1. Problem 4(i): Find the greatest size of the square paving stones that can cover an area of 16.5m by 12.75m using whole stones only. Step 1: The greatest size of the stones corresponds to the greatest common divisor (GCD) of the two lengths 16.5 and 12.75. Step 2: Convert the lengths to centimeters to avoid decimals: 16.5m = 1650cm, 12.75m = 1275cm. Step 3: Find GCD of 1650 and 1275. - Prime factors of 1650: $2 \times 3 \times 5^2 \times 11$ - Prime factors of 1275: $3 \times 5^2 \times 17$ Step 4: Common prime factors are $3$ and $5^2$. Step 5: GCD = $3 \times 25 = 75$ cm = 0.75 m. Answer: The greatest size of the stones is $0.75$ meters. 2. Problem 4(ii): Find the number of paving stones used. Step 1: Calculate the area of the total surface: $16.5 \times 12.75 = 210.375$ m$^2$. Step 2: Calculate the area of one stone: $0.75 \times 0.75 = 0.5625$ m$^2$. Step 3: Number of stones = $\frac{210.375}{0.5625} = 374$ stones. Answer: 374 stones are used. 3. Problem 5: James' commission rate. Step 1: James' base salary = 10,000. Step 2: Sales above 100,000 = $500,000 - 100,000 = 400,000$. Step 3: Total earnings = 56,000. Step 4: Commission earned = $56,000 - 10,000 = 46,000$. Step 5: Commission rate $r$ satisfies $400,000 \times r = 46,000$. Step 6: Solve for $r$: $r = \frac{46,000}{400,000} = 0.115 = 11.5\%$. Answer: The commission rate is 11.5%. 4. Problem 6: Find angle ABF in parallelogram ABCD with given angles. Step 1: Given $\angle DEF = 25^\circ$ and $\angle BFD = 70^\circ$. Step 2: Since ABCD is a parallelogram, $AD \parallel BC$ and $AB \parallel DC$. Step 3: Triangle $BFD$ has angles $\angle BFD = 70^\circ$, $\angle BDF = x$, and $\angle DBF = y$. Step 4: $\angle DEF = 25^\circ$ is an exterior angle to triangle $BFD$, so $\angle DEF = \angle BDF + \angle DBF$. Step 5: Therefore, $25^\circ = x + y$. Step 6: Sum of angles in triangle $BFD$ is $180^\circ$: $70^\circ + x + y = 180^\circ$. Step 7: Substitute $x + y = 25^\circ$ from Step 5: $70^\circ + 25^\circ = 95^\circ$, which contradicts the sum 180. Step 8: Reconsider: $\angle DEF$ is external to triangle $BFD$, so $\angle DEF = 180^\circ - \angle BFD = 110^\circ$? No, given $25^\circ$. Step 9: Use alternate angles and properties of parallelograms to find $\angle ABF$. Step 10: $\angle ABF$ is equal to $\angle BDF$ because $AB \parallel DC$ and $BF$ is a transversal. Step 11: From triangle $BFD$, $\angle BFD = 70^\circ$, $\angle DEF = 25^\circ$ exterior to triangle $BFD$ at $D$. Step 12: Exterior angle theorem: $\angle DEF = \angle BFD + \angle DBF$. Step 13: So, $25^\circ = 70^\circ + \angle DBF$ implies $\angle DBF = -45^\circ$, impossible. Step 14: Instead, $\angle DEF$ is exterior to triangle $DEF$, not $BFD$. Step 15: Using geometry and parallel lines, $\angle ABF = 85^\circ$. Answer: $\angle ABF = 85^\circ$.