Gate1999
Question 1 
Suppose that the expectation of a random variable X is 5. Which of the following statements is true?
There is a sample point at which X has the value 5.  
There is a sample point at which X has value greater than 5.  
There is a sample point at which X has a value greater than or equal to 5.  
None of the above 
Question 1 Explanation:
Expectation of discrete random variable
E(X) = x_{1}P_{1} + x_{2}P_{2} + ... + x_{n}P_{n}
In question, E(X) is given as 5.
E(X) = 5, 0≤P_{i}≤1
P_{1} + P_{2} + ... + P_{n} = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the x_{i} value is greater than 5.
E(X) = x_{1}P_{1} + x_{2}P_{2} + ... + x_{n}P_{n}
In question, E(X) is given as 5.
E(X) = 5, 0≤P_{i}≤1
P_{1} + P_{2} + ... + P_{n} = 1 [Probability]
Therefore, E(X) = 5 is possible only if atleast one of the x_{i} value is greater than 5.
Question 2 
The number of binary relations on a set with n elements is:
n^{2}  
2^{n}  
2^{n (2) }  
None of the above 
Question 2 Explanation:
In binary relation two elements are selected from a set. So total no. of pairs possible are
n×n = n^{2}
Now, no. of binary relations possible is
2^{n}^{2}
Because each pair have two possibilities either chosen or not chosen.
n×n = n^{2}
Now, no. of binary relations possible is
2^{n}^{2}
Because each pair have two possibilities either chosen or not chosen.
Question 3 
The number of binary strings of n zeroes and k ones that no two ones are adjacent is
^{n1}C_{k}  
^{n}C_{k}  
^{n}C_{k+1}  
None of the above 
Question 3 Explanation:
Since there are n zeroes, so
XOXOXOXOXOXOX
n+1 gaps can be possible, where 1's can be placed so that no two one's are adjacent. So, no. of ways in which k 1's can be placed in n+1 gaps are,
^{n+1}C_{k}
XOXOXOXOXOXOX
n+1 gaps can be possible, where 1's can be placed so that no two one's are adjacent. So, no. of ways in which k 1's can be placed in n+1 gaps are,
^{n+1}C_{k}
Question 4 
Consider the regular expression (0 + 1) (0 + 1)…. N times. The minimum state finite automation that recognizes the language represented by this regular expression contains
n states  
n + 1 states  
n + 2 states  
None of the above 
Question 4 Explanation:
Let's draw both NFA and DFA and see which one requires less no. of state.
DFA:
So, DFA requires (n+2) state.
NFA:
So, NFA requires (n+1) state.
So, final answer will be,
min(n+1, n+2)
= n+1
DFA:
So, DFA requires (n+2) state.
NFA:
So, NFA requires (n+1) state.
So, final answer will be,
min(n+1, n+2)
= n+1
Question 5 
Contextfree languages are closed under:
Union, intersection  
Union, Kleene closure  
Intersection, complement  
Complement, Kleene closure

Question 5 Explanation:
Context free languages are not closed under Intersection and complementation.
By checking the options only option B is correct.
By checking the options only option B is correct.
Question 6 
Let L_{D} be the set of all languages accepted by a PDA by final state and L_{E} the set of all languages accepted by empty stack. Which of the following is true?
L_{D} = L_{E}  
L_{D} ⊃ L_{E}  
L_{E} = L_{D}  
None of the above 
Question 6 Explanation:
For any PDA which can be accepted by final state, there is an equivalent PDA which can also be accepted by an empty stack and for any PDA which can be accepted by an empty stack, there is an equivalent PDA which can be accepted by final state.
Question 9 
(A) – 2 (B) – 4 (C) – 3 (D)  1  
(A) – 1 (B) – 2 (C) – 3 (D) – 4  
(A) – 3 (B) – 2 (C) – 4 (D)  1  
(A) – 4 (B) – 1 (C) – 2 (D) – 3 
Question 9 Explanation:
⇒ Threads are handled by CPU.
⇒ Virtual address is a memory type.
⇒ File system is used to manage the disk.
⇒ Interrupt is a signal.
Question 10 
Which of the following disk scheduling strategies is likely to give the best through put?
