KVPY 2019 SA Exam Previous Papers

Question : 8 of 80

Marks: +1, -0

Let

x_1, x_2, …., x_11

be 11 distinct positive integers. If we replace the largest of these integers by the median of the other 10 integers, then

Solution:

Let

x_1 < x_2 < x_3 ……x_11

median of

x_1, x_2 …..x_10 \is {x_5+x_6} /2

Now the new set of number are

x_1, x_2, ….x_5,{x_5+x_6} /2, x_6, ….x_10

Hence median is ${x_5+x_6} /2⇒ median decreases

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