10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

  • question_answer The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.

    Answer:

    Given, the houses in a row numbered consecutively from 1 to 49.
    Now, sum of numbers preceding the number X
                                        \[=\frac{X(X-1)}{2}\]
    And, sum of numbers following the number X
                                        \[=\frac{49(50)}{2}-\frac{X(X-1)}{2}-X\]
                                        \[=\frac{2450-{{X}^{2}}+X-2X}{2}\]
                                        \[=\frac{2450-{{X}^{2}}-X}{2}\]
    According to the given condition,
    Sum of no's preceding X = Sum of no's following X
                            \[\frac{X(X-1)}{2}=\frac{2450-{{X}^{2}}-X}{2}\]
                              \[{{X}^{2}}-X=2450-{{X}^{2}}-X\]
                                  \[2{{X}^{2}}=2450\]
                                    \[{{X}^{2}}=1225\]
                                      \[X=35\]
    Hence, at \[X=35\], sum of no. of houses preceding the house no. X is equal to sum of the no. of houses following X.


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