1. Problem 1: A human resource officer allocates employees to Accounts, Computer, and Human Resource Management departments in the ratio 3:4:5. Step 1: Let the total number of employees be $T$. Then employees in Accounts = $\frac{3}{3+4+5} T = \frac{3}{12}T = \frac{1}{4}T$. Step 2: Employees in Computer = $\frac{4}{12}T = \frac{1}{3}T$. Step 3: Employees in Human Resource Management = $\frac{5}{12}T$. If total employees $T$ is given, substitute to get actual numbers. 2. Problem 2: Five employees take 7 days to complete a task. How many employees are needed to finish in 3 days? Step 1: Total work $W$ is constant. Work done = employees $\times$ days. Step 2: $5 \times 7 = W = 35$ employee-days. Step 3: Let $x$ be the number of employees needed for 3 days: $$ x \times 3 = 35 \Rightarrow x = \frac{35}{3} \approx 11.67 $$ Step 4: Since employees must be whole numbers, at least 12 employees are needed. 3. Problem 3: Solve for $p$ and $q$ using matrix method: \[ 4p + 5q = 7 \\ 3p - 2q = 3 \] Step 1: Write system as $AX = B$ where $$ A = \begin{bmatrix}4 & 5 \\ 3 & -2\end{bmatrix}, X = \begin{bmatrix}p \\ q\end{bmatrix}, B = \begin{bmatrix}7 \\ 3\end{bmatrix} $$ Step 2: Find $A^{-1}$: $$ \det A = 4(-2) - 5(3) = -8 - 15 = -23 $$ $$ A^{-1} = \frac{1}{-23} \begin{bmatrix}-2 & -5 \\ -3 & 4\end{bmatrix} $$ Step 3: Compute $X = A^{-1}B$: $$ X = \frac{1}{-23} \begin{bmatrix}-2 & -5 \\ -3 & 4\end{bmatrix} \begin{bmatrix}7 \\ 3\end{bmatrix} = \frac{1}{-23} \begin{bmatrix}(-2)(7) + (-5)(3) \\ (-3)(7) + 4(3) \end{bmatrix} = \frac{1}{-23} \begin{bmatrix}-14 - 15 \\ -21 + 12 \end{bmatrix} = \frac{1}{-23} \begin{bmatrix}-29 \\ -9\end{bmatrix} $$ Step 4: Simplify: $$ p = \frac{-29}{-23} = \frac{29}{23}, \quad q = \frac{-9}{-23} = \frac{9}{23} $$ 4. Problem 4: Use Cramer's rule to solve for $x$ and $y$: \[ \frac{1}{3}x + \frac{1}{2}y = 7 \\ \frac{2}{3}x - y = -2 \] Step 1: Write coefficients matrix: $$ A = \begin{bmatrix}\frac{1}{3} & \frac{1}{2} \\ \frac{2}{3} & -1 \end{bmatrix} $$ Step 2: Calculate determinant $D$: $$ D = \frac{1}{3}(-1) - \frac{1}{2}\left(\frac{2}{3}\right) = -\frac{1}{3} - \frac{1}{3} = -\frac{2}{3} $$ Step 3: Calculate $D_x$ replacing first column with constants: $$ D_x = \begin{vmatrix}7 & \frac{1}{2} \\ -2 & -1 \end{vmatrix} = 7(-1) - \frac{1}{2}(-2) = -7 + 1 = -6 $$ Step 4: Calculate $D_y$ replacing second column with constants: $$ D_y = \begin{vmatrix}\frac{1}{3} & 7 \\ \frac{2}{3} & -2 \end{vmatrix} = \frac{1}{3}(-2) - 7\left(\frac{2}{3}\right) = -\frac{2}{3} - \frac{14}{3} = -\frac{16}{3} $$ Step 5: Find solutions: $$ x = \frac{D_x}{D} = \frac{-6}{-\frac{2}{3}} = -6 \times \frac{3}{-2} = 9 $$ $$ y = \frac{D_y}{D} = \frac{-\frac{16}{3}}{-\frac{2}{3}} = 8 $$ 5. Problem 5: Solve for $x, y, z$ using matrix method: \[ 3x - 7y + 2z = 43 \] Note: Only one equation provided; matrix method requires as many equations as variables. Additional equations needed for a unique solution. If user provides remaining equations, I can continue.