PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2017 Shift 2

36 Consider a binary code that consists of only four valid code words as given below:
00000,01011,10101,11110
Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q.
Then the values of p and q are

Given:
code1 00000
code2 01011
code3 10101
code4 11110
Hamming distance between code 1 and code 2 is 3
Hamming distance between code 1 and code 3 is 3
Hamming distance between code 1 and code 4 is 4
Hamming distance between code 2 and code 3 is 4
Hamming distance between code 2 and code 4 is 3
Hamming distance between code 3 and code 4 is 3
So, as per Hamming code, minimum Hamming distance of all code words is considered as Hamming distance i.e., 3 (p).
Now, the max number of erroneous bits that can be corrected by the Hamming code is 2d + 1
So,
So option A is correct.

37 A system shares 9 tape drives. The current allocation and maximum requirement of tape drives for three processes are shown below:

Which of the following best describes current state of the system ?

PID	Current allocation	Max Need	Available	Needed
P1	3	3	2	0
P2	1	6	-	5
P3	3	5	-	2

Hence, system is in safe state , because P1 and P3 anyone can execute first and after them P2 can be executed, no deadlocked option B is correct.

38 Two transactions T₁ and T₂ are given as:
T₁: r₁(X)w₁(X)r₁(Y)w₁(Y)
T₂ : r₂(Y)w₂(Y)r₂(Z)w₂(Z)
where r_i(V) denotes a read operation by transaction T_i on a variable V and w_i(V) denotes a write operation by transaction T_i on a variable V. The total number of conflict serializable schedules that can be formed by T₁ and T₂ is ______

Note: Numerical Type question

Conflict conditions RW WR WW
∴ 5 conflicts
T₁ - T₂
a b c d
r₁(X) w₁(X) r₁(Y) w₁(Y)
r₂(Y) w₂(Y) r₂(Z) r₂(Z)
1 2 3 4
Constraints:
a<b<c<d
1<2<3<4
d<1 (or) 2<c
only 1 way
Total = 70 - (12 + 5) = 53
Therefore total no of, 53 + 1 = 54

39 If w, x, y, z are Boolean variables, then which one of the following is INCORRECT?

A)LHS: wx + w(x + y) + x(x+y) = x + wy
RHS
⇒ wx + wx + wy + x + xy
⇒ wx + wy + x + xy [∵ xx = x]
⇒ x(w+y+1) + wy [∵1 + x = 1]
⇒ x + wy
⇒ L.H.S = R.H.S

L.H.S ≠ R.H.S
D) L.H.S.:(w + y)(wxy + wyz) = wxy + wyz
(w + y)(wxy + wyz)
wxy + wyz + wxy + wyz
wxy + wyz
L.H.S = R.H.S

40 Consider the following C Program.

(# include
#include
#int main ( ) {
char* c = “GATECSIT2017”;
char* p = c;
printf(“%d”, (int) strlen (c+p[2]-p[6]-1));
return 0;
})
The output of the program is ____.

Note – Numerical Type question

Assume starting address of string is 100 to make the whole expression is easy to understand.

Note: Whenever we have characters in the arithmetic expressions, we can replace those with their ASCII values
strlen (100 + x + 11 - x - 1)[assume x has the ASCII value of I]
⇒Strlen(110)
is printed
It gives address of second last character in the string. So it prints length 2.

PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2017 Shift 2

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