I have posted these solution keys here because all the books, available in the market, have lots of wrong solutions. Most of the students, like me, find themselves in dilemma and waste a lot of time in finding the correct keys. These solution keys are those which I had corrected while preparing for GATE Computer Science 2012 Exam. You will find most of these solution keys to be correct, except few for which I could not find the correct solution. Still I suggest you that please keep your mind open while using these keys.

## GATE 1998 CS Solution Keys

1.1. C
1.2. B 1.3. B 1.4. B 1.5. C 1.6. B 1.7. C |
1.8. C
1.9. D 1.10. A 1.11. B 1.12. C 1.13. D 1.14. C |
1.15. D
1.16. C 1.17. B 1.18. C 1.19. C 1.20. A 1.21. A |
1.22. B
1.23. D 1.24. C 1.25. C 1.26. C 1.27. C 1.28. B |
1.29. B
1.30. B 1.31. B 1.32. B 1.33. D 1.34. D 1.35. C |

2.1. D
2.2. B 2.3. B 2.4. D |
2.5. B
2.6. A 2.7. C 2.8. C |
2.9. D
2.10. A 2.11. A 2.12. C |
2.13. C
2.14. A 2.15. D 2.16. B |
2.17. B
2.18. A 2.19. A 2.20. C |

## Comments 7

1.17

342 in binary is 11100010

8085 numbers are in 2’s complement.

so 11100010 in 2’s complement is 00011110 = -30

so answer should be (d)

1.1 – B

The probability of getting odd or even number each time is = 1/2

3C1(1/2)(1/2)^2 = 3/8

1.5 – A,

“only if” are the only times when “if” does not introduce an antecedent, but rather a consequent.

so ‘I stay only if you go” becomes “If I stay then you go”

I->Y (I – I stay , Y- you go)

now converse is

Y->I

means If you go I stay or “I stay if you go”

ans for 1.24 is B

given combination of pre-order+in-order

OR

post-order+in-order

the tree can be uniquely constructed

2.6 – B,

for example a*b* is regular but subset of it a^nb^n is not regular

2.15 – C

faster access to non-local variables using an array of pointers to activation records,

the array is called a display.

2.20 – B,

As size of INT is 4..