2018 - WAEC Mathematics Past Questions and Answers - page 3
Solved the equation \(2x^2 - x - 6\) = 0
\(2x^2 - x - 6\) = 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
Factorise completely the expression
\((x + 2)^2\) - \((2x + 1)^2\)
\((x + 2)^2\) - \((2x + 1)^2\)
= \((x^2 + 4x + 4) - (4x^2 + 4x + 1)\)
= \(x^2 \) + 4x + 4 - 4 \(x^2 \) - 4x - 1
= -3 \(x^2 \) + 3
= 3 - 3 \(x^2 \)
= 3(1 - \(x^2 \))
= 3(1 + x)(1 - x)
Find the \(n^{th}\) term of the sequence 2 x 3, 4 x 6, 8 x 9, 16 x 12...
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
If 3x\(^o\) 4(mod 5), find the least value of x
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
Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16
\(Q_1 = \frac{1}{4}\)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
If x : y = \(\frac{1}{4} : \frac{3}{8}\) and y : z = \(\frac{1}{3} : \frac{4}{9}\), find x : z
\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\)
\(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\)
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
Expression 0.612 in the form \(\frac{x}{y}\), where x and y are integers and y \(\neq\) 0
0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\)
= \(\frac{612}{1000}\)
= \(\frac{153}{250}\)
The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of the tree.
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\(\sqrt{3m}\)
In the diagram, which of the following ratios is equal to \(\frac{|PN|}{|PQ|}\)?
If P and Q are two statements, under what condition would p|q be false?
From the truth table above, p \(\to\) q would be false if p is true and q is false