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    Subject

    Bank Exam

    Topic

    Bank Asia Ltd. - Trainee Officer - 2016

    13+23+3y=2312\frac{1}{3} + \frac{2}{3} + \frac{3}{y} = \frac{{23}}{{12}}, then y = ?

    ক)
    2
    খ)
    3
    গ)
    4
    ঘ)
    None

    Explanation

    Solution: 13+23+3y=2312\frac{1}{3} + \frac{2}{3} + \frac{3}{y} = \frac{{23}}{{12}} =>  3y=2312−(13+23)\frac{3}{y} = \frac{{23}}{{12}} - (\frac{1}{3} + \frac{2}{3})                 =>  3y=2312−33\frac{3}{y} = \frac{{23}}{{12}} - \frac{3}{3} => 3y=2312−1\frac{3}{y} = \frac{{23}}{{12}} - 1                                                          => 3y=23−1212\frac{3}{y} = \frac{{23 - 12}}{{12}} => 3y=1112\frac{3}{y} = \frac{{11}}{{12}}                                                  =>  11y = 36                  y = 3611\frac{{36}}{{11}}

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