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Mathematics
WAEC / WASSCE 1998
Indices, Logarithms & Surds
Solution
Simplify \(\frac{8^{\frac{2}{3}} \times 27^{\frac{-1}{3}}}{64^{\frac{1}{3}}}\)
A
-3
B
\(\frac{1}{9}\)
C
\(\frac{1}{3}\)
D
\(\frac{27}{8}\)
E
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{8^{\frac{2}{3}} \times 27^{\frac{-1}{3}}}{64^{\frac{1}{3}}}\)
Calculating each part:
1. \(8 = 2^3 \Rightarrow 8^{\frac{2}{3}} = 4\)
2. \(27 = 3^3 \Rightarrow 27^{\frac{-1}{3}} = \frac{1}{3}\)
3. \(64 = 4^3 \Rightarrow 64^{\frac{1}{3}} = 4\)
Now substituting these values:
\(\frac{4 \times \frac{1}{3}}{4} = \frac{1}{3}\)
Thus, the simplified expression is: \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #24003)
Mathematics
WAEC / WASSCE 1998
Linear, Simultaneous & Quadratic Equations
Solution
Solve the equation 5x2 - 4x - 1 = 0
A
-1, 1/5
B
-1, -1/5
C
1, 1/5
D
1, -1/5
E
-1, 5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
5x2 - 4x - 1 = 0
5x2 5x + x -1 = 0
5x(x - 1) + (x - 1) = 0
(5x + 1)(x -1) = 0
x = -1/5 or 1 = 0
Source: Myschool Past Questions Archive (Record #24004)
Mathematics
WAEC / WASSCE 1998
Plane Geometry, Polygons & Circle Theorems
p and q are two positive numbers such that p > 2q. Which one of the following statements is not true?
A
-p < -2q
B
-p > -2q
C
-q < 2p
D
q < 1/2p
E
p2 > 2q2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #24012)
Mathematics
WAEC / WASSCE 1998
Plane Geometry, Polygons & Circle Theorems
Solution
For what value of x is the expression x - 5/x2 - 2x - 3 not defined?
A
3, 1
B
-1, -3
C
-1, 3
D
3, -2
E
1, -3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x2 - 2x - 3 = 0
x2 - 3x + x - 3 = 0
x(x - 3) + (x - 3) = 0
(x + 1)(x - 3) = 0
x = -1 or 3
Source: Myschool Past Questions Archive (Record #24053)
Mathematics
WAEC / WASSCE 1998
Algebraic Expressions, Polynomials & Factorisation
Solution
Make S the subject of the formula: \(V = \frac{K}{\sqrt{T-S}}\)
A
\(T-\frac{K^2}{V^2} = S\)
B
\(T+\frac{K^2}{V^2} = S\)
C
\(T-\frac{K^2}{V} = S\)
D
\(T-\frac{K}{V} = S\)
E
\(T-\frac{K}{V^2} = S\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(V = \frac{K}{\sqrt{T-S}}\)
square both sides of the equation
\(V^2 = \frac{K^2}{T-S}\)
cross multiply
\(V^2 (T - S) = K^2\)
\(T - S = \frac{K^2}{V^2}\)
\(T = \frac{K^2}{V^2} + S\)
\(T = \frac{K^2}{V^2} + S\)
Source: Myschool Past Questions Archive (Record #24079)
Mathematics
WAEC / WASSCE 1998
Linear, Simultaneous & Quadratic Equations
Solution
Find (X-Y) if 4x - 3y = 7 and 3x - 2y = 5
A
4
B
3
C
2
D
-2
E
-3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
We are given the following simultaneous equations:
\(4x - 3y = 7 \quad \text{(Equation 1)}\)
\(3x - 2y = 5 \quad \text{(Equation 2)}\)
To eliminate \(y\), we multiply Equation 1 by 2 and Equation 2 by 3:
\(2(4x - 3y) = 2(7) \implies 8x - 6y = 14 \quad \text{(Equation 3)}\)
\(3(3x - 2y) = 3(5) \implies 9x - 6y = 15 \quad \text{(Equation 4)}\)
Now we subtract Equation 3 from Equation 4:
\((9x - 6y) - (8x - 6y) = 15 - 14\)
\(x = 1\)
Substituting \(x = 1\) into Equation 2:
\(3(1) - 2y = 5\)
\(3 - 2y = 5\)
\(-2y = 2 \implies y = -1\)
Now we have \(x = 1\) and \(y = -1\). Thus,
\(X - Y = x - y = 1 - (-1) = 1 + 1 = 2\)
Therefore, the value of \(X - Y\) is: 2.
Source: Myschool Past Questions Archive (Record #26335)
Mathematics
WAEC / WASSCE 1998
Inequalities & Linear Programming
Solution
solve the inequality \((Y-2)<\frac{y}{3}\)
A
y > -9
B
y < 3
C
y > 4
D
y < 9
E
y > 0
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((Y-2)<\frac{y}{3}\)
3y - 6 < y
2y< 6
y < \(\frac{6}{2}\)
y < 3.
