SETS AND VENN DIAGRAMS

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 $B$.

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$$\{6, 12\}$$

b. State $n(A)$ and $n(B)$ (the number of elements in each set).

[1]

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$n(A)=6$, $n(B)=5$

c. Is the statement $9 \in A$ true or false? Justify your answer.

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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 $P$.

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$$P = \{1, 2, 3, 4, 6, 8, 12, 24\}$$

b. State $n(P)$.

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$n(P) = 8$

c. Is $P$ a finite or infinite set? Explain.

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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).

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$\{a\}$: subsets are $\emptyset, \{a\}$ — 2 subsets. $\{a,b\}$: subsets are $\emptyset, \{a\}, \{b\}, \{a,b\}$ — 4 subsets.

b. List all the subsets of $\{a,b,c\}$ (3 elements). How many are there?

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$\emptyset, \{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}$ — 8 subsets.

c. Describe the pattern (2, 4, 8, ...) and predict the number of subsets for a set with 5 elements.

[2]

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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]

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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. State the correct value of $n(\{1,2,2,3\})$ and explain your reasoning using correct set notation.

[2]

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$$\{1,2,2,3\} = \{1,2,3\}, \text{ so } n(\{1,2,2,3\}) = 3$$

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).

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Using $n(M \cup A) = n(M)+n(A)-n(M\cap A) = 45+38-12=71$.

b. Find how many students take neither Music nor Art (i.e. they take only other electives).

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$$210 - 71 = 139 \text{ students}$$

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 $B$.

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$A=\{3,6,9,12,15,18\}$. $B=\{4,8,12,16,20\}$.

b. Find $A\cap B$, and explain what these elements represent in terms of multiples.

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$A\cap B=\{12\}$ — this is the multiple of BOTH 3 and 4, i.e. the multiple of LCM$(3,4)=12$ within the range.

c. Find $n(A\cup B)$, using the inclusion-exclusion formula.

[2]

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$n(A\cup B)=n(A)+n(B)-n(A\cap B)=6+5-1=10$.

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).

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$\emptyset$; $\{2\},\{4\},\{6\},\{8\}$; $\{2,4\},\{2,6\},\{2,8\},\{4,6\},\{4,8\},\{6,8\}$; $\{2,4,6\},\{2,4,8\},\{2,6,8\},\{4,6,8\}$; $\{2,4,6,8\}$ — 16 subsets total.

b. How many PROPER subsets does $S$ have (a proper subset excludes the set itself, but includes the empty set)?

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$16-1=15$ 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.

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$\{a,b\},\{a,c\},\{b,c\}$ — 3 subsets.

b. For $S=\{a,b,c,d\}$ (4 elements), list all subsets with exactly 2 elements.

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$\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\},\{c,d\}$ — 6 subsets.

c. Based on 3 elements?3 subsets and 4 elements?6 subsets, predict the number of 2-element subsets for 5 elements, and verify by listing them.

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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]

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$n(A)=3$, $n(B)=2$, $A\cap B=\emptyset$ so $n(A\cap B)=0$, $A\cup B=\{1,2,3,7,8\}$ so $n(A\cup B)=5$.

b. 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]

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$3+2-0=5$ ?, 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]

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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. Answer the student's question: is $(1,2)$ the same as $(2,1)$? Explain the key difference between SET notation $\{\}$ and ORDERED PAIR notation $()$.

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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]

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$n(M)+n(M')=142+58=200$ ?, matching the total surveyed — consistent.

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.

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$n(\text{car}\cup\text{bike})=142+95-40=197$. 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.

Sets