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Mathematics
WAEC / WASSCE 2019
Geometry and Trigonometry
Solution
From the top of a vertical cliff 20m high, a boat at sea can be sighted 75m away and on the same horizontal position as the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.
A
56\(^o\)
B
75\(^o\)
C
16\(^o\)
D
15\(^o\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Sin\(\theta = \frac{20}{75}\)
\(\theta = sin^{-1} (\frac{20}{75})\)
= 15\(^o\)
Source: Myschool Past Questions Archive (Record #57369)
Mathematics
WAEC / WASSCE 2019
Geometry and Trigonometry
Diagram
Solution
In the diagram, O is the centre of the circle of radius 18cm. If < zxy = 70\(^o\), calculate the length of arc ZY. [Take \(\pi = \frac{22}{7}\)]
Attached Figure / Diagram
A
11cm
B
22 cm
C
44 cm
D
80 cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Z\({O}\)T = 2 x 70\(^o\) 140\(^o\)
Length = \(\frac{\theta}{360}\) x 2\(\pi\)r
\(\frac{140^o}{360} \times 2 \times \frac{22}{7}\) x 18
= 44cm
Source: Myschool Past Questions Archive (Record #57378)
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. Find the value of y.
Attached Figure / Diagram
A
70\(^o\)
B
60\(^o\)
C
52\(^o\)
D
18\(^o\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
∠QRT = ∠SRT = 52 (opposite angles)
∠SRT = y = 52 (alternate angles)
y = 52\(^o\)
Source: Myschool Past Questions Archive (Record #57379)
Mathematics
WAEC / WASSCE 2019
Statistics and Probability
Solution
Calculate the variance of 2, 4, 7, 8 and 9
A
7.2
B
6.8
C
3.5
D
2.6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x
2
4
7
8
9
(\(\bar{x} - \bar{x}\))
-4
-2
1
2
4
(\(\bar{x} - \bar{x})^2\)
16
4
1
4
9
_________
\(\sum (x - x)^2\) = 34
\(\bar{x} = \frac{2 + 4 + 7 + 8 + 9}{5} = 6
variance = \(\frac{34}{5}\)
= 6.8
Source: Myschool Past Questions Archive (Record #57381)
Mathematics
WAEC / WASSCE 2019
Algebra
Solution
The fourth term of an Arithmetic Progression (A.P) is 37 and the first term is -20. Find the common difference.
A
3
B
57
C
19
D
17
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
a + 3d = 37
-20 + 3d = 37
3d = 37 + 20 = 57
d = \(\frac{57}{3}\)
= 19
Source: Myschool Past Questions Archive (Record #57382)
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 m.
Attached Figure / Diagram
A
130\(^o\)
B
105\(^o\)
C
75\(^o\)
D
55\(^o\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
50\(^o\) + m = 105\(^o\) (exterior angle of \(\triangle\)
m = 105\(^o\) - 50\(^o\)
m = 55\(^o\)
Source: Myschool Past Questions Archive (Record #57383)
Mathematics
WAEC / WASSCE 2019
Statistics and Probability
Solution
A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?
A
\(\frac{1}{6}\)
B
\(\frac{1}{4}\)
C
\(\frac{1}{3}\)
D
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Pr(green) = \(\frac{6}{5 + 6 + 7}\)
= \(\frac{6}{18}\)
= \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #57385)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Diagram
Solution
The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?
Attached Figure / Diagram
A
90
B
80
C
70
D
40
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No of fruits = x
\(\frac{80}{360}\) x \(x\) = 60
x = \(\frac{60 \times 360}{80}\) x 270
= 90
Source: Myschool Past Questions Archive (Record #57386)
Mathematics
WAEC / WASSCE 2019
Statistics and Probability
The following are scores obtained by some students in a test, Find the mode of the distribution
8
18
10
14
18
11
13
14
13
17
15
8
16
13
A
18
B
14
C
13
D
8
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #57391)
Mathematics
WAEC / WASSCE 2019
Statistics and Probability
Solution
The following are scores obtained by some students in a test. Find the median score
8
18
10
14
18
11
13
14
13
17
15
8
16
13
A
14.5
B
14.0
C
13.5
D
13.0
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{13 + 14}{2}\) = 13.5
Source: Myschool Past Questions Archive (Record #57394)
Mathematics
WAEC / WASSCE 2019
Algebra
Solution
(a) Find the equation of the line which passes through the points A(-2, 7) and B(2, -3)
