1. **Problem Statement:** Find the value of $k$ given that in the expansion of $ (3 - 4x)^5 $, the ratio of the coefficient of $x^2$ to that of $x$ is $-\frac{1}{3}k$. 2. **Recall the Binomial Expansion Formula:** $$ (a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r $$ where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. 3. **Identify terms:** Here, $a = 3$, $b = -4x$, and $n=5$. 4. **General term:** $$ T_{r+1} = \binom{5}{r} 3^{5-r} (-4x)^r = \binom{5}{r} 3^{5-r} (-4)^r x^r $$ 5. **Coefficient of $x$: when $r=1$** $$ C_1 = \binom{5}{1} 3^{4} (-4)^1 = 5 \times 81 \times (-4) = -1620 $$ 6. **Coefficient of $x^2$: when $r=2$** $$ C_2 = \binom{5}{2} 3^{3} (-4)^2 = 10 \times 27 \times 16 = 4320 $$ 7. **Given ratio:** $$ \frac{C_2}{C_1} = -\frac{1}{3} k $$ Substitute values: $$ \frac{4320}{-1620} = -\frac{1}{3} k $$ 8. **Simplify left side:** $$ \frac{4320}{-1620} = -\frac{4320}{1620} = -\frac{8}{3} $$ 9. **Equate and solve for $k$:** $$ -\frac{8}{3} = -\frac{1}{3} k \implies \frac{8}{3} = \frac{1}{3} k \implies k = 8 $$ **Final answer:** $k = 8$