ISRO2007
Question 1 
The Boolean expression Y = (A + B' + A'B)C' is given by
AC'  
BC'  
C'  
A 
Question 1 Explanation:
Y = (A + B' + A'B)C'
Y = AC' + B'C' + A'BC'
Y = (A + A'B)C' + B'C'
Y = (A + B)C' + B'C'
Y = AC' + BC' + B'C'
Y = AC' + C'(B + B') → B + B' = 1
Y = AC' + C'
Y = C'
Y = AC' + B'C' + A'BC'
Y = (A + A'B)C' + B'C'
Y = (A + B)C' + B'C'
Y = AC' + BC' + B'C'
Y = AC' + C'(B + B') → B + B' = 1
Y = AC' + C'
Y = C'
Question 2 
The circuit shown in the following figure realizes the function
(( A + B )’ +C ) ( D’E’ ))  
(( A + B )’ + C ) ( DE’ ))  
( A + ( B + C )’ ) ( D’E )  
( A + B + C’ ) ( D’E’ ) 
Question 2 Explanation:
The given function is equivalent to the following expression:
Y = (((A + B)' + C)' + ((D + E)')')'
Y = ((A + B)' + C)')' . (D + E)'
Y = ((A + B)' + C) (D'E')
Y = (((A + B)' + C)' + ((D + E)')')'
Y = ((A + B)' + C)')' . (D + E)'
Y = ((A + B)' + C) (D'E')
Question 3 
The circuit shown in the given figure is a
full adder  
full subtractor  
shift register  
decade counter 
Question 3 Explanation:
The above diagram is full subtractor. The equation is D=X⊕Y⊕Bin and B_{out=X'Bin+X'Y+YBin}
Question 4 
When two numbers are added in excess3 code and the sum is less than 9, then in order to get the correct answer it is necessary to
subtract 0011 from the sum  
add 0011 to the sum
 
subtract 0110 from the sum
 
add 0110 to the sum

Question 4 Explanation:
Subtract 0011 if there is no carry otherwise add 0011.
Example:
x+3
y+3

(x+y+6)
Here, sum is excess6. Hence, subtract 0011 to make it excess3.
Example:
x+3
y+3

(x+y+6)
Here, sum is excess6. Hence, subtract 0011 to make it excess3.
Question 5 
The characteristic equation of an SR flipflop is given by
Q_{n+1} = S + RQ_{n}  
Q_{n+1}= RQ^{’}_{n} + SQ_{n}  
Q_{n+1}= S^{’} + RQ^{n}  
Q_{n+1} = S + R^{’}Q_{n} 
Question 5 Explanation:
The characteristic table of an SR flipflop is:
So, by simplifying using kmaps:
characteristic equation of an SR flipflop = Q_{n+1} = S + R^{’}Q_{n}
So, by simplifying using kmaps:
characteristic equation of an SR flipflop = Q_{n+1} = S + R^{’}Q_{n}
Question 6 
A graph with n vertices and n1 edges that is not a tree, is
Connected  
Disconnected  
Euler  
A circuit 
Question 6 Explanation:
Consider a graph with two nodes(n1&n2) and number of edges are 1, There may be chance self edge with node n1 then graph is disconnected.
Consider the graph with three nodes(n1,n2&n3) and has 2 edges.
n1>n2 and n2>n1 then the graph is disconnected.
Consider the graph with three nodes(n1,n2&n3) and has 2 edges.
n1>n2 and n2>n1 then the graph is disconnected.
Question 7 
If a graph requires k different colours for its proper colouring, then the chromatic number of the graph is
1  
k  
k1  
k/2 
Question 7 Explanation:
The chromatic number of a graph is the smallest number of colors needed to color the vertices of so that no two adjacent vertices share the same color and if a graph requires k different colours for its proper colouring, then k is the chromatic number of the graph.
Question 8 
A read bit can be read
and written by CPU  
and written by peripheral  
by peripheral and written by CPU  
by CPU and written by the peripheral 
Question 8 Explanation:
The read and write functionality depends on the type of microcontroller peripheral. Generally, the status bits have a read only status and can be modified by peripherals only.
So, read bit can be read by CPU and written by the peripheral.
Question 9 
The term ‘aging’ refers to
booting up the priority of the process in multilevel of queue without feedback  
keeping track of the following a page has been in memory for the purpose of LRU replacement  
letting job reside in memory for a certain amount of time so that the number of pages required can be estimated accurately  
gradually increasing the priority of jobs that wait in the system for a long time to remedy infinite blocking  
In Operating systems, aging is a scheduling technique used to avoid starvation. In this, the priority of the jobs that have a longer waiting time is increased as compared to the newer processes, to avoid the starvation of older processes

