Answer:
Option C
Explanation:
Let C.P of each clock be Rs. $x$. Then, C.P pf 90 clocks = Rs. $90x$.
$\therefore$ $[(110$% of $40x)$ $+(120$% of $50x)]$ $-(115$% of $90x)$ $= 40$.
$\Rightarrow 44x+60x-103.5x$
$\Rightarrow 0.5x$ $=40$ $\Rightarrow x =80$.