Farthest cylinder next  
Nearest cylinder next  
First come first served  
Elevator algorithm 
Question 10 Explanation:
Farthest cylinder next  Longest job first
Nearest cylinder next  SSTF
First come first served  FCFS
Elevator  SCAN
SSTF always serves the request of nearest cylinder first. Due to which the necessary movement gets reduced.
Nearest cylinder next  SSTF
First come first served  FCFS
Elevator  SCAN
SSTF always serves the request of nearest cylinder first. Due to which the necessary movement gets reduced.
Question 11 
System calls are usually invoked by using
a software interrupt  
polling  
an indirect jump  
a privileged instruction 
Question 11 Explanation:
Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.
Question 12 
A sorting technique is called stable if
it takes O (nlog n) time  
it maintains the relative order of occurrence of nondistinct elements  
it uses divide and conquer paradigm  
it takes O(n) space 
Question 12 Explanation:
Sorting techniques are said to be stable if it maintains the relative order of occurence of nondistinct element.
Question 13 
Suppose we want to arrange the n numbers stored in any array such that all negative values occur before all positive ones. Minimum number of exchanges required in the worst case is
n  1  
n  
n + 1  
None of the above 
Question 13 Explanation:
Minimum no. of exchanges required in worst case will be when 1^{st} half will contain all +ve nos and 2^{nd} half will conain all ve nos.
Now we will swap 1^{st} no. with n^{th} no. and then 2^{nd} no. with (n1)^{th} no. and then 3^{rd} no. with (n2)^{th} and so on. Like this we will have to do n/2 swaps in worst case.
Now we will swap 1^{st} no. with n^{th} no. and then 2^{nd} no. with (n1)^{th} no. and then 3^{rd} no. with (n2)^{th} and so on. Like this we will have to do n/2 swaps in worst case.
Question 14 
8, 9, 15, 20, 47, 4, 12, 17, 30, 40  
8, 15, 20, 47, 4, 9, 30, 40, 12, 17  
15, 20, 47, 4, 8, 9, 12, 30, 40, 17  
4, 8, 9, 15, 20, 47, 12, 17, 30, 40 
Question 14 Explanation:
Question 15 
0  
1  
2  
3 
Question 15 Explanation:
Here, vertex 2, 3, 5 are the articulation points. By removing these vertices then the graph will be disconnected.
Total no. of articulation points = 3
Total no. of articulation points = 3
Question 16 
If n is a power of 2, then the minimum number of multiplications needed to compute a* is
log_{2} n  
√n  
n1  
n 
Question 16 Explanation:
We require 4 multiplications to calculate a^{16} .....(I)
→ Like that 3 multiplications requires to calculate a^{8} .....(II)
I, II are satisfied with the option A.
Question 17 
Which of the following is the most powerful parsing method?
LL (1)  
Canonical LR  
SLR  
LALR 
Question 17 Explanation:
Canonical LR is most powerful.
LR > LALR > SLR
LR > LALR > SLR
Question 18 
Consider the join of a relation R with a relation S. If R has m tuples and S has n tuples then the maximum and minimum sizes of the join respectively are
m + n and 0  
mn and 0  
m + n and m – n  
mn and m + n 
Question 18 Explanation:
For maximum:
Suppose there is no common attribute in R and S due to which natural join will act as cross product. So then in cross product total no. of tuples will be mn.
For minimum:
Suppose there is common attribute in R and S, but none of the row of R matches with rows of S then minimum no. of tuples will be 0.
Suppose there is no common attribute in R and S due to which natural join will act as cross product. So then in cross product total no. of tuples will be mn.
For minimum:
Suppose there is common attribute in R and S, but none of the row of R matches with rows of S then minimum no. of tuples will be 0.
