JAMB Mathematics Past Questions & Answers - Page 212

$\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}$ = $\frac{(3^2)^{\frac{1}{2}} \times (3^3)^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}$

= $\frac{3^{\frac{2}{3}} \times 3^{-\frac{3}{2}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}$

= $\frac{3^{-\frac{5}{6}}}{3^{-\frac{5}{6}}}$

= 1

log$_2$ 4 + log$_2$ 7 - log$_2$ n = 1

= log$_2$ (4 x 7) - log$_2$ n = 1

$\therefore$ log$_2$ 28 - log$_2$ n = 1

= $\frac{28}{n} = 2^1$

$\frac{28}{n}$ = 2

2n = 28

∴ n = 14

$(\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}$

$\frac{(\sqrt{5} - \sqrt{3}) - (\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})}$

= $\frac{\sqrt{5} - \sqrt{3} - \sqrt{5} - \sqrt{3}}{5 - \sqrt{15} + \sqrt{15} - 3}$

= $\frac{-2\sqrt{3}}{2}$

= $- \sqrt{3}$

$\therefore (\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}) \times \frac{1}{\sqrt{3}} = - \sqrt{3} \times \frac{1}{\sqrt{3}}$

= $-1$

Let the no. of years be y

24 - y = $\frac{1}{2}$(45 - y)

45 - y = 2(24 - y)

45 - y = 48 - 2y

2y - y = 48 - 45

∴ y = 3

The exact year = 1984

1984 - 3 = 1981

x - y = 6.......(i)

xy = 187.......(ii)

From equation (i), x(6 + y)

sub. for x in equation (ii) = y(6 + y)

= 187

y² + 6y = 187

y² + 6y - 187 = 0

(y + 17)(y - 11) = 0

y = -17 or y = 11

y cannot be negative, y = 11

Sub. for y in equation(i) = x - 11

= 16

x = 6 + 11

= 17

∴(x, y) = (17, 11)