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JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    A number x is chosen at random from the set {1, 2, 3, 4, ...., 100}. Define the event: A = the chosen number x satisfies\[\frac{\left( x-10 \right)\left( x-50 \right)}{\left( x-30 \right)}\ge 0\] Then P (A) is:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) 0.71                       

    B) 0.70

    C) 0.51                                       

    D) 0.20

    Correct Answer: A

    Solution :

                    Given \[\frac{(x-10)(x-50)}{(x-30)}\ge 0\] Let\[x\ge 10,x\ge 50\] equation will be true \[\forall x\ge 50\] as\[\left( \frac{x-50}{x-30} \right)\ge 0,\forall x\in [10,30)\] \[\frac{(x-10)(x-50)}{x-30}\ge 0,\forall x\in [10,30)\] Total value of x between 10 to 30 is 20. Total values of x between 50 to 100 including 50 and 100 is 51. Total values of x = 51 + 20 = 71 \[P(A)=\frac{71}{100}=0.71\]


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