Source: Myschool Past Questions Archive (Record #26336)
Mathematics
WAEC / WASSCE 1998
Plane Geometry, Polygons & Circle Theorems
Solution
If h(m+n) = m(h+r) find h in terms of m, n and r
A
\(h=\frac{mr}{2m+n}\)
B
\(h=\frac{mr}{n+m}\)
C
\(h=\frac{m+n}{n}\)
D
\(h=\frac{m+n}{m}\)
E
\(h=\frac{mr}{n}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(h(m+n) = m(h+r)\\
hm+hn=hm+mr\\
hn=mr\\
h=\frac{mr}{n}\)
Source: Myschool Past Questions Archive (Record #26337)
Mathematics
WAEC / WASSCE 1998
Mensuration (Areas, Perimeters & Volumes)
Solution
The radius of a geographical globe is 60cm. Find the length of the parallel of latitude 60oN
A
\(66\pi cm\)
B
\(60\pi cm\)
C
\(30\pi cm\)
D
\(15\pi cm\)
E
\(6\pi cm\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The radius of the globe (sphere), R = 60 cm.
At latitude 60^\circN, the radius of the parallel of latitude is given by:
r = R \(\times \cos(60^\circ)\)
\(\cos(60^\circ) = \frac{1}{2}\)
r = 60 \(\times \frac{1}{2} = 30 \text{ cm}\)
The length of the parallel is the circumference of this circle:
\(\text{Length} = 2\pi r = 2\pi \times 30 = 60\pi \text{ cm}\)
This is the actual distance around the globe along the 60^\circN parallel.
Source: Myschool Past Questions Archive (Record #26339)
Mathematics
WAEC / WASSCE 1998
Indices, Logarithms & Surds
Diagram
Solution
In the diagram above, O is the center if the two concentric circle of radii 13cm and 10cm respectively. Find the area of the shaded portion in the sector with angle 120^\circ at the center
Attached Figure / Diagram
A
\(\frac{1}{3}\pi cm^2\)
B
\(\pi cm^2\)
C
\(3\pi cm^2\)
D
\(23\pi cm^2\)
E
\(46\pi cm^2\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{120\pi}{360}(R^2 - r^2)\)
= \(\frac{1}{3}\times \pi (13^2 - 10^2)\)
= \(\frac{1}{3}\times \pi \times 69\) = 23\(\pi cm^2\)
Source: Myschool Past Questions Archive (Record #26340)
Mathematics
WAEC / WASSCE 1998
Fractions, Decimals, Percentages & Approximations
Solution
The following numbers represent at a set of scores for a class of 32 students, where the maximum score possible was 12, 6, 5, 9, 4, 4, 8, 7, 5, 6, 3, 2, 5, 4, 6, 9, 10, 4, 3, 2, 3, 4, 6, 8, 7, 4, 2, 1, 8, 7, 7, 6, 11. What is the percentage of the class, correct to the nearest whole number, scored above 6?
A
34%
B
35%
C
37%
D
39%
E
36%
F
38%
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\text{There are 33 scores in total. Hence, the class size is: } n = 33\)
\(\text{Scores greater than 6 are: } 7, 8, 9, 10, 11, 12.\)
\(\text{Number of students who scored above 6: } f = 12\)
\(\text{Fraction of students who scored above 6: } \frac{f}{n} = \frac{12}{33}\)
\(\text{Convert the fraction to percentage: } \frac{12}{33} \times 100 = 36.36\%\)
\(\text{Rounded to the nearest whole number: } 36\%\)
Source: Myschool Past Questions Archive (Record #26342)
Mathematics
WAEC / WASSCE 1998
Permutations, Combinations & Probability
Solution
A die with faces numbered 1 to 6 is rolled once. What is the probability of obtaining 4?
A
\(\frac{1}{36}\)
B
\(\frac{1}{6}\)
C
\(\frac{1}{3}\)
D
\(\frac{1}{2}\)
E
\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
When the die is rolled, occasion of getting 4
P(obtaining a 4) = \(\frac{1}{6}\)
= \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #26343)
Mathematics
WAEC / WASSCE 1998
Mensuration (Areas, Perimeters & Volumes)
Diagram
Solution
In the diagram above, the area of triangle ABC is 35cm\(^2\). Find the value of y
Attached Figure / Diagram
A
5cm
B
10cm
C
17.5cm
D
28cm
E
70cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area of triangle ABC = 35cm\(^2\).
Area of triangle ABC = \(\frac{1}{2}\) x B x H
= 0.5 x y x 7 = 35
= 3.5y = 35
y = \(\frac{35}{3.5}\) = 10 cm.
Source: Myschool Past Questions Archive (Record #26344)
Mathematics
WAEC / WASSCE 1998
Linear, Simultaneous & Quadratic Equations
Diagram
Solution
The graph is that of the quadratic expression
Attached Figure / Diagram
A
\(y=x^2 + x - 2\)
B
\(y=x^2 - x + 2\)
C
\(y=x^2 - x - 2\)
D
\(y=x + x + 2\)
E
\(y=x^2 - 2 x - 2\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The curve cuts the x-axis at x = -1 and x = 2
(x + 1)(x - 2) = 0
x\(^2\) - 2x + x - 2 = 0
x\(^2\) - x - 2 = 0
Source: Myschool Past Questions Archive (Record #26345)
Mathematics
WAEC / WASSCE 1998
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
The values of x when y = -1 are approximately
Attached Figure / Diagram
A
-10 and 2.0
B
-1.3 and 2.3
C
-0.6 and 2.6
D
-0.6 and 1.6
E
1.6 and 1.6
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Draw a straight line from y = -1 to meet the curve on both sides. From the point where the straight line touches the curve, draw a line from there to meet the x-axis, then read off the values.
So when y = -1, x = - 0.6 or 1.6
Source: Myschool Past Questions Archive (Record #26346)