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Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Diagram
Solution
(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108^\circ. Find < CDE.
(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate
\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
C\(B\)D = C\(D\)B = \(\frac{180^o - 108^o}{2}\)
= \(\frac{72^o}{2}\) = 36\(^o\)
< CDE = < BDE + 36\(^o\)
< CDE = 90\(^o\) + 36 = 126\(^o\)
(b) tan x = \(\sqrt{3}\)
y\(^2\) (\(\sqrt{3}\))\(^2\) - (l)\(^2\)
y\(^2\) 3 + 1, Cos x = \(\frac{1}{2}\)
y = \(\sqrt{4}\) = 2
sin x = \(\frac{\sqrt{3}}{2}\)
\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)
= \(\frac{(\frac{1}{2})^2 - \frac{\sqrt{3}}{2}} {(\frac{\sqrt{3}}{2})^2 + \frac{1}{2}}\)
= \(\frac{\frac{1}{4} - \frac{\sqrt{3}}{2}}{\frac{3}{4} + \frac{1}{2}}\)
= \(\frac{1 - 2\sqrt{3}}{4}\) \(\div\) \(\frac{5}{4}\)
= \(\frac{1 - 2 \sqrt{3}}{4}\) x \(\frac{4}{5}\)
= \(\frac{1 - 2\sqrt{3}}{5}\)
Source: Myschool Past Questions Archive (Record #58708)
Mathematics
WAEC / WASSCE 2020
Number and Numeration
Solution
The total surface area of a cone of slant height l cm and base radius r cm is 224\(\pi\) cm\(^2\). If r : l = 2:5, find:
(a) correct to one decimal place, the value of r
(b) correct to the nearest whole number, the volume of the cone [Take \(\pi\) = \(\frac{22}{7}\)]
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
TSA = 224\(\pi\) cm\(^2\)
r : l = 2: 5
\(\frac{r}{l} = \frac{2}{5}\)
r = \(\frac{2l}{5}\)
\(\frac{3r}{2}\) = l
TSA \(\pi\)rl + \(\pi ^2\)
TSA = \(\pi\)(rl + r\(^2\))
224\(\pi\) = \(\pi\)r(l + r)
224 = r (\(\frac{5r}{2} + \frac{r}{1}\))
224 = r(\(\frac{5r + 2r}{2}\))
224 = \(\frac{7r^2}{2}\)
r\(^2\) = \(\frac{448}{7}\)
r = \(\sqrt{448}{7}\)
r = \(\sqrt{64cm}\)
l = \(\frac{5 \times 8}{2}\)
r = 8cm
l = 20cm
(b) Volume of cone = \(\frac{1}{3}\)\(\pi ^2\)h
h\(^2\) = l\(^2\) - r\(^2\)
h\(^2\) = (20)\(^2\) - (8)\(^2\)
h\(^2\) = 400 - 64
h = \(\sqrt{336}\)
= 18.33cm
Volume = \(\frac{1}{3} \times \frac{22}{7}\) x 8\(^2\) x 18.33
= \(\frac{25,809.06}{21}\)
= 1229 cm\(^3\)
Source: Myschool Past Questions Archive (Record #58709)
Mathematics
WAEC / WASSCE 2020
Statistics and Probability
Solution
A die was rolled a number of times. The outcomes are as shown in the table
Number
1
2
3
4
5
6
Outcomes
32
m
25
40
28
45
If the probability of obtaining 2 is 0.15, find the:
(a) value of m;
(b) number of times the die was rolled;
(c) probability of obtaining an even number.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Pr(2) = 0.15
\(\frac{m}{32 + m + 25 + 40 + 28 + 45}\) = 0.15
\(\frac{m}{m + 170}\) = 0.15
m = 0.15(m + 170)
m = 0.15m + 25.5
m - 0.15m = 25.5
\(\frac{0.85m}{0.85} = \frac{25.5}{0.85}\)
m = 30
(b) Number of time
= 32 + 30 + 25 + 40 + 28 + 45
= 200 times
(c) Pr(even number) = \(\frac{30 + 40 + 45}{200}\)
= \(\frac{115}{200}\)
= \(\frac{23}{40}\)
Source: Myschool Past Questions Archive (Record #58710)
Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Solution
(a) Copy and complete the table of values for the relation y = 3 sin 2x.
x
o\(^o\)
15\(^o\)
30\(^o\)
45\(^o\)
60\(^o\)
75\(^o\)
90\(^o\)
105\(^o\)
120\(^o\)
135\(^o\)
\(^o\)
y
0.0
1.5
-2.6
(b) Using a scale of 2 cm to 15^\circ on the x-axis and 2cm to I unit on the y-axis, draw the graph of y = 3 sin 2x for 0^\circ \(\geq\) x \(\geq\) 150^\circ.
