Answer: (a)

\(\displaystyle \frac{x}{y}=\frac{3}{1}\) Þ \(\displaystyle \frac{{{{x}^{3}}}}{{{{y}^{3}}}}=\frac{{27}}{1}\)

\(\displaystyle \Rightarrow \) \(\displaystyle \frac{{{{x}^{3}}-{{y}^{3}}}}{{{{x}^{3}}+{{y}^{3}}}}=\frac{{27-1}}{{27+1}}\)

[By componendo  and dividendo]

= \(\displaystyle \frac{{26}}{{28}}=\frac{{13}}{{14}}=13:14\)

Answer: (c)

Let the fourth proportional be x Then, \(\displaystyle \frac{{0.12}}{{0.21}}=\frac{8}{x}\)

or x = \(\displaystyle 8\times \frac{{21}}{{12}}\)

or x = 14

Alternately  :

Fourth proportion = \(\displaystyle \frac{{bc}}{a}=\frac{{0.21\times 18}}{{0.12}}=14\)

Answer: (a)

Required ratio = \(\displaystyle \frac{{{{2}^{{1.5}}}}}{{{{2}^{{0.5}}}}}\)

= \(\displaystyle \frac{{{{2}^{{1.5-0.5}}}}}{1}\)

\(\displaystyle \frac{2}{1}=2:1\)

The Ratio and Proportion questions answer is (d)

A : D = \(\displaystyle \frac{A}{D}=\frac{A}{B}\times \frac{B}{C}\times \frac{C}{D}\)

= \(\displaystyle \frac{3}{4}\times \frac{5}{7}\times \frac{8}{9}=\frac{{10}}{{21}}=10:21\)

Answer: (d)

Since b is the mean proportional of a and c.

\(\displaystyle \frac{a}{b}=\frac{b}{c}=k\) (suppose)

a = bk, b = ck

\(\displaystyle \frac{{{{{(a-b)}}^{3}}}}{{{{{(b-c)}}^{3}}}}=\frac{{{{{(bk-b)}}^{3}}}}{{{{{(ck-c)}}^{3}}}}\)

\(\displaystyle \frac{{{{b}^{3}}{{{(k-1)}}^{3}}}}{{{{c}^{3}}{{{(k-1)}}^{3}}}}=\frac{{{{b}^{3}}}}{{{{c}^{3}}}}=\frac{{{{a}^{3}}}}{{{{b}^{3}}}}\)