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 a binary relation from the set two elements are chosen then n^{2} possible pairs happens.
Then finally no. of binary relations possible with the n^{2} pairing are to be 2^{n (2) }.
Then finally no. of binary relations possible with the n^{2} pairing are to be 2^{n (2) }.
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:
If n zeros are present then n+1 gaps are possible. So we have to fill (n+1) number of k's.
So, total possible ways is ^{(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 us consider the regular expression (0+1) (0+1) (0+1) ... 3 times. Then the corresponding DFA will be like
In DFA, requires N+2 states.
In NFA, requires N+1 states (eliminate dead state).
Minimum number is N+1 (i.e., min(N+1, N+2))
In DFA, requires N+2 states.
In NFA, requires N+1 states (eliminate dead state).
Minimum number is N+1 (i.e., min(N+1, N+2))
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 accept by final state, there is an equivalent PDA which can also be accepted by an empty stack and viceversa.
So, L_{1} = L_{2}.
So, L_{1} = L_{2}.
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:
Thread and processes are handled by CPU virtual address space belongs to memory type.
File system used for disk management Interrupt is a type of 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:
We need to arrange 'n' number stored in an array, then we have to swap all the positive numbers with the negative numbers then we require i/2 swaps only.
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:
If there is common attribute in R and S, and every row of R match with every row of S then total no. of tuples will be mn.
For minimum:
If there is no common attribute between R and S or if there is common attribute but none of the row of R matches with rows of S then output tuples will be 0.
If there is common attribute in R and S, and every row of R match with every row of S then total no. of tuples will be mn.
For minimum:
If there is no common attribute between R and S or if there is common attribute but none of the row of R matches with rows of S then output 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 having the two tuples such as t⋅B = 20
→ (Tuple having A=10) ∩ (Tuple having B=20) is equal to (Tuples having A=10 and B=20)
→ (Tuple having A=10) ∩ (Tuple having B=20) is equal to (Tuples having A=10 and B=20)
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:
Total set in cache = Total blocks in cache/Total blocks in set = 2c/2 = c
Cache set no. = (Main memory block no.) mod (Total set in cache)
= (k) mod c
= k mod c
Cache set no. = (Main memory block no.) mod (Total set in cache)
= (k) mod c
= 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:
This method is usually converge, provided this initial guess is close enough to the unknown zero, and that f'(x_{0}) != 0.
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:
R = (a, b, c, d, e, f)
c → f,
e → a,
ec → d,
a → b
Option B: EC
c → f (c, f)
e → a (c, f, e, a)
ec → d (c, f, e, d)
a → b (c, f, e, d, a, b)
Answer: Option B
c → f,
e → a,
ec → d,
a → b
Option B: EC
c → f (c, f)
e → a (c, f, e, a)
ec → d (c, f, e, d)
a → b (c, f, e, d, a, b)
Answer: Option B
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:
As our relation R on set L is Reflexive, Antisymmetric and Transitive, it is poset.
Since LUB and GLB exists for any two elements, so it is lattice.
Since LUB and GLB exists for any two elements, so it is lattice.
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 may enter infinite loop and hence terminates program may not be able to answer if the program does 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:
R1 ∩ R2 = ∅. This makes the decomposition lossless join, as all the attributes are keys, R1 ∩ R2 will be a key of the decomposed relations (lossless condition says the common attribute must be a key in atleast one of the decomposed relation). Now, even the original relation R isin 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).
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:
Since the figure is not clear. I assume there is a NOT gate just before taking Y making the final AND gate as NAND gate.
We have steady, means Y is not changing and remains 1 throughout.
So,
So, option (B) is true.
We have steady, means Y is not changing and remains 1 throughout.
So,
So, option (B) is true.
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) XOR and NOT gates can only make XOR and XNOR which are not functionally complete.
(B) 2×1 multiplexer is functinally complete provided we have external 1 and 0 available.
(C) XOR can be used to make a NOT gate (a⊕1=a') and (AND, NOT) is functionally complete. Again this required external 1.
(D) We cannot derive NOT gate here. So not functionally complete.
Hence, options (B) and (C) are true provided external 1 and 0 are available.
(B) 2×1 multiplexer is functinally complete provided we have external 1 and 0 available.