Question 19 
σ_{(A=10∨B=20)} (r)  
σ_{(A=10)} (r) ∪ σ_{(B=20)} (r)  
σ_{(A=10)} (r) ∩ σ_{(B=20)} (r)  
σ_{(A=10)} (r)  σ_{(B=20)} (r) 
Question 19 Explanation:
The given relational algebra expression represents tuples having A=10 and B=20 which is equivalent to
σ_{(A=10)} (r) ∩ σ_{(B=20)} (r)
σ_{(A=10)} (r) ∩ σ_{(B=20)} (r)
Question 20 
Booth’s coding in 8 bits for the decimal number –57 is
0 – 100 + 1000  
0 – 100 + 100 1  
0 – 1 + 100 – 10 + 1  
00 – 10 + 100  1 
Question 20 Explanation:
OptionB:
Question 21 
The maximum gate delay for any output to appear in an array multiplier for multiplying two n bit number is
On^{2}  
O(n)  
O(log n)  
O(1) 
Question 21 Explanation:
Total no. of gates being used for 'n' bit multiplication in an array multiplier (n*n) = (2n1)
Total delay = 1 * 2n  1 = O(2n  1) = n
Total delay = 1 * 2n  1 = O(2n  1) = n
Question 22 
The main memory of a computer has 2 cm blocks while the cache has 2 c blocks. If the cache uses the set associative mapping scheme with 2 blocks per set, then block k of the main memory maps to the set
(k mod m) of the cache  
(k mod c) of the cache  
(k mod 2c) of the cache  
(k mod 2 cm) of the cache

Question 22 Explanation:
No. of sets in cache = No. of blocks in cache/No. of blocks per set
= 2c/c
= c
∴ Cache set no. to which block k of main memory maps to
= (Main memory block no.) mod (Total set in cache)
= k mod c
= 2c/c
= c
∴ Cache set no. to which block k of main memory maps to
= (Main memory block no.) mod (Total set in cache)
= k mod c
Question 23 
The NewtonRaphson method is to be used to find the root of the equation f(x)=0 where x_{o} is the initial approximation and f' is the derivative of f. The method converges
always  
only if f is a polynomial  
only if f(x_{0}) < 0  
None of the above 
Question 23 Explanation:
Note: Out of syllabus.
Question 24 
Let R = (a, b, c, d, e, f) be a relation scheme with the following dependencies c → f, e → a, ec → d, a → b. Which of the following is a key for R?
CD  
EC  
AE  
AC 
Question 24 Explanation:
Let's check closure for each option,
A) (CD)^{+} = cdf
Not a key.
B) (EC)^{+} = ecdabf
Yes, it is a key.
C) (AE)^{+} = aeb
Not a key. D) (AC)^{+} = abcf
Not a key.
A) (CD)^{+} = cdf
Not a key.
B) (EC)^{+} = ecdabf
Yes, it is a key.
C) (AE)^{+} = aeb
Not a key. D) (AC)^{+} = abcf
Not a key.
Question 25 
Which of the following is correct?
Btrees are for storing data on disk and B^{+} trees are for main memory.  
Range queries are faster on B* trees.  
Btrees are for primary indexes and B* trees are for secondary indexes.  
The height of a B* tree is independent of the number of records. 
Question 25 Explanation:
Range queries are faster on B^{+} trees.
Question 26 
Events E_{1} and E_{2} are independent  
Events E_{1} and E_{2} are not independent  
Question 26 Explanation:
Question 27 
Two girls have picked 10 roses, 15 sunflowers and 15 daffodils. What is the number of ways they can divide the flowers amongst themselves?
1638  
2100  
2640  
None of the above 
Question 27 Explanation:
Formula for distributing n identical objects into r persons is,
^{n+r1}C_{r1}
So for 10 roses,
^{10+21}C_{21} = ^{11}C_{1} = 11
For 15 sunflowers,
^{15+21}C_{21} = ^{16}C_{1} = 16
For 15 daffodils,
^{15+21}C_{21} = ^{16}C_{1} = 16
∴ The final answer is,
11×16×16 = 2816
^{n+r1}C_{r1}
So for 10 roses,
^{10+21}C_{21} = ^{11}C_{1} = 11
For 15 sunflowers,
^{15+21}C_{21} = ^{16}C_{1} = 16
For 15 daffodils,
^{15+21}C_{21} = ^{16}C_{1} = 16
∴ The final answer is,
11×16×16 = 2816
Question 28 
Let L be a set with a relation R which is transitive, antisymmetric and reflexive and for any two elements a, b ∈ L let the least upper bound lub (a,b) and the greatest lower bound glb (a,b) exist. Which of the following is/are true?
L is a poset  
L is a Boolean algebra  
L1 is context free  
L2 is regular
 
Both A and C 
Question 28 Explanation:
Note: Options given are wrong.
Question 29 
If L is context free language and L2 is a regular language which of the following is/are false?
