MATHEMATICS PAPER 1 [40 Marks]

Since $4 = 2^2$, we can rewrite $4^x$ as $(2^2)^x$.

$$(2^2)^x = 2^{x+3}$$ $$2^{2x} = 2^{x+3}$$

$$2x - x = 3$$ $$x = 3$$

The value of $x$ is B. 3.

Question 2: Kwaku and Attah formed a company and agreed that their profit will be shared in the ratio 3:2 respectively. At the end of the year, Attah received GH¢5000.00 more than Kwaku. How much profit did the company make in the year?

Assumption: The problem states the ratio for Kwaku:Attah is 3:2, but then says Attah received GH¢5000.00 more than Kwaku. This implies Attah's share is larger. To reconcile this, we assume the ratio of shares is Kwaku:Attah = 2:3, which means Attah gets 3 parts and Kwaku gets 2 parts.

Let Kwaku's share be $2x$ and Attah's share be $3x$.
The difference in their shares is GH¢5000.00.

$$3x - 2x = 5000$$

$$\text{Total Profit} = 5x = 5 \times 5000$$ $$\text{Total Profit} = 25000$$

The company made a profit of B. GH¢25,000.00.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$.

$$\det(A) = (2)(4) - (-3)(1)$$ $$\det(A) = 8 - (-3)$$ $$\det(A) = 8 + 3$$ $$\det(A) = 11$$

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjugate is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$\text{adj}(A) = \begin{bmatrix} 4 & -(-3) \\ -1 & 2 \end{bmatrix}$$ $$\text{adj}(A) = \begin{bmatrix} 4 & 3 \\ -1 & 2 \end{bmatrix}$$

$$A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\ -1 & 2 \end{bmatrix}$$ $$A^{-1} = \begin{bmatrix} \frac{4}{11} & \frac{3}{11} \\ -\frac{1}{11} & \frac{2}{11} \end{bmatrix}$$

The inverse of the matrix is B. $\begin{bmatrix} \frac{4}{11} & \frac{3}{11} \\ -\frac{1}{11} & \frac{2}{11} \end{bmatrix}$.

Given $m = 3$ and $(x_1, y_1) = (-2, 0)$.

$$y - 0 = 3(x - (-2))$$ $$y = 3(x + 2)$$

$$y = 3x + 6$$ $$0 = 3x - y + 6$$

Or, multiplying by -1 to match option D:

$$y - 3x - 6 = 0$$

The equation of the line is D. $y - 3x - 6 = 0$.

We look for two numbers that multiply to $ac = 6 \times 5 = 30$ and add up to $b = -13$. These numbers are $-3$ and $-10$.
Rewrite the middle term using these numbers:

$$6x^2 - 3x - 10x + 5 = 0$$

$$3x(2x - 1) - 5(2x - 1) = 0$$ $$(3x - 5)(2x - 1) = 0$$

$$3x - 5 = 0 \implies 3x = 5 \implies x = \frac{5}{3}$$ $$2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$$

The solutions are $x = \frac{1}{2}$ and $x = \frac{5}{3}$. Note that $\frac{5}{3}$ can be written as $1\frac{2}{3}$.
The correct option is C. $x = \frac{1}{2}, x = 1\frac{2}{3}$.