Correct Answer: 45. (2)

Solution: 45. (2) I :

value $=\frac{1+3+5+7+9}{2+4+6+8}=\frac{25}{20}$

$=\frac{5}{4}=\frac{40}{32}$

II :

value $=\frac{1^{3}+3^{3}+5^{3}+7^{3}+9^{3}}{2^{3}+4^{3}+6^{3}+8^{3}}$

$=\frac{5^{2}\left(2 \times 5^{2}-1\right)}{2(4 \times 5)^{2}}=\frac{25 \times 49}{2 \times 400}=\frac{49}{32}$

$\therefore$ Quantity I $<$ Quantity II

Note :

(i) Sum of cubes of first $\mathrm{n}$ odd natural numbers $=n^{2}\left(2 n^{2}-1\right)$

(ii) Sum of cubes of first $n$ even natural numbers $=2[n(n+1)]^{2}$

(iii) Sum of first $n$ natural numbers $=\frac{n(n+1)}{2}$

(iv) Sum of first $n$ odd natural numbers $=n^{2}$

(v) Sum of first $n$ even natural numbers $=n(n+1)$