JEE Main 2026 — 24 January, Morning Shift — Question 6

$18 x^{2}-11 x+1>0$

$(2 \mathrm{x}-1)(9 \mathrm{x}-1)>0$

$\mathrm{x}<\frac{1}{9}$ or $\frac{1}{2}<\mathrm{x}$

Also $10 \mathrm{x}^{2}-17 \mathrm{x}+7>0$

$(\mathrm{x}-1)(10 \mathrm{x}-7)>0$

$\mathrm{x}<\frac{7}{10}$ or $1<\mathrm{x}$ &$10 \mathrm{x}^{2}-17 \mathrm{x}+7 \neq 1$

$x \in\left(-\infty, \frac{1}{9}\right) \cup\left(\frac{1}{2}, \frac{7}{10}\right) \cup(1, \infty)-\left\{\frac{6}{5}\right\}$

$90(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e})=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5}\right)$

$=10+45+63+90+108=316$