JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS[III]

(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: $$\begin{align*} &\frac{(1+\sqrt{2})^2-(1-\sqrt{2})^2}{2(1+\sqrt{2})(1-\sqrt{2})}\\ &=\frac{(1+2\sqrt{2}+2)-(3-2\sqrt{2})}{2(-1)}\\ &=\frac{1+2\sqrt{2}+2-3+2\sqrt{2}}{-2}\\ &=\frac{2\sqrt{2}+2\sqrt{2}}{-2}\\ &=\frac{4\sqrt{2}}{-2}\\ &=-2\sqrt{2} \end{align*}$$