Concurrency questions look like vocabulary tests, but operation order can change dirty read to lost update. One lock entry can settle the answer. This set has ten live, published bank questions with exact public deep links. Choose before reading each explanation. Exam and year labels are bank metadata only, not a claim that this set reproduces an official paper or answer key.
Uncontrolled concurrent updates: identify all consequences
Q1
Bank metadata: UGC NET, December 2019.
Two concurrent executing transactions 𝑇1 and 𝑇2 are allowed to update same stock item say ′A′ in an uncontrolled manner. In such scenario, following problems may occur:(a) Dirty read problem(b) Lost update problem(c) Transaction failure(d) Inconsistent database stateWhich of the following option is correct if database system has no concurrency module and allows concurrent execution of above two transactions?
(0)
(a), (b) and (c) only(1)
(c) and (d) only(2)
(a) and (b) only(3)
(a), (b) and (d) only
**Answer: option 3. Solved page**
Start with A = 100. For a dirty read, T1 writes 120 without committing, T2 reads 120, then T1 aborts, so T2 used a value that never became durable. For a lost update, both transactions read 100; T1 writes 110; T2 writes 125 from its stale copy. The final 125 erases T1's +10. These interleavings can also create an inconsistent state, but uncontrolled access does not itself force a transaction to fail. Hence (a), (b), and (d) only.
Read sets, write sets, and the isolation ceiling
Q2
Bank metadata: UGC NET, June 2015.
Let \(𝑃−𝑖\) and \(𝑃_𝑗\) two processes, R be the set of variables read from memory, and W be the set of variables written to memory. For the concurrent execution of two processes \(𝑝_𝑖\) and \(𝑃_𝑗\) which of the conditions are not true?
(0)
\(R(P_i) \cap W(P_j) = \Phi\)(1)
\(W(P_i) \cap R(P_j) = \Phi\)(2)
\(R(P_i) \cap R(P_j) = \Phi\)(3)
\(W(P_i) \cap W(P_j) = \Phi\)
**Answer: option 2. Solved page**
Take R(P_i) = {x, y}, W(P_i) = {z}, R(P_j) = {x}, and W(P_j) = {w}. Here R(P_i) ∩ R(P_j) = {x}, which is safe because both processes only read x. The read-write, write-read, and write-write intersections are all empty. Those three kinds of overlap could create interference, but a shared read cannot. Therefore the two read sets need not be disjoint. Option 2 is the condition that is not required for this concurrent execution.
Q3
Bank metadata: ISRO, 2013.
Which of the following is the highest isolation level in transaction management?
(0)
Serializable(1)
Repeated Read(2)
Committed Read(3)
Uncommitted Read
**Answer: option 0. Solved page**
Using the option wording, the order from weakest to strongest is Uncommitted Read, Committed Read, Repeated Read, then Serializable. Serializable requires a concurrent result to match a complete serial order, such as T1 then T2 or T2 then T1. It controls the schedule's overall effect, not just one repeated read, so its promise is stronger than the lower levels. Option 0 is therefore highest. Review the framework in Transactions and Concurrency Control in DBMS.
Timestamp ordering and Strict Two-Phase Locking
Q4
Bank metadata: DSSSB, 2021.
_____ is a unique identifier created by database management system to identify the relative starting time of a transaction.
(0)
Timestamp(1)
Spoofing(2)
Shadow paging(3)
Deadlock
**Answer: option 0. Solved page**
Suppose TS(T1) = 12 and TS(T2) = 27. The smaller value identifies T1 as the earlier transaction, and timestamp ordering uses this relative age when deciding whether an operation may proceed. Only relative order matters here, not elapsed time. Shadow paging is a recovery technique, deadlock is a circular waiting condition, and spoofing is unrelated to transaction start order. The requested DBMS-created identifier is therefore a timestamp, making option 0 correct.
Q5
Bank metadata: RPSC, 2024.
Which of the following problem(s) can occur in the Strict Two-Phase Locking (Strict 2PL) protocol?
(0)
Chances of deadlock(1)
Cascading Rollback(2)
Neither 1 nor 2(3)
Both 1 & 2(4)
Question not attempted
**Answer: option 0. Solved page**
T1 acquires X(A), T2 acquires X(B), T1 requests X(B) and waits, then T2 requests X(A) and waits. This circular wait proves that Strict 2PL can deadlock. However, it retains an exclusive lock until commit or abort, so another transaction cannot read that uncommitted write and later need a cascading rollback. Strict 2PL gives conflict serializability and blocks dirty reads of exclusive writes, but neither property gives deadlock freedom. Only problem 1 can occur, so option 0 is correct.
Deadlock prevention and multigranularity lock compatibility
Q6
Bank metadata: HTET, 2022.
In context of deadlock prevention in DBMS, wait-die and wound-wait schemes are ........ techniques.
