JAMB 1986 · UME · Q3

Write each number as a product of primes: $48 = 2^4 \times 3$, $64 = 2^6$ and $80 = 2^4 \times 5$.

The LCM takes the highest power of every prime: $2^6 \times 3 \times 5 = 960$.

The HCF takes the lowest power of the primes common to all three: only 2 is common, so the HCF is $2^4 = 16$.

So $960 \div 16 = 60$, option D.