2005 - WAEC Mathematics Past Questions and Answers - page 2
11
Convert 101101two to a number in base ten
A
61
B
46
C
45
D
44
correct option: c
1011012 = 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = 32 + 0 + 8 + 4 + 0 + 1 = 4510
Users' Answers & Comments12
Solve the equation \(2^7 = 8^{5-x}\)
A
\(\frac{5}{8}\)
B
\(\frac{8}{3}\)
C
\(\frac{3}{2}\)
D
\(\frac{15}{4}\)
correct option: b
\(2^7 = 2^{3(5-x)}\Rightarrow 7 = 3^{5-x} \Rightarrow 7 15 - 3x\
\Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3}\)
Users' Answers & Comments\Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3}\)
13
expand (2x-3y)(x-5y)
A
\(2x^2 + 13xy - 15y^2\)
B
\(2x^2 - 13xy - 15y^2\)
C
\(2x^2 + 13xy + 15y^2\)
D
\(2x^2 - 13xy + 15y^2\)
correct option: d
\((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2\
=2x^2 - 13xy + 15y^2\)
Users' Answers & Comments=2x^2 - 13xy + 15y^2\)
14
Make f the subject of the relation \(v = u + ft\)
A
\(\frac{v-u}{t}\)
B
\(\frac{u-v}{t}\)
C
\(t(v+u\)
D
\(\frac{v}{u}-t\)
15
The lengths of the adjacent sides of a right - angled triangle are xcm, (x-1)cm. If the length of the hypotenuse is \(\sqrt{13}cm\), find the value of x
A
2
B
3
C
4
D
5
correct option: b
\(x^2+(x-1)^2 = (\sqrt{13})^2 \Rightarrow x^2 + x^2 - 2x + 1 = 13\
2x^2 - 2x - 12 = 0\) dividing through by 2
\(x^2 - x- 6 = 0; (x-3)(x-2) = 0 \Rightarrow x = 3 or -2 \)
Users' Answers & Comments2x^2 - 2x - 12 = 0\) dividing through by 2
\(x^2 - x- 6 = 0; (x-3)(x-2) = 0 \Rightarrow x = 3 or -2 \)
16
The probability that John and James pass an examination are 3/4 and 3/5 respectively, find the probability of both boys failing the examination
A
\(\frac{1}{10}\)
B
\(\frac{3}{10}\)
C
\(\frac{9}{20}\)
D
\(\frac{11}{20}\)
correct option: a
Prob.(John pass)\(=frac{3}{4}\) prob.(John fail) \(=1-frac{3}{4}=\frac{1}{4}\)
Prob.(James pass) \(=frac{3}{5}\) Prob.(James fail) \(=1-frac{3}{5}=\frac{2}{5}\)
Prob(both boys fail)\(=frac{1}{4}\times \frac{2}{5}=\frac{1}{10}\)
17
In the diagram PQ and MN are straight lines. Find the value of x
A
13o
B
17o
C
28o
D
30o
correct option: b
The straight line MN = 180 = 2(x+30o) + 86o =>
180 = 2x + 60o + 86o => 180o = 2x + 146o => 2x = 180 - 46 = 34o => x = 34/2 = 17o
Users' Answers & Comments180 = 2x + 60o + 86o => 180o = 2x + 146o => 2x = 180 - 46 = 34o => x = 34/2 = 17o
18
In the diagram, PQRS is a parallelogram and ∠QRT = 30o. Find x
A
95o
B
100o
C
120o
D
150o
correct option: d
Since ∠QRS = 180o - 30o = 150o and ∠SPQ = ∠QRS (theorem opp. angles in a parallelogram are equal) ∴ x = 150o
Users' Answers & Comments19
From a point R, 300m north of P, man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degreeee
A
026o
B
045o
C
210o
D
240o
correct option: d
\(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}\
\theta = cos^{-1}(0.5000)=60^{\circ}\)
The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)
Users' Answers & Comments\theta = cos^{-1}(0.5000)=60^{\circ}\)
The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)
20
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)]
A
\(\frac{\sqrt{2}}{7}cm\)
B
\(3\frac{1}{2}cm\)
C
7cm
D
\(7\sqrt{2}cm\)
correct option: d
\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\
\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)
Users' Answers & Comments\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)