SQL Interview Question - Solution (Part - XVI) |

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Here are My profiles that will definitely help your preparation for data analyst or data engineer roles.
Medium: / mahendraee204
Github: github.com/mahendra204
Here are table create and insert statements:
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create table drivers (driver_id int, name varchar(20))
insert into drivers values (201, 'venkey'), (202, 'mahendra'), (210, 'mahi'),(203, 'maha')
create table riders (rider_id int, name varchar(20))
insert into riders values(10, 'sardar'), (11, 'kishore'),(12, 'dev')
create table trips (trip_id int, rider_id int, driver_id int,pickup_time time, dropoff_time time,fare_amount money, distance int);
insert into trips values
(1, 10, 201, '05:00', '06:30', 240.00, 15),
(2, 11, 202, '06:00', '07:30', 210.00, 12),
(3, 11, 201, '07:00', '08:30', 200.00, 10),
(4, 12, 202, '08:00', '09:30', 500.00, 20),
(5, 12, 203, '10:00', '12:30', 1000.00, 25),
(6, 11, 210, '09:00', '12:30', 750.00, 21);

[@krishnasai5806]

Is there any openings for freshers

[@MeanLifeStudies]

As of my knowledge there are very less openings for freshers.

[@kailashpatro5768]

Rider _name :
select top 1 r.name,
row_number() over(order by fare_amount/distance desc) as rider_per_km
from trips12 t inner join riders r on t.rider_id = r.rider_id
order by row_number() over(order by fare_amount/distance desc)
Driver_name
select top 1 d.name,
row_number() over(order by fare_amount/distance asc) as driver_per_km
from trips12 t inner join drivers d on t.driver_id = d.driver_id
order by row_number() over(order by fare_amount/distance asc)