Method-1:
Given data,
-- Set A={1, 2, 3, ........, 1000}
-- Set A shall be divisible by 3=?
-- Set A shall be divisible by 5=?
-- Set A shall be divisible by 3 and 5=?
Step-1: To find divisible by 3 numbers are
=⌊1000/3⌋
= 333
Step-2: To find divisible by 5 numbers are
=⌊1000/5⌋
= 200
Step-3: To find divisible by 3 and 5 numbers are
=⌊1000/(3*5)⌋
= 66
These 66 is already part of 333 and 200. So, we have to exclude it from the list.
Total= 333+200-66
= 467
Note: We are using floor because excluding fraction value.
Method-2:
The above problem is in the form of (AUB) = (A)+(B)-(A∩B)
A=1000/3
B=1000/5
(A∩B)=66
(AUB)=467
Note: Getting this idea in exam hall is very difficult. So, better follow method-1 ratherhan method-2

Method-1:
Here, they are clearly mentioned that “certain tree”
The tree contain maximum n-1 edges. Here, ‘n’ is number of vertices.
→ According to the handshaking lemma “sum of degrees of all vertices=2|e|”.
→ Two vertices of degree 4, we can write into (2*4)
→ One vertex of degree 3, we can write into (1*3)
→ One vertex of degree 2, we can write into (1*2).
→ Other vertices(X) have degree 1, we can write into (X*1)
Step-1: (2*4)+(1*3)+(1*2)+(X*1)
=2(X+4-1) [Note: ‘X’ value is not given ]
X=7
Step-2: To find total number of vertices, we are adding X+4 because they already given 4 vertices.
=X+4
=7+4
=11
Method-2:
Draw according to given constraints but it not suitable for very big trees

B, C, E, F, H, I, J degree = 1
D degree = 2
A degree = 4
K degree = 4
G degree = 3

Option-A: It is not complete graph because it won’t have n(n-1)/2 edges.
Option-B: It is not Bipartite Graph because it takes more than 2 colours. Bipartite Graph have exactly 2 colours.
Option-C: Hamiltonian path is a path in an undirected or directed graph that visits each vertex exactly once.

Hence, Option-C is correct answer.

Probability that the value k=1 appears in the printout at least once is
1-p(x=0)
We will apply the formula of binomial distribution
1-(^{​ 1000000​} C​ ₀​ (1/1000000)​ ⁰​ * (999999/1000000)​ ^{10^6}​ )
=1-(1*1*0.367879)
=0.632121
Hence answer is option C

Given f(x) = x​ ⁴​ , g(x) = √ x ² + 1 , h(x) = x​ 2​ + 72
for, ho(gof)
gof=g(f(x))
=g(x​ ⁴​ )
=√(x​ ⁸​ +1)
ho(gof)=h(gof)
=h(√(x​ ⁸​ +1))
=(√(x​ ⁸​ +1)​ 2​ +72
=x​ ⁸​ +1+72
=x​ ⁸​ +73
√ x ² + 1 , h(x) = x​ ²​ + 72
for, (hog)of,
hog=h(g(x))
=h(√(x​ ²​ +1)
=(√(x​ ²​ +1)​ 2​ +72
=x​ ²​ + 1+72
=x​ ²​ +73
(hog)of=(hog)(f(x))
=(hog)(x​ ⁴​ )
=(x​ ⁴​ )​ ²​ +73
=x​ ⁸​ +73
Hence, option-D is the correct answer

→ Each digit is represented by a 4-bit BCD code.
→ To add two 4-bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).
→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.
→ This can be done with 1 Half adder and 2 Full Adder
(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.)
→ Here, Half adder is used to add next significant bits.
Total 5 Full Adders and 2 Half Adders are needed

The Excess-3 decimal code is a self-complementing code because complement can be generated by inverting each bit pattern.

Given infix notation, we have to change infix notation into postfix notation.
after converting postfix notation the notations are ab* cde /* +

Total 5 PUSH and 4 POP operations performed.

The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is​ ​ 2 ^–129 to (1 – 2​ ^–23​ ) * 2^{​ 127}

The size of the ROM required to build an 8-bit adder/subtractor with mode control, carry input, carry output and two’s complement overflow output is given as 2​ ¹⁸​ * 10.

In this program we are using post increment. Post increment will send the value before updating the value.
printf(“\n%d”, 1 + x++); /* x=128 */
Here, 1+128=129

char *(*(*a[N]) ( )) ( ); /* an array of n pointers to function returning pointers to functions returning pointers to characters */