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Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
If P(2,3) and Q)2, 5) are points on a graph, calculate the length PQ
A
6 units
B
5 units
C
4 units
D
2 units
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\sqrt{(x_2 - x_1)^2 + (y_2 - Y_1)^2}\)
= \(\sqrt{(2 - 2)^2 + (5 - 3)^2}\)
= \(\sqrt{0^2 + 2^2}\)
= \(\sqrt{4}\)
= 2 units
Source: Myschool Past Questions Archive (Record #50002)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
In what number base was the addition 1 + nn = 100, where n > 0, done?
A
n - 1
B
n + 1
C
n
D
n + 2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Assuming the numbers 1 + nn =100 were added in base n + 1
n
n
1
1 0
0
n + 1 = n +1 and n + 1 in base n + 1 = 10, write 0 and carry 1, which will be added to the second n, resulting in another n + 1 in base n + 1 = 10.
So, putting the result together 1 + nn = 100 provided, the addition was carried out in base n + 1.
Source: Myschool Past Questions Archive (Record #50025)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
A
4
B
\(\sqrt{3} + 4\)
C
4 \(\sqrt{2}\)
D
4\(\sqrt{3} + 4\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)
2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4
= 4
Source: Myschool Past Questions Archive (Record #50027)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
Simplify the expression \(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\)
A
\(a^2 - b^2\)
B
\(b^2 - a^2\)
C
\(a^2b - ab^2\)
D
\(ab^2 - a^2b\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{a^2 b^4 - b^2 a^4}{ab(a + b)}\) = \(\frac{a^2 b^2(b^2 - a^2)}{ab(a + b)}\)
= \(\frac{ab [(b - a) (b + a)]}{a + b}\)
= ab(b - a)
= \(ab^2 - a^2b\)
Source: Myschool Past Questions Archive (Record #50112)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Consider the statements: p it is hot, q: it is raining
Which of the following symbols correctly represents the statement "It is raining if and only if it it is cold"?
A
p \(\iff\) \(\sim\)q
B
p \(\iff\) q
C
\(\sim\)p \(\iff\) \(\sim\)q
D
q \(\iff\) \(\sim\)p
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #50126)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
Given that t = \(2 ^{-x}\), find \(2 ^{x + 1}\) in terms of t.
A
\(\frac{2}{t}\)
B
\(\frac{t}{2}\)
C
\(\frac{1}{2t}\)
D
t
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
t = \(2^{-x} = \frac{1}{2^{x}}\)
\(\implies 2^{x} =\frac{1}{t}\)
\(2^{x+1} = 2^{x} \times 2^{1}\)
= \(\frac{1}{t} \times 2 = \frac{2}{t}\)
Source: Myschool Past Questions Archive (Record #50130)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
If the simple interest on a certain amount of money saved in a bank for 5 years at 2\(\frac{1}{2}\)% annum is N500.00, calculate the total amount due after 6 years at the same rate
A
N2,500.00
B
N2,600.00
C
N4,500.00
D
N4,600.00
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
P = \(\frac{100l}{RT} = \frac{100 \times 500}{5 \times 2.5} = \frac{50,000}{12.5}\)
= 4,000
Annual interest is \(\frac{500}{5}\) = 100
for 6 year = 4,000 + 100 + 100 + 100 + 100 + 100 + 100
= N 4,600
Source: Myschool Past Questions Archive (Record #50135)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
A circular pond of radius 4m has a path of width 2.5m round it. Find, correct to two decimal places, the area of the path. [Take\(\frac{22}{7}\)]
A
7.83\(m^2\)
B
32.29\(m^2\)
C
50.29\(m^2\)
D
82.50\(m^2\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of path = Area of (pond+path) - Area of pond
The area of the pond with the path: The radius = (4 + 2.5)m = 6.5m
Area = \(\pi \times r^{2}\) = \(\frac{22}{7} \times 6.5^{2} \approxeq 132.79m^{2}\)
Area of the pond = \(\frac{22}{7} \times 4^{2} \approxeq 50.29m^{2}\)
Area of the path = (132.79 - 50.29)m^{2} = 82.50m^{2}\)
Source: Myschool Past Questions Archive (Record #50142)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Diagram
Solution
Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n \(\oplus 4\))
Attached Figure / Diagram
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(n \(\oplus\) 4) \(\oplus\) 3 = 0 (mod 5)
(3 \(\oplus\) 4) \(\oplus\) 3
12 \(\oplus\) 3 = 15 (mod 5)
(5 x 3 + 0) = 0 (mod 5)
Source: Myschool Past Questions Archive (Record #50148)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Diagram
Solution
In the diagram, XY is a straight line.
