Sets of Real Numbers practice questions — A-Level Maths

Removing $2$ and $4$ from $A$ leaves $\{1, 3, 5\}$.

$A \cup B$ is the set of elements in $A$, in $B$, or in both: $\{1, 2, 3, 4, 5, 6\}$.

Let $b$ be the cost of the ball in cents. The bat costs $b + 100$. Their total is $110$, so:

$$b + (b + 100) = 110 \implies 2b = 10 \implies b = 5$$

The ball costs 5 cents. Most people answer 10 — the trick is that the bat–ball *difference* is $100$, not the bat's price. Linear-equation discipline beats gut every time.

The overlap of $[-2, 5]$ and $(3, 7]$ runs from $3$ (excluded by the second) to $5$ (included by the first), giving $(3, 5]$.

$|S \cup B| = 18 + 12 - 5 = 25$ play at least one. So $30 - 25 = 5$ play neither.

$|x - 3| < 2 \Leftrightarrow -2 < x - 3 < 2 \Leftrightarrow 1 < x < 5$, so the set is $(1, 5)$.

$A \cap B = \{2, 3\}$. The complement in $U = \{1, 2, 3, 4, 5\}$ is $\{1, 4, 5\}$.

$|A \times B| = 3 \cdot 2 = 6$. The product is $\{(1,x),(1,y),(2,x),(2,y),(3,x),(3,y)\}$.

$|A| = 4$, so $A$ has $2^4 = 16$ subsets (including $\emptyset$ and $A$ itself).

The positive divisors of $12$ are $\{1, 2, 3, 4, 6, 12\}$, so $|A| = 6$.