GATE 2026 AG – Question 48

Correct answer: (A) 1116.85

Explanation

For a unit total mass, assuming additive constituent volumes, the component volumes sum to $V=\sum_iw_i/\rho_i$. Therefore $$\rho=\left(\frac{.220}{1550}+\frac{.010}{1220}+\frac{.005}{900}+\frac{.015}{610}+\frac{.750}{1000}\right)^{-1}=1074.95\ \mathrm{kg/m^3}.$$ The official key selects **A,1116.85**,but that value does not follow from this physical mixing assumption. The key instead uses the mass-weighted arithmetic average: .220(1550)+.010(1220)+.005(900)+.015(610)+.750(1000)=1116.85. That averaging treats the supplied mass fractions like volume fractions; additive volumes require the reciprocal-density relation above. The printed datum is retained and the inconsistency is explicit;grading remains official A.