## GATE-IT-2004

Question 1 |

In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?

3/23 | |

6/23 | |

3/10 | |

3/5 |

**General Aptitude**

**GATE-IT-2004**

**Discuss it**

Question 1 Explanation:

x- total families. Given:

- 50% -3 Children -> (5*x/10) * 3=(1*x/2)*3
- 30% -2 Children-> (3*x /10) * 2
- Rear 20%- 1child -> (2*x/10)*1 =1*x/5

**So Answer is B**Question 2 |

In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course; 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization; 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses.How many students have not taken any of the three courses?

15 | |

20 | |

25 | |

35 |

**General Aptitude**

**GATE-IT-2004**

**Discuss it**

Question 2 Explanation:

P(AUBUC)=P(A)+P(B)+P(C)- P(A∩B)-P(A∩C)- P(B∩C)+P(A∩B∩C)

125+85+65-50−35−30+15 =175

No of students not taking any courses-> 200−175=25
**So C is the Answer.**

Question 3 |

Let a(x, y), b(x, y,) and c(x, y) be three statements with variables x and y chosen from some universe. Consider the following statement:

(∃x)(∀y)[(a(x, y) ∧ b(x, y)) ∧ ¬c(x, y)]
Which one of the following is its equivalent?
(∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)] | |

(∃x)(∀y)[(a(x, y) ∨ b(x, y)) ∧¬ c(x, y)] | |

¬ (∀x)(∃y)[(a(x, y) ∧ b(x, y)) → c(x, y)] | |

¬ (∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)] |

**Linear Algebra**

**GATE-IT-2004**

**Discuss it**

Question 4 |

Let R

_{1}be a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} and R_{2}be another relation from B to C = {1, 2, 3, 4} as defined below:- An element x in A is related to an element y in B (under R
_{1}) if x + y is divisible by 3. - An element x in B is related to an element y in C (under R
_{2}) if x + y is even but not divisible by 3.

_{1}R_{2}from A to C?R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)} | |

R1R2 = {(1, 2), (1, 3), (3, 2), (5, 2), (7, 3)} | |

R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)} | |

R1R2 = {(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)} |

**Set Theory & Algebra**

**GATE-IT-2004**

**Discuss it**

Question 4 Explanation:

R1 is a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} . Under R1, an element x in A is related to an element y in B if x + y is divisible by 3.

Thus, R1 = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)}

R2 is a relation from B = {2, 4, 6, 8} to C = {1, 2, 3, 4} Under R2, an element y in B is related to an element z in C if y + z is even but not divisible by 3.

Thus, R2 = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)}

Thus, R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}

Thus, option (C) is correct.

Please comment below if you find anything wrong in the above post.

Question 5 |

What is the maximum number of edges in an acyclic undirected graph with n vertices?

n-i | |

n | |

n + 1 | |

2n-1 |

**Graph**

**GATE-IT-2004**

**Discuss it**

Question 5 Explanation:

n * (n - 1) / 2 when not cyclic.
An acyclic graph with the maximum number of edges is actually a spanning tree and therefore, correct answer is n-1 edges.

**So Answer is A**Question 6 |

x=6,y=3,z=2 | |

x=12,y=3,z=—4 | |

x=6,y=6,z=—4 | |

x=12,y=—3,z=O |

**Linear Algebra**

**GATE-IT-2004**

**Discuss it**

Question 7 |

Which one of the following regular expressions is NOT equivalent to the regular expression (a +

*b*+*c)**?(a* + b* + c*)* | |

(a*b*c*)* | |

((ab)* + c*)* | |

(a*b* + c*)* |

**Regular languages and finite automata**

**GATE-IT-2004**

**Discuss it**

Question 7 Explanation:

**C**- (ab)* + c*)* will always give strings with

**"ab"**together.Whereas (a+b+c)* would generate language where a,b,c may not be always together.

**A,B,D**may generate same language as (a+b+c)*

**So, Answer is (C)**

Question 8 |

What is the minimum number of NAND gates required to implement a 2-input EXCLUSIVE-OR function without using any other logic gate?

3 | |

4 | |

5 | |

6 |

**Digital Logic & Number representation**

**GATE-IT-2004**

**Discuss it**

Question 8 Explanation:

Pic Courtesy: http://www.electronics-tutorials.ws/logic/logic_7.html

**Other way around:**x XOR y = x’y+xy’ = x’y+xy’+xx’+yy’ = (x+y) (x’+y’) Using NAND gates F= (x+y)(xy)’ = x. (xy)’ + y. (xy)’ Taking compliment F’= ( x. (xy)’ + y. (xy)’ )’ = (x. (xy)’)’. (y. (xy)’) Compliment again F=( (x. (xy)’)’. (y. (xy)’) )’**So Answer is B**Question 9 |

Which one of the following statements is FALSE?

There exist context-free languages such that all the context-free grammars generating them are ambiguous | |

An unambiguous context free grammar always has a unique parse tree for each string of the language generated by it. | |

Both deterministic and non-deterministic pushdown automata always accept the same set of languages | |

A finite set of string from one alphabet is always a regular language. |

**Context free languages and Push-down automata**

**GATE-IT-2004**

**Discuss it**

Question 9 Explanation:

A) For real-world programming languages, the reference CFG is often ambiguous, due to issues such as the dangling else problem. //Wikipedia

B) A string is ambiguous if it has two distinct parse trees;The grammar is unambiguous,if a string has distinct parse trees.

**C) Deterministic pushdown automata can recognize all deterministic context-free languages while nondeterministic ones can recognize all context-free languages**

**Therefore it's FALSE**

D)Properties of Regular Language:

- The set of regular languages over an alphabet is closed under operations union, concatenation and Kleene star.
- Finite languages are regular

**So Answer is C**

Question 10 |

What is the minimum size of ROM required to store the complete truth table of an 8-bit x 8-bit multiplier?

32 K x 16 bits | |

64 K x 16 bits | |

16 K x 32 bits | |

64 K x 32 bits |

**Computer Organization and Architecture**

**GATE-IT-2004**

**Discuss it**

Question 10 Explanation:

Input to ROM - 2 lines ,8 bit each.
Possible combinations in ROM - (2^8)x(2^8)
Size of truth table = (2^8)*(2^8)=2^16=

**64 KB**Maximum output size =**16 bit****So, Answer is B**
There are 90 questions to complete.