2005 - WAEC Mathematics Past Questions and Answers - page 1

\(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\
=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\
\frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\
\frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\
\frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)

\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\
=\frac{15}{2}-\left(\frac{5}{2}+\frac{3}{1}\right)\times\frac{2}{33}\
=\frac{15}{2}-\left(\frac{5+6}{2}\right)\times \frac{2}{33}=\frac{15}{2}-\frac{11}{2}\times \frac{2}{33}=\frac{15}{2}-\frac{1}{3}\
=\frac{45-2}{6}=\frac{43}{6}\)

\(log_{10}18 - log_{10}2.88+log_{10}16\
=log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\
=log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)

x² (sum of roots)x + (product of roots) = 0
Sum of roots \(2+-3\frac{1}{2} = -1\frac{1}{2}=-\frac{1}{2}\)
Product of roots \(=2 \times -3\frac{1}{2}=-7\
x^2-\left(\frac{-3}{2}\right)x+(-7)=0\Rightarrow 2x^2 + 3x - 14 = 0\)

The cost price of the whole mangoes = N5x
The sold amount of the mangoes = 3x * 220 = N6.60x
The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)

Using Pythagoras theorem
\(|PR|^2=|PO|^2+|OR|^2\
|PR|^2=5^2+12^2\
|PR|=\sqrt{25+144}=\sqrt{169}=13km\)

Probability of obtaining a 5 is P(5)=\(\frac{1}{6}\)
Probability of obtaining a 2 is P(2)=\(\frac{1}{6}\)
Probability of obatining either 2 or 5 = P(2∪5) = \(\frac{2}{6}\)
Probability of obtaining neither 5 or 2 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)

∠KPM = 360^o - (30 + 45)^o
= 360^o - 75^o = 285^o

180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle)
180 = ∠RPQ + ∠PRQ + 90 => 180^o = (3x-8)^o+(2y-7)^o+90^o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)