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Mathematics
WAEC / WASSCE 2017
Trigonometry, Elevation & Bearings
A bearing of 320\(^o\) expressed as a compass bearing is
A
N 50\(^o\) W
B
N 40\(^o\) W
C
N 50\(^o\) E
D
N 40\(^o\) E
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #50005)
Mathematics
WAEC / WASSCE 2017
Trigonometry, Elevation & Bearings
Solution
Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - 1}{1 - tan 30^o}\)
A
\(\sqrt{3 - 2}\)
B
2 - \(\sqrt{3}\)
C
\(\sqrt{3}\)
D
-2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
From the figure above, tan30º = \(\frac{1}{\sqrt3}\) and tan60º = \(\sqrt{3}\)
so \(\frac{tan60 - 1}{1 - tan30}\)
= \(\frac{\sqrt{3}}{1}\div(1 - (\frac{1}{\sqrt{3}}\)))
= \(\frac{\sqrt{3} - 1}{1}\div\frac{\sqrt{3} - 1}{\sqrt3}\)
= \(\frac{\sqrt{3} - 1}{1}\times \frac{\sqrt3}{\sqrt{3} - 1}\) = \(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #50014)
Mathematics
WAEC / WASSCE 2017
Plane Geometry, Polygons & Circle Theorems
Solution
In what number base was the addition 1 + nn = 100, where n > 0, done?
A
n - 1
B
n + 1
C
n
D
n + 2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Assuming the numbers 1 + nn =100 were added in base n + 1
n
n
1
1 0
0
n + 1 = n +1 and n + 1 in base n + 1 = 10, write 0 and carry 1, which will be added to the second n, resulting in another n + 1 in base n + 1 = 10.
So, putting the result together 1 + nn = 100 provided, the addition was carried out in base n + 1.
Source: Myschool Past Questions Archive (Record #50025)
Mathematics
WAEC / WASSCE 2017
Indices, Logarithms & Surds
Solution
Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
A
4
B
\(\sqrt{3} + 4\)
C
4 \(\sqrt{2}\)
D
4\(\sqrt{3} + 4\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)
2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4
= 4
Source: Myschool Past Questions Archive (Record #50027)
Mathematics
WAEC / WASSCE 2017
Plane Geometry, Polygons & Circle Theorems
Solution
Three exterior angles of a polygon are 30\(^o\), 40\(^o\) and 60\(^o\). If the remaining exterior angles are 46\(^o\) each, name the polygon.
A
decagon
B
nonagon
C
octagon
D
hexagon
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Sum of all exterior angles is 360\(^o\)
360\(^o\) (30\(^o\) - 40\(^o\))
360 - (130\(^o\))
230\(^o\)
remaining is 46\(^o\) = \(\frac{230}{46}\) = 5
5 + 3 = 8 sides; Octagon
Source: Myschool Past Questions Archive (Record #50029)
Mathematics
WAEC / WASSCE 2017
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify the expression \(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\)
A
\(a^2 - b^2\)
B
\(b^2 - a^2\)
C
\(a^2b - ab^2\)
D
\(ab^2 - a^2b\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\) = \(\frac{a^2 b^2(b^2 - a^2)}{ab(a + b)}\)
= \(\frac{ab [(b - a) (b + a)]}{a + b}\)
= ab(b - a)
= \(ab^2 - a^2b\)
Source: Myschool Past Questions Archive (Record #50112)
Mathematics
WAEC / WASSCE 2017
Sequences & Series (AP & GP)
Solution
Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),...
A
-\(\frac{1}{3}\)
B
-\(\frac{1}{5}\)
C
-\(\frac{1}{15}\)
D
\(\frac{1}{9}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)
= 7 - 10
= \(\frac{-3}{15}\)
d = - \(\frac{-1}{5}\)
T6 = a + 5d
= \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\)
= \(\frac{2}{3}\) - 1
= \(\frac{2 - 3}{3}\)
= \(\frac{-1}{3}\)
Source: Myschool Past Questions Archive (Record #50121)
Mathematics
WAEC / WASSCE 2017
Linear, Simultaneous & Quadratic Equations
Solution
The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation.
A
\(6x^2 - x + 2 = 0\)
B
\(6x^2 - x - 2 = 0\)
C
\(6x^2 + x - 2 = 0\)
D
\(6x^2 + x + 2 = 0\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\))
\(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\)
\(6x^2 - 4n + 3n - 2 = 0\)
\(6x^2 - x - 2 = 0\)
Source: Myschool Past Questions Archive (Record #50122)
Mathematics
WAEC / WASSCE 2017
Plane Geometry, Polygons & Circle Theorems
Consider the statements: p it is hot, q: it is raining
Which of the following symbols correctly represents the statement "It is raining if and only if it it is cold"?
