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NTSE Sample Paper NTSE SAT Practice Test-12

  • question_answer
    If \[x=\frac{a-b}{a+b},y=\frac{b-c}{b+c},\]\[z=\frac{c-a}{c+a}\]then what is the value of  \[\frac{1+x}{1-x}\times \frac{1+y}{1-y}\times \frac{1+z}{1-z}\]?

    A) 1                     

    B) 2

    C) 3                                 

    D) 4

    Correct Answer: A

    Solution :

    [a] \[\frac{1}{x}=\frac{a+b}{a-b}\] Or  \[\frac{1+x}{1-x}=\frac{a}{b}\] Also, \[\frac{1+y}{1-y}=\frac{b}{c}\]and \[\frac{1+z}{1-z}=\frac{c}{a}\] \[\therefore \,\,\,\left( \frac{1+x}{1-x} \right)\times \left( \frac{1+y}{1-y} \right)\times \left( \frac{1+z}{1-z} \right)\] \[=\frac{a}{b}\times \frac{b}{c}\times \frac{c}{a}=1\]    


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