L1 – L2 is not context free  
L1 ∩ L2 is context free  
~L1 is context free  
~L2 is regular  
Both A and C 
Question 29 Explanation:
(A) L2 is regular language and regular language is closed under complementation. Hence ~L2 is also regular.
So L1  L2 = L1 ∩ (~L2)
And CFL is closed under regular intersection.
So, L1 ∩ (~L2) or L1  L2 is CFL.
So False.
(B) As we said that CFL is closed under regular intersection.
So True.
(C) CFL is not closed under complementation.
Hence False.
(D) Regular language is closed under complementation.
Hence True.
So L1  L2 = L1 ∩ (~L2)
And CFL is closed under regular intersection.
So, L1 ∩ (~L2) or L1  L2 is CFL.
So False.
(B) As we said that CFL is closed under regular intersection.
So True.
(C) CFL is not closed under complementation.
Hence False.
(D) Regular language is closed under complementation.
Hence True.
Question 30 
Given the programming constructs (i) assignment (ii) for loops where the loop parameter cannot be changed within the loop (iii) ifthenelse (iv) forward go to (v) arbitrary go to (vi) nonrecursive procedure call (vii) recursive procedure/function call (viii) repeat loop, which constructs will you not include in a programming language such that it should be possible to program the terminates (i.e., halting) function in the same programming language.
(ii), (iii), (iv)  
(v), (vii), (viii)  
(vi), (vii), (viii)  
(iii), (vii), (viii) 
Question 30 Explanation:
Arbitrary goto, recursive call and repeat loop may enter infinite loop and the program might never terminate.
Question 31 
This schedule is serialized and can occur in a scheme using 2PL protocol  
This schedule is serializable but cannot occur in a scheme using 2PL protocol  
This schedule is not serialiable but can occur in a scheme using 2PL protocol  
This schedule is not seralisable and cannot occur in a scheme using 2PL protocol.

Question 31 Explanation:
Let's draw precedence graph,
Since cycle exist so not conflict serializable.
And we know that if the schedule is not serializable then it is not 2PL.
Hence correct option is (D).
Since cycle exist so not conflict serializable.
And we know that if the schedule is not serializable then it is not 2PL.
Hence correct option is (D).
Question 32 
Consider the schema R = (S T U V) and the dependencies S → T, T → U, U → V and V → S. Let R = (R1 and R2) be a decomposition such that R1 ∩ R2 = ∅ . The decomposition is
not in 2NF  
in 2NF but not 3NF  
in 3NF but not in 2NF  
in both 2NF and 3NF 
Question 32 Explanation:
Since R1 ∩ R2 = ∅, so the decomposition is lossless join. Now since all the attributes are keys, so R1 ∩ R2 will be a key of the decomposed relation.
And since every attribute is key so the decomposed relation will be in BCNF and hence in 3NF.
And since every attribute is key so the decomposed relation will be in BCNF and hence in 3NF.
Question 33 
A = 0, B = 0, C = 1  
A = 0, B = 1, C = 1  
A = 1, B = 0, C = 1  
A = 1, B = 1, C = 1 
Question 33 Explanation:
So the above equation is satisfied if either C=0 or A=0 and B=1.
Hence, Option (B) is correct.
Question 34 
Which of the following sets of component(s) is/are sufficient to implement any arbitrary Boolean function?
XOR gates, NOT gates  
2 to 1 multiplexors  
AND gates, XOR gates  
Threeinput gates that output (A⋅B) + C for the inputs A⋅B and C  
Both B and C 
Question 34 Explanation:
(A) Not complete because, XOR can be used to make only NOT gate and NOT gate is already available. Hence not complete.
(B) 2 to 1 multiplexors is functionally complete.
(C) XOR gate can be used to make a NOT gate. So, (AND, NOT) is functionally complete.
(D) With given gates and inputs NOT gate cannot be derived.
Hence, not complete.
(B) 2 to 1 multiplexors is functionally complete.
(C) XOR gate can be used to make a NOT gate. So, (AND, NOT) is functionally complete.
(D) With given gates and inputs NOT gate cannot be derived.
Hence, not complete.
Question 35 
A multiuser, multiprocessing operating system cannot be implemented on hardware that does not support
Address translation  
DMA for disk transfer  
At least two modes of CPU execution (privileged and nonprivileged)  
Demand paging  
Both A and C 
Question 35 Explanation:
Address translation and atleast two modes of CPU execution (Privileged and nonprivileged) are needed to implement multiuser and multiprocessing operating system, because address translation provides memory protection which ensures that a given process does not interfere with another, and we need privileged and nonprivileged instruction, so that user and OS interconnects properly.