Attached Figure / Diagram
A
60\(^o\)
B
90\(^o\)
C
100\(^o\)
D
120\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Source: Myschool Past Questions Archive (Record #50155)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Diagram
Solution
In the diagram, NQ//TS,
Attached Figure / Diagram
A
110\(^o\)
B
130\(^o\)
C
140\(^o\)
D
150\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< TSR = 180 - (80 + 50)
= 180 - (130)
= 50\(^o\)
< QPR = < TSR corresponding < s
< NPR + QPR = < NPR
180\(^o\) - < QPR = < NPR
180\(^o\) - 50 = < NPR
< NPR = 130\(^o\)
Source: Myschool Past Questions Archive (Record #50182)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
The diagonal of a square is 60 cm. Calculate its peremeter
A
20\(\sqrt{2}\)
B
40\(\sqrt{2}\)
C
90\(\sqrt{2}\)
D
120\(\sqrt{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(60^2 + x^2 + x^2\)
\(360^2 = 2x^2\)
\(x^2\) = 1800
x = \(\sqrt{1800}\)
x = 42.4264
x = 42.4264
perimeter = 4x
= 4 x 42.4264
= 169.7056
= 120\(\sqrt{2}\)
= 120\(\sqrt{2}\)
Source: Myschool Past Questions Archive (Record #50187)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Diagram
Solution
The graph of y = \(ax^2 + bx + c\) is shown oon the diagram. Find the minimum value of y
Attached Figure / Diagram
A
-2.0
B
-2.1
C
-2.3
D
-2.5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Find the minimum value of y is the point where the base of the curve corresponds to the value of y i.e. at -2.1
Source: Myschool Past Questions Archive (Record #50189)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
(a) A manufacturing company requires 3 hours of direct labour to process N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget at N18.25 per hour?
(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in the two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the:
(i) values of x and y
(ii) interest paid by the second bank.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Worth of raw materials = N30,450.00
Worth of raw material to get processed materials = N87.00
Amount of raw material to get processed material = \(\frac{30,450}{87}\)
Time required for direct labour = 3 hours
Amount that will be budgeted for direct labour at N18.25 per hour =
\(3 \times 18.25 \times \frac{30,450}{87}\)
= \(\frac{1667137.5}{87}\)
= N19,162.50
(b) Total amount invested \(N(x + y) = N8,000,000 ....(1)\)
Interest from bank M = \(\frac{x \times 4 \times 6}{100} = 0.24x\)
Interest from bank N = \(\frac{y \times 4 \times 8}{100} = 0.32y\)
\(0.24x + 0.32y = 2,320,000 .... (2)\)
(i) Multiplying (1) by 0.24 in order to eliminate x, we have
\(0.24(x + y) = 0.24(8,000,000.00)\)
\(0.24x + 0.24y = 1,920,000.00 .... (3)\)
(2) - (3) : \(0.32y - 0.24y = 2,320,000 - 1,920,000 = 400,000\)
\(0.08y = 400,000 \implies y = \frac{400,000}{0.08} = N5,000,000\)
\(x = 8,000,000 - y\)
= \(8,000,000 - 5,000,000\)
= \(N3,000,000\)
Therefore, x = N3,000,000 and y = N5,000,000.
(ii) Interest paid by bank N = 0.32y
= \(0.32 \times N5,000,000\)
= \(N1,600,000\)
Source: Myschool Past Questions Archive (Record #52338)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
(a) The operation (*) is defined on the set of real numbers, R, by \(x * y = \frac{x + y}{2}, x, y \in R\).
(i) Evaluate \(3 * \frac{2}{5}\).
(ii) If \(8 * y = 8\frac{1}{4}\), find the value of y.
(b) In \(\Delta ABC, \overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}\). If P is the midpoint of \(\overline{AB}\), express \(\overline{CP}\) as a column vector.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) (i) \(3 * \frac{2}{5} = \frac{3 + \frac{2}{5}}{2}\)
= \(\frac{\frac{17}{5}}{2}\)
= \(\frac{17}{10}\)
(ii) \(8 * y = 8\frac{1}{4}\)
\(\frac{8 + y}{2} = \frac{33}{4}\)
\(66 = 32 + 4y \implies 4y = 66 - 32 = 34\)
\(y = \frac{34}{4} = 8.5\)
(b) \(\overrightarrow{AP} = \frac{1}{2}(\overrightarrow{AB})\)
= \(\frac{1}{2} \begin{pmatrix} -4 \\ 6 \end{pmatrix}\)
= \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
Midpoint = \(\overrightarrow{AP} + \overrightarrow{PC} = \overrightarrow{AC}\)
\(\overrightarrow{PC} = \overrightarrow{AC} - \overrightarrow{AP}\)
= \(\begin{pmatrix} 3 \\ -8 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
= \(\begin{pmatrix} 5 \\ -11 \end{pmatrix}\)
\(\overrightarrow{CP} = -1 \times \begin{pmatrix} 5 \\ -11 \end{pmatrix}\)
= \(\begin{pmatrix} -5 \\ 11 \end{pmatrix}\)
Source: Myschool Past Questions Archive (Record #52423)