(C) XOR can be used to make a NOT gate (a⊕1=a') and (AND, NOT) is functionally complete. Again this required external 1.
(D) We cannot derive NOT gate here. So not functionally complete.
Hence, options (B) and (C) are true provided external 1 and 0 are available.
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 is needed to provide memory protection so that a given process does not interfere with another otherwise we must fix the number of processors to some limit and divide the memory space among them, which is not an efficient mechanism.
We also need atleast 2 nodes of execution to ensure user processes share resources properly and OS maintains controls.
Demand paging and DMA enhances the performances not necessarily.
We also need atleast 2 nodes of execution to ensure user processes share resources properly and OS maintains controls.
Demand paging and DMA enhances the performances not necessarily.
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:
Virtual memory provides an interface through which processes access the physical memory. So,
(A) False. Because in virtual memory conceptfirst virtual to physical address translation is required then only we can access MM. So here access will be slow.
(B) True. Because without virtual memory it is difficult to give protected address space to processes as they will be accessing physical memory directly  No protection mechanism can be done inside the physical memory as processes are dynamic and number of processes changes from time to time.
(C) Position independent can be produced even without virtual memory support.
(D) This is one primary use of virtual memory. So true.
(A) False. Because in virtual memory conceptfirst virtual to physical address translation is required then only we can access MM. So here access will be slow.
(B) True. Because without virtual memory it is difficult to give protected address space to processes as they will be accessing physical memory directly  No protection mechanism can be done inside the physical memory as processes are dynamic and number of processes changes from time to time.
(C) Position independent can be produced even without virtual memory support.
(D) This is one primary use of virtual memory. So 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:
Whenever any page table entry is referred for the first time it is temporarily saved in TLB. Every element of this memory has a tag. And whenever anything is searched it is compared against TLB and we can get that entry/data with less memory access.
And invalidation of TLB means resetting TLB which is necessary because a TLB entry may belong to any page table of any other process thus resetting ensures that the entry corresponds to the process that we are searching for.
Note: Invalidation is necessary, otherwise the new process might end up using the translation of the old process. Saving and reuse of TLB is not necessary but can lead to better performance.
And invalidation of TLB means resetting TLB which is necessary because a TLB entry may belong to any page table of any other process thus resetting ensures that the entry corresponds to the process that we are searching for.
Note: Invalidation is necessary, otherwise the new process might end up using the translation of the old process. Saving and reuse of TLB is not necessary but can lead to better performance.
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 LSB addition we do not need a full adder.
For addition of subsequent bits we need full adders since carry from previous addition has to be fed into the addition operation.
For addition of subsequent bits we need full adders since carry from previous addition has to be fed into the addition operation.
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 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, whereas microprogramming is software approach. So hardwire control should be faster than both microprogramming approaches.
Between horizontal and vertical microprogramming, horizontal is faster because in this control signals are not encoded.
Whereas in vertical microprogramming to save memory signals are encoded. So, it take less time in horizontal microprogramming because decoding of signals is not required.
Hence, correct option is (B).
Between horizontal and vertical microprogramming, horizontal is faster because in this control signals are not encoded.
Whereas in vertical microprogramming to save memory signals are encoded. So, it take less time in horizontal microprogramming because decoding of signals is not required.
Hence, correct option is (B).
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:
Pointer access requires indirect addressing which can be simulated with indexed addressing or register indirect addressing but not with direct and immediate addressing.
An array and record access needs a pointer access. So, options (A), (B) and (C) cannot be implemented on such a processor.
Now to handle recursive procedure we need to use stack. A local variable inside the stack will be accessed as *(SP+Offset) which is nothing but a pointer access and requires indirect addressing. Usually this is done by moving the SP value to the base register and then using base register addressing to avoid unnecessary memory access for indirect addressing but not possible with just direct and immediate addressing.
Hence all options are correct.
An array and record access needs a pointer access. So, options (A), (B) and (C) cannot be implemented on such a processor.
Now to handle recursive procedure we need to use stack. A local variable inside the stack will be accessed as *(SP+Offset) which is nothing but a pointer access and requires indirect addressing. Usually this is done by moving the SP value to the base register and then using base register addressing to avoid unnecessary memory access for indirect addressing but not possible with just direct and immediate addressing.
Hence all options are correct.
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.