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Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
Simplify: \(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\)
A
\(\frac{x - 7}{x + 7}\)
B
\(\frac{x + 7}{x - 7}\)
C
\(\frac{x - 2}{x + 4}\)
D
\(\frac{x + 2}{x - 2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{x^2 - 5x - 14}{x^2 - 9x + 14}\)
\(\frac{(x - 7)(x + 2)}{(x - 7)(x - 2)}\)
= \(\frac{x + 2}{x - 2}\)
Source: Myschool Past Questions Archive (Record #57221)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
Which of these values would make \(\frac{3p - 1}{p^{2} - p}\) undefined?
A
1
B
\(\frac{1}{3}\)
C
-\(\frac{1}{3}\)
D
-1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
P\(^2\) - p = 0
p(p - 1) = 0
p = 0 or p = 1
Source: Myschool Past Questions Archive (Record #57223)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
The total surface area of a solid cylinder 165cm\(^2\). Of the base diameter is 7cm, calculate its height. [Take \(\pi = \frac{22}{7}\)]
A
7.5 cm
B
4.5 cm
C
4.0 cm
D
2.0 cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
165 = 2 x \(\frac{22}{7}\) x \(\frac{7}{2}\) x h + 2 x \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\)
165 = 22h + 77
\(\frac{22h}{22} = \frac{88}{22}\)
height = 4cm
Source: Myschool Past Questions Archive (Record #57228)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
If 2\(^{a}\) = \(\sqrt{64}\) and \(\frac{b}{a}\) = 3, evaluate a\(^2 + b^{2}\)
A
250
B
160
C
90
D
48
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
2\(^a\) = \(\sqrt{64}\)
2\(^a\) = 8
2\(^a\) = 2\(^3\)
a = 3
b = 3\(^a\)
b = 3 x 3 = 9
a\(^2\) + b\(^2\) = 3\(^2\) + 9\(^2\)
= 9 + 81 = 90
Source: Myschool Past Questions Archive (Record #57238)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
If log\(_x\) 2 = 0.3, evaluate log\(_x\) 8.
A
2.4
B
1.2
C
0.9
D
0.6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
log\(_x\)2 = 0.3
log\(_x\)8 = log\(_x\)2\(^3\) = 3 log\(_x\)2
= 3 x 0.3
= 0.9
Source: Myschool Past Questions Archive (Record #57255)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
Make b the subject of the relation lb = \(\frac{1}{2}\) (a + b)h
A
\(\frac{ah}{2l - h}\)
B
\(\frac{2l - h}{al}\)
C
\(\frac{al}{2l - h}\)
D
\(\frac{al}{2 - h}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
lb = \(\frac{1}{2} (a + b)h
2lb = ah + bh
2lb - bh = ah
\(\frac{b(2l - h)}{2l - h} = \frac{ah}{2l - h}\)
b = \(\frac{ah}{2l - h}\)
Source: Myschool Past Questions Archive (Record #57261)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
Eric sold his house through an agent who charged 8% commission on the selling price. If Eric received $117,760.00 after the sale, what was the selling price of the house?
A
$130,000.00
B
$128,000.00
C
$125,000.00
D
$120,000.00
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{8}{100} \times x + 117,760 = x\)
x - \(\frac{8x}{100} = 117,760
\(\frac{92x}{100} = 117,760\)
x = \(\frac{117,760 \times 100}{92}\)
Selling price = N128,000
Source: Myschool Past Questions Archive (Record #57268)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
Factorize completely; (2x + 2y)(x - y) + (2x - 2y)(x + y)
A
4(x - y)(x + y)
B
(4(x - y)
C
2(x - y)(x + y)
D
2(x - y)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(2x + 2y)(x - y) + (2x - 2y)(x - y)
2(x + y)(x - y) + 2(x - y)(x + y)
4(x - y)(x + y)
Source: Myschool Past Questions Archive (Record #57320)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
The foot of a ladder is 6m from the base of an electric pole. The top of the ladder rest against the pole at a point 8m above the ground. How long is the ladder?
A
14m
B
12m
C
10m
D
7m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
H\(^2\) = 8\(^2\) + 6\(^2\) + 6\(^2\)
H\(^2\) = 64 + 36 = 100
H = \(\sqrt{100}\) = 10m
Source: Myschool Past Questions Archive (Record #57361)
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
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
Number and Numeration
Solution
The data shows the marks obtained by students in a biology test
52
56
25
56
68
73
66
64
56
48
20
39
9
50
46
54
54
40
50
96
36
44
18
97
65
21
60
44
54
32
92
49
37
94
72
88
89
35
59
34
15
88
53
16
84
52
72
46
60
42
(a) Construct a frequency distribution table using the class interval 0 - 9, 10 - 19, 19, 20, 29...
