hi reverse good morning this video is
for TRB teachers recruitment both
computer science and surkhi cushion
paper for the year 2006 to 7
ok let us see the Kushina answer this
question paper US teaches recruitment
boat special recruitment of computer
instructor in government Higher
Secondary School 2006 to 7 question
paper and the moths allotted or on 50
moths 150 questions one more for each
question and the time allotted will be 3
hovers you will be allotted with the
time period of 3 Hertz 150 moths in the
marking schemes now changed. 120 marks for computer.
remaining 40 from GK and education okay
let's see the let us either pushing
first question the most common error
detection code uses option yay party
boots B check bitsy binary bits D
generator bit stances yay option parity
bits among the other answer the parity
bits only used for Jared deduction code
the other codes not used the other
second one the other names for reflected
coders hacksaw baited gray excess K from
Google search you can know actually the
gray code is also called US or BC
reflected binary code the great Buddhist
also called us reflected binary code or
reflected code so the answer is the
other name for reflected coders gray
code and it is also called as reflected
binary code excursion
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readable x4 match next question the
manipulation of binary information is
done by logic circuit decoder truth
table star state a Bill Gates so the
answer is gates see here binary
information is carried by signals and
manipulation of binary information is
done by logic circuit called us gates so
this is wrong from the Google search so
third question answer gates the
manipulation of binary information is
done by the logic circuit gates next
question the flow of our just in eighty
eighty-five Earth unidirectional okay
see this is your 80-85 architecture 8080
fire architecture in this this is your
data verse which is bi-directional and
this is your address verse this
unidirectional so in eighty eight if I
address buses unidirectional that is one
way data transfer and data buses
bi-directional two-way data transfer and
next with pushing which register points
to the memory address from which the
next to bite us to be fetched so answer
program counter so program counter one
Li will point to the address of the next
instruction to be fetched so the stack
pointer is different based pointer is
different and index pointer is different
and next question how many address lines
are necessary to address the 2048 bytes
of memory so the answer is 11 so
actually the how to solve this
questioners so to power y n equal to
2048 so you have to find out each power
so try the answer for 2 to the power of
9 2 to the power of 10 2 to the power of
11 and 2 to the power of 12 so for this
question 2 to the power of 11 equal to 2
0 for 8 so the answer is a 11 so 11
address lines are necessary to address
the 2048 bytes of memory next question
if the memory chip size is 256 into four
birds how many chips are required to
make up to four 2048 bytes of memory so
the question read one more time is the
memory chip size 256 into four bits so
the question is given in terms of bits
here here they have given 256 into four
birds and they have asked for two 48
bytes of memory so let's solve this
Koshien a must word first write down the
numbers evolve faster chip size they
have given memory chip size equal to 256
in 2/4 so it is given in terms of births
now you convert the bits into bytes for
conversion you how to divide the two
sister 6 into 4 divided by H okay what
they have given in terms of bits so now
you convert bit into bytes by dividing
by H so how to divide by H so 256 into 4
divided by H divided by 8 so by solving
this equation we will get so for H so by
solving this we will get the answer
120 H so the answer will be 128 so 128
bytes so now read the question so we
have to require to make 2048 bytes of
data so the question is B how to convert
it into 2048 bytes of memory
so now 128 into some n number 128 into n
equal to 2048 we have to find out the y
n so n equal to 2048 divided by 128 so
the answer will be 16 otherwise you can
solve directly here also you can solve
128 into 16 equal to 2048 if you know
the sixteenth table directly you can
directly write 128 into 16 equal to 2048
so the answer is 16 here en equal to 16
so for 4
these type of pushing you how to first
step actually there are two steps are
there in this machine first step you how
to convert the bits into bytes by
dividing by H and the next the given X
bytes into some yen equal to given bytes
after that you have to find out that n
value so the N value will be the answer
so for the seventh machined answer is C
option 16 and next question the arches
and data bus an 80 85 microprocessor is
multiplex actually see the 80-85
architecture see this is your address
and data bus which is a d0 to a d7 the
lower order address on data bus or
multiplexed address versus
unidirectional data versus
bi-directional and address plus data
buses multiplexed so ninth question
which signal indicates that peripherals
such as DMA controller is requesting the
use of our json data bus and the signal
is hold h ho LD hold so for ninth
caution answer C option hold a next
question if the 80-85 has fetch to the
machine code located at the memory
location to 0 file EF h specify the
content of program counter so next it in
the cushion if the 80-85 has speech to
the machine code located at the memory
location to 0 Phi F H specify the
content of program counter so normally
program counters and store the content
of memory location plus 1 so this are
just plus 1 so this is calculation type
problem so we have to
calculate this so we can open em us so
first write down the number to Z zero
file yes
so the given number is 2:05 EF so write
the numbers with some space format to 0
Phi yes write the binary equivalent of
the number in four digit format
so actually as we know that for to the
binary equivalent is 1 0 so write it in
four digit format 0 0 1 0 4 0 we have to
write for number of zeros for 5 1 not 1
now write 0 1 not 1 for yes we know for
once for yes for once
now add 1 with the number we will get
the and search now we have to add 1 so
with answer we how to add the 1 1 + 1 0
carry is 1 1 + 1 0 sorry 1 plus 1 in
carry is 1 1 + 1 10 0 here and carry
will be 1 here 1 plus 1 n 0 will be here
1 will be here as a carry 1 + 1 10 0
will be here and 1 will be here as a
carry now 1 + 0 1 so now remaining
numbers we can write as the t's 1 will
be no number so 1 0 will be write write
down write it down ur 0 0 0 0 0 he write
the same number 0 1 0 0 so the addition
will be up to these digits so now we are
getting the answer this one 0 this one
sakes this one you are 0 again this one
you are 2 so the answer is 2 0 6 0 H
answer will be 2z
no six zero so the value of two zero
five hatch after this number if we add
one we will get the number two zero six
zero so next question what does dispatch
latency so dispatched latency means
actually is the right answer is B option
time to switch from one process to
another process so dispatch latency
means the time taken to switch from one
process to another process process
twelth machine time quantum is needed
for round-robin scheduling so answer is
d round-robin scheduling in round-robin
scheduling only we need time quantum
method and thirteenth pushing what is
the critical section problem so critical
section problem this problem of sharing
of data in code figments the correct
answer is what is critical section
problem the problem of sharing of data
in Code section and fourteenth question
which is not the condition for deadlock
actually we can Google search what are
all the conditions for deadlock actually
the conditions are mutual exclusion
Holden which no no preemption circular
which there are four main conditions for
deadlock so these four are the
conditions for deadlock so now read the
cushion circular rate it is their mutual
exclusion it is their Holden weight it
is their so the answer is swapping
swapping is not a condition for date
block now 15 question what does GPU
scheduling CPU scheduling the they have
given the answer select roses from ready
queue to CPU select process to the ready
queue select process from i/o buffer
fixing the time limit
so the correct answer is select crosses
from ready queue to CPU is called CPU
scheduling the number of process will be
available so the process which is in
radio Q will be assigned to CPU that is
called CPU scheduling thank you to
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11th the standard new syllabus computer
science lesson and a computer
application lessons if you want you can
use the materials and see the top
booster folder the top first folder is
TRB PS here only your T or B old pushing
paper and
we'll be uploaded thank you
