Given: the alloy has 25% Cd (75% Bi), and the eutectic has 40% Cd (60% Bi).
The alloy has less Cd (25%) than the eutectic (40%), so it has excess Bi. On cooling, the excess Bi separates out first, and the rest solidifies as eutectic. So the solid alloy is pure Bi + eutectic, and all the Cd is present in the eutectic.
Mass of Cd in 1 kg of alloy = 25% of 1000 g = 250 g.
\( \text{Mass of eutectic} = \dfrac{250 \times 100}{40} = 625\ \text{g} \)
Mass of eutectic = 625 g (0.625 kg), and the free Bi = 375 g.
Check: Bi in the eutectic = 0.60 × 625 = 375 g; 375 + 375 = 750 g = 75% of 1 kg (verified).