(b) Draw a cumulative frequency curve for the distribution
(c) Use the graph to estimate the;
(i) median
(ii) Percentage of students who scored at least 66 marks, correct to the nearest whole number.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Class
Intervals
Class
Boundaries
Frequency
Cumulative
Frequency
0 - 9
0.5 - 9.5
1
1
10 - 19
9.5 - 19.5
3
4
20 - 29
19.5 - 29.5
3
7
30 - 39
29.5 - 39.5
6
13
40 - 4921
39.5 - 49.5
8
21
50 - 59
49.5 - 59.5
12
33
60 - 69
59.5 - 69.5
6
39
70 - 79
69.5 - 79.5
3
42
80 - 89
79.5 - 89.5
4
46
90 - 99
89.5 - 99.5
4
50
\(\sum f = 50\)
(b)
(c) (I) Mean mark = 49.9
(ii) Percentage of at least 66 mark = \(\frac{13}{50} \times 100\)%
= 26%
Source: Myschool Past Questions Archive (Record #58053)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
A woman bought 130 kg of tomatoes for 52,000.00. She sold half of the tomatoes at a profit of 30%. The rest of the tomatoes began to go bad, she then reduced the selling price per kg by 12%. Calculate:
(a) the new selling price per kg;
(ii) the percentage profit on the entire sales if she threw away 5 kg of bad tomatoes.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Weight of full tomatoes = 130kg
Weight of half tomatoes = \(\frac{130}{2}\) = 65kg
Cost price of half tomatoes = \(\frac{N52,000}{2}\)
= N26,000
Profit = \(\frac{S.P - C.P}{C.P}\) x 100
30 = \(\frac{(S.P - 26,000)}{25,000}\) x 100
S.P = N7800 + N26,000
S.P = N7,800 + N26,000
S.P of half tomatoes = N33,800
Selling price per kg = \(\frac{33,800}{65}\)
= N520 per kg
Reduction in selling price = 12%
= \(\frac{12}{100}\) x 520 = N62.40k per kg
New selling price per kg
= N520 - N62.40k
= N457.60k
(b) 5kg is thrown away remaining 60kg
The selling price of entire 60kg
= 60 x N457.60k = N27,456
The entire selling price
N33,800 + N27,456
= N61,256
% profit = \(\frac{(61.256 - 50,000)}{52,000}\) x 100
= 17.8%
Source: Myschool Past Questions Archive (Record #58164)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
(a) The third and sixth terms of a Geometric Progression (G.P) are and \(\frac{1}{4}\) and \(\frac{1}{32}\) respectively.
Find:
(i) the first term and the common ratio;
(ii) the seventh term.
(b) Given that 2 and -3 are the roots of the equation ax\(^2\) \pm bx + c = 0, find the values of a, b and c.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) (i) ar\(^2\) = \(\frac{1}{4}\) ........(i)
ar\(^5\) = \(\frac{1}{32}\) ........(ii)
\(\frac{ar^5}{ar^2} = \frac{1}{32} \div \frac{1}{4}\)
r\(^3 = \frac{1}{32} \times \frac{4}{1}\)
r\(^3 = \frac{1}{32} \times \frac{4}{1}\)
r\(^3 = 2^{-3} = (\frac{1}{2})^3\)
r = \(\frac{1}{2}\)
Since ar\(^2\) = \(\frac{1}{4}\)
a(\(\frac{1}{2}\))\(^2\) = \(\frac{1}{4}\)
a = 1
The first term is 1 ad the common ratio is \(\frac{1}{2}\)
(ii) T\(_7\) = ar\(^6\) = 1 x (\(\frac{1}{2}\))\(^6\)
The seventh term = \(\frac{1}{64}\)
(b) x = 2 and x = -3
a(2)\(^2\) + b(2) + c = 0
4a + 2b + c = 0
4a + 2b + c = 0.....(i)
a(-3)\(^2\) + b(-3) + c = 0
9a - 3b + c = 0.......(ii)
Solving eq.(i) and (ii)
4a + 2b + c = 0
-(9a - 3b + c) = 0
-5a + 5b = 0
(x - 2)(x + 3) = 0
x\(^2\) + 3x - 2x - 6 = 0
x\(^2\) + x - 6 = 0
a = 1, b = 1, c = -6
Source: Myschool Past Questions Archive (Record #58166)
Mathematics
WAEC / WASSCE 2019
Number and Numeration
Solution
(a) Given that sin y = \(\frac{8}{17}\) find the value of \(\frac{tan y}{1 + 2 tan y}\)
(b) An amount of N300,000.00 was shared among Otobo, Ada and Adeola. Otobo received N60,000.00, Ada received \(\frac{5}{10}\) of the remainder, while the rest went to Adeola. In what ratio was the money shared?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x\(^2\) = 172\(^2\) - 8\(^2\)
x\(^2\) = 289 - 64
x\(^2\) = 225
x = \(\sqrt{225}\)
x = 15
tan y \(\frac{8}{15}\)
1 + 2 tan y = 1 + 2(\(\frac{8}{15}\))
= \(\frac{1}{1} + \frac{6}{15}\)
= \(\frac{15 + 16}{15} = \frac{31}{15}\)
\(\frac{\tan y}{1 + 2 \tan y}\)
= \(\frac{8}{15} \div \frac{31}{15}\)
= \(\frac{8}{15} \times \frac{15}{31} = \frac{8}{31}\)
(b) Otobo shares = N60,000
Remainder = N300,000 - N60,000
= N240,000
Ada's share = \(\frac{5}{12}\) x N240,0000
= N100,000
Adeola's share =;
= 300,000 - N160,000 = N140,000
Ratio of the money is
Otobo : Ada : Adeola
60,000 : 100,000 : 140,000
60 : 100 : 140
6 : 10 : 14
3 : 5 : 7
Source: Myschool Past Questions Archive (Record #58167)