The average of three numbers p, q and r is k. p is as much more than the average as q is less than the average. What is the value of r?

We are given that the average of p, q, and r is k. We recall from the analysis of PYQ 2017 and 2024 that the average behaves like a 'Balance Point' or fulcrum [1], [8]. \n\nThe question states: 'p is as much more than the average as q is less than the average.' \nIn the language of 'Symmetry in Numbers' observed in 2017 [13], this means p and q have equal and opposite deviations. If p is (k + d), then q is (k - d). \n\nWhen we look at the 'Net Deviation' of the group:\nDeviation of p = +d\nDeviation of q = -d\nThese two cancel each other out (Net deviation = 0).\n\nFor the overall average to remain k, the sum of all deviations must be zero. Since p and q already sum to zero, the deviation of the remaining number, r, must also be zero.\nDeviation of r = r - k = 0.\nTherefore, r = k.\n\nThis uses the 'Method of Deviation' [40] to solve by inspection, exactly as suggested for the 2017 problem regarding weights.