The given inequality is
$\large\frac{(2x-1)}{3} \geq \frac{(3x-2)} {4} - \frac{(2-x)}{5}$
=> $ \large\frac{(2x-1)}{3} \geq \frac{5(3x-2) -4(2-x)}{20}$
=> $ \large\frac{2x-1)}{3} \geq \frac{5(3x-2)-4(2-x)}{20}$
=> $ \large\frac{(2x-1)}{3} \geq \frac{19 x -18}{20}$
Multiplying both sides if the inequality by a positive number 60.
=> $20 (2x-1) \geq 3( 19 x -18)$
=> $40 x -20 \geq 57 x -54$
Adding 54 and -40 on both sides of the inequality
$-20 +54 \geq 57 x -40 x$
$34 \geq 17 x$
$2 \geq x$
Step 2:
All numbers which are less than or equal to 2 satisfy the inequality .
The solution set is $(-\infty,2]$
Hence B is the correct answer.