(b) Given that \(\frac{5b - a}{8b + 3a} = \frac{1}{5}\) = find, correct to two decimal places, the value \(\frac{a}{b}\)
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) A(-2, 7), B(2, -3)
m = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 7}{2 - (-2)} = \frac{10}{4}\)
\(\frac{-5}{2} = \frac{y - 7}{x + 2}\)
-5(x + 3) = 2(y - 7)
-5x - 10 = 2y - 14
2y - 14 + 10 + 5x = 0
2y - 14 + 10 + 5x = 0
2y + 5x - 4 = 0
(b) \(\frac{5b - a}{8b + 3a} = \frac{1}{5}\)
5(5b - a) = 1(1)(8b + 3a)
25b - 5a = 8b + 3a
25b - 8b = 3a + 5a
17b = 8a
\(\frac{a}{b} = \frac{17}{8}\)
\(\frac{a}{b}\) = 2.13 to 2 d.p
Source: Myschool Past Questions Archive (Record #57871)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
(a) Ali and Yusif shared N420.000.00 in the ratio 3.,; 5 : 8 respectively. Find the sum of Ali and Yusuf's shares
(b) Solve: 2(\(\frac{1}{8}\))\(^x\) = 32\(^{x - 1}\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Total ratio = 3 + 5 + 8 = 26
Ali's share is \(\frac{3}{16}\) x N420,000
= N78,000
Yusuf's share is \(\frac{8}{16}\) x N420,000
= N210,000
Sum of Ali's share + Yusuf's share
= N78,750 + N210,000
= N288,750
(b) 2(\(\frac{1}{8}\))\(^2\) = 32\(^{x - 1}\)
2\(^{1}\) x 2\(^{3x}\) = 2\(^{5x - 5}\)
2\(^{1 - 3x}\) = 2\(^{5x - 5}\)
1 - 3x = 5x - 5
-3x - 5x = -5 -1
-8x = -6
x = \(\frac{-6}{-8}\)
= \(\frac{3}{4}\)
Source: Myschool Past Questions Archive (Record #57950)
Mathematics
WAEC / WASSCE 2019
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQRS is a quadrilateral, < PQR = < PRS = 90\(^o\), |PQ| =3cm, |QR| = 4cm and |PS| = 13 cm. Find the area of the quadrilateral.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
In \(\triangle\)PQR
|PR|\(^2\) = |PQ|\(^2\) + |QR|\(^2\)
= 3\(^2\) + 4\(^2\)
|PR|\(^2\) = 9 + 16
|PR| = \(\sqrt{25}\) = 5cm
In \(\triangle\)PRS
|PS|\(^2\) = |PR|\(^2\) + |RS|\(^2\)
13\(^2\) = 5\(^2\) + |RS|\(^2\)
169 =25 + |RS|\(^2\)
|RS|\(^2\) = 169 - 25
= 144
|RS| = \(\sqrt{144}\) = 12cm
Area of quadrilateral PQRS = Area of \(\triangle\)PQR + Area of \(\triangle\)PQR + Area of \(\triangle\)PRS
= (\(\frac{1}{2} \times 3 \times 4\)) + (\(\frac{1}{2} \times 12 \times 5\))
= 6cm\(^2\) + 30cm\(^2\)
= 36cm\(^2\)
Source: Myschool Past Questions Archive (Record #58025)
Mathematics
WAEC / WASSCE 2019
Statistics and Probability
Solution
Three red balls, five green balls, and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is \(\frac{1}{6}\) find;
(a) The number of blue balls in the sack
(b) the probability of picking a green ball
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Let number of blue balls = n
Total number of balls = 3 + 5 + n
= 8 + n
p(red ball) = \(\frac{\text{no. of red balls}}{\text{Total no. of balls}}\)
\(\frac{1}{6} = \frac{3}{8 + n}\)
8 + n = 18
n + 18 - 8 = 10
number of blue balls = 10
(b) P(green ball) = \(\frac{\text{No. of green balls}}{\text{Total no. of balls}}\)
= \(\frac{5}{3 + 5 + 10}\)
= \(\frac{5}{18}\)
= 0.2778
Source: Myschool Past Questions Archive (Record #58027)
Mathematics
WAEC / WASSCE 2019
Geometry and Trigonometry
Solution
The force of attraction F, between two bodies, varies directly as the product of their masses, \(m_1\) and m\(_2\) and inversely as the square of the distance, d, between them. Given that F = 20N, when m\(_1\) = 25kg, m\(_2\) = 10kg and d = 5m, find:
(I) an expression for F in terms of m\(_1\), m\(_2\) and d;
(ii) the distance, d for F = 30N, m\(_1\) = 7.5kg and m\(_2\) = 4kg
(b) Find the value of x in the diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
F \(\infty\) \(\frac{m_1m_ 2}{d^2}\)
F = \(\frac{km_1m_2}{d^2}\)
20 = \(\frac{k(25)(10)}{(25)}\)
\(\frac{10k}{10} = \frac{20}{10}\)
k = 2
(i) F = \(\frac{2m_1m_2}{d_2}\)
(ii) Since F = \(\frac{2m_1m_2}{d_2}\)
30 = \(\frac{2(7.5)(4)}{d^2}\)
30 \(\frac{60}{d^2}\)
d\(^2\) = \(\frac{60}{30}\)
d\(^2\) = 2
d = \(\sqrt{2}\)m
(b) Sum of angles = (n - 2) 180
= (5 - 2) 180
= 540\(^o\)
x + x + 20 + x + 60 + x + 40 + x + 80 = 540
5x + 200 = 540
5x = 540 - 200
\(\frac{5x}{5} = \frac{340}{5}\)
x = 68\(^o\)
Source: Myschool Past Questions Archive (Record #58032)