SETS AND VENN DIAGRAMS

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).

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$$C \cap D = \{4,5,6\}$$

b. Find $C \cup D$ (the union).

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$$C \cup D = \{1,2,3,4,5,6,7,8,9\}$$

c. State $n(C \cup D)$.

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$n(C\cup D) = 9$

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 $F$.

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$E=\{4,8,12,16,20,24,28\}$. $F=\{6,12,18,24,30\}$.

b. Find $E \cap F$, and explain what these numbers represent in terms of multiples.

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$E\cap F = \{12,24\}$ — 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 $n(A\cup B)$.

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$n(A)=5$, $n(B)=5$, $A\cap B=\{3,4,5\}$ so $n(A\cap B)=3$, $A\cup B=\{1,2,3,4,5,6,7\}$ so $n(A\cup B)=7$.

b. Calculate $n(A)+n(B)-n(A\cap B)$ and compare it to $n(A\cup B)$.

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$5+5-3=7$, which matches $n(A\cup B)=7$ exactly.

c. Explain why adding $n(A)+n(B)$ directly (without subtracting the intersection) would give the wrong count for $n(A\cup B)$.

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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.

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$A\cap B=\{3\}$ (only the shared element). $A\cup B=\{1,2,3,4,5\}$ (all elements from either set). These are clearly different sets.

b. Explain, in your own words, the conceptual difference between intersection and union.

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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.

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$n(S\cup T)=n(S)+n(T)-n(S\cap T) \Rightarrow 89 = 64+41-n(S\cap T) \Rightarrow n(S\cap T)=105-89=16$.

b. The club has 120 members in total. How many play NEITHER soccer nor tennis?

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$$120-89=31 \text{ members}$$

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 $B$.

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$A=\{1,2,3,4,6,9,12,18,36\}$. $B=\{1,2,3,4,6,8,12,16,24,48\}$.

b. Find $A\cap B$, and explain the connection between this set and HCF$(36,48)$.

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$A\cap B=\{1,2,3,4,6,12\}$. 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 $n(R\cup S)$.

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$$40-8=32$$

b. Find $n(R\cap S)$, the number of students who play BOTH sports.

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$n(R\cup S)=n(R)+n(S)-n(R\cap S) \Rightarrow 32=22+19-n(R\cap S) \Rightarrow n(R\cap S)=41-32=9$.

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 $B$).

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$A\cap B=\{4,5,6\}$. $A$ only (elements in $A$ but not $B$): $\{1,2,3\}$.

b. Compare $n(A\text{ only})$ to $n(A)$ and $n(A\cap B)$. Write a general formula connecting these three quantities.

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$n(A\text{ only})=3$, $n(A)=6$, $n(A\cap B)=3$. Since $6-3=3$, the formula is: $$n(A\text{ only}) = n(A) - n(A\cap B)$$

c. 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.

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$n(B\text{ only})=n(B)-n(A\cap B)=5-3=2$. 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 $A\cap(B\cup C)$.

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$B\cup C=\{3,4,5,6\}$. $A\cap(B\cup C)=\{3,4\}$.

b. Now find $A\cap B$ and $A\cap C$ separately, then find $(A\cap B)\cup(A\cap C)$.

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$A\cap B=\{3,4\}$. $A\cap C=\{4\}$. $(A\cap B)\cup(A\cap C)=\{3,4\}$.

c. Compare your two results. Do they match, supporting the distributive law $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$?

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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 $A\cap B$.

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$A\cup B=\{1,2,3\}$ and $A\cap B=\{1,2,3\}$ — they are EXACTLY THE SAME, not one bigger than the other.

b. Explain precisely under what condition $n(A\cup B)=n(A\cap B)$ can happen, correcting the student's overgeneralization.

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$n(A\cup B)=n(A\cap B)$ 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.

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$n(C\cup D)=38+27-15=50$. Neither: $60-50=10$ applicants.

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.

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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.

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$n(F\cup G)=250$ (since everyone visited at least one). $n(F\cap G)=n(F)+n(G)-n(F\cup G)=180+165-250=95$.

b. The carnival organizers want to know how many people visited ONLY the Food stalls (not Games), to help plan next year's food stall capacity. Find this number, and explain why this figure (rather than the total $n(F)=180$) is the more USEFUL number for planning food-only queue capacity.

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Food only $=n(F)-n(F\cap G)=180-95=85$. This is more useful than the total $n(F)=180$ because the 95 people who visited BOTH areas likely split their time/spending between food and games, while the 85 'food only' visitors represent people whose FULL visit was concentrated at food stalls — better reflecting genuine food-stall-focused demand for capacity planning.

Intersection and union