2013 - JAMB Mathematics Past Questions and Answers - page 1

\(\begin{array}{c|c}
3 & \text{27 rem 0} \\ \hline
3 & \text{ 9 rem 0} \\ \hline
3 & \text{ 3 rem 0} \\ \hline
3 & \text{ 1 rem 1}\\ \hline
& 0
\end{array}\)

Hence the correct answer is 1000₃

The sum, S of ratio is S = 5 + 3 + 2 = 10.

But highest share = \(\frac{5}{10} \times T\), where T is the total number of apples.

Thus, \(40 = \frac{5}{10} \times T\),

given 40 x 10 = 5T,

\(T = \frac{40 \times 10}{5} = 80\)

Hence the smallest share = \(\frac{2}{10} \times 80\)

= 16 apples

\(\frac{1.25 \times 0.025}{0.05}\)

\( = \frac{125 \times 10^{-2} \times 25 \times 10^{-3}}{5 \times 10^{-2}}\)

= 125 x 5 x 10^-3

= 625 x 0.001

= 0.625

= 0.6 Approx to 1 d.p.

Using \(S.I =\frac{P \times T \times R}{100}\)

\(600 =\frac{3000 \times T \times 8}{100}\)

\(T =\frac{600 \times 100}{3000 \times 8}\)

\(\frac{20}{8}\)

= 2\(\frac{1}{2}\) years

\(\frac{3^{-5n}}{9^{1-n}} \times 27^{n + 1}\)

\(\frac{3^{-5n}}{3^{2(1-n)}} \times 3^{3(n + 1)}\)

\(3^{-5n} \div 3^{2(1-n)} \times 3^{3(n + 1)}\)


\(3^{-5n - 2(1-n) + 3(n + 1)}\)


\(3^{-5n - 2 + 2n + 3n + 3}\)

\(3^{-5n + 5n + 3 - 2}\)

\(3^{1}\)

= 3

log₁₀4^1/3 = 1/3 log₁₀4

= 1/3 x 0.6021

= 0.2007

\(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\)

\(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\)

\(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\)

\(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\)

= 5

S = \(\sqrt{t^2 - 4t + 4}\)

S² = t² - 4t + 4

t² - 4t + 4 - S² = 0

Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting, we have;

Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)

\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)

\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)

\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)

\(t = \frac{2(2 \pm S)}{2}\)

Hence t = 2 + S or t = 2 - S

Let f(x) = x² - x - k
Then by the factor theorem,

(x - 4): f(4) = (4)² - (4) - k = 0

16 - 4 - k = 0

12 - k = 0

k = 12