Subject

    Quantitative Aptitude

    Topic

    Average

    Find the average of ten consecutive multiples of 12, starting from the fifth multiple.

    ক)
    124
    খ)
    1140
    গ)
    114
    ঘ)
    126
    ঙ)
    1260

    Explanation

    We know that, Average = sum of multiplesno. of multiples\frac{{{\bf{sum}}\ {\bf{of}}\ {\bf{multiples}}}}{{{\bf{no}}.\ {\bf{of}}\ {\bf{multiples}}}} =(12×5)+(12×6)+……….+(12×14)10= \frac{{\left( {12 \times 5} \right) + \left( {12 \times 6} \right) + \ldots \ldots \ldots . + \left( {12 \times 14} \right)}}{{10}} =12/10 (5+6+ ………+14) =12/10 (1+2+3+ ………+14-(1+2+3+4)) =1210(14×152−4×52)= \frac{{12}}{{10}}\left( {\frac{{14 \times 15}}{2} - \frac{{4 \times 5}}{2}} \right) ∵ Sum of first n numbers = n(n+1)2\frac{{n\left( {n + 1} \right)}}{2} ∴ Required average =1210×(105−10)=114= \frac{{12}}{{10}} \times \left( {105 - 10} \right) = 114

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