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Mathematics
WAEC / WASSCE 2020
Number and Numeration
Solution
The interquartile range of distribution is 7. If the 25th percentile is 16, find the upper quartile.
A
35
B
30
C
23
D
9
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Q\(_3\) - Q\(_1\) = 7
Q\(_3\) - 16 = 7
Q\(_3\) - 7 + 16
= 23
Source: Myschool Past Questions Archive (Record #58406)
Mathematics
WAEC / WASSCE 2020
Number and Numeration
Solution
If (x + 2) is a factor of x\(^2\) +px - 10, find the value of P.
A
3
B
-3
C
7
D
-7
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x + 2 = 0
x -2
x\(^2\) = px - 10 = 0
(-2)\(^2\) - 2p - 10 = 0
4 - 2p - 10 = 0
\(\frac{-2p}{-2} = \frac{6}{-2}\) \(\to\) p = -3
Source: Myschool Past Questions Archive (Record #58417)
Mathematics
WAEC / WASSCE 2020
Geometry and Trigonometry
Diagram
Solution
In the diagram, O is the centre of the circle. If < NLM = 74\(^o\), < LMN = 39\(^o\) and < LOM = x, find the value of x.
Attached Figure / Diagram
A
134\(^o\)
B
126\(^o\)
C
113\(^o\)
D
106\(^o\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
L\(N\)M = 180\(^o\) - (74\(^o\) + 39\(^o\))
180\(^o\) - 113\(^o\)
= 67\(^o\)
L\(^O\)
M = x = 2 x 67\(^o\)
= 134\(^o\)
Source: Myschool Past Questions Archive (Record #58465)
Mathematics
WAEC / WASSCE 2020
Algebra
Solution
Find the least value of x which satisfies the equation 4x = 7(mod 9)
A
7
B
6
C
5
D
4
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
4x = 7 (mod 9)
4x = 7 + 9 (mod 9)
\(\frac{4x}{4} = \frac{16}{4}\) (mod 9)
x = 4
Source: Myschool Past Questions Archive (Record #58619)
Mathematics
WAEC / WASSCE 2020
Number and Numeration
Solution
(a) If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of \(\mu\) = {x:1 \(\leq\) x \(\leq\) 10} find A ′ \(\cap\) B ′ \(\cap\) C′
(b) Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
\(\mu\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 4, 6, 8, 10}
B = {3,6,9}
C = {1, 2, 3, 6}
A` ={1, 3, 5, 7, 9}
B` = {1, 2, 4, 5, 7, 8, 10}
C` = {4, 5, 7, 8, 9, 10}
a` \(\cap\) B` C` = { 5 and 7 }
(b) bulk cost $80
1 will cost \(\frac{$80}{5}\) = $16
for 5 gentlemen each cost $18.50
total cost = 18.50 x 5
= $92.50
5 cost for $16 each = 5 x $16
= $(92.50 - 80.00)
each will save \(\frac{$12.50}{5}\)
= $2.50
Source: Myschool Past Questions Archive (Record #58706)
Mathematics
WAEC / WASSCE 2020
Number and Numeration
Solution
(a) Given that P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
(i) Make Q the subject of the relation;
(ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that \(\frac{x + 2y}{5}\) = x - 2y, find x : y
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
P\(^{\frac{3}{2}}\) = \(\frac{rk}{Q}\) - ms
P\(^{\frac{3}{2}}\) + ms = \(\frac{rk}{Q}\)
Q = \(\frac{rk}{p^{\frac{3}{2}}ms}\)
(a)(ii)
Q = \(\frac{10 \times 4}{3\frac{3}{2} + 15 (0.2)}\)
Q = \(\frac{40}{5.1958 + 3}\)
Q = \(\frac{40}{8.1958}\)
= 4.88 to 2 d.p
(b) \(\frac{x + 2y}{5} = x - 2y\)
x + 2y = 5(x - 2y)
x + 2y = 5x - 10y
x - 5x = -10y - 2y
-4x = -12y
\(\frac{x}{y} = \frac{12}{4}\)
x : y = 3 : 1
Source: Myschool Past Questions Archive (Record #58707)
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)