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SSC Sample Paper Mock Test-10 SSC CGL Tear-II Paper-1

  • question_answer
    AB and CD are two parallel lines. The points B and C are joined such that \[\angle ABC=65{}^\circ .\]A line CE is drawn making angle of \[35{}^\circ \]with the line CB, EF is drawn parallel to AB, as shown in figure, then \[\angle CEF\]is equal to                                

    A)  \[160{}^\circ \]

    B)  \[155{}^\circ \]

    C)  \[150{}^\circ \]

    D)  \[145{}^\circ \]

    Correct Answer: C

    Solution :

    \[\angle ABC=\angle BCD\] as \[AB||CD\] \[\angle BCD=65{}^\circ \angle ECD=65{}^\circ -\angle BCE=65{}^\circ -35{}^\circ \]\[=30{}^\circ \] \[\angle CEF+\angle ECD=180{}^\circ \] \[\angle CEF=180{}^\circ -30{}^\circ =150{}^\circ \]


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