actually fit together logically).
(b)
Check 1: re-verify all arithmetic steps carefully, since simple calculation errors are the most common cause. Check 2: verify the ORIGINAL given data is internally consistent — e.g. that no individual set's size is smaller than a claimed overlap involving that set, and that all given totals genuinely add up correctly across all four regions of a 2-set Venn diagram (or eight regions for 3 sets).
QUESTION 11
6 marks
Criterion C
Hard
A market research problem states: 'Of 80 respondents, 55 like Brand X, 48 like Brand Y, and 60 like at least one of the two brands.' A student is asked to communicate a full solution finding how many like both brands.
a.
Write a complete, well-organized solution (clearly defining your sets/notation, showing all working, and stating a final conclusion in a full sentence) to find how many respondents like BOTH brands.
[4] b.
Explain why clearly DEFINING your sets and notation at the start of a solution (as done in part a) is considered good mathematical communication practice, even though the final NUMBER would be the same without it.
[2] Show complete worked solution
(a)
Let $X=\{\text{respondents who like Brand X}\}$ and $Y=\{\text{respondents who like Brand Y}\}$, with $n(X)=55$, $n(Y)=48$, and $n(X\cup Y)=60$. Using the formula $n(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$: $$60=55+48-n(X\cap Y) \Rightarrow n(X\cap Y)=103-60=43$$ Therefore, 43 respondents like both Brand X and Brand Y.
(b)
Clearly defining sets and notation at the start makes the ENTIRE solution understandable to someone reading it independently — it removes ambiguity about what each symbol represents, allows the reader to follow the logical structure of the argument, and demonstrates that the solver understands the underlying mathematical framework being used, not just performing calculations by rote.
QUESTION 12
6 marks
Criterion C
Hard
A student solves a survey problem involving 95 students. Of these, 50 joined Music, 45 joined Sport, and 15 joined both. The student writes only: '50 + 45 - 15 = 80; 95 - 80 = 15.'
a.
Explain why the student's unlabeled calculation would lose mathematical-communication credit even though its two numerical results are correct.
[3] b.
Rewrite the calculation with set notation, labels and a conclusion that answers the survey question.
[3] Show complete worked solution
(a)
The numbers are not identified, and neither operation is connected to a set rule. A reader cannot tell that 80 is the union or that 15 is the number in neither set. The work needs set labels and a conclusion in context.
(b)
Let $M$ be Music and $S$ be Sport.
$$n(M\cup S)=n(M)+n(S)-n(M\cap S)=50+45-15=80.$$
Therefore 80 students joined at least one activity.
$$n(\text{neither})=95-80=\boxed{15}.$$
Hence 15 students joined neither Music nor Sport.
QUESTION 13
8 marks
Criterion D
Hard
A hospital emergency department tracks 300 patients in one week: 130 required X-rays, 95 required blood tests, and 45 required BOTH.
a.
Find the number of patients who required AT LEAST one of the two procedures.
[3] b.
The hospital wants to estimate STAFFING needs: X-ray technicians can process 8 patients per hour, and phlebotomists (blood test staff) can process 12 patients per hour, during an 8-hour shift. Using the TOTAL number needing each procedure (not just 'only' each), determine whether current staffing of 2 X-ray technicians and 1 phlebotomist would be sufficient to handle a full week's procedures within a single 8-hour shift, showing your reasoning and identifying any bottleneck.
[5] Show complete worked solution
(a)
$n(X\cup B)=130+95-45=180$.
(b)
X-ray capacity needed: 130 patients. 2 technicians $\times8\text{h}\times8\text{ patients/h}=128$ patients — INSUFFICIENT (130 needed, only 128 capacity, short by 2). Blood test capacity needed: 95 patients. 1 phlebotomist $\times8\text{h}\times12\text{ patients/h}=96$ patients — sufficient (barely, with 1 to spare). The BOTTLENECK is X-ray staffing, which falls just short of meeting demand within a single 8-hour shift; the hospital should consider adding staff or extending X-ray hours specifically, while blood test staffing is adequate.
QUESTION 14
7 marks
Criterion D
Hard
A university's 400 first-year students were surveyed: 220 are enrolled in a Maths course, 180 in a Science course, and 60 in NEITHER.
a.
Find the number enrolled in BOTH Maths and Science.
[3] b.