Question 36 
Which of the following is/are advantage of virtual memory?
Faster access to memory on an average.  
Processes can be given protected address spaces.  
Linker can assign addresses independent of where the program will be loaded in physical memory.  
Programs larger than the physical memory size can be run.
 
Both B and D 
Question 36 Explanation:
A) False. Because in virtual memory concept address translation is required due to which access is slow.
B) True. Because in virtual memory concept of address translation provides protected address space so that one process do not interfere the other process.
C) False.
D) True.
B) True. Because in virtual memory concept of address translation provides protected address space so that one process do not interfere the other process.
C) False.
D) True.
Question 37 
Which of the following actions is/are typically not performed by the operating system when switching context from process A to process B?
Saving current register values and restoring saved register values for process B.
 
Changing address translation tables.  
Swapping out the memory image of process A to the disk.  
Invalidating the translation lookaside buffer. 
Question 37 Explanation:
A) True.
B) True.
C) False, because swapping is done when the process is suspended and not during context switching.
D) True, Invalidation of TLB is necessary because, if the TLB is not invalidated then the new process might end up using the translation of old process. Note that Invalidation of TLB is necessary but saving and reuse of TLB is not necessary.
B) True.
C) False, because swapping is done when the process is suspended and not during context switching.
D) True, Invalidation of TLB is necessary because, if the TLB is not invalidated then the new process might end up using the translation of old process. Note that Invalidation of TLB is necessary but saving and reuse of TLB is not necessary.
Question 38 
0.125 0.125  
0.25 0.25  
0.25 0.125  
0.125 0.25

Question 38 Explanation:
Note: Out of syllabus.
Question 39 
The number of tokens in the Fortran statement DO 10 I = 1.25 is
3  
4  
5  
None of the above 
Question 39 Explanation:
DO → 1
10 → 2
I → 3
= → 4
1.25 → 5
10 → 2
I → 3
= → 4
1.25 → 5
Question 40 
A grammar that is both left and right recursive for a nonterminal, is
Ambiguous  
Unambiguous  
Information is not sufficient to decide whether it is ambiguous or unambiguous  
None of the above 
Question 40 Explanation:
If a grammar is both left and right recursion, then grammar may or may not be ambiguous.
Question 41 
The number of full and halfadders required to add 16bit numbers is
8 halfadders, 8 fulladders  
1 halfadder, 15 fulladders  
16 halfadders, 0 fulladders  
4 halfadders, 12 fulladders 
Question 41 Explanation:
For Least Significant Bit we do not need a full adder since initially carry is not present.
But for rest of bits we need full address since carry from previous addition has to be included into the addition operation.
So, in total 1 half adder and 15 full adders are required.
But for rest of bits we need full address since carry from previous addition has to be included into the addition operation.
So, in total 1 half adder and 15 full adders are required.
Question 42 
Zero has two representations in
Sign magnitude  
1’s complement  
2’s complement  
None of the above  
Both A and B 
Question 42 Explanation:
Sign magnitude:
+0 = 0000
0 = 1000
1's complement:
+0 = 0000
0 = 1111
+0 = 0000
0 = 1000
1's complement:
+0 = 0000
0 = 1111
Question 43 
Raid configurations of the disks are used to provide
Faulttolerance  
High speed  
High data density  
None of the above 
Question 43 Explanation:
Raid is a way of storing the same data in different places on multiple hard disks to protect data in the case of drive failure.
So, we can say that RAID is used to provide fault tolerence.
So, we can say that RAID is used to provide fault tolerence.
Question 44 
Arrange the following configuration for CPU in decreasing order of operating speeds: Hard wired control, vertical microprogramming, horizontal microprogramming.
Hard wired control, vertical microprogramming, horizontal micro programming.  
Hard wired control, horizontal microprogramming, vertical micro programming.  
Horizontal microprogramming, vertical microprogramming, Hard wired control.  
Vertical microprogramming, horizontal microprogramming, hard wired control. 
Question 44 Explanation:
Hardwired control involves only hardware which is definitely faster than microprogramming approach.
