Question : What is the value of $\frac{(45 \times 3 \div 27 \times 4)(81 \div 5 \times 85 \div 51)}{(108 \div 18 \times 7\div21)(133 \times 3 \div 38 \times 4)} ?$
Option 1: $\frac{46}{7}$
Option 2: $\frac{45}{7}$
Option 3: $\frac{54}{7}$
Option 4: $\frac{39}{7}$
Correct Answer: $\frac{45}{7}$
Solution :
$\frac{(45 \times 3 \div 27 \times 4)(81 \div 5 \times 85 \div 51)}{(108 \div 18 \times 7\div 21)(133 \times 3 \div 38 \times 4)}$
$=\frac{(45 \times \frac{3}{27} \times 4)(\frac{81}{5} \times \frac{85}{51})}{(\frac{108}{18 }\times \frac{7}{21})(133 \times \frac{3}{38} \times 4)}$
$=\frac{(45 \times \frac{1}{9} \times 4)(\frac{81}{5} \times \frac{5}{3})}{(2)(133 \times \frac{3}{19} \times 2)}$
$=\frac{20\times 27}{2\times21 \times 2}$
$=\frac{45}{7}$
Hence, the correct answer is $\frac{45}{7}$.
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