GATE 1998 CS Solution Keys

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

anon

February 4, 2013 at 3:05 PM

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)

Niharika

February 6, 2013 at 10:46 AM

1.1 – B
The probability of getting odd or even number each time is = 1/2
3C1(1/2)(1/2)^2 = 3/8

Niharika

February 6, 2013 at 11:17 AM

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”