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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
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
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
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
UTME / JAMB 2019
Number and Numeration
Solution
A binary operation \(\otimes\) is defined by \(m \otimes n = mn + m - n\) on the set of real numbers, for all m, n \(\in\) R. Find the value of 3 \(\otimes\) (2 \(\otimes\) 4).
A
6
B
25
C
15
D
18
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(m \otimes n = mn + m - n\)
3 \(\otimes\) (2 \(\otimes\) 4)
2 \(\otimes\) 4 = 2(4) + 2 - 4 = 6
3 \(otimes\) 6 = 3(6) + 3 - 6 = 15
Source: Myschool Past Questions Archive (Record #54945)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
In a class of 50 students, 40 students offered Physics and 30 offered Biology. How many offered both Physics and Biology?
A
42
B
20
C
70
D
54
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
n(Total) = 50
n(Physics) = 40
n(Biology) = 30
Let n(Physics and Biology) = x
n(Physics only) = 40 -x
n(Biology only) = 30 - x
40 - x + 30 - x + x = 50
70 - x = 50
x = 20
Source: Myschool Past Questions Archive (Record #54947)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
Rationalize \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}\)
A
\(-5 - 2\sqrt{6}\)
B
\(-5 + 3\sqrt{2}\)
C
\(5 - 2\sqrt{3}\)
D
\(5 + 2\sqrt{6}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}\)
= \((\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}})(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}})\)
= \(\frac{2 + \sqrt{6} + \sqrt{6} + 3}{2 - \sqrt{6} + \sqrt{6} - 3}\)
= \(\frac{5 + 2\sqrt{6}}{-1}\)
= \(- 5 - 2\sqrt{6}\)
Source: Myschool Past Questions Archive (Record #54951)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
\(\frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} - \frac{1}{3}}\)
A
\(\frac{31}{50}\)
B
\(\frac{20}{31}\)
C
\(\frac{31}{20}\)
D
\(\frac{50}{31}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} - \frac{1}{3}}\)
\(\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4}\)
= \(\frac{5}{6}\)
\(\frac{1}{4} + \frac{3}{5} - \frac{1}{3} = \frac{15 + 36 - 20}{60}\)
= \(\frac{31}{60}\)
\(\therefore \frac{\frac{2}{3} \div \frac{4}{5}}{\frac{1}{4} + \frac{3}{5} - \frac{1}{3}} = \frac{5}{6} \div \frac{31}{60}\)
= \(\frac{5}{6} \times \frac{60}{31}\)
= \(\frac{50}{31}\)
Source: Myschool Past Questions Archive (Record #54958)
Mathematics
UTME / JAMB 2019
Number and Numeration
If \(6x^3 + 2x^2 - 5x + 1\) divides \(x^2 - x - 1\), find the remainder.
A
9x + 9
B
2x + 6
C
6x + 8
D
5x - 3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #54960)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
In how many ways can the word MATHEMATICIAN be arranged?
A
6794800 ways
B
2664910 ways
C
6227020800 ways
D
129729600 ways
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
MATHEMATICIAN = 13 letters with 2M, 3A, 2T, 2I.
Hence, the word MATHEMATICIAN can be arranged in \(\frac{13!}{2! 3! 2! 2!}\)
= 129729600 ways
Source: Myschool Past Questions Archive (Record #54962)
Mathematics
UTME / JAMB 2019
Number and Numeration
Diagram
Solution
In the figure above, |CD| is the base of the triangle CDE. Find the area of the figure to the nearest whole number.
Attached Figure / Diagram
A
56 cm\(^2\)
B
24 cm\(^2\)
C
42 cm\(^2\)
D
34 cm\(^2\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of rectangle ABCD = length x breadth
= 7 x 4
= 28 cm\(^2\)
Area of triangle CDE = \(\frac{1}{2}\) base x height
= \(\frac{1}{2} \times 3 \times 4\)
= 6 cm\(^2\)
Area of the figure = 28 cm\(^2\) + 6 cm\(^2\)
= 34 cm\(^2\)
Source: Myschool Past Questions Archive (Record #55004)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
The marks scored by 30 students in a Mathematics test are recorded in the table below:
Scores (Mark)
0
1
2
3
4
5
No of students
4
3
7
8
6
2
What is the total number of marks scored by the children?
A
82
B
15
C
63
D
75
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Scores (Mark)
0
1
2
3
4
5
No of students
4
3
7
8
6
2
fx
0
3
14
24
24
10
75
Source: Myschool Past Questions Archive (Record #55016)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
A
\(\frac{\mathrm d y}{\mathrm d x} = 15x^2 - 4x^{-2} - 3x^{-2}\)
B
\(\frac{\mathrm d y}{\mathrm d x} = 6x + 4x^{-1} - 3x^{-4}\)
C
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)
D
\(\frac{\mathrm d y}{\mathrm d x} = 12x^2 + 4x^{-1} - 3x^{-2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(y = 6x^3 + 2x^{-2} - x^{-3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)
Source: Myschool Past Questions Archive (Record #55023)
Mathematics
UTME / JAMB 2019
Number and Numeration
Solution
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) is equal to
A
\(\frac{12x - 2}{4x^2}\)
B
\(\frac{43x^2 - 2x}{7x}\)
C
\(\frac{4x^2 - 2}{7x + 6}\)
D
\(\frac{12x^2 - 2}{4x^3 - 2x}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)
Let u = 4x\(^3\) - 2x.
\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)
\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\)
\(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\)
= \(\frac{12x^2 - 2}{4x^3 - 2x}\)
Source: Myschool Past Questions Archive (Record #55035)