Below are carefully selected WASSCE-style SHS Mathematics questions with worked answers. The questions cover algebra, indices, logarithms, sequences, geometry, trigonometry, statistics, probability and financial mathematics.
Question 1: Indices
Simplify:
$$ \frac{2^5 \times 2^{-2}}{2^1} $$
Solution
Using the law:
$$ a^m \times a^n=a^{m+n} $$
We have:
$$ \frac{2^{5+(-2)}}{2^1} = \frac{2^3}{2^1} $$ $$ =2^{3-1} $$ $$ \boxed{4} $$
Answer: 4
Question 2: Logarithms
Evaluate:
$$ \log_{10}1000+\log_{10}0.01 $$
Solution
Since:
$$ 1000=10^3 $$
and
$$ 0.01=10^{-2} $$
Therefore:
$$ \log_{10}1000=3 $$ $$ \log_{10}0.01=-2 $$
Hence:
$$ 3+(-2)=1 $$ $$ \boxed{1} $$
Answer: 1
Question 3: Quadratic Equations
Solve:
$$ x^2-7x+12=0 $$
Solution
Factorising:
$$ x^2-7x+12=(x-3)(x-4) $$
Therefore:
$$ (x-3)(x-4)=0 $$
Hence:
$$ x-3=0 \quad \text{or} \quad x-4=0 $$ $$ x=3 \quad \text{or} \quad x=4 $$ $$ \boxed{x=3,\;4} $$
Answer: \(x=3\) or \(x=4\)
Question 4: Simultaneous Equations
Solve the simultaneous equations:
$$ 2x+y=11 $$ $$ x-y=1 $$
Solution
From the second equation:
$$ x-y=1 $$ $$ y=x-1 $$
Substitute into the first equation:
$$ 2x+(x-1)=11 $$ $$ 3x-1=11 $$ $$ 3x=12 $$ $$ x=4 $$
Therefore:
$$ y=4-1=3 $$ $$ \boxed{x=4,\;y=3} $$
Question 5: Arithmetic Progression
The first term of an arithmetic progression is \(5\) and the common difference is \(3\). Find the 20th term.
Solution
The formula for the \(n\)th term is:
$$ T_n=a+(n-1)d $$
Where:
$$ a=5,\quad d=3,\quad n=20 $$
Therefore:
$$ T_{20}=5+(20-1)(3) $$ $$ =5+19(3) $$ $$ =5+57 $$ $$ \boxed{62} $$
Answer: 62
Question 6: Sum of an Arithmetic Progression
Find the sum of the first 25 terms of the arithmetic progression:
$$ 4,7,10,13,\ldots $$
Solution
We have:
$$ a=4,\quad d=3,\quad n=25 $$
The sum is given by:
$$ S_n=\frac{n}{2}[2a+(n-1)d] $$
Therefore:
$$ S_{25} = \frac{25}{2}[2(4)+(25-1)(3)] $$ $$ = \frac{25}{2}[8+72] $$ $$ = \frac{25}{2}(80) $$ $$ \boxed{1000} $$
Answer: 1000
Question 7: Sets
In a class of 50 students, 30 study Mathematics, 25 study Physics and 15 study both Mathematics and Physics.
Find the number of students who study at least one of the two subjects.
Solution
Using:
$$ n(M\cup P)=n(M)+n(P)-n(M\cap P) $$
Therefore:
$$ n(M\cup P)=30+25-15 $$ $$ \boxed{40} $$
Answer: 40 students
Question 8: Probability
A fair die is thrown once. Find the probability of obtaining a number greater than 4.
Solution
The possible outcomes are:
$$ \{1,2,3,4,5,6\} $$
Numbers greater than 4 are:
$$ \{5,6\} $$
There are 2 favourable outcomes out of 6 possible outcomes.
$$ P(\text{number}>4)=\frac{2}{6} $$ $$ =\boxed{\frac{1}{3}} $$
Question 9: Trigonometry
A ladder \(10\) metres long leans against a vertical wall. The ladder makes an angle of \(60^\circ\) with the ground. Find the height reached by the ladder.
Solution
Using:
$$ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} $$
Therefore:
$$ \sin60^\circ=\frac{h}{10} $$
Since:
$$ \sin60^\circ=\frac{\sqrt3}{2} $$
Then:
$$ \frac{\sqrt3}{2}=\frac{h}{10} $$ $$ h=10\left(\frac{\sqrt3}{2}\right) $$ $$ \boxed{5\sqrt3\text{ m}} $$
Approximately:
$$ h\approx8.66\text{ m} $$
Answer: \(5\sqrt3\) m or approximately 8.66 m
Question 10: Bearings
A ship travels \(20\) km on a bearing of \(060^\circ\). Find the distance travelled eastwards.
Solution
The bearing \(060^\circ\) is measured clockwise from north.
The angle between the direction of travel and east is:
$$ 90^\circ-60^\circ=30^\circ $$
Therefore:
$$ \text{Eastward distance}=20\cos30^\circ $$ $$ =20\left(\frac{\sqrt3}{2}\right) $$ $$ \boxed{10\sqrt3\text{ km}} $$
Approximately:
$$ \boxed{17.32\text{ km}} $$
Question 11: Statistics
The marks obtained by five students are:
$$ 12,\;15,\;18,\;10,\;20 $$
Find the mean mark.
