(1) A container has 30 gold medals,22 silver medals and 18 bronze medals.if one medal is selected at random from the container, what is the probability that it is not a gold medal.?
A.$\frac{4}{7}$**
B.$\frac{3}{7}$
C.$\frac{11}{35}$
D.$\frac{12}{35}$
Solution:The formula for the probability that it is not a gold medal is $p(not a gold medal)=1-p(gold medal)$

so we first compute \begin{align*}
&\text{p(gold medal)}=\frac{\text{Number of possible outcome}}{\text{Number of total outcome}}\\
&=\frac{30}{30+22+18}\\
&=\frac{30}{70}\\
&=\frac{3}{7}\\
&\text{Now we proceed to}\\
&\text{p(not a gold medal)}=1-\frac{3}{7}=\frac{4}{7}
\end{align*}
(2) the sum of the first n terms of an A.P is 252. if the first term is -16 and the last term is 72. find the number of terms in the series.
A.7
B.9**
C.6
D.8
Solution: The formula for the sum of an A.P involving a last term is $s_n=\frac{n}{2}(a+l)$
where a=first term, n=number of terms in the A.P,l=last term and $s_n=\text{sum of the A.P}$ so $s_n=252$, a=-16,l=72 substituting these values into formula we get
\begin{align*}
&252=\frac{n}{2}(-16+72)\\
&252=\frac{n}{2}(56)\\
&252=28n\\
&n=9
\end{align*}
(3) Factorize completely $4abx-2axy-12b^2x+6bxy$
A.$2x(3b-a)(2b-y)$
B.$2x(a-3b)(b-2y)$
C.$2x(2b-a)(3b-y)$
D.$2x(a-3b)(2b-y)$**
Solution: \begin{align*}
&4abx-2axy-12b^2x+6bxy\\
&=4abx-12b^2x-2axy+6bxy\\
&=4bx(a-3b)-2xy(a-3b)\\
&=(4bx-2xy)(a-3b)\\
&=2x(2b-y)(a-3b)
\end{align*}\
(4) If $\frac{9^{2x-1}}{27^{x+1}}=1$ find value of x.
A.2
B.8
C.5**
D.3
Solution:
\begin{align*}
&\frac{9^{2x-1}}{27^{x+1}}=1\\
&=\frac{3^{2(2x-1)}}{3^{3(x+1)}}=1\\
&=\frac{2(2x-1)}{3(x+1)}=1\\
&2(2x-1)=3(x+1)\\
&4x-2=3x+3\\
&4x=3x+5\\
&x=5
\end{align*}
(5) find $(\frac{1}{0.06}\div{\frac{1}{0.042}})^{-1}$
A.4.42
B.3.14
C.1.53
D.1.43**
Solution: \begin{align*}
&(\frac{1}{0.06}\div{\frac{1}{0.042}})^{-1}\\
&=(\frac{1}{0.06}\times{\frac{0.042}{1}})^{-1}\\
&=(\frac{0.042}{0.06})^{-1}\\
&=(0.7)^{-1}\\
&=\frac{1}{0.7}=1.43
\end{align*}
(6) In a class of 40 students,32 offer maths, 24 offer physics and 4 offer neither maths nor physics. How many offer both maths and physics.
