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Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Solution
Find the equation whose roots are \(\frac{3}{4}\) and -4
A
4x2 - 13x + 12 = 0
B
4x2 - 13x - 12 = 0
C
4x2 + 13x - 12 = 0
D
4x2 + 13x + 12 = 0
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let x = \(\frac{3}{4}\) or x = -4
i.e. 4x = 3 or x = -4
(4x - 3)(x + 4) = 0
therefore, 4x2 + 13x - 12 = 0
Source: Myschool Past Questions Archive (Record #45707)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Solution
Adding 42 to a given positive number gives the same result as squaring the number. Find the number
A
14
B
13
C
7
D
6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let the given positive number be x
Then x + 42 = \(x^2\)
0 = \(x^2\) - x - 42
or \(x^2\) - x - 42 = 0
\(x^2\) - 7x + 6x - 42 = 0
x(x - 7) + 6(x - 7) = 0
= (x + 6)(x - 7) = 0
x = -6 or x = 7
Hence, x = 7
Source: Myschool Past Questions Archive (Record #45711)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Solution
Ada draws the graph of y = x2 - x - 2 and y = 2x - 1 on the same axes. Which of these equations is she solving?
A
x2 - x - 3 = 0
B
x2 - 3x - 1 = 0
C
x2 - 3x - 3 = 0
D
x2 + 3x - 1 = 0
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Given; y = x2 - x - 2, y = 2x - 1
Using y = y, gives
x2 - x - 2 = 2x - 1
x2 - 3x - 2 + 1 = 0
therefore, x2 - 3x - 1 = 0
Source: Myschool Past Questions Archive (Record #45712)
Mathematics
WAEC / WASSCE 2015
Mensuration (Areas, Perimeters & Volumes)
Solution
The volume of a cone of height 3cm is 38\(\frac{1}{2}\)cm3. Find the radius of its base. [Take \(\pi = \frac{22}{7}\)]
A
3.0cm
B
3.5cm
C
4.0cm
D
4.5cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Using V = \(\frac{3}{1} \pi r^2h\),
so, 38\(\frac{1}{2} = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 3\)
\(\frac{77}{2} = \frac{22}{7} \times r^2\)
r2 = \(\frac{77 \times 7}{2 \times 22}\)
r2 = \(\frac{49}{4}\)
Hence, r = \(\sqrt{\frac{49}{4}}\)
= 3\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #45724)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Solution
The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take \(\pi = \frac{22}{7}\)]
A
14cm
B
7m
C
3\(\frac{1}{2}\)m
D
1\(\frac{3}{4}\)m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Volume of rectangular tank = L x B x H
= 2 x 7 x 11
= 154cm3
volume of cylindrical tank = \(\pi r^2h\)
154 = \(\frac{22}{7} \times r^2 \times 4\)
r2 = \(\frac{154 \times 7}{22 \times 4}\)
= \(\frac{49}{4}\)
r = \(\sqrt{\frac{49}{4}} = \frac{7}{2}\)
= 3\(\frac{1}{2}\)m
Source: Myschool Past Questions Archive (Record #45741)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, O is the centre. If PQ//RS and ∠ONS = 140^\circ, find the size of ∠POM.