(0)
both preemptive(1)
both non-preemptive(2)
preemptive and non-preemptive respectively(3)
non-preemptive and preemptive respectively
**Answer to teach after correction or adjudication: option 3. Solved page**
Let TS(T_old) = 5 and TS(T_young) = 20. In wait-die, an older requester may wait for a younger holder, while a younger requester aborts if the older transaction holds the lock. The holder is not forced out, so wait-die is non-preemptive. In wound-wait, an older requester forces a younger holder to roll back, so it is preemptive. The standard classification is option 3. Editorial warning: the live record stores option 0 and reverses these age rules. Correct or adjudicate it before publication; no official HTET key is claimed here.
Q7
Bank metadata: UGC NET, December 2025.
If a transaction ta holds a lock on some data item d in shared and intention-exclusive (SIX) mode, then which of the following is true for another transaction tb? tb can lock d in shared (S) mode tb can lock d in exclusive (X) mode tb can lock d in intention-exclusive (IX) mode tb can lock d in intention-shared (IS) mode
(0)
1(1)
2(2)
3(3)
4
**Answer: option 3, meaning statement 4. Solved page**
Check every mode against the SIX lock on the same object d. SIX combines shared access at d with an intention to take exclusive locks below it. Another transaction's S lock conflicts with that exclusive intention, X conflicts with both parts, and IX conflicts with the shared part. IS only announces an intention to take shared locks lower in the hierarchy, so it is compatible with SIX at d. Only statement 4 is true, and the option naming statement 4 is option 3.
Temporary updates, read-write conflicts, and phantom rows
Q8
Bank metadata: UGC NET, June 2012.
The problem that occurs when one transaction updates a database item and then the transaction fails for some reason is ________.
(0)
Temporary Select Problem(1)
Temporary Modify Problem(2)
Dirty Read Problem(3)
None
**Answer from the exact wording and choices: option 3. Solved page**
Set balance = 500. T1 writes 450 and then fails before commit. This is commonly called a temporary update, but that term is absent from the choices. A dirty read would additionally require T2 to read the uncommitted 450 before T1 aborts, and the stem gives no such read. Temporary Select Problem and Temporary Modify Problem are not the standard label. Among the listed choices, None is therefore defensible. This explanation does not claim an official answer key.
Q9
Bank metadata: DSSSB, 2018.
Which of the following concurrency anomalies is also known as a read-write conflict?
(0)
Dirty read(1)
Lost update problem(2)
Unrepeatable read(3)
Uncommitted dependency
**Answer: option 2. Solved page**
Let X = 100. T1 reads 100; T2 writes X = 120 and commits; T1 reads X again and sees 120. The two reads disagree because T2's intervening write creates a read-write conflict. That is an unrepeatable read, so option 2 fits. By contrast, a write followed by another transaction reading the uncommitted value is a dirty read. Two transactions reading the same old value and then writing can produce a lost update. Here the defining order is read, committed write, reread.
Q10
Bank metadata: MPPSC, 2025.
What is phantom read in the context of transaction isolation levels?
(0)
A transaction reads a set of rows that were inserted by another transaction after it started(1)
A transaction reads data that was committed by another transaction(2)
A transaction reads data that was changed by another transaction but not yet committed(3)
None of the above
**Answer: option 0. Solved page**
T1 runs SELECT id FROM accounts WHERE balance >= 1000 and gets {11, 18}. T2 inserts account 23 with balance 1400 and commits. T1 reruns the predicate and gets {11, 18, 23}. Row 23 is the phantom because the repeated query finds a new qualifying row, not merely a changed value in an existing row. An uncommitted changed value would instead indicate a dirty read. Option 0 captures the changed result set caused by a later insertion.
The exam traps behind these ten answers
Do not memorise an anomaly label without rebuilding its operation order:
W1(A=120), R2(A=120)before T1 commits is a dirty read.R1(A=100), R2(A=100), W1(A=110), W2(A=125)is a lost update.R1(X=100), W2(X=120), R1(X=120)is an unrepeatable read.A repeated predicate changing from
{11, 18}to{11, 18, 23}is a phantom read.
For locks, remember that Strict 2PL prevents cascading rollback but can deadlock, a timestamp orders transactions but is not a lock, and SIX is compatible with IS at the same node among Q7's modes. Continue with DBMS Transaction MCQs for adjacent practice.
Short version and next practice step
Keep five rules ready: the same old value then two writes means lost update; uncommitted write then another transaction's read means dirty read; read, other write, then reread means unrepeatable read; a changed result set means phantom; strict locking can still deadlock.
Redo the questions without notes. Mark each anomaly WR, RW, or WW, draw the two-node Strict 2PL wait-for cycle, and revisit answers you cannot justify in two sentences.
Next, use GATE Guidance by Sanchit Sir for the DBMS sequence, GATE Test Series for timed practice, or compare options in the GATE CS Exam category.
