CAT 2022 Slot 2 Paper

Maximum number of widgets delivered in a day is 100 units $(C)+70$ units $(A)+60$ units $(B)+$ 50 units $(D)$, 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. $A - (70,50), B - (60,40), C-(100,70) \text{ and } D - (50,30)$ 20 units can be decreased from $A, B$ or $D$ . Demand at location $C$ will be 100 units and supplier first visits $C$. In the question, it is also given that the route ends at $B$ . $C(100$ units),_, , $B$ If B's demand is 60 units, $\text{D's}$ 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 $B$ . This implies demand at location $A$ is 70 units and at location $D$ is 50 units. Order will be $C(100 \text{ units}) - A(70 \text{ units}) - D(50 \text{ units}) - B(40 \text{ units})$. It is given, $C - 100 \text{ units} - 70\%$ $A - 70 \text{ units} - 60\%$ $D - 50 \text{ units} - 60\%$ $B - 40 \text{ units} - 30\%$ Required value $=0.7 \times 0.6 \times 0.6 \times 0.3 = 0.0756 = 7.56\%$ The answer is option D.