JAMB MATHEMATICS PAST QUESTIONS AND SOLUTIONS

JAMB Past Questions And Solutions

(1) factorize completely $ac-2bc-a^2+4b^2$
A. $(a-2b)(c+a-2b)$
B. $(a-2b)(c-a-2b)$****
C. $(a-2b)(c+a+2b)$
D. $(a-2b)(c-a+2b)$

Solution:
$$\begin{equation*} ac-2bc-a^2+4b^2 \end{equation*}$$
\begin{equation*}
= ac-2bc-(a^2-4b^2)

\end{equation*}
expand $a^2-4b^2=(a+2b)(a-2b)$
$$\begin{equation*} =ac-2bc-(a+2b)(a-2b) \end{equation*}$$
$$\begin{equation*} =c(a-2b)-(a+2b)(a-2b) \end{equation*}$$
$$\begin{equation*} =(a-2b)[c-(a+2b)] \end{equation*}$$
$$\begin{equation*} =(a-2b)(c-a-2b) \end{equation*}$$

(2) The length $L$ of a simple pendulum varies directly as the square of its period $T$. If a pendulum with period $4sec$ is $64cm$ long, find the length of a pendulum whose period is $9sec$.
A.$36cm$
B.$96cm$
C.$144cm$
D.$324cm$***

Solution: The relation can be written as $L\alpha{T^2}$.
which is $L=KT^2$  where K=constant of proportionality.
Now we use the first set of given data to obtain our K. i.e $T=4$ and $L=64$ so $L=KT^2\Rightarrow{64}=K(4^2)$
$=16K=64$ and hence $K=4$. We use the value K to obtain the new length at $T=9sec$. $L=KT^2\Rightarrow{L}=4(9^2)=324cm$

(3) y is inversely proportional to $x$ and $y=4$ when $x=\frac{1}{2}$. find $x$ when $y=10$
A.$\frac{1}{10}$
B.$\frac{1}{5}$***
C.$2$
D.$10$

Solution: $y\alpha\frac{1}{x}$ this implies $y=\frac{k}{x}$, where k=constant. and $k=yx\Rightarrow{K}=4(\frac{1}{2})=2$.
To find $x$ when $y=10$,make x subject of formula,so that $x=\frac{k}{y}$ and since $k=2$ and $y=10$ then $x=\frac{2}{10}=\frac{1}{5}$

(4) Find the remainder when $3x^3+5x^2-11x+2$ is divided by $x+3$
A.4
B.1
C.-1***
D.-4

Solution: Use polynomial long division
$$\begin{equation} \frac{3x^3+5x^2-11x+2}{x+3} \end{equation}$$
$$\begin{equation} \sqrt[x+3]{3x^3+5x^2-11x+2}=3x^2-4x+1 \end{equation}$$
and the remainder will be $-1$

(5) Given that the first and fourth terms of a G.P are 6 and 162 respectively,find the sum of the first 3 terms of the progression.
A.8
B.27
C.48
D.78***

Solution: the first a=6, and $4th=ar^3=162$ subst. a=6 into the 4th term. $ar^3=162$ and a=6 then $6r^3=162$ and $r^3=\frac{162}{6}=27$ sum of first 3 terms are $S=\frac{a(r^n-1)}{r-1}$
$S=\frac{6(3^3-1}{3-1}=\frac{6(26)}{2}=78$

(6) find the sum to infinity of the series $\frac{1}{2},\frac{1}{6},\frac{1}{18}$
A.1
B.$\frac{3}{4}$***
C.$\frac{2}{3}$
D.$\frac{1}{3}$

Solution: the first term $a=\frac{1}{2}$ and common ration $r=\frac{1}{3}$
$$\begin{equation} S_{\infty}=\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{3}} \end{equation}$$
$$\begin{equation} =\frac{1}{2}\times\frac{3}{2}=\frac{3}{4} \end{equation}$$

(7) The nth terms of two sequences are $Q_n=3.2^{n-2}$ and $U_m=3.2^{2m-3}$
Find the product of $Q_2$ and $U_2$
A.3
B.6***
C.12
D.18

Solution: for $Q_2$ it means $n=2$ and $Q_2=3.2^{2-2}=3$ and $U_2=3.2^{2m-3}=3.2^{2(2)-3}=6$ then $Q_2\times{U_2}=3\times6=18$

(8) Find the values of $x$ and $y$ respectively if $3x-5y+5=0$ and $4x-7y+8=0$
A.-4,-5
B.-5,-4
C.5,4***
D.4,5

Solution: rearrange the equation
$$\begin{align*} 3x-5y=-5\\......... (i) 4x-7y=-8.......... (ii) \end{align*}$$
you can use any method of simultaneous equation such as elimination and substitution method to solve the linear system. but i am going to use the substitution method. from (i) let $3x-5y=-5$ so that $x=\frac{-5+5y}{3}$ subst. x into (ii)
$$\begin{equation} 4(\frac{-5+5y}{3})-7y=-8 \end{equation}$$
$$\begin{equation} =\frac{-20+20y-21y}{3}=-8 \end{equation}$$
$$\begin{equation} \frac{-20-y}{3}=-8\Rightarrow{-20-y}=-24 \end{equation}$$
$$\begin{equation} y=4 \end{equation}$$
now substitute y=4 into (i)
$$\begin{align} 3x-5(4)=-5\\ =3x-20=-5\\ =3x=15\\ x=5 \end{align}$$
Hence x=5 and y=4

(9) Simplify
$$\begin{equation} \frac{(\sqrt{98}-\sqrt{50})}{\sqrt{32}} \end{equation}$$
A.$\frac{1}{2}$***
B.$\frac{1}{4}$
C.1
D.3

Solution:
$$\begin{align*} \frac{(\sqrt{98}-\sqrt{50})}{\sqrt{32}}\\ =\frac{\sqrt{2\times{49}}-\sqrt{2\times{25}}}{\sqrt{2\times{16}}}\\ =\frac{7\sqrt{2}-5\sqrt{2}}{4\sqrt{2}}\\ =\frac{\sqrt{2}(7-5)}{4\sqrt{2}}\\ =\frac{2}{4}\\ =\frac{1}{2} \end{align*}$$

(10) Find the mean of the given set of data
$$\begin{align*} 7,-3,4,-2,5,-9,\\ 4,8,-6,12 \end{align*}$$
A.1
B.2***
C.3
D.4

Solution: $$\begin{align*}\frac{7+(-3)+4+(-2)+5+(-9)+ 4+8+(-6)+12}{10}=\frac{20}{2}=2 \end{align*}$$