NIMCET 2019 Question Paper with solutions

We have, first 30 natural numbers.
Two number are selected a and b such that ${\mathit{a}}^2-{\mathit{b}}^2$
is divisible by 2
$\mathit{n}\left(\mathit{S}\right){=}^{30}{\mathit{C}}_2$
$\because {\mathit{a}}^2-{\mathit{b}}^2$ is divisible by 3
$\therefore \left(\mathit{a}+\mathit{b}\right)\left(\mathit{a}-\mathit{b}\right)=3\mathit{m}$
If $\left(\mathit{a}-\mathit{b}\right)$ is multiple of 3, then ${\mathit{a}}^2-{\mathit{b}}^2$ is multiple of 3
∴ Such numbers are
(1,4,7,....,28),(2,5,....,29),(3,6,9,...,30) are possible are
$\therefore {}^{10}{\mathit{C}}_2{+}^{10}{\mathit{C}}_2{+}^{10}{\mathit{C}}_2=45\times 3=135$
o c 5) is multiple of 3 then ${\mathit{a}}^2-{\mathit{b}}^2$ is multiple of 3
Such numbers are possible
Total$=10\times 10=100$
Total numbers are $135+100=235$
∴ Required probability$=\frac{235}{{}_{}^{30}{\mathit{C}}_2^{}}=\frac{235\times 2}{30\times 25}=\frac{47}{87}$