JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Binomial Theorem

The general term of ${\left(7^{1/5}-3^{1/10}\right)}^{60}$ is $\left(\genfrac{}{}{0ex}{}{60}{r}\right)7^{(60-r)/5}(-1{)}^r{3}^{r/10}$, which is rational only when both $\frac{60-r}{5}$ and $\frac{r}{10}$ are integers. That needs $r$ a multiple of $10$, giving $r=0,10,20,30,40,50,60$ $-$ seven rational terms. Out of $61$ terms in all, the remaining $61-7=54$ are irrational.