SETS AND VENN DIAGRAMS

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 $A$).

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$$A' = \{1,3,5,7,9,11,13,15\}$$

b. State $n(A')$, and verify that $n(A) + n(A') = n(U)$.

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$n(A')=8$. 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 $B$.

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$$B = \{3,6,9,12,15,18\}$$

b. List the elements of $B'$.

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$$B' = \{1,2,4,5,7,8,10,11,13,14,16,17,19,20\}$$

c. Find $n(B')$.

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$n(B')=14$

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 $A'$).

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$A' = \{2,4,6\}$. $(A')' = U \setminus A' = \{1,3,5\}$.

b. Compare $(A')'$ to the original set $A$. What do you notice?

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$(A')' = A$ — they are identical.

c. Explain why this makes sense conceptually (what does 'the complement of the complement' actually mean?).

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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.

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$n(A') = n(U)-n(A) = 30-18=12$. The student's answer is correct.

b. Explain, in words, why this rule ($n(A')=n(U)-n(A)$) is always true for any set $A$ within a universal set $U$.

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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.

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$$n(R') = 150-97=53$$

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?

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$$60\% \times 53 = 31.8 \approx 32 \text{ bike racks (rounding up to ensure enough capacity)}$$

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 $P$.

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$P=\{2,3,5,7,11,13,17,19,23,29\}$ — 10 primes under 30.

b. List the elements of $P'$ (numbers from 1 to 30 that are NOT prime), and state $n(P')$.

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$P'=\{1,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30\}$ — 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 $n(B)$.

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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. Find $n(A\cap B)$ (multiples of LCM$(6,8)=24$), then find $n((A\cup B)')$ — the number of elements in NEITHER set.

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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 $(A\cap B)'$.

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$A\cap B=\{3,4,5\}$. $(A\cap B)'=\{1,2,6,7,8,9,10\}$.

b. Now find $A'$ and $B'$ separately, then find $A'\cup B'$.

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$A'=\{6,7,8,9,10\}$. $B'=\{1,2,8,9,10\}$. $A'\cup B' = \{1,2,6,7,8,9,10\}$.

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.

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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 $(A\cup B)'$.

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$A\cup B=\{1,2,3,4,5,6,7\}$. $(A\cup B)'=\{8,9,10\}$.

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 $(A\cup B)'$.

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$A'\cap B' = \{8,9,10\}$ — 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.

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$A'=\{6,7,8,9,10\}$, $B'=\{1,2,8,9,10\}$. $A'\cap B'=\{8,9,10\}$. This is clearly DIFFERENT from $(A\cap B)'=\{1,2,6,7,8,9,10\}$ — the student's claim is false.

b. 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.

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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.

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$$500-85=415$$

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.

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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.

Complement of a set