2018 - WAEC Mathematics Past Questions and Answers - page 3

\(2x^2 - x - 6\) = 0

\(2x^2 - 4x + 3x - 6\) = 0

2x(x - 2) + 3(x - 2) = 0

(2x + 3) (x - 2) = 0

Either; 2x + 3 = 0 or x - 2 = 0

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

\((x + 2)^2\) - \((2x + 1)^2\)

= \((x^2 + 4x + 4) - (4x^2 + 4x + 1)\)

= \(x^2 \) + 4x + 4 - 4 \(x^2 \) - 4x - 1

= -3 \(x^2 \) + 3

= 3 - 3 \(x^2 \)

= 3(1 - \(x^2 \))

= 3(1 + x)(1 - x)

2 x 3, 4 x 6, 8 x 9, 16 x 12,...

2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,.... 2\(^n\) x 3n

3x \(\equiv\) 4(mod 5)

In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x.

Thus; 3x = 4 + 5 = 9

x = \(\frac{9}{3}\)

x = 3

\(Q_1 = \frac{1}{4}\)Nth

\(\frac{1}{4} \times 11^{th}\) no.

= 2\(\frac{3}{4}th\) (\(\cong\) 4)

\(Q_3 = \frac{3}{4}\)  Nth

= \(\frac{3}{4}\) x 11th no.

= 8\(\frac{1}{4}\) no. (\(\cong\) 11) (\(\cong\) 11)

Hence, interquartile range

= \(Q_3 = Q_1\)

= 11 - 4

= 7

\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\)

\(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\)

But,

x = \(\frac{2}{5}T_1\), y = \(\frac{3}{5}T_1\)

y = \(\frac{3}{7}T_2\), z = \(\frac{4}{7}T_2\)

Using y = y

\(\frac{3}{5}T_1\) = x = \(\frac{3}{7}T_2\)

 \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) = \(\frac{15}{21}\)

\(T_1 = 15\) and \(T_2 = 21\)

Thus , x = \(\frac{2}{5}\) x 15 = 6

y = \(\frac{3}{5}\) x 15 = 9

y = \(\frac{3}{7}\) x 21 = 9

z = \(\frac{4}{7}\) x 21 = 12

Hence; x : z = 6 : 12

= 1 : 2

0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\)

= \(\frac{612}{1000}\)

= \(\frac{153}{250}\)

From the diagram,

tan 30\(^o\) = \(\frac{h}{27}\)

h = 27 tan 30\(^o\)

= 27 x \(\frac{1}{\sqrt{3}}\)

= \(\frac{27}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{27 \sqrt{3}}{3}\)

= 9\(\sqrt{3m}\)

\(\frac{|PN|}{|PQ|}\) = \(\frac{|PM|}{|PR|}\)

From the truth table above, p \(\to\) q  would be false if p is true and q is false