ISRO-2007

Question 1
The Boolean expression Y = (A + B' + A'B)C' is given by
A
AC'
B
BC'
C
C'
D
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'
Question 2
The circuit shown in the following figure realizes the function
A
(( A + B )’ +C ) ( D’E’ ))
B
(( A + B )’ + C ) ( DE’ ))
C
( A + ( B + C )’ ) ( D’E )
D
( 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')
Question 3
The circuit shown in the given figure is a
A
full adder
B
full subtractor
C
shift register
D
decade counter
Question 3 Explanation: 
The above diagram is full subtractor. The equation is D=X⊕Y⊕Bin and Bout=X'Bin+X'Y+YBin
Question 4
When two numbers are added in excess-3 code and the sum is less than 9, then in order to get the correct answer it is necessary to
A
subtract 0011 from the sum
B
add 0011 to the sum
C
subtract 0110 from the sum
D
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 excess-6. Hence, subtract 0011 to make it excess-3.
Question 5
The characteristic equation of an SR flip-flop is given by
A
Qn+1 = S + RQn
B
Qn+1= RQn + SQn
C
Qn+1= S + RQn
D
Qn+1 = S + RQn
Question 5 Explanation: 
The characteristic table of an SR flip-flop is:

So, by simplifying using k-maps:
characteristic equation of an SR flip-flop = Qn+1 = S + RQn
Question 6
A graph with n vertices and n-1 edges that is not a tree, is
A
Connected
B
Disconnected
C
Euler
D
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.
Question 7
If a graph requires k different colours for its proper colouring, then the chromatic number of the graph is
A
1
B
k
C
k-1
D
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
A
and written by CPU
B
and written by peripheral
C
by peripheral and written by CPU
D
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
A
booting up the priority of the process in multi-level of queue without feedback
B
keeping track of the following a page has been in memory for the purpose of LRU replacement
C
letting job reside in memory for a certain amount of time so that the number of pages required can be estimated accurately
D
gradually increasing the priority of jobs that wait in the system for a long time to remedy infinite blocking
E
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 r1, r2….rn to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput?
A
Round Robin
B
Shortest job first
C
Highest response ratio next
D
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.
Question 11
Consider a job scheduling problem with 4 jobs J1, J2, J3, J4 and with corresponding deadlines: ( d1, d2, d3, d4) = (4, 2, 4, 2).
Which of the following is not a feasible schedule without violating any job schedule?
A
J2, J4, J1, J3
B
J4, J1, J2, J3.
C
J4, J2, J1, J3.
D
J4, J2, J3, J1
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
Question 12
By using an eight-bit optical encoder the degree of resolution that can be obtained is (approximately)
A
1.8o
B
3.4o
C
2.8o
D
1.4o
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, 8-bit optical encoder will have 28 increments Resolution = 360/2n = 360/28 = 1.4o
Question 13
The principal of the locality of reference justifies the use of
A
virtual memory
B
interrupts
C
main memory
D
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.
Question 14
Consider the following pseudo-code
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 pseudo-code?
A
4
B
5
C
6
D
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.
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
A
B
B
C
C
D
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
A
FIFO
B
Shortest job first
C
Shortest remaining time
D
Longest remaining time
Question 16 Explanation: 
FIFO is when implemented in preemptive version, it acts like round-robin algorithm.
Question 17
The number of digit 1 present in the binary representation of 3 × 512 + 7 × 64 + 5 × 8 + 3
A
8
B
9
C
10
D
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.
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
A
109
B
216
C
300
D
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 8-bit 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
A
5 kHz
B
10 kHz
C
47 kHz
D
57 kHz
Question 19 Explanation: 
Bandwidth = Highest frequency - Lowest frequency.
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.
Question 20
An Ethernet hub
A
functions as a repeater
B
connects to a digital PBX
C
connects to a token-ring network
D