A.16
B.4
C.20**
D.8
Solution: from the data above,let M=32, P=24, $(M\cap{P})^c=4$
$M\cap{P}=x$, $n(M)=32-x$,$n(P)=24-x$ and
\begin{align*}
&n(M\cap{P})=32-x+x+24-x+4=40\\
&=32+24+4-x=40\\
&\Rightarrow{60}-x=40\\
&x=20
\end{align*}
(7) Given \begin{align*}
&U=\{\text{Even numbers between 0 and 30}\}\\
&P=\{\text{Multiples of 6 between 0 and 30}\}\\
&Q=\{\text{Multiples of 4 between 0 and 30}\}
\end{align*}
find $(P\cup{Q})^c$
A.$\{0,2,6,22,26\}$
B.$\{2,4,14,22,26\}$
C.$\{2,10,14,22,26\}$**
D.$\{0,10,14,22,26\}$
Solution: \begin{align*}
&U=\{2,4,6,8,10,12,14,...,30\}\\
&P=\{6,12,18,24,30\}\\
&Q=\{4,8,12,16,20,24,28\}\\
&P\cup{Q}=\{4,6,8,12,16,18,20,24,28,30\}\\
&(P\cup{Q})^c=\{2,10,14,22,26\}
\end{align*}       
(8) If \begin{equation*}
N=
\begin{vmatrix}
3 & 5 & -4\\
6 & -3 & -5\\
-2 & 2 & 1
\end{vmatrix}
\end{equation*}
find $|N|$
A.91
B.65
C.23
D.17**
Solution: $|N|$ means we should find N determinants
\begin{equation*}
|N|=3
\begin{vmatrix}
-3 & -5\\
2 & 1
\end{vmatrix}
-5\begin{vmatrix}
6 & -5\\
-2 & 1
\end{vmatrix}
-4\begin{vmatrix}
6 & -3\\
-2 & 2
\end{vmatrix}
\end{equation*}
\begin{align*}
&3[-3(1)-(-5(2)]-5[(6(1)-(-5(-2)]-4[6(2)-(-3(2)]\\
&=21+20-24\\
&=17
\end{align*}
(9) The binary operation is defined on the set of integers p and q by $p\star{q}=pq+p+q$. find $2(3\star{4})$
A.19
B.38**
C.59
D.67
Solution: since $p\star{q}=pq+p+q$ then
\begin{align*}
&2(3\star{4})\\
&=2(3(4)+3+4))\\
&=2(12+7)\\
&=38
\end{align*}
(10) A bucket is 12cm in diameter at the top, 8cm in diameter at the bottom and 4cm deep. calculate its volume.
A.$144\pi{cm^3}$
B.$\frac{304}{3}\pi{cm^3}$**
C.$72\pi{cm^3}$
D.$128\pi{cm^3}$
Solution: A bucket is a frustrum of a cone so we will use the formula for calculating volume of a frustrum.
let $R=\frac{d}{2}=\frac{12}{2}=6cm$ and $r=\frac{d}{2}=\frac{8}{2}=4cm$
now volume is
\begin{align*}
&Volume=\frac{1}{3}\pi{s}(R^2+r^2+Rr)\\
\text{where s=slant height}\\
&=\frac{1}{3}\pi{4}(6^2+4^2+6(4))\\
&=\frac{4}{3}\pi(76)=\frac{304}{3}\pi
\end{align*}
(11)  find the number of ways of selecting 8 subjects from 12 subjects for an examination.
A.498
B.496
C.495
D.490
Solution: using combinatorial analysis
\begin{align*}
&{12}C_{8}=\frac{12!}{8!(12-8)!}\\
&=\frac{12!}{8!4!}\\
&=\frac{12\times{11}\times{10}\times{9}\times{8!}}{8!4!}\\
&=\frac{11880}{24}=495
\end{align*}
(12) Given $p=1+\sqrt{2}$ and $q=1-\sqrt{2}$ evaluate $\frac{p^2-q^2}{2pq}$\\
A.$-2(2+\sqrt{2})$
B.$2(2+\sqrt{2})$
C.$-2\sqrt{2}$**
D.$2\sqrt{2}$
Solution: \begin{align*}
&\frac{(1+\sqrt{2})^2-(1-\sqrt{2})^2}{2(1+\sqrt{2})(1-\sqrt{2})}\\
&=\frac{(1+2\sqrt{2}+2)-(3-2\sqrt{2})}{2(-1)}\\
&=\frac{1+2\sqrt{2}+2-3+2\sqrt{2}}{-2}\\
&=\frac{2\sqrt{2}+2\sqrt{2}}{-2}\\
&=\frac{4\sqrt{2}}{-2}\\
&=-2\sqrt{2}
\end{align*}