The university is planning a joint Maths-Science study support session and can only accommodate 50 students in the room available. Using the number who take BOTH subjects (the most likely group to benefit from a JOINT session), determine what percentage of this dual-enrolled group could be accommodated, and suggest ONE fair method for selecting which students get a spot if not everyone can attend.
[4] Show complete worked solution
(a)
$n(M\cup S)=400-60=340$. $n(M\cap S)=220+180-340=60$.
(b)
Dual-enrolled: 60 students. Room capacity: 50. Percentage accommodated: $\frac{50}{60}\times100\approx83.3\%$. Since not all 60 can fit, a fair selection method could be a random lottery/draw among the 60 dual-enrolled students, or a first-come-first-served registration system — either approach avoids favoritism and gives every eligible student an equal, transparent chance at one of the 50 spots.
QUESTION 15
4 marks
Criterion B
Hard
Investigate the maximum and minimum possible values of $n(A\cap B)$, given only $n(A)=18$ and $n(B)=25$ (with no universal set size specified).
a.
What is the LARGEST possible value of $n(A\cap B)$? (Hint: the intersection can't have more elements than the smaller of the two sets.)
[2] b.
What is the SMALLEST possible value of $n(A\cap B)$? Explain your reasoning.
[2] Show complete worked solution
(a)
Maximum $n(A\cap B)=18$ — this occurs if EVERY element of $A$ is also in $B$ (i.e. $A$ is entirely a subset of $B$); the overlap can never exceed the size of the smaller set.
(b)
Minimum $n(A\cap B)=0$ — this occurs if $A$ and $B$ share no elements at all (disjoint sets); an intersection can never be negative, so 0 is always achievable and is the theoretical floor.
QUESTION 16
3 marks
Criterion C
Hard
A student solving a Venn diagram problem writes only the final answer '$n(A\cap B)=17$' with no working shown at all.
a.
Explain why, even if $17$ happens to be correct, providing NO working significantly weakens the mathematical communication of the solution.
[3] Show complete worked solution
(a)
Without shown working, there is no way to verify HOW the answer was reached, whether the correct method was used, or whether the student genuinely understands the underlying relationship (versus guessing or recalling a memorized number) — clear working demonstrates the LOGICAL PATH to a solution, which is central to communicating mathematics effectively, not just stating a result.
QUESTION 17
5 marks
Criterion D
Hard
A charity runs two donation drives: Drive A collected from 84 total donors, of whom 38 gave ONLY to Drive A, 29 gave ONLY to Drive B, and the rest gave to both.
a.
Find the number of donors who gave to BOTH drives.
[2] b.
The charity wants to send a special thank-you gift to donors who gave to BOTH drives, at a cost of \$12 per gift, but has budgeted only \$150. Determine whether the budget is sufficient, and if not, state exactly how many donors would need to be excluded to fit the budget.
[3] Show complete worked solution
(a)
$84-38-29=17$ donors.
(b)
Cost for all 17 dual donors: $17\times12=\$204$, which EXCEEDS the \$150 budget by $204-150=\$54$. Maximum donors affordable: $150\div12=12.5\to12$ donors. So $17-12=5$ donors would need to be excluded from receiving the special gift to stay within budget.
QUESTION 18
6 marks
Criterion D
Hard
A tech company surveyed 200 employees on remote-work tool usage: 120 use Tool X.
a.
Given that 45 employees use ONLY Tool X (not any other main tool), and the remaining Tool X users also use Tool Y, find how many employees use BOTH Tool X and Tool Y.
[2] b.
IT support wants to prioritize training resources for employees using BOTH tools (since they need to manage more complexity), estimating 30 minutes of training per dual-tool user. Calculate the total training TIME (in hours) needed, and discuss whether this seems like a realistic amount of total training time for a company to schedule within a single week (assume a standard 40-hour work week per trainer, and 1 trainer available).
[4] Show complete worked solution
(a)
Both $=120-45=75$ employees.
(b)
Total training time: $75\times30\text{ min}=2250$ minutes $=37.5$ hours. With 1 trainer and a 40-hour week, this is JUST barely feasible ($37.5<40$), leaving only 2.5 hours of slack — this is a tight but technically achievable schedule, though it leaves little room for other trainer duties, breaks, or unexpected delays, suggesting the company might benefit from a second trainer or spreading the training across more than one week for a more comfortable margin.