The average atomic mass of an element is calculated as the weighted mean of the masses of its isotopes, with each isotope's mass weighted by its fractional abundance. The concept accounts for the fact that in any macroscopic sample, both isotopes are present in their natural proportions. Average atomic mass = (mass of X-35 × fractional abundance) + (mass of X-37 × fractional abundance) = (35 × 0.75) + (37 × 0.25) = 26.25 + 9.25 = 35.5 u. Option 36.0 u would result from a simple arithmetic mean of 35 and 37, ignoring the different abundances. Option 35.0 u would be correct only if X-35 were 100% abundant. Option 36.5 u would correspond to a 50:50 mixture of X-35 and X-37. The calculated value of 35.5 u identifies element X as chlorine, whose actual atomic mass is 35.45 u due to the 75.77% : 24.23% abundance ratio. This is a standard NCERT problem on atomic mass and isotopes, fundamental to the JEE syllabus. Plausibility check: the average must lie between 35 and 37; with X-35 being three times more abundant than X-37, the average must be closer to 35 than to 37, and 35.5 satisfies this constraint.