Question 10 
Consider a set of n tasks with known runtimes r_{1}, r_{2}….r_{n} to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput?
Round Robin  
Shortest job first  
Highest response ratio next  
first come first served 
Question 10 Explanation:
Throughput means total number of tasks executed per unit time i.e. sum of waiting time and burst time.
Shortest job first scheduling is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
Thus, in shortest job first scheduling, shortest jobs are executed first. This means CPU utilization is maximum. So, maximum number of tasks are completed.
Shortest job first scheduling is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
Thus, in shortest job first scheduling, shortest jobs are executed first. This means CPU utilization is maximum. So, maximum number of tasks are completed.
Question 11 
Consider a job scheduling problem with 4 jobs J_{1}, J_{2}, J_{3}, J_{4} and with corresponding deadlines: ( d_{1}, d_{2}, d_{3}, d_{4}) = (4, 2, 4, 2).
Which of the following is not a feasible schedule without violating any job schedule?
Which of the following is not a feasible schedule without violating any job schedule?
J_{2}, J_{4}, J_{1}, J_{3}  
J_{4}, J_{1}, J_{2}, J_{3}.  
J_{4}, J_{2}, J_{1}, J_{3}.  
J_{4}, J_{2}, J_{3}, J_{1} 
Question 11 Explanation:
→ Feasible schedule is completing all the jobs within deadline.
→ From the dead line, we can deduce that Job J2 & J4 will complete by time “2” whereas remaining two requires time “4”.
→ So the order of completion of Jobs are Either J2 or J4 and followed by either J1 or J3.
From the given options , Option A,C & D gives the solution because after completion of Jobs J2 and J4 then only jobs J1 and J3 is going to complete.
→ But option B , order of completing jobs is J4,J1,J2 ,J3 which is not possible and it is not feasible schedule
→ From the dead line, we can deduce that Job J2 & J4 will complete by time “2” whereas remaining two requires time “4”.
→ So the order of completion of Jobs are Either J2 or J4 and followed by either J1 or J3.
From the given options , Option A,C & D gives the solution because after completion of Jobs J2 and J4 then only jobs J1 and J3 is going to complete.
→ But option B , order of completing jobs is J4,J1,J2 ,J3 which is not possible and it is not feasible schedule
Question 12 
By using an eightbit optical encoder the degree of resolution that can be obtained is (approximately)
1.8^{o}  
3.4^{o}  
2.8^{o}  
1.4^{o} 
Question 12 Explanation:
An optical encoder is an electromechanical device which has an electrical output in digital form proportional to the angular position of the input shaft.
Optical encoders enable an angular displacement to be converted directly into a digital form.
Encoder resolution is often referred to in bits, which are binary units: a 16 bit resolution rotary encoder will have 65,536 (216) increments per turn, or PPR.
In the given question, 8bit optical encoder will have 2^{8} increments Resolution = 360/2^{n} = 360/2^{8} = 1.4^{o}
Optical encoders enable an angular displacement to be converted directly into a digital form.
Encoder resolution is often referred to in bits, which are binary units: a 16 bit resolution rotary encoder will have 65,536 (216) increments per turn, or PPR.
In the given question, 8bit optical encoder will have 2^{8} increments Resolution = 360/2^{n} = 360/2^{8} = 1.4^{o}
Question 13 
The principal of the locality of reference justifies the use of
virtual memory  
interrupts  
main memory
 
