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Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify: \((\frac{16}{81})^{\frac{1}{4}}\)
A
8/27
B
1/3
C
4/9
D
2/3
E
-4/3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\((\frac{16}{81})^{\frac{1}{4}}\)
= \(((\frac{2}{3})^{4})^{\frac{1}{4}}\)
= \(\frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #23145)
Mathematics
WAEC / WASSCE 1992
Indices, Logarithms & Surds
Solution
Evaluate \(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)
A
0.2
B
2
C
100
D
409
E
490
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)
= \(\log_{10} (\frac{25 \times 32}{8})\)
= \(\log_{10} 100 \)
= 2
Source: Myschool Past Questions Archive (Record #23146)
Mathematics
WAEC / WASSCE 1992
Linear, Simultaneous & Quadratic Equations
Solution
Construct a quadratic equation whose roots are \(-\frac{1}{2}\) and 2.
A
3x2-3x+2=0
B
3x2+3x-2=0
C
2x2+3x-2=0
D
2x2-3x+2=0
E
2x2-3x-2=0
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If x = \(-\frac{1}{2}\) and 2; then
\(x + \frac{1}{2} = 0\) and \(x - 2 = 0\)
\(\implies (x + \frac{1}{2})(x - 2) = 0\)
\(x^2 - 2x + \frac{1}{2}x - 1 = 0\)
\(x^2 - \frac{3}{2}x - 1 = 0\)
\(2x^2 - 3x - 2 = 0\)
Source: Myschool Past Questions Archive (Record #23148)
Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Solution
Write as a single fraction \(\frac{1}{1 - x} + \frac{2}{1 + x}\)
A
\(\frac{x + 3}{1 - x^2}\)
B
\(\frac{3 - x}{(1 - x)^2}\)
C
\(\frac{3 - x}{1 + x^2}\)
D
\(\frac{3 - x}{(1 + x)^2}\)
E
\(\frac{3 - x}{1 - x^2}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(\frac{1}{1 - x} + \frac{2}{1 + x}\)
= \(\frac{(1 + x) + 2(1 - x)}{(1 - x)(1 + x)}\)
= \(\frac{1 + x + 2 - 2x}{1 - x^2}\)
= \(\frac{3 - x}{1 - x^2}\)
Source: Myschool Past Questions Archive (Record #23149)
Mathematics
WAEC / WASSCE 1992
Linear, Simultaneous & Quadratic Equations
Solution
What must be added to the expression x\(^2\) - 18x to make it a perfect square?
A
3
B
9
C
36
D
72
E
81
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
x\(^2\) - 18x to be a perfect square.
\((\frac{b}{2})^2\) is added to ax\(^2\) + bx + c in order to make it a perfect square.
\(x^2 - 18x + (\frac{-18}{2})^2\)
= \(x^2 - 18x + 81\)
Source: Myschool Past Questions Archive (Record #23150)
Mathematics
WAEC / WASSCE 1992
Linear, Simultaneous & Quadratic Equations
Solution
Solve the equation \(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)
A
-3
B
-2
C
2
D
3
E
4
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)
\(\frac{m}{3} - \frac{m}{4} = \frac{3}{4} - \frac{1}{2}\)
\(\frac{m}{12} = \frac{1}{4}\)
\(4m = 12 \implies m = 3\)
Source: Myschool Past Questions Archive (Record #23151)
Mathematics
WAEC / WASSCE 1992
Mensuration (Areas, Perimeters & Volumes)
Solution
The angle of a sector of a circle of radius 8cm is 240^\circ. This sector is bent to form a cone. Find the radius of the base of the cone.
