Subject

    Quantitative Aptitude

    Topic

    Simplification

    Simplify:4+12  of  102  ÷243+382  ×210  (60%  of  400÷8)÷10  \frac{{4 + \frac{1}{2}\;of\;{{10}^2}\; \div \sqrt {243 + 382\;} \times {2^{10}}\;}}{{\left( {60\% \;of\;400 \div 8} \right) \div 10\;}}

    ক)
    678
    খ)
    649
    গ)
    684
    ঘ)
    650
    ঙ)
    698

    Explanation

    Follow BODMAS rule to solve this question, as per the order given below, Step-1-Parts of an equation enclosed in 'Brackets' must be solved first, Step-2-Any mathematical 'Of' or 'Exponent' must be solved next, Step-3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, Step-4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated. The given expression is:4+12  of  102  ÷243+382  ×210  (60%  of  400÷8)÷10\frac{{4 + \frac{1}{2}\;of\;{{10}^2}\; \div \sqrt {243 + 382\;} \times {2^{10}}\;}}{{\left( {60\% \;of\;400 \div 8} \right) \div 10}} =  4  +  12  of  102  ÷  625    ×  210  (60%  of  400÷8)÷10= \;\frac{{4\; + \;\frac{1}{2}\;of\;{{10}^2}\; \div \;\sqrt {625\;} \; \times \;{2^{10}}\;}}{{\left( {60\% \;of\;400 \div 8} \right) \div 10}} =  4  +  12  ×  100  ÷25  ×1024  (240÷8)÷10= \;\frac{{4\; + \;\frac{1}{2}\; \times \;100\; \div 25\; \times 1024\;}}{{\left( {240 \div 8} \right) \div 10}} =  4  +  50  ÷25  ×  102430÷10= \;\frac{{4\; + \;50\; \div 25\; \times \;1024}}{{30 \div 10}} =  4  +  2  ×  10243= \;\frac{{4\; + \;2\; \times \;1024}}{3} =  4  +  20483= \;\frac{{4\; + \;2048}}{3} =    20523= \;\frac{{\;2052}}{3} = 684

    Find Tutors

    View all Tutors