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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-3

  • question_answer
    A and B together can do a work in 12 days. B and C together do it in 15 days. If A's efficiency is twice that of C, then the days required for B alone to finish the work is

    A)  60                               

    B)  30

    C)  20                     

    D)         15

    Correct Answer: C

    Solution :

     [A + B]?s 1 day's work             \[=\frac{1}{12}\]                                               ? (i) [B + C]?s 1 day's work             \[=\frac{1}{15}\]                                               ? (ii) \[\therefore \]Difference between A and C?s 1 day's work             \[=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}\] If A alone completes the work in \[x\] days, C will do the same in \[2x\] days. \[\therefore \]      \[\frac{1}{x}-\frac{1}{2x}=\frac{1}{60}\] \[\Rightarrow \]   \[\frac{2-1}{2x}=\frac{1}{60}\] \[\Rightarrow \]   \[\frac{1}{2x}=\frac{1}{60}\] \[\Rightarrow \]   \[x=30\] \[\therefore \]B's 1 day?s work             \[=\frac{1}{12}-\frac{1}{30}\]         [From equation (1)]             \[=\frac{5-2}{60}=\frac{3}{60}=\frac{1}{20}\] Hence, B alone will complete the work in 20 days.


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