2. Sample Papers
  3. Banking

Banking Sample Paper IBPS PO (Main) Sample Test Paper-3

  • question_answer
    Direction: In each of the following questions read the given statements and compare the given two quantities on its basis.
    \[{{({{x}^{b}})}^{e}}={{x}^{e}}\] and \[\frac{{{x}^{3c}}}{{{x}^{b}}}=({{x}^{7b}})\times ({{x}^{y}})\times ({{x}^{2c}})\] Also, \[e\ne o\] Quantity I. c Quantity II. y

    A)  Quantity I > Quantity II

    B)  Quantity I \[\ge \] Quantity II

    C)  Quantity I \[\le \] Quantity II

    D)  Quantity I < Quantity II

    E)  No relation between Quantity I and II

    Correct Answer: A

    Solution :

    \[{{({{x}^{b}})}^{e}}={{x}^{e}}\] \[\Rightarrow \]\[be=e\] \[\Rightarrow \]\[b=1\,\,(\because \,e\ne o)\] Now, \[\frac{{{x}^{3c}}}{{{x}^{b}}}=({{x}^{7b}})\times ({{x}^{y}})\times ({{x}^{2e}})\] Putting the value of b = 1 in equation (i), we get or, \[\frac{{{x}^{3c}}}{{{x}^{1}}}={{x}^{7}}\times {{x}^{y}}\times {{x}^{2c}}\] or, \[3c-1=7+y+2c\] or, \[3c-2c-y=7+1\] or, \[c-y=8\] or, \[c=8+y\] Hence \[c>y\]


You need to login to perform this action.
You will be redirected in 3 sec spinner