## UGC NET CS 2014 Dec-Paper-2

Question 1 |

Consider a set A = {1, 2, 3, ........, 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5 ?

533 | |

599 | |

467 | |

66 |

Question 1 Explanation:

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

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

Question 2 |

A certain tree has two vertices of degree 4, one vertex of degree 3 and one vertex of degree 2. If the other vertices have degree 1, how many vertices are there in the graph ?

5 | |

n-3 | |

20 | |

11 |

Question 2 Explanation:

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

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

Question 3 |

Consider the Graph shown below :

This graph is a __________.

This graph is a __________.

Complete Graph | |

Bipartite Graph | |

Hamiltonian Graph | |

All of the above |

Question 3 Explanation:

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.

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.

Question 4 |

A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k=1 appears in the printout at least once ?

0.5 | |

0.704 | |

0.632121 | |

0.68 |

Question 4 Explanation:

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-(

=1-(1*1*0.367879)

=0.632121

Hence answer is option C

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

Question 5 |

If we define the functions f, g and h that map R into R by :
f(x) = x

^{ 4} , g(x) = √ x^{ 2}+ 1 , h(x) = x^{ 2} + 72, then the value of the composite functions ho(gof) and (hog)of are given asx ^{8} – 71 and x ^{8} – 71 | |

x ^{8} – 73 and x^{8} – 73 | |

x ^{8} + 71 and x ^{8} + 71 | |

x ^{8} + 73 and x ^{8} + 73 |

Question 5 Explanation:

Given f(x) = x

for, ho(gof)

gof=g(f(x))

=g(x

=√(x

ho(gof)=h(gof)

=h(√(x

=(√(x

=x

=x

√ x

for, (hog)of,

hog=h(g(x))

=h(√(x

=(√(x

=x

=x

(hog)of=(hog)(f(x))

=(hog)(x

=(x

=x

Hence, option-D is the correct answer

^{4} , g(x) = √ x^{2}+ 1 , h(x) = x 2 + 72for, 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} + 72for, (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} +73Hence, option-D is the correct answer

Question 6 |

The BCD adder to add two decimal digits needs minimum of

6 full adders and 2 half adders | |

5 full adders and 3 half adders | |

4 full adders and 3 half adders | |

5 full adders and 2 half adders |

Question 6 Explanation:

→ 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

→ 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

Question 7 |

The Excess-3 decimal code is a self-complementing code because

The binary sum of a code and its 9’s complement is equal to 9. | |

It is a weighted code | |

Complement can be generated by inverting each bit pattern | |

The binary sum of a code and its 10’s complement is equal to 9 |

Question 7 Explanation:

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

Question 8 |

How many PUSH and POP operations will be needed to evaluate the following expression by reverse polish notation in a stack machine (A * B) + (C * D/E) ?

4 PUSH and 3 POP instructions | |

5 PUSH and 4 POP instructions | |

6 PUSH and 2 POP instructions | |

5 PUSH and 3 POP instructions |

Question 8 Explanation:

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.

after converting postfix notation the notations are ab* cde /* +

Total 5 PUSH and 4 POP operations performed.

Question 9 |

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 ^{-128} to (1 – 2^{ –23} ) * 2^{ 127} | |

(1 – 2 ^{ –23} ) * 2 ^{ -127} to 2 ^{128} | |

(1 – 2 ^{–23} ) * 2^{ –127} to 2^{ 23>} | |

2 ^{–129} to (1 – 2^{ –23} ) * 2^{ 127} |

Question 9 Explanation:

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}Question 10 |

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 ^{16} * 8 | |

2 ^{18} * 10 | |

2 ^{16} * 10 | |

2 ^{18} * 8 |

Question 10 Explanation:

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.Question 11 |

What will be the output of the following ‘C’ code ?

main( )

{

int x=128;

printf (“\n%d”, 1 + x++);

}

main( )

{

int x=128;

printf (“\n%d”, 1 + x++);

}

128 | |

129 | |

130 | |

131 |

Question 11 Explanation:

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

printf(“\n%d”, 1 + x++); /* x=128 */

Here, 1+128=129

Question 12 |

What does the following expression means ?
char *(*(*a[N]) ( )) ( );

a pointer to a function returning array of n pointers to function returning character pointers. | |

a function return array of N pointers to functions returning pointers to characters | |

an array of n pointers to function returning pointers to characters | |

an array of n pointers to function returning pointers to functions returning pointers to characters. |

Question 12 Explanation:

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