Here are the solutions to the problems: 20. 2 boys can load a lorry of cement in 6hrs. How long will it take 5 such boys to load the same lorry of cement? Step 1: Recognize this as an inverse proportion problem. More boys will take less time to load the same amount of cement. Let $B$ be the number of boys and $T$ be the time in hours. The product $B \times T$ is constant. $$B_1 T_1 = B_2 T_2$$ Step 2: Substitute the given values. $B_1 = 2$ boys, $T_1 = 6$ hours $B_2 = 5$ boys, $T_2 = ?$ $$2 \times 6 = 5 \times T_2$$ Step 3: Solve for $T_2$. $$12 = 5 T_2$$ $$T_2 = \frac{12}{5}$$ $$T_2 = 2.4$$ It will take 5 boys $\boxed{\text{2.4 hours}}$ to load the same lorry of cement. 21. A tank of water of capacity 5000 litres serves a family for 16 days. How long will it take a water tank of 8000 litres to serve the same family? Step 1: Recognize this as a direct proportion problem. A larger capacity tank will serve the family for more days. Let $C$ be the capacity in litres and $D$ be the number of days. The ratio $\frac{C}{D}$ is constant. $$\frac{C_1}{D_1} = \frac{C_2}{D_2}$$ Step 2: Substitute the given values. $C_1 = 5000$ litres, $D_1 = 16$ days $C_2 = 8000$ litres, $D_2 = ?$ $$\frac{5000}{16} = \frac{8000}{D_2}$$ Step 3: Solve for $D_2$. $$5000 D_2 = 8000 \times 16$$ $$5000 D_2 = 128000$$ $$D_2 = \frac{128000}{5000}$$ $$D_2 = \frac{128}{5}$$ $$D_2 = 25.6$$ A water tank of 8000 litres will serve the same family for $\boxed{\text{25.6 days}}$. 22. 9 men working at the same rate and working 8 hrs in a day take 12 days to complete a given task. How long will it take six such men working 10 hrs in a day to complete the same task? Step 1: Recognize this as a compound proportion problem. The total work done is constant. Let $M$ be the number of men, $H$ be the hours worked per day, and $D$ be the number of days. The total work units can be represented as $M \times H \times D$. $$M_1 H_1 D_1 = M_2 H_2 D_2$$ Step 2: Substitute the given values. $M_1 = 9$ men, $H_1 = 8$ hours/day, $D_1 = 12$ days $M_2 = 6$ men, $H_2 = 10$ hours/day, $D_2 = ?$ $$9 \times 8 \times 12 = 6 \times 10 \times D_2$$ Step 3: Calculate the total work units for the first scenario. $$9 \times 8 \times 12 = 72 \times 12 = 864$$ Step 4: Set up the equation for the second scenario and solve for $D_2$. $$864 = 60 D_2$$ $$D_2 = \frac{864}{60}$$ $$D_2 = \frac{144}{10}$$ $$D_2 = 14.4$$ It will take six men working 10 hours a day $\boxed{\text{14.4 days}}$ to complete the same task. 23. What is the order of the matrix below? $$\begin{pmatrix} 7 & -1 \\ -2 & 3 \\ 9 & 4 \end{pmatrix}$$ Step 1: Count the number of rows. There are 3 rows. Step 2: Count the number of columns. There are 2 columns. Step 3: State the order as (number of rows) $\times$ (number of columns). The order of the matrix is $\boxed{3 \times 2}$. 24. Are the matrices below compatible? $$\begin{pmatrix} 2 & 3 \\ 4 & 9 \end{pmatrix} - \begin{pmatrix} 3 & 9 & -2 \\ 4 & 6 & 4 \\ 1 & 3 & 8 \end{pmatrix}$$ Step 1: Determine the order of the first matrix. It has 2 rows and 2 columns, so its order is $2 \times 2$. Step 2: Determine the order of the second matrix. It has 3 rows and 3 columns, so its order is $3 \times 3$. Step 3: Check for compatibility for subtraction. For matrix addition or subtraction, the matrices must have the exact same order. Since the orders ($2 \times 2$ and $3 \times 3$) are different, the matrices are not compatible for subtraction. The matrices are $\boxed{\text{not compatible}}$. 25. Work out $$\begin{pmatrix} 6 & 6 \\ 3 & -7 \\ 4 & 10 \end{pmatrix} + \begin{pmatrix} 0 & 9 \\ 1 & 6 \\ -2 & -8 \end{pmatrix}$$ Step 1: Perform matrix addition by adding corresponding elements. $$\begin{pmatrix} 6+0 & 6+9 \\ 3+1 & -7+6 \\ 4+(-2) & 10+(-8) \end{pmatrix}$$ Step 2: Simplify each element. $$\begin{pmatrix} 6 & 15 \\ 4 & -1 \\ 2 & 2 \end{pmatrix}$$ The result of the addition is $\boxed{\begin{pmatrix} 6 & 15 \\ 4 & -1 \\ 2 & 2 \end{pmatrix}}$. 26. Given $A=B-C$ and $$A = \begin{pmatrix} 2 & 0 & 3 \\ 4 & -1 & 7 \\ 6 & 3 & 9 \end{pmatrix}$$ $$B = \begin{pmatrix} 3 & 6 & 9 \\ -1 & 0 & 4 \\ 4 & 2 & 6 \end{pmatrix}$$ Step 1: Rearrange the equation to solve for $C$. Given $A = B - C$, we can write $C = B - A$. Step 2: Substitute the given matrices $A$ and $B$ into the equation for $C$. $$C = \begin{pmatrix} 3 & 6 & 9 \\ -1 & 0 & 4 \\ 4 & 2 & 6 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 3 \\ 4 & -1 & 7 \\ 6 & 3 & 9 \end{pmatrix}$$ Step 3: Perform matrix subtraction by subtracting corresponding elements. $$C = \begin{pmatrix} 3-2 & 6-0 & 9-3 \\ -1-4 & 0-(-1) & 4-7 \\ 4-6 & 2-3 & 6-9 \end{pmatrix}$$ Step 4: Simplify each element. $$C = \begin{pmatrix} 1 & 6 & 6 \\ -5 & 1 & -3 \\ -2 & -1 & -3 \end{pmatrix}$$ The matrix $C$ is $\boxed{\begin{pmatrix} 1 & 6 & 6 \\ -5 & 1 & -3 \\ -2 & -1 & -3 \end{pmatrix}}$. That's 2 down. 3 left today — send the next one.