The requirement to retrieve the details of vehicles whose total service cost exceeds ₹50,000 is a common database query problem in a DBMS and C++ Lab. It involves storing vehicle information, maintaining service records, calculating the total service cost for each vehicle, and displaying only those vehicles for which the accumulated service expenditure is greater than ₹50,000. This type of query demonstrates the practical use of relational tables, aggregate functions, grouping, joins, and the `HAVING` clause in SQL. A suitable database design can contain two related tables: `Vehicle` and `Service`. The `Vehicle` table stores basic information about each vehicle, while the `Service` table stores individual service transactions. For example, the `Vehicle` table may contain attributes such as `Vehicle_ID`, `Vehicle_No`, `Vehicle_Model`, `Owner_Name`, and `Vehicle_Type`. The `Service` table may contain `Service_ID`, `Vehicle_ID`, `Service_Date`, `Service_Type`, and `Service_Cost`. The `Vehicle_ID` acts as the primary key in the `Vehicle` table and as a foreign key in the `Service` table. An example structure can be represented as follows: ```sql CREATE TABLE Vehicle ( Vehicle_ID INT PRIMARY KEY, Vehicle_No VARCHAR(20), Vehicle_Model VARCHAR(50), Owner_Name VARCHAR(50), Vehicle_Type VARCHAR(30) ); CREATE TABLE Service ( Service_ID INT PRIMARY KEY, Vehicle_ID INT, Service_Date DATE, Service_Type VARCHAR(50), Service_Cost DECIMAL(10,2), FOREIGN KEY (Vehicle_ID) REFERENCES Vehicle(Vehicle_ID) ); ``` The service table can contain multiple records for the same vehicle because a vehicle may be serviced many times. For example, a vehicle with `Vehicle_ID = 101` might have service costs of ₹15,000, ₹20,000, and ₹25,000 on different dates. Its total service cost would therefore be ₹60,000. Since ₹60,000 is greater than ₹50,000, the vehicle must be included in the result. The SQL query to retrieve the required vehicle details is: ```sql SELECT v.Vehicle_ID, v.Vehicle_No, v.Vehicle_Model, v.Owner_Name, v.Vehicle_Type, SUM(s.Service_Cost) AS Total_Service_Cost FROM Vehicle v JOIN Service s ON v.Vehicle_ID =