Railways Quantitative Aptitude Power, Surds And Indices Sample Paper

done Surds Indices Sample Test Paper-4 Total Questions - 20    [Download]

• question_answer1)
  The value of \[{{\left( \frac{-1}{216} \right)}^{-\,\frac{2}{3}}}\]  is :
  A)
   \[\frac{1}{36}\]
  done clear
  B)
          \[-\frac{1}{36}\]
  done clear
  C)
   \[-36\]              
  done clear
  D)
   36
  done clear
  View Answer play_arrow

• question_answer2)
  The value of \[{{\left( \frac{1}{4} \right)}^{-2}}\] is :
  A)
   2                    
  done clear
  B)
   \[-\frac{1}{2}\]
  done clear
  C)
   \[-\frac{1}{16}\]           
  done clear
  D)
    16
  done clear
  View Answer play_arrow

• question_answer3)
  Simplify: \[{{13}^{\frac{1}{5}}}{{.17}^{\frac{1}{5}}}\]
  A)
   221                 
  done clear
  B)
   \[\sqrt{221}\]
  done clear
  C)
   \[\sqrt[5]{221}\]
  done clear
  D)
   \[\frac{1}{5}\]
  done clear
  View Answer play_arrow

• question_answer4)
  Simplify: \[{{\left( \frac{{{2}^{a}}}{{{2}^{b}}} \right)}^{a+b}}{{\left( \frac{{{2}^{b}}}{{{2}^{c}}} \right)}^{b+c}}{{\left( \frac{{{2}^{c}}}{{{2}^{a}}} \right)}^{c+a}}\]
  A)
    0                   
  done clear
  B)
   1          
  done clear
  C)
    2       
  done clear
  D)
   \[{{(2)}^{a+b+c}}\]
  done clear
  View Answer play_arrow

• question_answer5)
  Show that: \[\frac{{{x}^{a(b-c)}}}{{{x}^{b(a-c)}}}\div {{\left( \frac{{{x}^{b}}}{{{x}^{a}}} \right)}^{c}}=?\]
  A)
   0                    
  done clear
  B)
   1          
  done clear
  C)
   2                    
  done clear
  D)
   \[{{2}^{(a+b+c)}}\]
  done clear
  View Answer play_arrow

• question_answer6)
  If \[{{\left[ {{\left\{ {{\left( \frac{1}{{{7}^{2}}} \right)}^{-2}} \right\}}^{-1/3}} \right]}^{\frac{1}{4}}}={{7}^{m}},\] then find the value of m.
  A)
   \[m=1\]           
  done clear
  B)
   \[m=\frac{1}{3}\]
  done clear
  C)
   \[m=-\frac{1}{3}\]       
  done clear
  D)
   \[m=-7\]
  done clear
  View Answer play_arrow

• question_answer7)
  When simplified the product \[\left( 1+\frac{1}{2} \right)\left( 1+\frac{1}{3} \right)\left( 1+\frac{1}{4} \right).....\left( 1+\frac{1}{n} \right)\] becomes
  A)
   \[n\]                
  done clear
  B)
   \[\frac{n-1}{2}\]
  done clear
  C)
   \[\frac{n+1}{2}\]         
  done clear
  D)
          \[\frac{n}{2}\]
  done clear
  View Answer play_arrow

• question_answer8)
  Evaluate \[\sqrt[3]{{{\left( \frac{1}{64} \right)}^{2}}}\]
  A)
    4                   
  done clear
  B)
   16        
  done clear
  C)
   \[\frac{1}{4}\] 
  done clear
  D)
   \[\frac{1}{16}\]
  done clear
  View Answer play_arrow

• question_answer9)
  \[\frac{{{2}^{n+2}}-2({{2}^{n}})}{{{2}^{(2n+2)}}}\] when simplified is
  A)
   \[1-2({{2}^{n}})\]      
  done clear
  B)
   \[{{2}^{n+3}}-\frac{1}{4}\]
  done clear
  C)
   \[\frac{1}{{{2}^{n+1}}}\]     
  done clear
  D)
   \[\frac{1}{{{2}^{n-1}}}\]
  done clear
  View Answer play_arrow

