$$2\frac{1.5}{5}$$ + 2$$\frac{1}{6}$$ - $$1\frac{3.5}{15}$$ = $$\left( {\frac{{{{(?)}^{\frac{1}{3}}}}}{4}} \right)$$  + $$1\frac{7}{30}$$

Solution(By Examveda Team)

$$\eqalign{ & 2\frac{{1.5}}{5} + 2\frac{1}{6} - 1\frac{{3.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + 1\frac{7}{{30}} \cr & \Rightarrow \frac{{11.5}}{5} + \frac{{13}}{6} - \frac{{18.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr} $$
⇒ L.C.M. of 5, 6 and 15 is 30
$$\eqalign{ & \Rightarrow \frac{{69 + 65 - 37}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{97}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{{x^{\frac{1}{3}}}}}{4} = \frac{{97}}{{30}} - \frac{{37}}{{30}} \cr & \Rightarrow {x^{\frac{1}{3}}} = \frac{{60}}{{30}} \times 4 \cr & \Rightarrow {x^{\frac{1}{3}}} = 8 \cr & \Rightarrow x = {\left( 8 \right)^3} \cr & \Rightarrow x = 512 \cr} $$
Hence, the number is 512