2007 - JAMB Mathematics Past Questions and Answers - page 8

If the line 3y = 4x – 1 is parallel t[ line qy = x + 3

Implies gradient of 3y = 4x – 1

Y = ⁴/₃x - ¹/₃

∴Gradient = ⁴/₃

Gradient of line qy = x + 3

Y = ¹/_qx + ³/_q

∴ Gradient = ¹/_q
⁴/₃ = ¹/_q

4q = 3

Q = ³/₄

x = 30^o (alternate ∠s b/c PQ//Sr)

Y = 50^o (vertical opposite ∠s)

r = 30^o (alternate ∠s b/c PQ//SR)

But = y + q + t = 180^o (∠s on a straight line)

50 + q + 30 = 180

q + 80 = 180

q = 180 -80

q = 100^o

Volume of bowl (\frac{2}{3}\pi r^2\

718\frac{2}{3}=\frac{2}{3}\pi r^2\

\frac{2156}{3}=\frac{2}{3} \times \frac{22}{7} \times r^3

∴r^3 = \frac{2156 \times 3 \times 7}{3 \times 2 \times 22}\

r^3 = 343\

r = \sqrt[3]{343}\

r= 7.0cm)

If the point of intersection is (1, -2), it implies that x = 1 and y = -2 when the two equation are solved simultaneously.

∴ substitute x = 1 and y = -2 in any of the equations

2y - k x + 2 = 0

2(-2) - k(1) + 2 = 0

-4 - k + 2 = 0

-4 + 2 = k

-2 = k

(Tan 30 = \frac{h}{40}\

\frac{1}{\sqrt{3}}=\frac{h}{40}\

h\sqrt{3}=40\

h = \frac{40}{\sqrt{3}}\

h = \frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\

h = \frac{40\sqrt{3}}{3})

Locus of point P equidistant from y - 5 = 0 and y - 3 = 0 is y = 4 i.e y - 4 = 0

(1-k+2) / 2 = - k and -4 + k + 1 = k

3-k = -2k and -3 + k = 2k

K = -3 and k = -3

Each Exterior ∠ = ³⁶⁰/_n

= ³⁶⁰/₈

= 45^o

(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}= \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}\

=\left(\frac{\sqrt{3}\sqrt{3}-1}{\sqrt{3}}\right)\div \left(\frac{\sqrt{3}\sqrt{3}+1}{\sqrt{3}}\right)\

=\frac{(3-1)}{(3+1)}\

=\frac{2}{4}=\frac{1}{2})