Question : The square root of $\left(\frac{1}{4}\right) \times\left(\frac{1}{9}\right) \times\left(\frac{1}{25}\right) \times\left(\frac{1}{49}\right) \div\left(\frac{36}{121}\right)$ is:
Option 1: $\frac{11}{1260}$
Option 2: $\frac{1260}{11}$
Option 3: $\frac{11}{12.60}$
Option 4: $\frac{1}{1260}$
Correct Answer: $\frac{11}{1260}$
Solution : $\sqrt{\left(\frac{1}{4}\right) \times\left(\frac{1}{9}\right) \times\left(\frac{1}{25}\right) \times\left(\frac{1}{49}\right) \div\left(\frac{36}{121}\right)}$ = $\sqrt{\left(\frac{1}{4}\right) \times\left(\frac{1}{9}\right) \times\left(\frac{1}{25}\right) \times\left(\frac{1}{49}\right) \times \left(\frac{121}{36}\right)}$ = $\left(\frac{1}{2}\right) \times\left(\frac{1}{3}\right) \times\left(\frac{1}{5}\right) \times\left(\frac{1}{7}\right) \times\left(\frac{11}{6}\right)$ = $\frac{11}{1260}$ Hence, the correct answer is $\frac{11}{1260}$.
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