1. **Problem c)**: Given the proportion $\frac{7}{9} : \frac{4}{7} = \frac{m}{n} : 729$, find $m$ and $n$. 2. **Step 1:** Understand the proportion. The ratio $\frac{7}{9}$ to $\frac{4}{7}$ equals the ratio $\frac{m}{n}$ to $729$. This means: $$\frac{\frac{7}{9}}{\frac{4}{7}} = \frac{\frac{m}{n}}{729}$$ 3. **Step 2:** Simplify the left side: $$\frac{\frac{7}{9}}{\frac{4}{7}} = \frac{7}{9} \times \frac{7}{4} = \frac{49}{36}$$ 4. **Step 3:** Set up the equation: $$\frac{49}{36} = \frac{\frac{m}{n}}{729} = \frac{m}{n} \times \frac{1}{729} = \frac{m}{729n}$$ 5. **Step 4:** Cross-multiply: $$49 \times 729 n = 36 m$$ 6. **Step 5:** Express $m$ in terms of $n$: $$m = \frac{49 \times 729}{36} n$$ 7. **Step 6:** Simplify the fraction: $$\frac{49 \times 729}{36} = \frac{49 \times 729}{36}$$ Calculate numerator and denominator: $$49 = 7^2, \quad 729 = 3^6, \quad 36 = 6^2 = (2 \times 3)^2 = 2^2 \times 3^2$$ Simplify: $$\frac{7^2 \times 3^6}{2^2 \times 3^2} = 7^2 \times 3^{6-2} \times \frac{1}{2^2} = 49 \times 3^4 \times \frac{1}{4} = 49 \times 81 \times \frac{1}{4}$$ Calculate: $$49 \times 81 = 3969$$ So: $$\frac{3969}{4}$$ 8. **Step 7:** So: $$m = \frac{3969}{4} n$$ 9. **Step 8:** To have $m$ and $n$ as integers, choose $n=4$: $$m = 3969$$ 10. **Final answer:** $$m = 3969, \quad n = 4$$ --- 1. **Problem Q2 a):** Resolve the expression $$\frac{4x^2 - 1x - 4}{6x^2 - 25x - 46x - 24}$$ into partial fractions. 2. **Step 1:** Simplify the denominator: $$6x^2 - 25x - 46x - 24 = 6x^2 - 71x - 24$$ 3. **Step 2:** Factor the denominator if possible. 4. **Step 3:** Factor numerator and denominator and express as sum of partial fractions. (Note: The problem is incomplete or ambiguous; please clarify if needed.) --- 1. **Problem Q2 (ii):** The cubic polynomial $f(x)$ has roots $1, k, k^2$ and leading coefficient 1. Given $f(x)$ leaves remainder 7 when divided by $x-2$, show: $$k^3 - 2k^2 - 2k - 3 = 0$$ 2. **Step 1:** Write $f(x)$ as: $$f(x) = (x-1)(x-k)(x-k^2)$$ 3. **Step 2:** Use the Remainder Theorem: $$f(2) = 7$$ 4. **Step 3:** Calculate: $$f(2) = (2-1)(2-k)(2-k^2) = 7$$ 5. **Step 4:** Simplify: $$(1)(2-k)(2-k^2) = 7$$ 6. **Step 5:** Expand: $$(2-k)(2-k^2) = 4 - 2k^2 - 2k + k^3 = 7$$ 7. **Step 6:** Rearrange: $$k^3 - 2k^2 - 2k + 4 = 7$$ 8. **Step 7:** Subtract 7: $$k^3 - 2k^2 - 2k - 3 = 0$$ 9. **Step 8:** To find $k$, test possible roots or use methods like Rational Root Theorem. 10. **Step 9:** Check $k=3$: $$3^3 - 2(3)^2 - 2(3) - 3 = 27 - 18 - 6 - 3 = 0$$ 11. **Step 10:** So $k=3$ is a root. 12. **Step 11:** Factor polynomial and show no other real roots exist. --- 1. **Problem Q3 (a):** A class has 4 boys and $g$ girls. Every Sunday, 5 students including at least 3 boys go for a picnic. Each girl in the group receives a doll. Total dolls distributed are 85. Find $g$. 2. **Step 1:** Possible groups have 3 boys + 2 girls or 4 boys + 1 girl. 3. **Step 2:** Number of groups with 3 boys and 2 girls: $$\binom{4}{3} \times \binom{g}{2} = 4 \times \binom{g}{2}$$ 4. **Step 3:** Number of groups with 4 boys and 1 girl: $$\binom{4}{4} \times \binom{g}{1} = 1 \times g = g$$ 5. **Step 4:** Total dolls given: $$2 \times 4 \times \binom{g}{2} + 1 \times g = 85$$ 6. **Step 5:** Simplify: $$8 \binom{g}{2} + g = 85$$ 7. **Step 6:** Recall: $$\binom{g}{2} = \frac{g(g-1)}{2}$$ 8. **Step 7:** Substitute: $$8 \times \frac{g(g-1)}{2} + g = 85$$ 9. **Step 8:** Simplify: $$4g(g-1) + g = 85$$ 10. **Step 9:** Expand: $$4g^2 - 4g + g = 85$$ 11. **Step 10:** Combine like terms: $$4g^2 - 3g = 85$$ 12. **Step 11:** Rearrange: $$4g^2 - 3g - 85 = 0$$ 13. **Step 12:** Solve quadratic: $$g = \frac{3 \pm \sqrt{9 + 1360}}{8} = \frac{3 \pm \sqrt{1369}}{8} = \frac{3 \pm 37}{8}$$ 14. **Step 13:** Possible values: $$g = \frac{40}{8} = 5 \quad \text{or} \quad g = \frac{-34}{8} = -4.25$$ 15. **Step 14:** Since $g$ must be positive integer, $g=5$. --- 1. **Problem Q3 (b):** Show by induction that: $$10^n + 3 \cdot 4^{n+2} + 5$$ is divisible by 9 for all $n \geq 1$. 2. **Step 1:** Base case $n=1$: $$10^1 + 3 \cdot 4^{3} + 5 = 10 + 3 \times 64 + 5 = 10 + 192 + 5 = 207$$ 3. **Step 2:** Check divisibility: $$207 \div 9 = 23$$ remainder 0, so base case holds. 4. **Step 3:** Assume true for $n=k$: $$9 \mid 10^k + 3 \cdot 4^{k+2} + 5$$ 5. **Step 4:** Prove for $n=k+1$: $$10^{k+1} + 3 \cdot 4^{k+3} + 5$$ 6. **Step 5:** Express: $$10^{k+1} + 3 \cdot 4^{k+3} + 5 = 10 \times 10^k + 3 \times 4 \times 4^{k+2} + 5 = 10 \times 10^k + 12 \times 4^{k+2} + 5$$ 7. **Step 6:** Rewrite as: $$= (10^k + 3 \cdot 4^{k+2} + 5) + 9 \times 10^k + 9 \times 4^{k+2}$$ 8. **Step 7:** By induction hypothesis, $10^k + 3 \cdot 4^{k+2} + 5$ is divisible by 9. 9. **Step 8:** The terms $9 \times 10^k$ and $9 \times 4^{k+2}$ are clearly divisible by 9. 10. **Step 9:** Sum of multiples of 9 is divisible by 9, so the expression for $n=k+1$ is divisible by 9. 11. **Final conclusion:** By induction, the expression is divisible by 9 for all $n \geq 1$.