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Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
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
Geometry and Trigonometry
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
Geometry and Trigonometry
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
Geometry and Trigonometry
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
Geometry and Trigonometry
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
Number and Numeration
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)
Mathematics
WAEC / WASSCE 2015
Statistics and Probability
Solution
Score
0 - 4
5 - 9
10 - 14
Frequency
2
1
2
The table shows the distribution of the scores of some students in a test. Calculate the mean scores.
A
5.6
B
6.2
C
6.6
D
7.0
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
To calculate the mean of grouped data,
- First step to determine the midpoint (x) of each interval or class.
0 - 4 ►2
5 - 9 ► 7
10 - 14 ►12
These midpoints must then be multiplied by the frequencies of the corresponding classes:
2 X 2 = 4
1 X 7 = 7
2 X 12 = 24
Mean = ( 24 + 7 + 4) \div ( 2 + 1 + 2 )
: Mean = 7
Source: Myschool Past Questions Archive (Record #45822)
Mathematics
WAEC / WASSCE 2015
Statistics and Probability
Solution
The probability that kebba, Ebou and Omar will hit a target are \(\frac{2}{3}\), \(\frac{3}{4}\) and \(\frac{4}{5}\) respectively. Find the probability that only Kebba will hit the target.
A
\(\frac{2}{5}\)
B
\(\frac{7}{60}\)
C
\(\frac{1}{30}\)
D
\(\frac{1}{60}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Pr(Ebou did not hit the target) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\) = Pr(E'), Pr(Omar did not hit the target) = 1 - \(\frac{4}{5}\) = \(\frac{1}{5}\) = Pr(O')
Hence the probability that only Kebba will hit the target
= Pr(K) x Pr(E') x Pr(O')
= \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\)
= \(\frac{1}{30}\)
Source: Myschool Past Questions Archive (Record #45827)
Mathematics
WAEC / WASSCE 2015
Number and Numeration
Diagram
Solution
In the diagram, VW//YZ, |WX| = 6cm, |XY| = 16cm, |YZ| = 20cm and |ZX| = 12cm. Calculate |VX|
Attached Figure / Diagram
A
3cm
B
4cm
C
6cm
D
8cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
In the diagram, \(\frac{16}{12} = \frac{VX}{6}\) (similar \(\Delta\)s)
VX = \(\frac{16 \times 6}{12}\)
= 8cm
Source: Myschool Past Questions Archive (Record #45834)
Mathematics
WAEC / WASSCE 2015
Number and Numeration
Solution
Tom will be 25 years old in n years' time. If he is 5 years younger than Bade, find Bade's present age.
A
(30 - n)years
B
(20 - n)years
C
(25 - n)years
D
(30 + n)years
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Let Tom's present age be x.
Then x = 25 - n
If Tom is 5 years younger than Bade, then Bade's present age is x + 5 = 25 - n + 5
= (30 - n)
Source: Myschool Past Questions Archive (Record #45852)
Mathematics
WAEC / WASSCE 2015
Number and Numeration
Solution
If \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) is simplified as m + n\(\sqrt{6}\), find the value of (m + n)
A
\(\frac{1}{3}\)
B
\(\frac{2}{3}\)
C
1\(\frac{1}{3}\)
D
1\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{\sqrt{6} + 3}{3}\)
= \(\frac{3 + \sqrt{6}}{3}\)
= Hence, (m + n) = 1 + \(\frac{1}{3}\)
= 1\(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #45859)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the given diagram, \(\bar{QT}\) and \(\bar{PR}\) are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.
Attached Figure / Diagram
A
35o
B
40o
C
55o
D
60o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In the diagram, QOR + 2m(vertically opposite angles)
So, m + 90^\circ + 2m = 180^\circ
(angles on str. line)
3m = 180^\circ - 90^\circ
3m = 90^\circ
m = \(\frac{90^o}{3}\)
= 30^\circ
substituting 30^\circ for m in
2m + 4n = 200^\circ gives
2 x 30^\circ + 4n = 200^\circ
60^\circ + 4n = 200^\circ
4n = 200^\circ - 60^\circ
= 140^\circ
n = \(\frac{140^\circ}{4}\)
= 35^\circ
Source: Myschool Past Questions Archive (Record #45861)
Mathematics
WAEC / WASSCE 2015
Number and Numeration
Solution
Make K the subject of the relation T = \(\sqrt{\frac{TK - H}{K - H}}\)
A
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
B
K = \(\frac{HT}{(T - 1)^2}\)
C
K = \(\frac{H(T^2 + 1)}{T}\)
D
K = \(\frac{H(T - 1)}{T}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
T = \(\sqrt{\frac{TK - H}{K - H}}\)
Taking the square of both sides, give
T2 = \(\frac{TK - H}{K - H}\)
T2(K - H) = TK - H
T2K - T2H = TK - H
T2K - TK = T2H - H
K(T2 - T) = H(T2 - 1)
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
Source: Myschool Past Questions Archive (Record #45862)
Mathematics
WAEC / WASSCE 2015
Statistics and Probability
Which of the following is used to determine the mode of a grouped data?
A
Bar chart
B
Frequency polygon
C
Ogive
D
Histogram
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 #45864)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Solution
The area of a rhombus is 110 cm\(^2\). If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.
A
5.0
B
4.0
C
3.0
D
2.5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Diagonal |AC| = (2x + 1)cm
In the diagram,area of \(\Delta\)ABC
is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB|
55 = \(\frac{1}{2}\) x (2x + 1) x 10
55 = (2x + 1)5
55 = 10x + 5
55 - 5 = 10x
50 = 10x
x = \(\frac{50}{10}\)
= 5.0
Source: Myschool Past Questions Archive (Record #45900)