Solution
The mean is:
$$ \bar{x}=\frac{\sum x}{n} $$
Therefore:
$$ \bar{x}= \frac{12+15+18+10+20}{5} $$ $$ =\frac{75}{5} $$ $$ \boxed{15} $$
Answer: 15
Question 12: Median
Find the median of:
$$ 14,\;8,\;21,\;10,\;17,\;12,\;19 $$
Solution
Arrange the numbers in ascending order:
$$ 8,\;10,\;12,\;14,\;17,\;19,\;21 $$
There are seven values. The middle value is therefore the fourth value.
$$ \boxed{14} $$
Answer: 14
Question 13: Simple Interest
A student invests GH¢5,000 at a simple interest rate of \(8\%\) per annum for 3 years. Find the interest earned.
Solution
Using:
$$ I=\frac{PRT}{100} $$
Where:
$$ P=5000,\quad R=8,\quad T=3 $$
Therefore:
$$ I=\frac{5000\times8\times3}{100} $$ $$ =1200 $$ $$ \boxed{\text{GH¢1,200}} $$
Answer: GH¢1,200
Question 14: Compound Interest
Find the amount obtained when GH¢4,000 is invested at \(10\%\) compound interest per annum for 2 years.
Solution
Using:
$$ A=P\left(1+\frac{r}{100}\right)^n $$
Therefore:
$$ A=4000\left(1+\frac{10}{100}\right)^2 $$ $$ =4000(1.1)^2 $$ $$ =4000(1.21) $$ $$ \boxed{\text{GH¢4,840}} $$
Question 15: Variation
If \(y\) varies directly as \(x\), and \(y=12\) when \(x=4\), find \(y\) when \(x=10\).
Solution
Since \(y\) varies directly as \(x\):
$$ y=kx $$
When \(y=12\) and \(x=4\):
$$ 12=4k $$ $$ k=3 $$
Therefore:
$$ y=3x $$
When \(x=10\):
$$ y=3(10) $$ $$ \boxed{30} $$
Question 16: Mensuration
A cylinder has radius \(7\) cm and height \(10\) cm. Find its volume. Take \(\pi=\frac{22}{7}\).
Solution
The volume of a cylinder is:
$$ V=\pi r^2h $$
Therefore:
$$ V=\frac{22}{7}\times7^2\times10 $$ $$ =\frac{22}{7}\times49\times10 $$ $$ =1540 $$ $$ \boxed{1540\text{ cm}^3} $$
Question 17: Differentiation
Given:
$$ y=3x^3-5x^2+4x-7 $$
Find:
$$ \frac{dy}{dx} $$
Solution
Differentiate each term:
$$ \frac{dy}{dx} = 9x^2-10x+4 $$ $$ \boxed{\frac{dy}{dx}=9x^2-10x+4} $$
Question 18: Gradient of a Straight Line
Find the gradient of the line passing through the points \(A(2,5)\) and \(B(6,13)\).
Solution
The gradient is:
$$ m=\frac{y_2-y_1}{x_2-x_1} $$
Therefore:
$$ m=\frac{13-5}{6-2} $$ $$ =\frac{8}{4} $$ $$ \boxed{2} $$
Question 19: Matrices
Given:
$$ A= \begin{pmatrix} 2&3\\ 1&4 \end{pmatrix} $$
Find the determinant of \(A\).
Solution
For:
$$ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix} $$
The determinant is:
$$ |A|=ad-bc $$
Therefore:
$$ |A|=(2)(4)-(3)(1) $$ $$ =8-3 $$ $$ \boxed{5} $$
Question 20: WASSCE Word Problem
A trader bought an article for GH¢800 and sold it for GH¢920. Calculate the percentage profit.
Solution
Profit:
$$ \text{Profit}=920-800 $$ $$ =120 $$
Percentage profit is:
$$ \text{Percentage Profit} = \frac{\text{Profit}}{\text{Cost Price}}\times100 $$
Therefore:
$$ =\frac{120}{800}\times100 $$ $$ =15\% $$ $$ \boxed{15\%} $$
WASSCE Mathematics Preparation Tips
- Practise Mathematics every day rather than waiting until examination week.
- Learn the meaning and application of important formulas.
- Work through past WASSCE questions under timed conditions.
- Show all important working in your answers.
- Pay attention to units in mensuration, statistics and word problems.
- Use your calculator correctly and check whether your answer is reasonable.
- Practise topics you find difficult instead of avoiding them.
- Review mistakes after every practice test.
- Learn how to interpret graphs, tables and diagrams.
- Develop good examination time management.
Final Challenge
Do not simply read the solutions. Cover the answers and attempt each question yourself first.
For every question, ask yourself:
- What information has been given?
- What am I required to find?
- Which formula or mathematical principle applies?
- Have I shown my working clearly?
- Does my final answer make sense?
Consistent practice is one of the most effective ways to prepare for WASSCE Mathematics. Start with the topics you find difficult, practise regularly, and learn from every mistake.
