Solution
Let the two numbers be $$12x$$ and $$12y$$ respectively where $$x$$ and $$y$$ are co-primes
Product of numbers = $$(12x) \times (12y)=2160$$
=> $$xy=\frac{2160}{144}=15$$
Possible pairs of $$x$$ and $$y$$ whose H.C.F. is 1 = $$(3,5)$$
$$\therefore$$ Required numbers = $$(12 \times 3),(12 \times 5)$$
= 36 , 60
=> Ans - (C)
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