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Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
A pole of length L leans against a vertical wall so that it makes an angle of 60o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.
A
\(16\sqrt{3}m\)
B
\(4\sqrt{3}m\)
C
\(\frac{\sqrt{3}}{16}\)
D
\(\frac{16\sqrt{3}}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\)
When we rationalize gives \(\frac{16\sqrt{3}}{3}\)
Source: Myschool Past Questions Archive (Record #26403)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
The height of a pyramid on square base is 15cm. If the volume is 80cm\(^3\), find the length of the side of the base
A
3.3cm
B
5.3cm
C
4.0cm
D
8.0cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Volume = \(\frac{1}{3} b^2 h\)
\(\therefore b = \sqrt{\frac{3v}{h}}\)
\(b = \sqrt{\frac{3(80)}{15}} \)
\(b = \sqrt{16} = 4.0 cm\)
Source: Myschool Past Questions Archive (Record #26405)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
Two numbers 2\(4_x\) and 31\(_y\) are equal in value when converted to base ten. Find the equation connecting x and y
A
2x = 3(y - 1)
B
4x - y = 1
C
3y + 2x = 3
D
3y = 2 (x + 3)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
24\(_x\) = 31\(_y\)
\(2 \times x^1 + 4 \times x^0 = 3 \times y^1 + 1 \times y^0\)
\(2x + 4 = 3y + 1 \implies 2x = 3y + 1 - 4\)
\(2x = 3y - 3 = 3(y - 1)\)
Source: Myschool Past Questions Archive (Record #26407)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed
A
\(\frac{2x + 2y}{3}kmh^{-1}\)
B
\(\frac{x + y}{3}kmh^{-1}\)
C
\(\frac{x + 2y}{3}kmh^{-1}\)
D
\(\frac{2x + y}{3}kmh^{-1}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Travelled x km/h for 1 hour \(\therefore\) traveled x km in the first hour.
Traveled y km/h for 2 hours \(\therefore\) traveled 2y km in the next 2 hours.
Average speed = \(\frac{x + 2y}{1 + 2}\)
= \(\frac{x + 2y}{3} kmh^{-1}\)
Source: Myschool Past Questions Archive (Record #26408)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm\(^2\), find the length of the base.
A
24cm
B
36cm
C
72cm
D
144cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area = \(\frac{1}{2} \times base \times height\)
\(height : base = 1 : 3\)
\(\implies base = 3 \times height\)
Let height = h;
Area = \(\frac{1}{2} \times 3h \times h = 216\)
\(3h^2 = 216 \times 2 = 432\)
\(h^2 = \frac{432}{3} = 144\)
\(h = \sqrt{144} = 12.0 cm\)
\(\therefore base = 3 \times 12 = 36 cm\)
Source: Myschool Past Questions Archive (Record #26409)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
Water flows from a tap into cylindrical container at the rate 5\picm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.
A
2cm
B
5cm
C
8cm
D
15cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Volume of water after 9 seconds = \(5\pi \times 9 = 45\pi cm^3\)
Volume of cylinder = \(\pi r^2 h\)
\(\therefore \pi r^2 h = 45\pi\)
\(\pi \times 3^2 \times h = 45\pi\)
\(\implies 9h = 45 \)
\(h = 5 cm\)
(where h = height of the water after 9 secs)
Source: Myschool Past Questions Archive (Record #26410)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
If \(x^2 +15x + 50 = ax^2 + bx + c = 0\). Which of the following statement is not true?
A
x = -5
B
x = 10
C
x + 10 = 0
D
bc = 750
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(x^2 + 15x + 50 = 0\)
\(x^2 + 5x + 10x + 50 = 0\)
\(x(x + 5) + 10(x + 5) = 0\)
\((x + 5)(x + 10) = 0\)
x + 5 = 0 or x + 10 = 0.
Comparing \(x^2 + 15x + 50\) with \(ax^2 + bx + c\), b = 15 and c = 50.
\(\therefore\) bc = 750.
Source: Myschool Past Questions Archive (Record #26413)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Diagram
Solution
In the diagram, DE||BC, |AD| = x cm and |DB| = |AE| = ycm. Find |CE| in terms of x and y
Attached Figure / Diagram
A
x
B
\(\frac{x^2}{y}\)
C
\(\frac{y}{x^2}\)
D
\(\frac{y^2}{x}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{x}{x+y}=\frac{y}{x+CE}\\
xy + xCE = xy + y^2\\
∴xCE = y^2\\
CE=\frac{y^2}{x}\)
Source: Myschool Past Questions Archive (Record #26417)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Diagram
Describe the locus L shown in the diagram below
Attached Figure / Diagram
A
Locus of points equidistant from X and Y
B
Locus of points equidistant from X and Z
C
Locus of points equidistant from \(\bar{ZX}\) and \(\bar{ZY}\)
D
Locus of points equidistant from \(\bar{XY}\) and \(\bar{ZY}\)
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 #26418)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Diagram
Which of the following statements is true from the diagram above?
