2008 - WAEC Mathematics Past Questions and Answers - page 2

4y - 9 > y + 3x; 4y - y > 3x + 9

3y > 3(x + 3); y = > \(\frac{3(x + 3)}{3}\)

y > x + 3; y - 3 > x

y is greater than x

p = \(\frac{3}{5} \sqrt{\frac{q}{r}}; \frac{5p}{3} = \sqrt{\frac{q}{r}}\)

= (\(\frac{5}{3}p\))²

= \(\frac{q}{r}\)

= \(\frac{25p^2}{9} = \frac{q}{r}\)

q = \(\frac{25}{9} p^2 r\)

\(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) = \(\frac{3x^2 - 8x + 3x - 12}{(2x)^2 - 3^2}\)

= \(\frac{32(x - 4) + 3(x - 4)}{(2x - 3)(2x + 3)} - \frac{(x - 4) + (2x + 3)}{(2x - 3) (2x + 3)}\)

= \(\frac{x - 4}{2x - 3}\)

Given q² = 25 - r².....(1)

from pythagora's theorem

P² = q² + r².......(2)

put (1) into (2)

p² = 25 - p² + x²

p² = 25

p = \(\sqrt{25}\)

= 5

Volume of a cube = (side)³

= (side)³

(side)³ = 343

side = 3\(\sqrt{343}\)

side = 7cm

Sum of the angles in a quadrilateral = 360^o

(x + 10^o) + 2y + 90^o + 100 - y = 360^o

x + y + 200 = 360^o

y = 360 - 200 - x

y = 160 - x

obtuse angle \(\to\) 90^o < x < 180^o

tan x = 1; x = tan^-1(10 = 45^o

sin x + cos x

= sin 45^o + cos 45^o

= \(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\)

= \(\frac{\sqrt{2} + \sqrt{2}}{2}\)

= \(\frac{2\sqrt{2}}{2}\)

= \(\sqrt{2}\)

cos(x + 25^o) = sin 45^o

using cos \(\theta\) = sin(90 - \(\theta\))

cos(x + 25^o) = cos(90 - 45)

cos(x + 25^o) = cos 45

x + 25 = 45

x = 45 - 25

x = 20^o

2ⁿ = 128

2ⁿ = 2⁷

n = 7

(2^{n - 1})(5^{n - 2}) = (2^{n - 2}.2)(5^{n - 2}) put n = 7

(2^{n - 1})(5^{n - 2}) = 2(2^{n - 2} x 5^{n - 2})

= 2(2 x 5)^{n - 2}

= 2(10^{n - 2}) put n = 7

(2^n-1)(5^n-2) = 2(10⁵)