CAT 2015 Quantitative Ability

Since LCM (c. a) and LCM (b, c) is$2^4\times 5^3$ and LCM (a, b) = $2^3\times 5^3$ so the factor $2^4$ must be present in c.

Hence $\mathit{c}=2^4\times 5^{\mathit{x}}$, where x ranges from 0 to 3 Therefore, there are four possible values of C.

Since, HCF of (a, b) = $\mathit{K}\times 5^3$, it means
$\mathit{a}=2^{\mathit{y}}\times 5^3$
$\mathit{b}=2^{\mathit{z}}\times 5^3$
x = 0 to 3, y = 0, then z = 3 → 4 cases,
x = 0 to 3, y = 1, then z = 3 → 4 cases,
x = 0 to 3, y = 2, then z = 3 → 4 cases,
x = 0 to 3, y = 3, then z = 3 → 4 cases,
x = 0 to 3, y = 3, then z = 2 → 4 cases,
x = 0 to 3, y = 3, then z = 1 → 4 cases,
x = 0 to 3, y = 3, then z = 0 → 4 cases.
Hence, total cases = 28.