RDBMS Questions and Answers - SQL Queries - III

This set of RDBMS Multiple Choice Questions & Answers (MCQs) focuses on “Sql Queries – III”.

1. Find all the bank customers having a loan, an account, or both at the bank.
View Answer

Answer: select customer_name from depositor
union all
select customer_name from borrower;

2. Find all the bank customers having both a loan and an account at the bank
View Answer

Answer: select customer_name from depositor
intersect all
select customer_name from borrower;

3. Find all customers who have an account but no loan at the bank.
View Answer

Answer: select customer_name from depositor
except
select customer_name from borrower;

Aggregate Functions (avg,min,max,sum,count) / Group By
4. Find the average account balance at the Perryridge branch.
View Answer

Answer: select avg(balance) from account where branch_name=’Perryridge’;

5. Find the average account balance at each branch.
View Answer

Answer: select avg(balance),branch_name
from account natural join depositor
group by branch_name;

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6. Find the number of depositors for each branch (Use distinct).
View Answer

Answer: select count(Distinct customer_name),branch_name
from account natural join depositor
group by branch_name;

7. Find those branches where the average accounts balance is more than Rs. 1200.
View Answer

Answer: select branch_name from account
group by branch_name
having avg(balance)>1200

8. Find the number of branches of the bank.
View Answer

Answer: select count(distinct branch_name) from branch;

9. Find the average balance for each customer who lives in Harrison and has at least three accounts.
View Answer

Answer: /*select avg(balance) from account where account_number in
(
select account_number from depositor
where customer_name =
(select customer_name from depositor group by customer_name
having count(account_number) >= 2
intersect
select customer_name from customer where
customer_city = ‘Harrison’));*/

select avg(balance), customer_name from
depositor natural join account natural join customer
where customer_city = ‘Harrison’ group by customer_name
having count(customer_name)>=2;

Null values
10. Find all loan numbers that appear in the loan relation with null values for amount.
View Answer

Answer: select loan_number from loan where amount is null.

Sanfoundry Global Education & Learning Series – RDBMS.

To practice all areas of RDBMS, here is complete set of 1000+ Multiple Choice Questions and Answers.

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