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JEE Main & Advanced Sample Paper JEE Main Sample Paper-29

  • question_answer
    A bag contains \[(2n+1)\] coins. It is known that n of these coin have a head on both sides, whereas the remaining \[(n+1)\] coins are fair. A coin is selected at random from the bag and tossed once. If the probability the toss results in a head is \[31/42\], then n is equal to

    A) 10                                          

    B) 11  

    C) 12                                          

    D) 13

    Correct Answer: A

    Solution :

    \[\frac{31}{42}=\left( \frac{n}{2n+1} \right).1+\frac{(n+1)}{(2n+1)}.\frac{1}{2}\] \[\Rightarrow (31).2(2n+1)=42(2n+n+1)\]                            \[\Rightarrow 31(2n+1)=21(3n+1)\] \[\Rightarrow 62n+31=63n+21\]               \[\Rightarrow n=10\]


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