JAMB Mathematics Past Questions & Answers - Page 14

Mean = $\frac{\sum fx}{\sum f} = \frac {820.5}{107}$ = 7.7 (1 d.p)

$lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}$ = $lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}$

$lim_{x\to2} \frac {x + 6}{x}$

$\frac {2 + 6}{2} = \frac {8}{2}$ = 4

Amount (A) = Principal (P) + Interest (I) i.e. A = P + I

1 = $\frac {PTR}{100}$

A = 3P; T = 10 years (Given)

Since A = P + I

$\implies 3P = P + \frac {P\times 10 \times R}{100}$

$\implies 3P = P + \frac {10PR}{100}$

$\implies 3P = P + \frac {PR}{10}$

$\implies 3P - P = \frac {PR}{10}$

$\implies 2P = \frac {PR}{10}$

$\implies \frac {2P}{1} = \frac {PR}{10}$

$\implies$ PR = 20P

$\therefore$ R = 20%

Since the rate stays the same

A = 5P; R = 20%;T =?; A  =P + I

$\implies 5P = P + \frac {p \times T \times 20}{100}$

$\implies 5P - P = \frac {2PT}{10}$

$\implies 4P = \frac {2PT}{10}$

$\implies 2P = \frac {PT}{10}$

$\implies$ PT = 20P

$\therefore$ T = 20 years

Let the given points be:

P(-3, -14) = (x$_1$, y$_1$)

Q(t, -5) = (x$_2$, y$_2$)

PQ = 9 units (given)

Using the distance formula,

d = √ [ $(x_2 - x_1)^2 + (y_2 - y_1)^2$]

PQ = √ [ $(t - (-3))^2 + (-5 + 14)^2$]

$\implies$ √ [ $(t + 3)^2 + 81$] = 9

Squaring on both sides,

⇒ (t + 3)$^2$ + 81 = 81

⇒ (t + 3)$^2$  = 0

⇒ t + 3 = 0

∴ t = -3

$\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3$

Rationalize

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

= $\frac {(3 - \sqrt 3)}{(2 + \sqrt 3)} \frac {(2 - \sqrt 3)}{(2 - \sqrt 3)}$

= $\frac {6 - 3 \sqrt 3 - 2 \sqrt 3 + (\sqrt 3)^2}{4 - 2 \sqrt 3 + 2 \sqrt 3 - (\sqrt 3)^2}$

= $\frac {6 - 5 \sqrt 3 + 3}{4 - 3}$

= $\frac {9 - 5 \sqrt 3}{1} = 9 - 5 \sqrt 3$

= 9 + (-5) $\sqrt 3$

$\therefore a = 9, b = - 5$