CAT 2020 Exam Shift 2 Solved Paper

$2\mathit{x}+5\mathit{y}=99$
When $\mathit{y}=-19,\mathit{x}=97;$ since $\mathit{x}\ge \mathit{y};$ the maximum value of $\mathit{y}$ is 13 and corresponding value of $\mathit{x}$ is 17
We know that the solutions of $\mathit{y}$ are in arithmetic progression with common difference 2
Here $\mathit{a}=-19,\mathit{d}=2,\mathit{t}\mathit{n}=13$
${\mathit{t}}_{\mathit{n}}=\mathit{a}+\left(\mathit{n}-1\right)\mathit{d}$
$-19+\left(\mathit{n}-1\right)\left(2\right)=13$
$\left(\mathit{n}-1\right)2=32\Rightarrow \mathit{n}=17$
Hence number of pairs integers is 17