CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    0.5737373 ... is equal to :

    A)  \[\frac{284}{497}\]                                        

    B)  \[\frac{284}{495}\]

    C)  \[\frac{568}{999}\]                                        

    D)  \[\frac{567}{990}\]

    Correct Answer: B

    Solution :

    Let                          \[0.5737373=y\]                                ... (i) \[\therefore \]  \[10y=5.737373\]                                             ... (ii) (on multiply equation (1) by 10) Again multiply equation (ii) by 100 Then,                 \[1000y=573.737373\]                    ... (iii) Now, subtracting equation (ii) from (iii) we get                 \[990y=568\]                 \[y=\frac{568}{990}=\frac{284}{495}\] \[\therefore \]  \[0.5737373=\frac{284}{495}\]


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