• question_answer10)
  Simplify: \[{{\left[ 5{{\left( {{8}^{\frac{1}{3}}}+{{27}^{\frac{1}{3}}} \right)}^{3}} \right]}^{\frac{1}{4}}}\]
  A)
   0                    
  done clear
  B)
   1         
  done clear
  C)
   5                    
  done clear
  D)
   2
  done clear
  View Answer play_arrow

• question_answer11)
  Simplify: \[\sqrt[3]{2}+\sqrt[4]{64}+\sqrt[4]{2500}+\sqrt[6]{8}\]
  A)
   \[\sqrt{2}\]                   
  done clear
  B)
   \[2\sqrt{2}\]
  done clear
  C)
   \[11\sqrt{2}\]   
  done clear
  D)
   \[9\sqrt{2}\]
  done clear
  View Answer play_arrow

• question_answer12)
  If abc = 1, then \[\left( \frac{1}{1+a+{{b}^{-1}}}+\frac{1}{1+b+{{c}^{-1}}}+\frac{1}{1+c+{{a}^{-1}}} \right)=?\]
  A)
   0                    
  done clear
  B)
   1         
  done clear
  C)
   \[\frac{1}{ab}\]            
  done clear
  D)
   ab
  done clear
  View Answer play_arrow

• question_answer13)
  \[\frac{{{(243)}^{\frac{n}{5}}}\times {{3}^{2n+1}}}{{{9}^{n}}\times {{3}^{n-1}}}=?\]
  A)
   1                    
  done clear
  B)
   3        
  done clear
  C)
   9                    
  done clear
  D)
   \[{{3}^{n}}\]
  done clear
  View Answer play_arrow

• question_answer14)
  If \[{{27}^{k}}=\frac{9}{{{3}^{k}}},\] then value of \[\frac{1}{{{k}^{2}}}\] is
  A)
   \[\frac{1}{4}\]             
  done clear
  B)
   4        
  done clear
  C)
   \[\frac{1}{2}\] 
  done clear
  D)
   2
  done clear
  View Answer play_arrow

• question_answer15)
  If \[\frac{{{3}^{x}}}{1+{{3}^{x}}}=\frac{1}{9},\] the value of \[\frac{{{9}^{x}}}{1+{{9}^{x}}}\]
  A)
   \[\frac{1}{27}\]            
  done clear
  B)
   \[\frac{1}{64}\]
  done clear
  C)
   \[\frac{1}{65}\]            
  done clear
  D)
   None of these
  done clear
  View Answer play_arrow

• question_answer16)
  If \[a={{x}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}}\]  then \[{{a}^{3}}-3a=\]
  A)
   \[x-{{x}^{-1}}\]         
  done clear
  B)
   \[2x\]
  done clear
  C)
   \[x+{{x}^{-1}}\]       
  done clear
  D)
   0
  done clear
  View Answer play_arrow

• question_answer17)
  On simplification \[{{\left[ \frac{{{x}^{\frac{a}{a-b}}}}{{{x}^{\frac{a}{a+b}}}}\div \frac{{{x}^{\frac{b}{b-a}}}}{{{x}^{\frac{b}{b+1}}}} \right]}^{a+b}}\]reduces to
  A)
   1                    
  done clear
  B)
   \[-1\]       
  done clear
  C)
   0        
  done clear
  D)
   None of these.
  done clear
  View Answer play_arrow

• question_answer18)
  If \[{{4}^{\sqrt{x}}}^{^{\sqrt{x}}}=256\] then the value of x is
  A)
   2                    
  done clear
  B)
   16        
  done clear
  C)
   4                    
  done clear
  D)
   \[\sqrt{2}\]
  done clear
  View Answer play_arrow

• question_answer19)
  If \[{{3}^{2x}}^{^{2}}-{{2.3}^{{{x}^{2}}+x+6}}+{{3}^{2(x+6)}}=0\] then the values of x are
  A)
   \[x=-3,-2\]       
  done clear
  B)
   \[x=3,2\]
  done clear
  C)
   \[x=-3,2\]        
  done clear
  D)
   \[x=3,-2\]
  done clear
  View Answer play_arrow

• question_answer20)
  Value of \[\frac{991\times 991\times 991+9\times 9\times 9}{991\times 991-991\times 9+9\times 9}\] is
  A)
   991                 
  done clear
  B)
   9        
  done clear
  C)
   1000               
  done clear
  D)
   \[991\times 9\]
  done clear
  View Answer play_arrow