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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    If the median and the range of four numbers                 \[\{x,y,2x+y,x-y\},\] where                 \[0<y<x<2y\] are 10 and 28  respectively, then the mean of the numbers is :     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                   18

    B)                                                          10

    C)                                          5                                           

    D)                                          14

    Correct Answer: D

    Solution :

                      Since                 \[0<y<x<2y\]                 \[\therefore\]                 \[y>\frac{x}{2}\Rightarrow x-y<\frac{x}{2}\]                 \[\therefore\]                 \[x-y<y<x<2x+y\] Hence median                 \[=\frac{y+x}{2}=10\]                 \[\Rightarrow\]                 \[x+y=20\] ?(i) And range                 \[=(2x+y)-(x-y)=x+2y\] But range =28                 \[\therefore\]                 \[x+2y=28\]                                                                 ...(ii) From equations (i) and (ii),                 \[\therefore\] Mean                 \[=\frac{(x-y)+y+x+(2x+y)}{4}=\frac{4x+y}{4}\]                 \[=x+\frac{y}{4}=12+\frac{8}{4}=14\]


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