Attached Figure / Diagram
A
40o
B
50o
C
60o
D
80o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In the diagram above,
∠MNO = 140^\circ and angles on a straight line is 180^\circ
: ∠NMO = (180 - 140)^\circ = 40^\circ
Hence; ∠POM = 40^\circ ( alternate angle ∠S)
Source: Myschool Past Questions Archive (Record #45754)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PTR is a tangent to the centre O. If angles TON = 108^\circ, Calculate the size of angle PTN
Attached Figure / Diagram
A
132o
B
126o
C
108o
D
102o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram; 108^\circ + x + x = 180^\circ (sum of angle in a triangle)
108^\circ + 2x = 180^\circ
x = 180^\circ - 108^\circ
= 72^\circ
x = \(\frac{72^o}{2}\)
= 36^\circ
(Angle between tangent and a chord through the point of contact)
Hence, angle PTN = 90 + 36
= 126^\circ
Source: Myschool Past Questions Archive (Record #45757)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PQ//RT, QR//SU,
Attached Figure / Diagram
A
134o
B
132o
C
96o
D
48o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram; a = b = 48º (alternate < S)
x = 180 ^\circ - b (angles on a str. line)
x = 180º - 48º
= 132º
Source: Myschool Past Questions Archive (Record #45767)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and
Attached Figure / Diagram
A
73o
B
67o
C
57o
D
37o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the diagram given, < PRT = 3\(^o\) (Change in same segment)
< TPR = 90\(^o\) (angle in a semicircle)
Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)
= 180\(^o\) - 123\(^o\)
= 57\(^o\)
Source: Myschool Past Questions Archive (Record #45775)
Mathematics
WAEC / WASSCE 2015
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, O is the centre of the circle, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and < TOS = 76\(^o\). Using the diagram, find the value of angle PTR.
Attached Figure / Diagram
A
76o
B
71o
C
57o
D
33o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
PTR = 180º - (90 + 33)º
= 180º - 123
= 57º
Source: Myschool Past Questions Archive (Record #45785)
Mathematics
WAEC / WASSCE 2015
Coordinate Geometry
Solution
\(\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}\)
The table given shows some values for a linear graph. Find the gradient of the line
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let: (\(x_1, y_1\)) = (0, 3)
(\(x_2, y_2\)) = (\(\frac{5}{4}, \frac{11}{2}\))
Using gradient, m = \(\frac{y_2 - y_2}{x_2 - x_1}\)
= \(\frac{\frac{11}{2} - 3}{\frac{5}{4} - 0}\)
= \(\frac{11 - 6}{2} \div \frac{5}{4}\)
= \(\frac{5}{2} \div \frac{5}{4}\)
= \(\frac{5}{2} \times \frac{4}{5}\)
= 2
Source: Myschool Past Questions Archive (Record #45788)
Mathematics
UTME / JAMB 2015
Indices, Logarithms & Surds
Solution
Given that tan x = \(\frac{2}{3}\), where 0o d" x d" 90o, Find the value of 2sinx.
A
\(\frac{2\sqrt{13}}{13}\)
B
\(\frac{3\sqrt{13}}{13}\)
C
\(\frac{4\sqrt{13}}{13}\)
D
\(\frac{6\sqrt{13}}{13}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)
thus m2 = 22 + 32
= 4 + 9 = 13
m = \(\sqrt{13}\)
Hence, 2sin x = 2 x \(\frac{2}{m}\)
2 x\(\frac{2}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)
= \(\frac{4\sqrt{13}}{13}\)
Source: Myschool Past Questions Archive (Record #45797)
Mathematics
WAEC / WASSCE 2015
Coordinate Geometry
Solution
PQRS is a square. If x is the midpoint of PQ, calculate correct to the nearest degree,
A
53o
B
55o
C
63o
D
65o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the diagram given,
tan\(\alpha\) = \(\frac{1}{0.5}\) = 2
\(\alpha\) = tan - 1(2) = 63.43º
= 63º
Source: Myschool Past Questions Archive (Record #45809)
Mathematics
WAEC / WASSCE 2015
Trigonometry, Elevation & Bearings
Solution
The angle of elevation of an aircraft from a point K on the horizontal ground 30\(\alpha\). If the aircraft is 800m above the ground, how far is it from K?
A
400.00m
B
692.82m
C
923.76m
D
1,600.00m
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
In \(\bigtriangleup\)KBC, sin 30 = \(\frac{800}{IKCI}\)
IKCL = \(\frac{800}{sin30^o}\)
= \(\frac{800}{0.5}\)
= 1600m
Source: Myschool Past Questions Archive (Record #45810)
Mathematics
WAEC / WASSCE 2015
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.
A
62
B
63
C
64
D
65
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Third quartile Q3 = \(\frac{3}{4}\)Nth
Alternatively:
Split the score into equal quarters
44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75.
Third quartile Q3 = [63 + 65] / 2 = 64
Source: Myschool Past Questions Archive (Record #45817)