CurrentGK Papers GAT(Graduate Aptitude Test In Engineering) GATE Sample Exam Papers

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**Q.1.** **Data Structure and Algorithms**

In a box there are random number of white and black marbles. At a time two marbles are taken out at random and if

A) Both Black: Discard both and insert a white

B) Both White: Discard one and retain one

C) One Black and One White: Discard White and retain Black.

If initially there are nb black and nw white marbles then determine the color of the only marble remaining at the end.

**Options**

1. white if nb is even

2. white if nw is odd

3. black if nb is even

4. black if nw is odd

Ans: **1**

**Explanation:**

Here parity of black is preserved and the parity of white is neglected.

Lets consider each case

A) Both Black: Discard both and insert a white Parity of back is preserved by discarding both, i.e if there are nb marbles then nb-2 remain. Thus if nb is even, nb-2 is also even. For nw, it becomes nw+1 and hence parity is changed(or is not conserved).

B) Both White: Discard one and retain one Again no change to nb and nw parity is neglected.

C) One Black and One White: Discard White and retain Black. Same applies here also.

Thus, the color of last marble depends on nb and not on nw.

Hence, we get the following table:

nb | nw | remaining |
---|---|---|

even | even | white |

even | odd | white |

odd | even | black |

odd | odd | black |

which can be further reduce to :

nb | nw | remaining |
---|---|---|

even | X | white |

odd | X | black |

**Q.3.** **Databases**

Consider three tables with following number of tuples in each

S(a,b,c) = 100

R(a,d,e) = 80

T(x,d,f) = 90

Tuples in S and R with same value of attribute "a" = 60

Tuples in R and T with same value of attribute "d" = 70

Maximum and minimum number of tuples in ** (S left outer join R) full outer join T** is:

**Options**

1. 130 120

2. 150 140

3. 140 130

4. 140 120

Ans: **3**

**Explanation:**

Consider, * X = ( S left outer join R ) * This will have 100 rows, with *d = null* for 40(100-60) rows.

*X full outer join T*: Here, the common values of d in tables X and T can be between 50 and 60.

**Case 1:** *Minimum value of common values of d between X and T*

Out of 80 values of d in R, 20 were not found in X (since X has only 60 non-null values of d). These 20 can all be the ones that were common between S and T. In this case the common values between X and T will be 50.

**Case 2**: *Maximum value of common values of d between X and T*

Now assume that 20 values of d that were lost, none was common between S and T. But this is not possible, because S has 80 tuples and 70 values are common with T. So at least 10 common values will get lost. Thus max. value of common d's will b 60.

Now full outer join (X + T - common values of d in X and T) will be either

100 + 90 - 50 or 100 + 90 - 60.

Thus the answer is 140,130. Hence 3

**Q.4.** **Data Structures - Hashing**

A hash table implementation uses function of (% 7) and linear probing to resolve collision. What is the ratio of numbers in the following series with out collision and with collision if 7 buckets are used:

32, 56, 87, 23, 65, 26, 93

**Options**

1. 2,5

2. 3,4

3. 4,3

4. 5,2

Ans: **3**

**Explanation:**

Since we're using **mod 7** hash function, we've a hash table with 7 buckets numbered from 0 to 6. Any number **x** will be mapped to ** x % 7** bucket.

Lets look at the insertion of sequence into hash table:

1. **32** is inserted into *32 % 7 = 4* i.e. 4th bucket.

0 | 1 | 2 | 3 | 4 |
5 | 6 |

32 |

2. **56** is inserted into *56 % 7 = 0* i.e. 0th bucket.

0 |
1 | 2 | 3 | 4 | 5 | 6 |

56 |
32 |

3. **87** is inserted into *87 % 7 = 3* i.e. third bucket.

0 | 1 | 2 | 3 |
4 | 5 | 6 |

56 | 87 |
32 |

3A

4. **23** is inserted into *23 % 7 = 2*

0 | 1 | 2 |
3 | 4 | 5 | 6 |

56 | 23 |
87 | 32 |

5. **65** is inserted *65 % 7 = 2*. Since bucket 3 also contains 23, its a collision. We'll put 65 into the next empty bucket i.e. 5

0 | 1 | 2 |
3 | 4 | 5 |
6 |

56 | 23 |
87 | 32 | 65 |

6. **26** is inserted into *26 % 7 = 5*. However, since bucket 5 is also full, its a collision and 26 is inserted into the next empty bucket i.e. 6.

0 | 1 | 2 | 3 | 4 | 5 |
6 |

56 | 23 | 87 | 32 | 65 |
26 |

7. **93** is inserted into *93 % 7 = 2*. However, since bucket 2 is also full, its a collision and 93 is inserted into the next empty bucket i.e. 1.

0 | 1 |
2 |
3 | 4 | 5 | 6 |

56 | 93 |
23 |
87 | 32 | 65 | 26 |

Final position of buckets will be:

Bucket # | Element | Remark |
---|---|---|

0 | 56 | |

1 | 93 | |

2 | 23 - 65 - 93 | |

65 ( move to next empty - 5) | Collision | |

93 ( move to next empty - 1) | Collision | |

3 | 87 | |

4 | 32 | |

5 | 65 - 26 | |

26 (move to next empty - 6) | Collision | |

6 | 26 |

**Q.5.** **Computer Organization**

In a certain machine data references constitute 40% of the mix, and that the ideal CPI of the pipelined machine is 1. Assume that the machine with structural hazards has a clock rate that is 1.05 times higher than the clock rate of machine without hazard. What is effective speedup with pipelining without structural hazard?

**Options**

1. 1.05

2. 1.3

3. 1

4. none of above

Ans: **2**

**Explanation:**

We major the average instruction time on both the machines to access speedup/slowdown.

Avg. instruction time = CPI * clock cycle time

Since there are no stalls/hazards, the average instruction time for ideal machine is the clock cycle times.

The average clock cycle time for the machine with 40% structural hazard is M

= (1+0.4 x 1) x Clock cycle time ideal/1.05 = 1.3 clock cycle time ideal

Machine without structural hazard is 1.3 times faster.

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