Attached Figure / Diagram
A
m + n = 90o
B
m = n
C
n > 45o
D
m o
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 #26419)
Mathematics
UTME / JAMB 1999
Number and Numeration
Diagram
the shaded portion in the graph is represented by
Attached Figure / Diagram
A
y + x - x \(\leq\) 0, y - x \(\leq\) 0
B
y - x + x, 3 \(\leq\) 0, y - x \(\geq\) 0
C
y + x - 3 \(\geq\) 0, y + x \(\leq\) 0
D
y - x + x3 \(\geq\) 0, y + x \(\leq\) 0
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 #33084)
Mathematics
UTME / JAMB 1999
Number and Numeration
Solution
The sectoral allocations to various ministries in a state budget are as follows:
Agriculture - N25,000,000.00
Education - N20,000,000.00
Women affairs - N35,000,000.00
Commercial and Industries - N20,000,000.00
Use a pie chart to represent this information. The corresponding angle to Agriculture is
A
90\(^0\)
B
72\(^0\)
C
60\(^0\)
D
45\(^0\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Total allocation = ₦25m + ₦20m + ₦35m + ₦20m = ₦100,000,000
Fraction for Agriculture = ₦25,000,000 / ₦100,000,000 = 0.25 (or 25%)
Angle for Agriculture = 0.25 \times 360^\circ = 90^\circ
Other sectors:
- Education: (20/100) \times 360^\circ = 72^\circ
- Women Affairs: (35/100) \times 360^\circ = 126^\circ
- Commercial and Industries: (20/100) \times 360^\circ = 72^\circ
Source: Myschool Past Questions Archive (Record #33088)
Mathematics
UTME / JAMB 1999
Number and Numeration
Diagram
Solution
The grades of 36 students in a class test are as shown in the pie chart. How many student has excellent?
Attached Figure / Diagram
A
7
B
8
C
9
D
12
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
degree of excellent in the pie-chart = 360o - (90o + 120o + 80o) = 360o - 290o = 70o
= 360o = 36 students
70 = \(\frac{70^o}{360^o}\) x 36 = 7
Source: Myschool Past Questions Archive (Record #33090)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
(a) Simplify \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
(b) Evaluate \(\log_{5} (\frac{3}{5}) + 3 \log_{5} (\frac{5}{2}) - \log_{5} (\frac{81}{8})\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
= \((\frac{25}{4})^{\frac{1}{2}} \times 2^{4} \div (\frac{2}{15})^{2}\)
= \(\frac{5}{2} \times 16 \times \frac{225}{4} = 2250\)
(b) \(\log_{5} (\frac{3}{5}) + 3\log_{5} (\frac{15}{2}) - \log_{5} (\frac{81}{8})\)
= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 15 - 3\log_{5} 2 - \log_{5} 81 + \log_{5} 8\)
= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} (3 \times 5) - 3\log_{5} 2 - \log_{5} 3^{4} + \log_{5} 2^{3}\)
= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 3 + 3\log_{5} 5 - 3\log_{5} 2 - 4\log_{5} 3 + 3\log_{5} 2\)
= \(2\log_{5} 5 \)
= \(2\)
Source: Myschool Past Questions Archive (Record #54376)
Mathematics
WAEC / WASSCE 1999
Number and Numeration
Solution
(a) If \(\varepsilon\) is the set \({1, 2, 3,..., 19, 20}\) and A, B and C are subsets of \(\varepsilon\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;
(b) Find : (i) \(A \cap B\) ; (ii) \(A \cap C\) ; (iii) \(B \cup C\).
(c) Using your results in (b), show that \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) A = {5, 10, 15, 20}
(ii) B = {4, 8, 12, 16, 20}
(iii) C = {3, 6, 9, 12, 15, 18}
(b) (i) \(A \cap B = {20}\)
(ii) \(A \cap C = {15}\)
(iii) \(B \cap C = {12}\)
(c) \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\)
\((A \cap B) \cup (A \cap C) = {15, 20}\)
\(A \cap (B \cup C) = {5, 10, 15, 20} \cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\)
= \({15, 20}\)
\(\therefore (A \cap B) \cup (A \cap C) = A \cap (B \cup C)\)
Source: Myschool Past Questions Archive (Record #54414)