Set Language and Absolute Value Worked ExamplesIGCSE Maths 0580

(a) $\cap$ means intersection — elements in both sets.

(b) $\cup$ means union — everything in either set, listed once.

(c) $A'$ is the complement — everything in $\mathscr{E}$ that is not in $A$.

(d) $A \cap B'$ is in $A$ but not in $B$. $n(\;)$ asks for how many, not which.

Always fill the intersection first and work outwards — this is what prevents double-counting.

(a) French only is the French total minus the overlap.

German only, likewise.

(b) Subtract everyone inside the circles from the class total.

Check all four regions sum to $40$: $12 + 6 + 8 + 14 = 40$. ✓

(a) 'Play football' means every region inside the football circle — four regions in all.

Add them.

(b) Add all seven regions.

Each element is either in a subset or not — two choices per element, independently.

(a) Here $n = 4$.

(b) A proper subset excludes the set itself, so subtract one. The empty set still counts as a proper subset.

List the sets within $\mathscr{E}$ first.

(a) Every multiple of $6$ is a multiple of $3$, so it cannot be one more than a multiple of $3$. No element is in both sets: the intersection is the empty set.

(b) The odd elements of $B$ are $13$, $19$ and $25$. $n(\;)$ asks how many.

(c) $A \cup C$ contains the multiples of $6$ and all the odd numbers. Its complement is every element of $\mathscr{E}$ not in that union: the even numbers that are not multiples of $6$.

List the smaller sets. $M = \{3, 6, 9, 12, 15, 18, 21, 24\}$ and $F = \{1, 2, 3, 4, 6, 8, 12, 24\}$.

(a) Numbers in all three sets: even multiples of $3$ that are factors of $24$.

(b) The complement is everything else in $\mathscr{E}$, and $n(\mathscr{E}) = 24$.

(c) There are $12$ even numbers and $8$ multiples of $3$, and $4$ of them ($6, 12, 18, 24$) are in both.

(d) The shaded region is inside $E$ and inside $M$ but outside $F$. Of $6, 12, 18, 24$, only $18$ is not a factor of $24$.

(a) Set $A$ is shaded between $-2$ and $4$. Both circles are open, so neither end value is included: use $<$ at both ends.

(b) The circle at $4$ is open, so $4$ is not in $A$.

(c) $A \cap B$ is the part shaded on both number lines. $B$'s left-hand part ($x \leqslant 1$) overlaps $A$ from $-2$ up to $1$; $B$'s right-hand part starts at $6$, which is outside $A$.

Take the end symbols from the sets they come from: $-2$ is an open circle in $A$, so $<$; $1$ is a filled circle in $B$, so $\leqslant$.

(d) $A \cup B$ covers everything below $4$ (from $B$ up to $1$, then from $A$ up to $4$) and everything from $6$ upwards. The only gap is from $4$ to $6$.

Check the ends: $4$ is not in $A$ or $B$, so it belongs to the complement and gets $\leqslant$. $6$ is in $B$, so it is not in the complement and gets $<$.

(a) List each set first. Remember $1$ is not prime, and $1 = 1^2$ is a square number.

Place each number in the region for exactly the sets it belongs to. The odd primes go in $P \cap O$ only; the odd squares go in $S \cap O$ only.

$2$, $4$ and $15$ are already placed. The remaining numbers belong to none of the sets, so they go outside all three circles.

Check: $1 + 5 + 2 + 1 + 1 + 5 = 15$ numbers placed, which is $n(\mathscr{E})$. ✓ The regions $P \cap S$ and $P \cap S \cap O$ stay empty.

(b) $O \cap P'$ means odd numbers that are not prime.

(c) A prime has exactly two factors, but a square number greater than $1$ has an odd number of factors (at least three), and $1$ has only one factor. So no number is both.

(d)(i) False: $4 \in S$ but $4 \notin O$. (ii) True: $3$ and $5$ are both odd primes. (iii) False: $9$ is a square number, so $9 \in S \cup P$.

(a) A point belongs to a set of this type exactly when its coordinates satisfy the equation. Substitute $x = 4$, $y = 7$ into each equation.

So the point lies on line $A$ but not on line $B$.

(b) $A \cap B$ is the set of points on both lines, so solve the equations simultaneously. Substitute $y = 2x - 1$ into $x + y = 8$.

The answer is a set containing one point. On the diagram the two lines cross at $(3, 5)$. ✓

(c) Lines $A$ and $C$ both have gradient $2$ but different $y$-intercepts ($-1$ and $3$), so they are parallel and never meet. Trying to solve them gives a contradiction.

So there are no points in common.

(d) $B \cap E$ is the points on $x + y = 8$ whose coordinates are both positive integers. List them systematically, starting from $x = 1$; $x = 8$ would give $y = 0$, which is not positive.