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SSC Quantitative Aptitude Algebra Question Bank Elementary Algebra (II)

  • question_answer
    If \[a=0.25,\] \[b=0.05\]and \[c=0.5,\]then the value of \[\frac{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}-2bc}{{{a}^{2}}+{{b}^{2}}-2ab-{{c}^{2}}}\]is

    A) \[\frac{8}{7}\]

    B) \[\frac{14}{17}\]

    C) \[1\]

    D) \[\frac{25}{16}\]

    Correct Answer: A

    Solution :

    [a] \[\frac{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}-2abc}{{{a}^{2}}+{{b}^{2}}-2ab-{{c}^{2}}}\] \[=\frac{{{a}^{2}}-({{b}^{2}}+{{c}^{2}}+2abc)}{({{a}^{2}}+{{b}^{2}}-2ab)-{{c}^{2}}}\] \[=\frac{{{a}^{2}}-{{(b+c)}^{2}}}{{{(a\,-b)}^{2}}-{{c}^{2}}}\] Now, \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\] \[\therefore \]      \[\frac{(a+b+c)(a-b-c)}{(a-b-c)(a-b+c)}=\frac{a+b+c}{a-b+c}\] \[=\frac{0.25+0.05+0.5}{0.25-0.05+0.5}=\frac{0.8}{.07}=\frac{8}{7}\]


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