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    The value of (0.29‾÷0.32‾)×6.60.95‾÷1.05‾×1.2‾\frac{(0.\overline{29} \div 0.\overline{32}) \times 6.6}{0.\overline{95} \div 1.\overline{05} \time
    Question

    The value of (0.29÷0.32)×6.60.95÷1.05×1.2\frac{(0.\overline{29} \div 0.\overline{32}) \times 6.6}{0.\overline{95} \div 1.\overline{05} \times 1.\overline{2}}​ is

    A.

    4.5

    B.

    5.2

    C.

    5.4

    D.

    6.3

    Correct option is C

    Given:

    (0.29÷0.32)×6.60.95÷1.05×1.2\frac{(0.\overline{29} \div 0.\overline{32}) \times 6.6}{0.\overline{95} \div 1.\overline{05} \times 1.\overline{2}}​​

    Formula Used:
    For a two-digit recurring decimal:

    0.ab=ab990.\overline{ab}=\frac{ab}{99}​​

    Solution:

    (0.29÷0.32)×6.60.95÷1.05×1.2 =(2999÷3299)×6.69599÷105199×1219 =(2932)×661095104×119 =2932×335×10495×911 =101791900 5.4\frac{(0.\overline{29} \div 0.\overline{32}) \times 6.6}{0.\overline{95} \div 1.\overline{05} \times 1.\overline{2}} \\ \ \\ = \frac{(\frac{29}{99} \div \frac{32}{99}) \times 6.6}{\frac{95}{99} \div \frac{105 -1}{99} \times \frac{12-1}{9}}\\ \ \\ = \frac{(\frac{29}{32} ) \times \frac{66}{10}}{\frac{95}{104} \times \frac{11}{9}}\\ \ \\ = \frac{29}{32} \times \frac{33}{5} \times \frac{104} {95} \times \frac{9}{11} \\ \ \\ = \frac{10179}{1900} \\ \ \\ \approx 5.4​​

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