If $\sqrt{4096} = 64$, then the value of $\sqrt{40.96} + \sqrt{0.4096} + \sqrt{0.004096} + \sqrt{0.00004096}$ up to two places of decimals is

Concept & Logic Evaluate a sequence of decimal square roots by applying the provided base root and dynamically adjusting the decimal places for each term before summing them. Step-by-Step Solution * **Given:** The core integer root is $\sqrt{4096} = 64$. * **Calculation:** Determine the precise value of each decimal term: * $\sqrt{40.96} = 6.4$ (2 places converts to 1 place) * $\sqrt{0.4096} = 0.64$ (4 places converts to 2 places) * $\sqrt{0.004096} = 0.064$ (6 places converts to 3 places) * $\sqrt{0.00004096} = 0.0064$ (8 places converts to 4 places) * Add the evaluated results together: $6.4 + 0.64 + 0.064 + 0.0064 = 7.1104$ * The question asks for the answer up to two decimal places: $7.1104 \approx 7.11$ Exam Strategy & Shortcut To save time, only sum the terms that affect the first two decimal places. $6.4 + 0.64 + 0.064 = 7.104$. We know the final term ($0.0064$) will push this to $7.1104$. Rounding to two decimal places yields $7.11$. Common Pitfall Failing to properly carry over values during decimal addition, or rounding prematurely before all necessary terms have been combined. Final Answer **Therefore, the correct answer is 7.11.**