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Mathematics
WAEC / WASSCE 2005
Indices, Logarithms & Surds
Solution
Simplify the expression \(log_{10}18 - log_{10}2.88+log_{10}16\)
A
31.12
B
3.112
C
2
D
1
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(log_{10}18 - log_{10}2.88+log_{10}16\\
=log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\\
=log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)
Source: Myschool Past Questions Archive (Record #26673)
Mathematics
WAEC / WASSCE 2005
Linear, Simultaneous & Quadratic Equations
Solution
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
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
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\(^2\) - \(\frac{-3}{2}\)x + (-7) = 0 ⇒ 2x\(^2\) + 3x - 14 = 0
Source: Myschool Past Questions Archive (Record #26674)
Mathematics
WAEC / WASSCE 2005
Fractions, Decimals, Percentages & Approximations
Solution
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
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The cost price of the whole mangoes = N5x
The sold amount of the mangoes = 3x * 220 = N6.60x ( 660x converted to naira by dividing by hundred.)
The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)
Source: Myschool Past Questions Archive (Record #26675)
Mathematics
WAEC / WASSCE 2005
Indices, Logarithms & Surds
Solution
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
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Using Pythagoras theorem
\(|PR|^2=|PO|^2+|OR|^2\\
|PR|^2=5^2+12^2\\
|PR|=\sqrt{25+144}=\sqrt{169}=13km\)
Source: Myschool Past Questions Archive (Record #26676)
Mathematics
WAEC / WASSCE 2005
Permutations, Combinations & Probability
Solution
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
\(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
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 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #26677)
Mathematics
WAEC / WASSCE 2005
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, KL//MN, ∠LKP = 30o and ∠NMP = 45o. Find the size of the reflex ∠KPM.
Attached Figure / Diagram
A
285o
B
255o
C
225o
D
210o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
∠KPM = 360o - (30 + 45)o
= 360o - 75o = 285o
Source: Myschool Past Questions Archive (Record #26678)
Mathematics
WAEC / WASSCE 2005
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PR is a diameter, ∠PRQ = (3x-8)o and ∠RPQ = (2y-7)o. s x in terms of y
Attached Figure / Diagram
A
\(x=\frac{75-2y}{3}\)
B
\(x=\frac{105-3y}{2}\)
C
\(x=\frac{105-2y}{3}\)
D
\(x=\frac{75-3y}{2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
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 =>\(\frac{105-2y}{3}=x\)
Source: Myschool Past Questions Archive (Record #26679)
Mathematics
WAEC / WASSCE 2005
Number Bases & Modular Arithmetic
Solution
Convert 101101two to a number in base ten
A
61
B
46
C
45
D
44
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
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
Source: Myschool Past Questions Archive (Record #26680)
Mathematics
WAEC / WASSCE 2005
Indices, Logarithms & Surds
Solution
Solve the equation 2\(^7\) = 8\(^{5-x}\)
A
\(\frac{5}{8}\)
B
\(\frac{8}{3}\)
C
\(\frac{3}{2}\)
D
\(\frac{15}{4}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
2\(^7\) = 8\(^{5-x}\)
2\(^7\) = 2\(^{3(5 - x)}\)
7 = 15 - 3x
7 - 15 = - 3x
-8 = -3x
x = \(\frac{8}{3}\)
Source: Myschool Past Questions Archive (Record #26681)
Mathematics
WAEC / WASSCE 2005
Linear, Simultaneous & Quadratic Equations
Solution
expand (2x-3y)(x-5y)
A
\(2x^2 + 13xy - 15y^2\)
B
\(2x^2 - 13xy - 15y^2\)
C
\(2x^2 + 13xy + 15y^2\)
D
\(2x^2 - 13xy + 15y^2\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2\\
=2x^2 - 13xy + 15y^2\)
Source: Myschool Past Questions Archive (Record #26682)
Mathematics
WAEC / WASSCE 2005
Algebraic Expressions, Polynomials & Factorisation
Solution
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
\(\frac{v}{u}-t\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(v=u+ft \Rightarrow v-u=ft\Rightarrow f=\frac{v-u}{t}\)
Source: Myschool Past Questions Archive (Record #26683)
Mathematics
WAEC / WASSCE 2005
Indices, Logarithms & Surds
Solution
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
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Using Pythagoras' theorem
\(x^2+(x-1)^2 = (\sqrt{13})^2\)
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
x = 3 or -2
Source: Myschool Past Questions Archive (Record #26684)
Mathematics
WAEC / WASSCE 2005
Permutations, Combinations & Probability
Solution
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
\(\frac{11}{20}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
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)\(=frac{1}{4}\times \frac{2}{5}=\frac{1}{10}\)
Source: Myschool Past Questions Archive (Record #26685)
Mathematics
WAEC / WASSCE 2005
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram PQ and MN are straight lines. Find the value of x
Attached Figure / Diagram
A
13o
B
17o
C
28o
D
30o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The straight line MN = 180 = 2(x+30o) + 86o =>
180 = 2x + 60o + 86o => 180o = 2x + 146o => 2x = 180 - 46 = 34o => x = 34/2 = 17o
Source: Myschool Past Questions Archive (Record #26686)
Mathematics
WAEC / WASSCE 2005
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PQRS is a parallelogram and ∠QRT = 30o. Find x
Attached Figure / Diagram
A
95o
B
100o
C
120o
D
150o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Since ∠QRS = 180o - 30o = 150o and ∠SPQ = ∠QRS (theorem opp. angles in a parallelogram are equal) ∴ x = 150o
Source: Myschool Past Questions Archive (Record #26687)