These are the solutions for the WASSCE 2026 Private Candidates (PC1) Core Mathematics Paper 1. The questions without the solutions may be found at WASSCE 2026 (PC1) Core Mathematics Paper 1.
- Given that μ = {x: 0 x < 10, where x is a whole number}, A is {1, 2, 3, 5, 6} and B = {2, 4, 6, 8}, find n(A B’).
A. 2
B. 3
C. 4
D. 5
Answer: B
Solution
B’ = {0, 1, 3, 5, 7, 9}.
Hence, A B’= {1, 3, 5}. Thus, n(A B’) = 3 as there are three elements in the set {1, 3, 5}.
2. Given that 213six = Qfour, find the value of Q.
A. 1101
B. 1111
C. 1010
D. 1001
Answer: A
Solution
213six = 2 × 62 + 1 × 61 + 3 × 60 = 72 + 6 + 3 = 81.
The powers of four are 40 = 1, 41 = 4, 42 = 16, 43 = 64, and so on.
Since 81 = 64 + 16 + 1, 213six = 1101four. Thus, Q = 1101.
We could also use the usual division algorithm where we repeatedly divide 81 by 4, noting the remainder at each step and gathering them together to get the result.
3. Find the value of \( \displaystyle \frac{\log 32}{\log{6} – \log{3}} \)
A. 2
B. 3
C. 4
D. 5
Answer: D
Solution
This question invites us to apply the laws of logarithms to simplify the expression.
Applying the subtraction law, we get
$$ \begin{aligned} \frac{\log 32}{\log{6} – \log{3}} = \frac{\log 32}{\log \frac{6}{3}} = \frac{\log 32}{\log 2} \end{aligned} $$
Noticing that 32 = 25 and applying the power law of logarithms and simplifying, we get
$$ \begin{aligned} \frac{\log 32}{\log 2} = \frac{\log 2^5}{\log 2} = \frac{5 \log 2}{\log 2} = 5 \end{aligned} $$
This question may also be solved by just punching the figures into a calculator and getting a result. Thus, students can obtain the right answer without understanding the concepts the examiners may have wanted to test, making this an ineffective question for testing understanding. Will the examiners be bold enough to replace the figures with letters?
4. If \( 3^x \times 3^{x + 1} = 2187 \), find the value of \( x \).
A. 2
B. 3
C. 4
D. 5
Answer: B
Solution
Since the left hand side contains powers of 3, we would like to try to write the right hand side, 2187, as a power of 3.
Dividing by 3 gives 729. Noticing that 729 = 81 × 9 = 34 × 32 = 36, we conclude that 2187 = 37.
Thus, we have
$$ \begin{align} 3^x \times 3^{x + 1} &= 2187 \\ 3^x \times 3^{x + 1} &= 3^7 \end{align} $$
Applying the addition law of indices, we simplify the left hand side to get
$$ \begin{align} 3^{x + x + 1} &= 2187 \\ 3^{2x + 1} &= 3^7 \end{align} $$
Equating indices, we get \( 2x + 1 = 7 \) or x = 3.
5. Correct 10.10057 to four significant figures.
A. 10.1001
B. 10.105
C. 10.10
D. 10.101
Answer: C
Solution
Starting from the first non-zero digit and counting four digits ahead, we get 10.10. The digit that comes after these first four digits is 0, hence, it rounds down, giving the answer as 10.10.
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