CAT 2025 Slot 3 · Quant · Question 9

Break every base into primes: 12={2}^{2}\cdot 3, 4={2}^{2}, 8={2}^{3}, 20={2}^{2}\cdot 5 and 243={3}^{5}.

The left side becomes {2}^{24x}\cdot{3}^{12x}\times{2}^{48x+24}\times{5}^{2y}={2}^{72x+24}\cdot{3}^{12x}\cdot{5}^{2y}.

The right side becomes {2}^{12z}\times{2}^{24x}\cdot{5}^{12x}\times{3}^{15x-30}={2}^{12z+24x}\cdot{3}^{15x-30}\cdot{5}^{12x}.

Matching the powers of 3 gives 12x=15x-30, so 3x=30 and x=10.

Matching the powers of 5 gives 2y=12x=120, so y=60.

Matching the powers of 2 gives 72x+24=12z+24x, that is 744=12z+240.

So 12z=504 and z=42.

Hence x+y+z=10+60+42=112.

Shortcut. Start with the prime that appears least often. Only one factor on each side carries a 3, so that comparison gives x on its own, and y and z then fall out one at a time.