A
p \(\iff\) \(\sim\)q
B
p \(\iff\) q
C
\(\sim\)p \(\iff\) \(\sim\)q
D
q \(\iff\) \(\sim\)p
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 #50126)
Mathematics
WAEC / WASSCE 2017
Indices, Logarithms & Surds
Solution
Given that t = \(2 ^{-x}\), find \(2 ^{x + 1}\) in terms of t.
A
\(\frac{2}{t}\)
B
\(\frac{t}{2}\)
C
\(\frac{1}{2t}\)
D
t
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
t = \(2^{-x} = \frac{1}{2^{x}}\)
\(\implies 2^{x} =\frac{1}{t}\)
\(2^{x+1} = 2^{x} \times 2^{1}\)
= \(\frac{1}{t} \times 2 = \frac{2}{t}\)
Source: Myschool Past Questions Archive (Record #50130)
Mathematics
WAEC / WASSCE 2017
Permutations, Combinations & Probability
Solution
Two bottles are drawn with replacement from a crate containing 8 Coke, 12 Fanta, and 4 Sprite bottles. What is the probability that the first is coke and the second is not coke?
A
\(\frac{1}{12}\)
B
\(\frac{1}{6}\)
C
\(\frac{2}{9}\)
D
\(\frac{3}{8}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total = 8 + 12 + 4
= 24
\(\frac{8}{24} \times (\frac{12}{24} + \frac{4}{24}\))
= \(\frac{1}{3} \times (\frac{1}{2} + \frac{1}{6}\))
= \(\frac{1}{3} \times \frac{3 + 1}{6}\)
\(\frac{1}{3} \times \frac{4}{6} = \frac{2}{9}\)
Source: Myschool Past Questions Archive (Record #50132)
Mathematics
WAEC / WASSCE 2017
Fractions, Decimals, Percentages & Approximations
Solution
If the simple interest on a certain amount of money saved in a bank for 5 years at 2\(\frac{1}{2}\)% annum is N500.00, calculate the total amount due after 6 years at the same rate
A
N2,500.00
B
N2,600.00
C
N4,500.00
D
N4,600.00
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
P = \(\frac{100l}{RT} = \frac{100 \times 500}{5 \times 2.5} = \frac{50,000}{12.5}\)
= 4,000
Annual interest is \(\frac{500}{5}\) = 100
for 6 year = 4,000 + 100 + 100 + 100 + 100 + 100 + 100
= N 4,600
Source: Myschool Past Questions Archive (Record #50135)
Mathematics
WAEC / WASSCE 2017
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9
A
2.2
B
3.4
D
4.0
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x = \(\frac{2 + 3 + 3 + 4 + 5 + 5+ 5+ 7 + 7 + 9}{10}\)
=\(\frac{50}{10}\)
= 5
Variance = \(\frac{\sum{(x - x})^2}{N}\)
\(\frac{9 + 4+ 4+ 1 + 4 + 4 + 16}{10}\)
= \(\frac{42}{10}\)
= 4.2
Source: Myschool Past Questions Archive (Record #50138)
Mathematics
WAEC / WASSCE 2017
Fractions, Decimals, Percentages & Approximations
Solution
A circular pond of radius 4m has a path of width 2.5m round it. Find, correct to two decimal places, the area of the path. [Take\(\frac{22}{7}\)]
A
7.83\(m^2\)
B
32.29\(m^2\)
C
50.29\(m^2\)
D
82.50\(m^2\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of path = Area of (pond+path) - Area of pond
The area of the pond with the path: The radius = (4 + 2.5)m = 6.5m
Area = \(\pi \times r^{2}\) = \(\frac{22}{7} \times 6.5^{2} \approxeq 132.79m^{2}\)
Area of the pond = \(\frac{22}{7} \times 4^{2} \approxeq 50.29m^{2}\)
Area of the path = (132.79 - 50.29)m^{2} = 82.50m^{2}\)
Source: Myschool Past Questions Archive (Record #50142)
Mathematics
WAEC / WASSCE 2017
Number Bases & Modular Arithmetic
Diagram
Solution
Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n \(\oplus 4\))
Attached Figure / Diagram
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(n \(\oplus\) 4) \(\oplus\) 3 = 0 (mod 5)
(3 \(\oplus\) 4) \(\oplus\) 3
12 \(\oplus\) 3 = 15 (mod 5)
(5 x 3 + 0) = 0 (mod 5)
Source: Myschool Past Questions Archive (Record #50148)