CAT 2022 Slot 2 Paper

Maximum number of widgets delivered in a day is 100 units $\left(\mathrm{C}\right)+70$ units $\left(\mathrm{A}\right)+60$ units $\left(\mathrm{B}\right)+$ 50 units $\left(\mathrm{D}\right)$, i.e. 280 units
Given, total number of widgets delivered is 260 units. This implies 20 units must be decreased from any one of the locations.
$\mathrm{A}-\left(70,50\right),\mathrm{B}-\left(60,40\right),\mathrm{C}-\left(100,70\right)\mathrm{and}\mathrm{D}-\left(50,30\right)$
20 units can be decreased from $\mathrm{A},\mathrm{B}$ or $\mathrm{D}$ .
Demand at location $\mathrm{C}$ will be 100 units and supplier first visits $\mathrm{C}$.
In the question, it is also given that the route ends at $\mathrm{B}$ .
$\mathrm{C}(100$ units),_, , $\mathrm{B}$
If B's demand is 60 units, $\mathrm{D}\prime \mathrm{s}$ demand should be more than 60 units which is not possible. Therefore, B's demand should be 40 units.
20 units is decreased at location $\mathrm{B}$ . This implies demand at location $\mathrm{A}$ is 70 units and at location $\mathrm{D}$ is 50 units.
Order will be $\mathrm{C}\left(100\mathrm{units}\right)-\mathrm{A}\left(70\mathrm{units}\right)-\mathrm{D}\left(50\mathrm{units}\right)-\mathrm{B}\left(40\mathrm{units}\right)$.
It is given,
$\mathrm{C}-100\mathrm{units}-70\%$
$\mathrm{A}-70\mathrm{units}-60\%$
$\mathrm{D}-50\mathrm{units}-60\%$
$\mathrm{B}-40\mathrm{units}-30\%$
Required value $=0.7\times 0.6\times 0.6\times 0.3=0.0756=7.56\%$
The answer is option D.