# GATE IT 2006

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Question 1 |

In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?

0.4 | |

0.6 | |

0.8 | |

0.9 |

**GATE IT 2006**

**Discuss it**

Question 1 Explanation:

Question 2 |

For the set N of natural numbers and a binary operation f : N x N → N, an element z ∊ N is called an identity for f, if f (a, z) = a = f(z, a), for all a ∊ N. Which of the following binary operations have an identity?

- f (x, y) = x + y - 3
- f (x, y) = max(x, y)
- f (x, y) = x
^{y}

I and II only | |

II and III only | |

I and III only | |

None of these |

**Set Theory & Algebra**

**GATE IT 2006**

**Discuss it**

Question 2 Explanation:

I f(x,y) = x+y-3 = x= y+x-3 => y=3 Here identity elements is 3
II f(x,y) = max(x,y)=x=max(y,x) => y=1 Here identity elements is 1
(III f(x,y) =x^y is not same as f(y,x) = y^x. So no identity element.

Question 3 |

In the automaton below, s is the start state and t is the only final state.
Consider the strings u = abbaba, v = bab, and w = aabb. Which of the following statements is true?

The automaton accepts u and v but not w | |

The automaton accepts each of u, v, and w | |

The automaton rejects each of u, v, and w | |

The automaton accepts u but rejects v and w |

**Regular languages and finite automata**

**GATE IT 2006**

**Discuss it**

Question 3 Explanation:

For the acceptance and rejection of any string we can simply check for the movement on each input alphabet between states. A string is accepted if we stop at any final state of the DFA.
For string u=abbaba the string ends at t (final state) hence it is accepted by the DFA.
For string v=bab the string ends at s (non-final state) and hence rejected by the DFA.
For string w=aabb the string ends at s (non-final state) and hence rejected by the DFA.
This solution is contributed by

**Yashika Arora**.Question 4 |

In the context-free grammar below, S is the start symbol, a and b are terminals, and ϵ denotes the empty string
S → aSa | bSb | a | b | ϵ
Which of the following strings is NOT generated by the grammar?

aaaa | |

baba | |

abba | |

babaaabab |

**Regular languages and finite automata**

**GATE IT 2006**

**Discuss it**

Question 4 Explanation:

The given Language is Palindrome.
A,C and D follow the language but

**baba**is not a palindrome , so B is the answerQuestion 5 |

Which regular expression best describes the language accepted by the non-deterministic automaton below?

(a + b)* a(a + b)b | |

(abb)* | |

(a + b)* a(a + b)* b(a + b)* | |

(a + b)* |

**Regular languages and finite automata**

**GATE IT 2006**

**Discuss it**

Question 6 |

Given a boolean function f (x

_{1}, x_{2}, ..., x_{n}), which of the following equations is NOT truef (x1, x2, ..., xn) = x1'f(x1, x2, ..., xn) + x1f(x1, x2, ..., xn) | |

f (x1, x2, ..., xn) = x2f(x1, x2, …, xn) + x2'f(x1, x2, …,xn) | |

f (x1, x2, ..., xn) = xn'f(x1, x2, …, 0) + xnf(x1, x2, …,1) | |

f (x1, x2, ..., xn) = f(0, x2, …, xn) + f(1, x2, .., xn) |

**C Functions**

**Set Theory & Algebra**

**GATE IT 2006**

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Question 6 Explanation:

**Option A**: f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn)

**Case 1**: taking x1=0 RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn).

**Case 2**: taking x1=1 RHS = 0.f(x1, x2, …, xn) + 1.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). In both cases RHS=LHS, so, (A) is true

**Option B**: f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn)

**Case 1**: taking x2=0 RHS= 0.f(x1, x2, …, xn) + 1.f(x1, x2…,xn) RHS =f(x1, x2, …, xn).

**Case 2**: taking x2=1 RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn) RHS =f(x1, x2, …, xn). In both cases RHS=LHS, so, (B) is true.

**Option C**: f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1)

**Case 1**: taking xn=0 RHS= 1.f(x1, x2, …, 0) + 0.f(x1, x2, …, 1) RHS =f(x1, x2, …, 0)

**Case 2**: taking xn=1 RHS = 0.f(x1, x2, …, 0) + 1.f(x1, x2, …, 1) RHS =f(x1, x2, …, 1)In both cases RHS=LHS, so, (C) is true.

**Option D**: f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn) Here, no way to equate LHS and RHS so

**‘NOT true’**. NO term depends on value of ‘x1’. This solution is contributed by

**S**

**andeep pandey.**

Question 7 |

The addition of 4-bit, two's complement, binary numbers 1101 and 0100 results in

0001 and an overflow | |

1001 and no overflow | |

0001 and no overflow | |

1001 and an overflow |

**Digital Logic & Number representation**

**Number Representation**

**GATE IT 2006**

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Question 7 Explanation:

Its -3+4=1, so no overflow
So Answer is C.

Question 8 |

Which of the following DMA transfer modes and interrupt handling mechanisms will enable the highest I/O band-width?

Transparent DMA and Polling interrupts | |

Cycle-stealing and Vectored interrupts | |

Block transfer and Vectored interrupts | |

Block transfer and Polling interrupts |

**OS Input Output Systems**

**Computer Organization and Architecture**

**GATE IT 2006**

**Secondary memory and DMA**

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Question 9 |

In a binary tree, the number of internal nodes of degree 1 is 5, and the number of internal nodes of degree 2 is 10. The number of leaf nodes in the binary tree is

10 | |

11 | |

12 | |

15 |

**Binary Trees**

**GATE IT 2006**

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Question 9 Explanation:

In a binary tree, the number of leaf nodes is always 1 more than number of internal nodes with 2 children, refer http://www.geeksforgeeks.org/handshaking-lemma-and-interesting-tree-properties/
So,
Number of Leaf Nodes = Number of Internal nodes with 2 children + 1
Number of Leaf Nodes = 10 + 1
Number of Leaf Nodes = 11

Question 10 |

A problem in NP is NP-complete if

It can be reduced to the 3-SAT problem in polynomial time | |

The 3-SAT problem can be reduced to it in polynomial time | |

It can be reduced to any other problem in NP in polynomial time | |

some problem in NP can be reduced to it in polynomial time |

**Analysis of Algorithms**

**NP Complete**

**GATE IT 2006**

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Question 10 Explanation:

A problem in NP becomes NPC if all NP problems can be reduced to it in polynomial time. This is same as reducing any of the NPC problem to it. 3-SAT being an NPC problem, reducing it to a NP problem would mean that NP problem is NPC.
Please refer: http://www.geeksforgeeks.org/np-completeness-set-1/

There are 86 questions to complete.