Let $a = (4 \div 3) \div 3 \div 4, b = 4 \div (3 \div 3) \div 4, c = 4 \div 3 \div (3 \div 4)$. The maximum value among the above three is

Concept & Rule This question tests the strict application of the order of operations (BODMAS/PEMDAS) in sequential division. Division is NOT associative. When encountering a string of division operations like $$x \div y \div z$$, you must evaluate them strictly from left to right unless parentheses dictate otherwise. Step-by-Step Solution * **Evaluate expression $a$:** $$a = (\frac{4}{3}) \div 3 \div 4$$ $$a = (\frac{4}{3} \times \frac{1}{3}) \div 4$$ $$a = \frac{4}{9} \times \frac{1}{4} = \frac{1}{9}$$ (Approx $0.11$) * **Evaluate expression $b$:** $$b = 4 \div (3 \div 3) \div 4$$ Resolve parentheses first: $$(3 \div 3) = 1$$ $$b = 4 \div 1 \div 4$$ Evaluate left to right: $$(4 \div 1) = 4$$, then $$4 \div 4 = 1$$ $$b = 1$$ * **Evaluate expression $c$:** $$c = 4 \div 3 \div (3 \div 4)$$ Resolve parentheses first: $$(3 \div 4) = \frac{3}{4}$$ $$c = (4 \div 3) \div \frac{3}{4}$$ $$c = \frac{4}{3} \div \frac{3}{4}$$ $$c = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$$ (Approx $1.77$) * Compare the final values: $$\frac{1}{9}$$, $$1$$, and $$\frac{16}{9}$$. The largest value is $$\frac{16}{9}$$, which corresponds to $c$. Exam Strategy & Shortcut Avoid calculating exact decimals. Keep everything in fractional form to make comparisons obvious. Once you see that $a$ is a proper fraction ($<1$), $b$ is exactly $1$, and $c$ is an improper fraction ($>1$), you can instantly identify $c$ as the maximum without further simplification. Common Pitfall The most common trap is assuming that $4 \div 3 \div 3 \div 4$ allows you to arbitrarily cancel the 4s and 3s. Without parentheses, $x \div y \div z$ is always evaluated as $(x/y)/z$, which is mathematically equivalent to $\frac{x}{y \times z}$. Final Answer Therefore, the correct answer is **c**.