(c) Use the graph to find the truth set of;
(i) 3 sin 2x + 2 = 0;
(ii ) \(\frac{3}{2}\) sin 2x = 0.25.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x
o\(^o\)
15\(^o\)
30\(^o\)
45\(^o\)
60\(^o\)
75\(^o\)
90\(^o\)
105\(^o\)
120\(^o\)
135\(^o\)
\(^o\)
y
0.0
1.5
2.6
2.1
2.6
1.5
0
- 1.5
- 2.6
- 3
-2.6
y = 35m\(^2\)x
when x = 15\(^o\)
y = 3 sin = (15\(^o\))
y = sin 30\(^o\)
y = 3 x \(\frac{1}{2}\) = 1.5
When x = 30\(^o\)
y = 3 sin 2(30\(^o\))
y = 3 sin 60\(^o\) = 2.6
When x = 45\(^o\)
y = 3sin45\(^o\)
= 2.1
When x = 60\(^o\)
y = 3 sin 2(60\(^o\))
= 2 sin 120\(^o\)
= 2.6
When x = 75\(^o\)
y = 3 sin 2(75\(^o\))
= 3 sin 150\(^o\)
= 1.5
When x = 90\(^o\)
y = 3 sin 2(90\(^o\))
= 3 sin 180\(^o\)
= 0
When x = 105\(^o\)
y = 3 sin 2(105\(^o\))
= sin 210\(^o\)
= -1.5
When x = 120\(^o\)
y = 3 sin 2(120\(^o\))
= 3 sin 240\(^o\)
= -2.6
When x = 135\(^o\)
y = 3 sin 2(135\(^o\))
y = 3 sin 270\(^o\)
y = -3
(b
SEE THE GRAPH ABOVE...
(c)(i) 113
(ii) 3\(^o\) and 82\(^o\)
Source: Myschool Past Questions Archive (Record #58711)
Mathematics
WAEC / WASSCE 2020
Number and Numeration
Diagram
Solution
(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius is 14. Calculate, correct to three significant figures, the surface area of the structure, [Take \(\pi = \frac{22}{7}\)]
(b) Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Surface Area of the structure = curved surface area of cone + surface area of hemisphere = \(\pi rl + 2 \pi r^2\)
But slant height l = l\(^2\) = 48\(^2\) + 14\(^2\)
l\(^2\) = 2500
l = \(\sqrt{2500}\) = 50
Surface Area of the structure = \(\pi rl + 2 \pi r^2\) = \(\frac{22}{7}\) x 14 x 50 + 2 x \(\frac{22}{7}\) x 14 x 14
= 3432.0m\(^2\) ≈ 3430m\(^2\) to 3 s.f
(b) Let Musah age be M
Let Sesay be S
M - 5 = 2(S - 5) ......(i)
M + S = 100 ......(ii)
From eqn ii) M = 100 - S
By substitution, 100 - S - 5 = 2(S - 5)
95 - S = 2S - 10
3S = 105
\(\frac{3S}{3} = \frac{105}{3}\)
S = 35
Therefore, Sesay's present age is 35 years
Source: Myschool Past Questions Archive (Record #58712)
Mathematics
WAEC / WASSCE 2020
Algebra
Solution
(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at an open market and \(\frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had N222,000.00 left, find:
(i) her monthly income.
(ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A. P.) is 4m - 2n. If the ninth term of the progression is 2m - 8n. find the common difference in terms of m and n.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Let the salary be x
Shopping mall = \(\frac{1}{4}\) x \(x = \frac{x}{4}\)
Open market = \(\frac{1}{3} \times x = \frac{x}{3}\)
The remainder x - [\(\frac{x}{4} + \frac{x}{3}\)]
= x - [\(\frac{3x + 4x}{12}\)]
= \(\frac{x}{1} - \frac{7x}{12}\)
= \(\frac{12x - 7x}{12}\)
= \(\frac{5x}{12}\)
mechanic = \(\frac{2}{5} \times \frac{5x}{12} = \frac{x}{6}\)
\(\frac{x}{6} + \frac{x}{3} + \frac{x}{4}\) + 225,00 = x
\(\frac{2x + 4x + 3x}{12}\) + 225,000
\(\frac{9x}{12}\) + 225,000 = x
225,000 = \(\frac{x}{1} = \frac{9x}{12}\)
225,000 = \(\frac{12x - 9x}{12}\)
225,000 = \(\frac{3x}{12}\)
\(\frac{225,000 \times 12}{3}\) = x
N900,000 = x
(i) Her monthly income = N900,000
(ii) Amount in open market
= \(\frac{N900,000}{3}\)
= N300,000
(b) T\(_3\) = 4m - 2n
T\(_9\) = 2m - 8n
a + 2d = 4m - 2n .....(i)
a + 8d = 2m - 8n ......(ii)
a + 8d = 2m - 8n
-(a + 2d = 4m - 2n
\(\overline{6d = -2m - 6n}\)
6d = -2(m + 3n)
d = -\(\frac{m + 3n}{3}\)
Common difference = -\(\frac{m + 3n}{3}\)
Source: Myschool Past Questions Archive (Record #58713)
Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Solution
(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049^\circ and cyclist Y travels at the rate of 9 km/h on a bearing of 319^\circ.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate. correct to the nearest whole number, the:
(i) distance between cyclist X and Y;
(ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) diagram above
(b)(i) m\(^2\) = 5\(^2\) + 9\(^2\)
m\(^2\) = 25 + 81
m\(^2\) = 106
m = \(\sqrt{106}\) = 10.296 km/hr
Speed = \(\frac{distance}{times}\)
distance = speed x time
= 10.296 x 2
= 21km
(b)(ii)
tan\(\theta\) = \(\frac{9km/hr}{5km/hr}\)
tan\(\theta\) = 1.8
\(\theta\) = tan\(^{-1}\)
\(\theta\) = 60.9\(^o\)
the bearing of cyclist x from y
90\(^o\) + 19.9\(^o\) = 109.9\(^o\)
= 110\(^o\)
(c) The average speed = \(\frac{distance }{time}\)
= \(\frac{21}{4}\)km
= 5km/her
Source: Myschool Past Questions Archive (Record #58714)
Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Solution
The table shows the distribution of marks obtained by students in an examination.
Marks (%)
0 - 9
10 - 19
20 - 29
30 - 39
40 - 49
50 - 59
60 - 69
70 - 79
80 - 89
90 - 99
Frequency
7
11
17
20
29
34
30
25
21
6
(a) Construct a cumulative frequency table for the distribution.
(b) Draw the cumulative frequency curve for the distribution.
(c) Using the curve, find correct to one decimal place, the:
(i) median mark;
(ii) lowest mark for the distinction if 5% of the students passed with distinction
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Marks
Upper
Class
Boundary
Frequency (F)
Cumulative
Frequency (F)
(CF)
0 - 9
10 - 19
20 - 29
30 - 39
40 - 49
50 - 59
60 - 69
70 - 79
80 - 89
90 - 99
9.5
19.5
29.5
39.5
49.5
59.5
69.5
79.5
89.5
99.5
7
11
17
20
29
34
30
25
21
6
7
18
35
55
84
118
148
173
194
200
(b)
(c) (I) Median Mark = 54.5
(ii) Lowest mark, if 5% of the students passed = 14
Source: Myschool Past Questions Archive (Record #58715)
Mathematics
WAEC / WASSCE 2020
Algebra
Diagram
Solution
(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50^\circ. Find:
(I) < MNP
(ii) < POQ
(b) Find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12)
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
(i)
< NPQ = 180\(^o\) - 50\(^o\) = 130\(^o\)
< MNP = 130\(^o\) - 90\(^o\) = 40\(^o\)
< NMP = 40\(^o\) (Isolceles triangle)
< MNP = 180\(^o\) - (40\(^o\) + 40\(^o\))
= 180\(^o\) - 80\(^o\) = 100\(^o\)
(ii) < OPQ = 90\(^o\)(50\(^o\) - 40\(^o\)) = 80\(^o\)
< OPQ = 80\(^o\) (Isoseles triangle)
< POQ = 180\(^o\) (80\(^o\) + 80\(^o\))
180\(^o\) - 160\(^o\) = 20\(^o\)
(b) 8y + 4x = 24
\(\frac{8y}{8} = \frac{-4y}{8} + \frac{24}{8}\)
y = \(\frac{-1}{2}x + 6\)
m = \(\frac{-1}{2}\)
The equation will give
\(\frac{y - 12}{x + 8} = \frac{-1}{2}\)
2y - 24 = x - 8
2y = - x - 8 + 24
2y = -x + 16
Source: Myschool Past Questions Archive (Record #58716)
Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Diagram
Solution
(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40^\circ, find: < ODE.
(b) ABCD is a parallelogram in which |\(\overline{CD}\)| = 7 cm, I\(\overline{AD}\)I = 5 cm and < ADC= 125^\circ.
(i) Illustrate the information in a diagram.
(ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\)). Evaluate (2x\(^2\) - 2x).
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
< AOB = 108\(^o\) - (90\(^o\) + 40\(^o\))
= 180\(^o\) - 130\(^o\) = 50\(^o\)
< DCO = 40\(^o\) (Alternate angle)
< ODE = \(\frac{180^o - 80^o}{2}\) = \(\frac{100^o}{2}\)
= 50\(^o\)
(b)(i)
(b)(ii) Area of the parallelogram
= ab sin \(\theta\)
= 5cm x 7cm sin 125\(^3\)
= 35 sin 125
= 28.7cm\(^2\)
(c)
x = \(\frac{1}{2}(1 - \sqrt{2})\)
evaluate (2x\(^2\) - 2x)
= 2[\(\frac{1}{2}(1 - \sqrt{2})^2\)] - 2[\(\frac{1}{2} (1 - \sqrt{2}\))]
= 2[\(\frac{1}{4}(1 - 2\sqrt{2} + 2)\)] - 1 + \(\sqrt{2}\)
= 2[\(\frac{1}{4}(3 - 2\sqrt{2})\) - 1 + \(\sqrt{2}\)]
2[\(\frac{3}{4} - \frac{1}{2} \sqrt{2} - 1 + \sqrt{2}\)]
= \(\frac{3}{2} - \sqrt{2} - 2 + 2\sqrt{2}\)
= \(\frac{1}{2} + \sqrt{2}\)
Source: Myschool Past Questions Archive (Record #58717)
Mathematics
WAEC / WASSCE 2019
Geometry and Trigonometry
Diagram
Solution
In the diagram, RT is a tangent to the circle at R, < PQR = 70\(^o\), < QRT = 52\(^o\), < QSR and < PRQ = x. Calculate the value of x.
Attached Figure / Diagram
A
70\(^o\)
B
58\(^o\)
C
55\(^o\)
D
48\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x = 180\(^o\) - (52\(^o\) + 70\(^o\))
= 180\(o\) - 122\(^o\)
= 58\(^o\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #57380)
Mathematics
WAEC / WASSCE 2019
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQ is parallel to RS, < QFG = 105\(^o\) and < FEG = 50\(^o\). Find the value of n
Attached Figure / Diagram
A
40\(^o\)
B
55\(^o\)
C
75\(^o\)
D
130\(^o\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
E\(G\)R = 50\(^o\) (Alternative angles)
n + 55\(^o\) + 50\(^o\) = 180\(^o\) (Sum of angles on a straight line)
n = 180\(^o\) - 105\(^o\)
n = 75\(^o\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #57384)
Mathematics
WAEC / WASSCE 2019
Fractions, Decimals, Percentages & Approximation
Solution
Evaluate: \(\frac{0.42 \div 2.5}{0.5 \times 2.05}\), leaving the answer in the standard form.
A
1.639 x 10\(^2\)
B
1.639 x 10\(^1\)
C
1.639 x 10\(^{-1}\)
D
1.639 x 10\(^{-2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{0.42 \div 2.5}{0.5 \times 2.05} = \frac{0.168}{1.025}\)
= 1.639 x 10\(^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #57686)
Mathematics
WAEC / WASSCE 2019
Sets & Set Operations
Solution
If T = {prime numbers} and M = {odd numbers} are subsets of \(\mu\) = {x : 0 < x ≤ 10} and x is an integer, find (T\(^{\prime}\) n M\(^{\prime}\)).
A
{4, 6, 8, 10}
B
{1. 4, 6, 8, 10}
C
{1, 2, 4, 6, 8, 10}
D
{1, 2, 3, 5, 7, 8, 9}
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
T = {2, 3, 5, 7}
M = {1, 3, 5, 7, 9}
\(\mu\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
T\(^{\prime}\) = = {1, 4, 6, 8, 9, 10}
M\(^{\prime}\) = {2, 4, 6, 8, 10}
(T\(^{\prime}\) \(\cap\) M\(^{\prime}\)) = {4, 6, 8, 10}
Source: WAEC / JAMB / NECO Classroom Archive (Record #57122)
Mathematics
WAEC / WASSCE 2019
Triangles and Polygons
Diagram
Solution
In XYZ, |YZ| = 32cm, < YXZ 53\(^o\) and XZY = 90\(^o\). Find, correct to the nearest centimetre, |XZ|
Attached Figure / Diagram
A
31 cm
B
25 cm
C
20 cm
D
13 cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Tan 52\(^o\) = \(\frac{32}{\text{|XZ|}}\)
|XZ| = \(\frac{32}{Tan 52^o}\)
= 25cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #57253)