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:

$$b+(b+100)=110⟹2b=10⟹b=5$$

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 $(3,5]$.

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

$\mathrm{\mid }x-3\mathrm{\mid }<2\iff -2<x-3<2\iff 1<x<5$, so the set is $(1,5)$.

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

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

$\mathrm{\mid }A\mathrm{\mid }=4$, so $A$ has $2^4=16$ subsets (including $\mathrm{\varnothing }$ and $A$ itself).

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