cache memory 
Question 13 Explanation:
Spatial Locality of reference – this says that there is chance that element will be present in the close proximity to the reference point and next time if again searched then more close proximity to the point of reference.
Temporal Locality of reference – In this Least recently used algorithm will be used. Whenever there is page fault occurs within word will not only load word in main memory but complete page fault will be loaded because spatial locality of reference rule says that if you are referring any word next word will be referred in
its register that’s why we load complete page table so complete block will be loaded.
Principle of locality of reference justifies the use of cache.
Temporal Locality of reference – In this Least recently used algorithm will be used. Whenever there is page fault occurs within word will not only load word in main memory but complete page fault will be loaded because spatial locality of reference rule says that if you are referring any word next word will be referred in
its register that’s why we load complete page table so complete block will be loaded.
Principle of locality of reference justifies the use of cache.
Question 14 
Consider the following pseudocode
x:=1;
i:=1;
while (x <= 1000)
begin
x:=2^x;
i:=i+1;
end;
What is the value of i at the end of the pseudocode?
x:=1;
i:=1;
while (x <= 1000)
begin
x:=2^x;
i:=i+1;
end;
What is the value of i at the end of the pseudocode?
4  
5  
6  
7 
Question 14 Explanation:
Initialisation: x = 1, i = 1;
Loop: x i
21 2
22 3
24 4
216 5
After this condition becomes false.
Loop: x i
21 2
22 3
24 4
216 5
After this condition becomes false.
Question 15 
The five items: A, B, C, D, and E are pushed in a stack, one after other starting from A. The stack is popped four items and each element is inserted in a queue. The two elements are deleted from the queue and pushed back on the stack. Now one item is popped from the stack. The popped item is
A  
B  
C  
D 
Question 15 Explanation:
When five items: A, B, C, D, and E are pushed in a stack: Order of stack becomes: A, B, C, D, and E (A at the bottom and E at the top.) stack is popped four items and each element is inserted in a queue: Order of queue: B, C, D, E (B at rear and E at the front) Order of stack after pop operations = A. Two elements deleted from the queue and pushed back stack: New order of stack = A, E, D(A at the bottom, D at the top) As D is on the top so when pop operation occurs D will be popped out. So, correct option is (D).
Question 16 
Round Robin schedule is essentially the preemptive version of
FIFO  
Shortest job first  
Shortest remaining time  
Longest remaining time 
Question 16 Explanation:
FIFO is when implemented in preemptive version, it acts like roundrobin algorithm.
Question 17 
The number of digit 1 present in the binary representation of 3 × 512 + 7 × 64 + 5 × 8 + 3
8  
9  
10  
12 
Question 17 Explanation:
3 × 512 + 7 × 64 + 5 × 8 + 3
= (2 + 1)× 512 + (4 + 2 + 1)× 64 + (4 + 1)× 8 + 2 + 1
= 1024 + 512 + 64 x 4 + 64 x 2 + 64 + 32 + 8 + 2 + 1
= 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 + 1
As 1024 has ten 0’s followed by 1, 512 has nine 0’s followed by 1 and so on..
So, the expression will contain total nine 1’s and will be be represented as 11111101011.
= (2 + 1)× 512 + (4 + 2 + 1)× 64 + (4 + 1)× 8 + 2 + 1
= 1024 + 512 + 64 x 4 + 64 x 2 + 64 + 32 + 8 + 2 + 1
= 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 + 1
As 1024 has ten 0’s followed by 1, 512 has nine 0’s followed by 1 and so on..
So, the expression will contain total nine 1’s and will be be represented as 11111101011.
Question 18 
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an synchronous serial line at 2400 baud rate, and with two stop bits is
109  
216  
300  
219 
Question 18 Explanation:
Synchronous communication requires that the clocks in the transmitting and receiving devices are synchronized – running at the same rate – so the receiver can sample the signal at the same time intervals used by the transmitter. No start or stop bits are required. For this reason “synchronous communication permits more information to be passed over a circuit per unit time.
2400 baud means that the serial port is capable of transferring a maximum of 2400 bits per second.
Number of 8bit characters that can be transmitted per second = 2400/8 = 300
2400 baud means that the serial port is capable of transferring a maximum of 2400 bits per second.
Number of 8bit characters that can be transmitted per second = 2400/8 = 300
Question 19 
If the bandwidth of a signal is 5 kHz and the lowest frequency is 52 kHz, what is the highest frequency
5 kHz  
10 kHz  
47 kHz  
57 kHz 
Question 19 Explanation:
Bandwidth = Highest frequency  Lowest frequency.
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.
Question 20 
An Ethernet hub
functions as a repeater  
connects to a digital PBX  
connects to a tokenring network  