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SSC Sample Paper Mock Test-13 SSC CGL Tear-II Paper-1

  • question_answer
    \[\frac{{{(3.06)}^{3}}-{{(1.98)}^{3}}}{{{(3.06)}^{2}}+3.06\times 1.98+{{(1.98)}^{2}}}\]is equal to

    A) 1.08     

    B)                     5.04    

    C)  2.16    

    D)  1.92

    Correct Answer: A

    Solution :

    Let 3.06 = a and 1.98 = b \[\therefore \] Expression\[=\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}}\] \[=\frac{(a-b)({{a}^{2}}+ab+{{b}^{2}})}{{{a}^{2}}+ab+{{b}^{2}}}\] \[=a-b=3.06-1.98=1.08\]


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