2005 - WAEC Mathematics Past Questions and Answers - page 1
1
Correct 0.04945 to two significant figures
A
0.040
B
0.049
C
0.050
D
0.49
2
Simplify \(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}\)
A
\(\frac{1}{6}(5\sqrt{3}-3\sqrt{2}\)
B
\(\frac{1}{6}(15\sqrt{3}-6\sqrt{2}\)
C
\(\frac{1}{6}(3\sqrt{2}-\sqrt{3}\)
D
\(\frac{1}{6}(10\sqrt{3}-9\sqrt{2}\)
correct option: d
\(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\
=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\
\frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\
\frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\
\frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)
Users' Answers & Comments=\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})\)
3
Evaluate, correct to the nearest whole number \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\)
A
33
B
8
C
7
D
o
correct option: c
\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\
=\frac{15}{2}-\left(\frac{5}{2}+\frac{3}{1}\right)\times\frac{2}{33}\
=\frac{15}{2}-\left(\frac{5+6}{2}\right)\times \frac{2}{33}=\frac{15}{2}-\frac{11}{2}\times \frac{2}{33}=\frac{15}{2}-\frac{1}{3}\
=\frac{45-2}{6}=\frac{43}{6}\)
Users' Answers & Comments=\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}\)
4
Simplify the expression \(log_{10}18 - log_{10}2.88+log_{10}16\)
A
31.12
B
3.112
C
2
D
1
correct option: c
\(log_{10}18 - log_{10}2.88+log_{10}16\
=log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\
=log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)
Users' Answers & Comments=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\)
5
Find the equation whose roots are 2 and \(-3\frac{1}{2}\)
A
2x2 + 3x + 14 = 0
B
2x2 + 5x + 7 = 0
C
2x2 + 5x - 7 = 0
D
2x2 + 3x - 14 = 0
correct option: d
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 \(=2 \times -3\frac{1}{2}=-7\
x^2-\left(\frac{-3}{2}\right)x+(-7)=0\Rightarrow 2x^2 + 3x - 14 = 0\)
Users' Answers & CommentsSum of roots \(2+-3\frac{1}{2} = -1\frac{1}{2}=-\frac{1}{2}\)
Product of roots \(=2 \times -3\frac{1}{2}=-7\
x^2-\left(\frac{-3}{2}\right)x+(-7)=0\Rightarrow 2x^2 + 3x - 14 = 0\)
6
A man bought 220 mangoes at N5x. He sold each for 3x kobo and made a gain of N8. Find the value of x
A
2
B
5
C
6
D
10
correct option: b
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 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)
Users' Answers & CommentsThe sold amount of the mangoes = 3x * 220 = N6.60x
The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)
7
From a point P, R is 5km due west and 12km due south. Find the distance between P and R
A
5km
B
12km
C
13km
D
17km
correct option: c
Using Pythagoras theorem
\(|PR|^2=|PO|^2+|OR|^2\
|PR|^2=5^2+12^2\
|PR|=\sqrt{25+144}=\sqrt{169}=13km\)
Users' Answers & Comments\(|PR|^2=|PO|^2+|OR|^2\
|PR|^2=5^2+12^2\
|PR|=\sqrt{25+144}=\sqrt{169}=13km\)
8
A fair die is tossed once, what is the probability of obtaining neither 5 or 2
A
\(\frac{5}{6}\)
B
\(\frac{2}{3}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{6}\)
correct option: b
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 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)
Users' Answers & CommentsProbability 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 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)
9
In the diagram, KL//MN, ∠LKP = 30o and ∠NMP = 45o. Find the size of the reflex ∠KPM.
A
285o
B
255o
C
225o
D
210o
10
In the diagram, PR is a diameter, ∠PRQ = (3x-8)o and ∠RPQ = (2y-7)o. s x in terms of y
A
\(x=\frac{75-2y}{3}\)
B
\(x=\frac{105-3y}{2}\)
C
\(x=\frac{105-2y}{3}\)
D
\(x=\frac{75-3y}{2}\)
correct option: c
180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle)
180 = ∠RPQ + ∠PRQ + 90 => 180o = (3x-8)o+(2y-7)o+90o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)
Users' Answers & Comments180 = ∠RPQ + ∠PRQ + 90 => 180o = (3x-8)o+(2y-7)o+90o; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)