Now between horizontal and vertical microprogramming, the signals in vertical microprogramming are in encoded form which has to be decoded first using decoder. So, horizontal microprogramming is faster than vertical microprogramming.
Now between horizontal and vertical microprogramming, the signals in vertical microprogramming are in encoded form which has to be decoded first using decoder. So, horizontal microprogramming is faster than vertical microprogramming.
Question 45 
The minimum number of record movements required to merge five files A (with 10 records), B (with 20 records), C (with 15 records), D (with 5 records) and E (with 25 records) is:
165  
90  
75  
65 
Question 45 Explanation:
Always merge two files with minimum no. of records.
10, 20, 15, 5, 25
Merge 5 & 10:
5+10=15 movements
Now the list is
15, 20, 15, 25
Merge 15 & 15:
15+15=30 movements
Now the list is
30, 20, 25
Merge 20 & 25:
20+25=45 movements
Now the list is
30, 45
Merge 30 & 45:
30+45=75 movements
∴ Total no. of movements
= 15+30+45+75
= 165
10, 20, 15, 5, 25
Merge 5 & 10:
5+10=15 movements
Now the list is
15, 20, 15, 25
Merge 15 & 15:
15+15=30 movements
Now the list is
30, 20, 25
Merge 20 & 25:
20+25=45 movements
Now the list is
30, 45
Merge 30 & 45:
30+45=75 movements
∴ Total no. of movements
= 15+30+45+75
= 165
Question 46 
M – W N – V O – U P  X  
M – W N – U O – X P  V  
M – V N – W O – X P  U  
None of the above 
Question 46 Explanation:
(M) T(n) = Sum of first n matural nos. = n(n+1)/2 = O(n^{2})
(N) Apply Master's theorem
T(n) = θ(n) = O(n)
(O) Apply Master's theorem
T(n) = θ(n logn) = O(n logn)
(P) Here we are adding the log of firstn natural numbers.
So,
T_{n} = log1 + log2 + log3 + ... + logn
= log (1×2×...n)
= log(n!)
= θ(n logn)
(N) Apply Master's theorem
T(n) = θ(n) = O(n)
(O) Apply Master's theorem
T(n) = θ(n logn) = O(n logn)
(P) Here we are adding the log of firstn natural numbers.
So,
T_{n} = log1 + log2 + log3 + ... + logn
= log (1×2×...n)
= log(n!)
= θ(n logn)
Question 47 
The main differences(s) between a CSIC and A RISC processor is/are that a RISC processor typically
has fewer instructions  
has fewer addressing modes  
has more registers  
is easier to implement using hardwired control logic
 
All the above 
Question 47 Explanation:
All are properties of RISC processor.
Question 48 
A certain processor supports only the immediate and the direct addressing modes. Which of the following programming language features cannot be implemented on this processor?
Pointers  
Arrays  
Records  
Recursive procedures with local variable  
All the above 
Question 48 Explanation:
A) Cannot be implemented because pointers need indirect addressing mode.
B) Cannot be implemented because arrays need Register indexing.
C) Records also needs pointers which needs indirect addressing modes, so this also cannot be implemented.
D) Recursive procedures needs stack, and so it needs stack pointers which inturn needs indirect addressing. So this also cannot be implemented.
B) Cannot be implemented because arrays need Register indexing.
C) Records also needs pointers which needs indirect addressing modes, so this also cannot be implemented.
D) Recursive procedures needs stack, and so it needs stack pointers which inturn needs indirect addressing. So this also cannot be implemented.
Question 49 
Finds the maximum of a, b, and c  
Finds the minimum of a, b and c  
Finds the middle number of a, b, c  
None of the above

Question 49 Explanation:
Try for (3,2,2), it will go for infinite loop.
Question 50 
Which of the following is/are correct?
An SQL query automatically eliminates duplicates  
An SQL query will not work if there are no indexes on the relations  
SQL permits attribute names to be repeated in the same relation  
None of the above 
Question 50 Explanation:
→ SQL won't remove duplicates like relational algebra projection, we have to remove it explicitly by distinct.
→ If there are no indexes on the relation SQL, then also it works.
→ SQL does not permit 2 attributes to have same name in a relation.
→ If there are no indexes on the relation SQL, then also it works.
→ SQL does not permit 2 attributes to have same name in a relation.
There are 50 questions to complete.