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1.  

Directions: The next 4 questions are based on the table ACC of this chapter:

AccNo Name Branch Type Balance Nominee
101 Ravi Salem SB 45000 Lakshmi
102 Meena Madurai CA 120000 NULL
103 Arun Salem SB 8000 Kumar
104 Divya Coimbatore FD 250000 Ravi
105 Karthik Madurai SB 30000 NULL
106 Priya Salem CA 75000 Anand
107 Suresh Coimbatore SB 15000 NULL
SELECT SUM(Balance) FROM ACC WHERE Branch = 'Salem' AND Type = 'SB'; returns:

Correct Answer: B [53000]

Explanation:

Salem SB rows: Ravi 45000 + Arun 8000 = 53000.

2.  
SELECT COUNT(DISTINCT Type) FROM ACC; returns:

Correct Answer: D [3]

Explanation:

Types SB, CA, FD → 3.

3.  
SELECT Name FROM ACC WHERE Name LIKE '%i' ORDER BY Name; returns:

Correct Answer: E [Ravi]

Explanation:

Names ending in 'i': only Ravi (Divya and Priya end in 'a').

4.  
SELECT Branch FROM ACC GROUP BY Branch HAVING SUM(Balance) > 150000 ORDER BY Branch; returns:

Correct Answer: A [Coimbatore]

Explanation:

Totals: Salem 128000, Madurai 150000 (not > 150000), Coimbatore 265000 → only Coimbatore.

5.  
Consider the following statements:
I. Every candidate key is a super key.
II. A table can have more than one primary key.
III. A foreign key may contain duplicate values in its own table.
Which of the statements given above is/are correct?

Correct Answer: C [Only I and III]

Explanation:

I true (minimal super key); II false – one primary key per table; III true – many accounts can refer to one customer.

6.  
Consider the following statements:
I. TRUNCATE is a DML command.
II. DROP removes both the data and the structure of a table.
III. DELETE can remove selected rows with a WHERE clause.
Which of the statements given above is/are correct?

Correct Answer: D [Only II and III]

Explanation:

I false – TRUNCATE is DDL; II and III true.

7.  
A transaction debits ₹ 5,000 from one account, but the system fails before the credit to the other account. The DBMS undoes the debit. This is ensured by:

Correct Answer: B [Atomicity]

Explanation:

All or nothing = atomicity.

8.  
A table has candidate keys {EmpID}, {PAN} and {Email}. EmpID is chosen as the primary key. How many alternate keys does it have?

Correct Answer: A [2]

Explanation:

Alternate keys = 3 candidate keys − 1 primary = 2.

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