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Mathematics
WAEC / WASSCE 2002
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram O is the centre of the circle. Which of the following is/are not true? I. a = b II. b + c = 180o III. a + b = c
Attached Figure / Diagram
A
I and II only
B
II and III only
C
II only
D
III only
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
i.a and b are 90( angles in same segment are equal)
ii. angle b+c=180 (opposite angles in a cyclic quadrilateral)
iii. a=90 and b=90 .:a+b=180 which is not equal to c
Source: Myschool Past Questions Archive (Record #26550)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
Simplify \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\)
A
1
B
2
C
4
D
16
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\\=\frac{2^{\frac{1}{2}}\times 2^{\frac{3}{2}}}{2^2}\\=2^{\frac{1}{2}}+2^{\frac{3}{2}}-2=2^0 = 1\)
Source: Myschool Past Questions Archive (Record #26551)
Mathematics
WAEC / WASSCE 2002
Plane Geometry, Polygons & Circle Theorems
Diagram
In the diagram, |SR| = |RQ| and ∠PRQ = 58o ∠VQT = 19o, PQT, SQV and PSR are straight lines. Find ∠QPS
Attached Figure / Diagram
A
42o
B
39o
C
38o
D
30o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #26552)
Mathematics
WAEC / WASSCE 2002
Trigonometry, Elevation & Bearings
Solution
The angle of depression of a point Q from a vertical tower PR, 30m high, is 40^\circ. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, |PQ|.
A
35.75m
B
25.00m
C
22.98m
D
19.28
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\tan 40^\circ = \frac{30}{|PQ|}\)
\(|PQ| = \frac{30}{\tan 40^\circ} = \frac{30}{0.839}\)
= 35.75m
Source: Myschool Past Questions Archive (Record #26553)
Mathematics
WAEC / WASSCE 2002
Mensuration (Areas, Perimeters & Volumes)
Solution
The volume of a cylinder of radius 14cm is 210cm3. What is the curved surface area of the cylinder?
A
15cm2
B
30cm2
C
616cm2
D
1262cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(V = \pi r^2 h\)
210 = \(\frac{22}{7} \times 14^2 \times\) h
h = \(\frac{210}{22 \times 28}\)
Curved surface area = 2\(\pi\) rh
= 2 \(\times \frac{22}{7} \times 14 \times \frac{210}{22 \times 26} = 30cm^2\)
Source: Myschool Past Questions Archive (Record #26554)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)
A
\(7\sqrt{3}\)
B
\(10\sqrt{3}\)
C
\(14\sqrt{3}\)
D
\(18\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)
= \(3(\sqrt{4 \times 3}) + 10\sqrt{3} - (\frac{6}{\sqrt{3}})(\frac{\sqrt{3}}{\sqrt{3}})\)
= \(6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\)
= \(14\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #26555)
Mathematics
WAEC / WASSCE 2002
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize m(2a-b)-2n(b-2a)
A
(2a-b)(2n-m)
B
(2a+b)(m-2n)
C
(2a-b)(m+2n)
D
(2a-b)(m-2n)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
m(2a - b) - 2n(b - 2a)
= m(2a - b) - (-2n)(2a - b)
= m(2a - b) + 2n(2a - b)
= (m + 2n)(2a - b)
Source: Myschool Past Questions Archive (Record #26556)
Mathematics
WAEC / WASSCE 2002
Fractions, Decimals, Percentages & Approximations
Solution
If q oranges are sold for t Naira, how many oranges can be bought for p naira?
A
\(\frac{p}{2}t\)
B
\(\frac{qt}{p}\)
C
\(\frac{q}{pt}\)
D
\(\frac{pq}{t}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
q oranges = t naira
1 naira = \(\frac{q}{t}\)
p naira = \(p(\frac{q}{t})\)
= \(\frac{pq}{t}\) oranges
Source: Myschool Past Questions Archive (Record #26557)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Diagram
Solution
In the diagram, QRT is a straight line. If angle PTR = 90^\circ, angle PRT = 60^\circ, angle PQR = 30^\circ and |PQ| = \(6\sqrt{3}cm\), calculate |RT|
Attached Figure / Diagram
A
0.3cm
B
\(\frac{\sqrt{3}}{2}cm\)
C
3cm
D
\(3\sqrt{3}cm\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In \(\Delta\) QPT,
\(\frac{PT}{6\sqrt{3}} = \sin 30^\circ\)
PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\)
In \(\Delta\) RPT,
\(\frac{PT}{RT} = \tan 60^\circ\)
\(\frac{3\sqrt{3}}{RT} = \tan 60^\circ\)
\(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)
Source: Myschool Past Questions Archive (Record #26558)
Mathematics
WAEC / WASSCE 2002
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, calculate the value of x
Attached Figure / Diagram
A
35o
B
80o
C
100o
D
115o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
From the diagram above.
115º = A (corresponding angles)
t + 115º = 180^\circ ( sum of
Source: Myschool Past Questions Archive (Record #26559)
Mathematics
WAEC / WASSCE 2002
Plane Geometry, Polygons & Circle Theorems
Diagram
In the diagram, \(\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}\) are two altitudes of ∆PQR. Which of the following is equal to ∠RQT?
Attached Figure / Diagram
A
∠PQT
B
∠SRP
C
∠PQR
D
∠SPR
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #26560)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Diagram
Solution
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30^\circ and angle PQR = 45^\circ. Calculate in meters |QS|
Attached Figure / Diagram
A
\(10(1+\sqrt{3})\)
B
\(20\sqrt{3}\)
C
\(10\sqrt{3}\)
D
\((10+\sqrt{3})\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In \(\Delta\) PQR,
\(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\)
= 10m
In \(\Delta\) PRS,
\(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\)
= \(\frac{10}{\frac{1}{\sqrt{3}}\)
= \(10\sqrt{3}\)
PS = \(10 + 10\sqrt{3}\)
= \(10(1 + \sqrt{3}) cm\)
Source: Myschool Past Questions Archive (Record #26561)
Mathematics
WAEC / WASSCE 2002
Fractions, Decimals, Percentages & Approximations
Solution
Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4
A
2.0
B
2.5
C
6.6
D
6.7
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Measured value = 0.375
Approximation = 0.4
Error = 0.4 - 0.375 = 0.025
Error% = \(\frac{0.025}{0.375}\) x 100% = 6.67% = 6.7%
Source: Myschool Past Questions Archive (Record #26562)
Mathematics
WAEC / WASSCE 2002
Plane Geometry, Polygons & Circle Theorems
Solution
To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?
A
60km/hr
B
80km/hr
C
90km/hr
D
108km/hr
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(speed = \frac{distance}{time}\\
72 = \frac{60}{time}\\
t = \frac{60}{72} = \frac{5}{6}hr\)
time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\)
Time required for the journey
\(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\
speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)
Source: Myschool Past Questions Archive (Record #26563)
Mathematics
WAEC / WASSCE 2002
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} \). Find the value of x
Attached Figure / Diagram
A
92o
B
68o
C
67o
D
23o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Sum of the angles in a triangle = 180^\circ
3x - 180^\circ + 92^\circ + x = 180^\circ
4x - 88^\circ = 180^\circ
4x = 268^\circ
x = 67^\circ
Source: Myschool Past Questions Archive (Record #26564)