Elementary Set Theory - SS1 Mathematics Past Questions and Answers - page 1

Given $U=\{4,6,8,10,12,15,19\},$

$$C=\{4,6,8\},$$

$$D=\{10,12,6,15\}$$

$$E=\{4,15,19,6,10\}$$

(a) $C\cap D=\{6\}$

(b) $D\cup E=\{4,6,10,12,15,19\}$

(c) $C-D=\{4,8\}$

(d) $D^{{}^{\prime }}=\{4,8,19\}$

(e) $P(C)=2^3=8=\varnothing ,\left\{4\right\},\left\{6\right\},\left\{8\right\},\{4,6\},\{4,8\},\{6,8\},\{4,6,8\}$

The universal set $U$ is $100\mathrm{\%}$ of students at the Interhouse sports competition

$$n(U)=100\mathrm{\%}$$

$n(A)=80\mathrm{\%}$, percentage of students at the athletic competition.

$n(F)=60\mathrm{\%}$, percentage of students at the football competition.

$n(A\cap F)=x$, percentage that attended both.

$$n(U)=n(A)+n(F)-n(A\cap F)$$

$$100\mathrm{\%}=80\mathrm{\%}+60\mathrm{\%}-x$$

$$100\mathrm{\%}=140\mathrm{\%}-x$$

$$100\mathrm{\%}-140\mathrm{\%}=-x$$

$$-40\mathrm{\%}=-x$$

$x=40\mathrm{\%}$