A
16/3cm
B
15/3cm
C
16/5cm
D
8/3cm
E
16/10cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
L = \theta/360 x 2\piR = 240/360 x 2 x 22/7 x 8/1 ... (1)
This must be equal to the circumference of the circle which is 2\pir = 44R/7 .... (2)
equate (1) and (2)
r = 16/3 = 5\(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #23152)
Mathematics
WAEC / WASSCE 1992
Mensuration (Areas, Perimeters & Volumes)
Solution
The curved surface area of a cylindrical tin is 704cm\(^2\). Calculate the height when the radius is 8cm. [Take \pi = 22/7]
A
3.5cm
B
7cm
C
14cm
D
28cm
E
32cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Curved surface area of a cylindrical tin = \(2\pi rh\)
\(\therefore 2\pi rh = 704cm^2\)
\(2 \times \frac{22}{7} \times 8 \times h = 704\)
\(h = \frac{704 \times 7}{2 \times 22 \times 8}\)
\(h = 14cm\)
Source: Myschool Past Questions Archive (Record #23153)
Mathematics
WAEC / WASSCE 1992
Mensuration (Areas, Perimeters & Volumes)
Solution
The volume of a cone of height 9cm is 1848cm\(^3\). Find its radius. [Take \pi = 22/7]
A
7cm
B
14cm
C
28cm
D
98cm
E
196cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
1/3 \pir\(^2\) x 9 = 1848
3 x \pir2 = 1848 r2 = 1848/3 x 7/22 r = 14
\(\text{Volume of a cone} = \frac{1}{3} \pi r^2 h\)
\(\frac{1}{3} \times \frac{22}{7} \times r^2 \times 9 = 1848\)
\(r^2 = \frac{1848 \times 7}{22 \times 3}\)
\(r^2 = 196 \therefore r = 14cm\)
Source: Myschool Past Questions Archive (Record #23154)
Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Solution
The area of a parallelogram is 513cm\(^2\) and the height is 19cm. Calculate the base.
A
13.5cm
B
25cm
C
27cm
D
54cm
E
108cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of parallelogram = base \(\times\) height.
\(513 = base \times 19 \implies base = \frac{513}{19}\)
= 27 cm
Source: Myschool Past Questions Archive (Record #23155)
Mathematics
WAEC / WASSCE 1992
Mensuration (Areas, Perimeters & Volumes)
Diagram
Solution
The diagram above shows the shaded segment of a circle of radius 7cm. if the area of the triangle OXY is 12\(\frac{1}{4}\)cm\(^2\), calculate the area of the segment
[Take \pi = 22/7]
Attached Figure / Diagram
A
5/12cm2
B
7/12cm2
C
11/6cm2
D
2 1/3cm2
E
6 1/6cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area of \(\Delta OXY\) = \(12\frac{1}{4} cm^2\)
Area of sector OXY = \(\frac{30}{360} \times \frac{22}{7} \times 7 \times 7\)
= \(\frac{77}{6} = 12\frac{5}{6} cm^2\)
Area of the shaded portion = \(12\frac{5}{6} - 12\frac{1}{4}\)
= \(\frac{7}{12} cm^2\)
Source: Myschool Past Questions Archive (Record #23156)
Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above PQT is a tangent to the circle QRS at Q. Angle QTR = 48^\circ and ∠QRT = 95^\circ. Find ∠QSR
Attached Figure / Diagram
A
48o
B
45o
C
37o
D
32o
E
30o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
∠RQT = 180^\circ - (95^\circ + 48^\circ) = 37^\circ
∠OQR = 90^\circ - 37^\circ = 53^\circ
∠QOR = 180^\circ - (53^\circ + 53^\circ) = 74^\circ
QSR = 74^\circ/2 = 37^\circ
Source: Myschool Past Questions Archive (Record #23157)
Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above, O is the center of the circle and ∠POR = 126^\circ. Find ∠PQR
Attached Figure / Diagram
A
234o
B
117o
C
72o
D
63o
E
54o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Reflex < POR = 360^\circ - 126^\circ = 234^\circ
\(\therefore\) < PQR = \(\frac{1}{2} \times 234^\circ\)
= 117^\circ
Source: Myschool Past Questions Archive (Record #23158)
Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above, TR = TS and ∠TRS = 60o. Find the value of x
Attached Figure / Diagram
A
30
B
45
C
60
D
120
E
150
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
∠TRS = ∠RST = 60o
∠PSR + ∠PQR = 180o, 120o + xo = 180o
x = 60o
Source: Myschool Past Questions Archive (Record #23159)
Mathematics
WAEC / WASSCE 1992
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above, O is the center of the circle PQRS and ∠QPS = 36^\circ. Find ∠QOS
Attached Figure / Diagram
A
36o
B
72o
C
108o
D
144o
E
288o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< QOS = 2 < QPS (angle subtended at the centre)
\(\therefore\) < QOS = 2 x 36^\circ = 72^\circ
Source: Myschool Past Questions Archive (Record #23160)