Which of the following is equal to 1?

### Concept & Logic For a fraction involving powers of matching base significant digits to equal $1$, the total power of $10$ in the numerator must balance out perfectly with the total power of $10$ in the denominator. ### Step-by-Step Solution * Let us verify option (d) by breaking it down using powers of $10$ relative to the base integer $11$: $$ \text{Numerator} = (0.11)^2 = (11 \times 10^{-2})^2 = 11^2 \times 10^{-4} $$ $$ \text{Denominator} = 11^2 \times 0.01 = 11^2 \times 10^{-2} $$ Let's recheck option (b): $$ \text{Numerator} = (1.1)^2 = (11 \times 10^{-1})^2 = 11^2 \times 10^{-2} $$ $$ \text{Denominator} = 11^2 \times (0.01)^2 = 11^2 \times (10^{-2})^2 = 11^2 \times 10^{-4} $$ Let's compute the exact value of option (a): $$ \frac{11^2 \times 10^{-4}}{(11 \times 10^{-1})^2 \times 10^{-1}} = \frac{11^2 \times 10^{-4}}{11^2 \times 10^{-2} \times 10^{-1}} = \frac{10^{-4}}{10^{-3}} = 0.1 $$ Let's recalculate option (d) directly: $$ \frac{(0.11)^2}{11^2 \times 0.01} = \frac{0.0121}{121 \times 0.01} = \frac{0.0121}{1.21} = 0.01 $$ Let's carefully check option (c): $$ \text{Numerator} = (0.011)^2 = (11 \times 10^{-3})^2 = 11^2 \times 10^{-6} $$ $$ \text{Denominator} = (1.1)^2 \times (0.01)^2 = (11 \times 10^{-1})^2 \times (10^{-2})^2 = 11^2 \times 10^{-2} \times 10^{-4} = 11^2 \times 10^{-6} $$ $$ \text{Fraction Value} = \frac{11^2 \times 10^{-6}}{11^2 \times 10^{-6}} = 1 $$ ### Exam Strategy & Shortcut Count total decimal places on both sides of each option, squaring shifts appropriately: Option (c): Top has $3 \times 2 = 6$ decimal places. Bottom has $(1 \times 2) + (2 \times 2) = 2 + 4 = 6$ decimal places. Since the integer base elements $11$ cancel out completely, matching decimal counts ensure the ratio equals $1$. ### Common Pitfall Miscounting decimal counts during squaring (e.g., assuming $(0.01)^2$ has $4$ decimal places but adding them incorrectly to other indices). ### Final Answer **Therefore, the correct answer is $\frac{(0.011)^2}{(1.1)^2 \times (0.01)^2}$.**