1. Write IEEE floating point representation of the following decimal number 3.75

3.75 = 011.11 011.11 = 01.111 x 2^1 → exp – 127 = 1 exp = 128 = 10000000 → 1.111 x 2^1 = 11100000000000000000000 → 3.75 = positive = 0 = 01000000011100000000000000000000 #

2. Write IEEE floating point representation of the following decimal number -55 23/64

110111010111 = 1.10111010111 x 2^5 → exp – 127 = 5 exp = 5 + 127 = 132 = 10000100 → fraction = 10111010111000000000000 → – 55 23/64 = negative = 1 = 11000010010111010111000000000000 #

3. Write IEEE floating point representation of the following decimal number 64000

32768 16384 8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1 1 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 01111101000000000 = 01.111101000000000 x 2^15 → exp – 127 = 15 exp = 15 + 127 = 142 = 10001110 → Sign = positive = 0 = 01000111011110100000000000000000 #

4. Write the decimal equivalent for this IEEE floating point number 01000000000000000000000000000000

→ Sign = positive = 0 → exp = 10000000 = 128 → Fraction = 00000000000000000000000000 = 0 → N = (-1)^s x 1. fraction x 2^(exp – 127) = (-1)^0 x 1.0 x 2^(128 – 127) = 1 x 1.0 x 2^(1) = 1 x 2 = 2 #

5. Write the decimal equivalent for this IEEE floating point number 11000001100010000000000000000000

→ Sign = negative = 1 → exp = 10000011 = 131 → Fraction = 00010000000000000000000 = 0.0625 → N = (-1)^s x 1.fraction x 2^(exp – 127) = (-1)^1 x 1.0625 x 2^(131 – 127) = -1 x 1.0625 x 2^4 = -1.0625 x 16 = -17 #

6. Write the decimal equivalent for this IEEE floating point number 01111111100000000000000000000000

→ Sign = positive = 0 → exp = 11111111 = 255 → Fraction = 00000000000000000000000000 = 0 → N = + - ∞ = + ∞ #

7. Write the decimal equivalent for this IEEE floating point number 11000000010010000000000000000000

→ Sign = negative = 1 → exp = 10000000 = 128 → Fraction = 10010000000000000000000 = 0.5625 → N = (-1)^s x 1.fraction x 2^(exp – 127) = (-1)^1 x 1.5625 x 2^(128 – 127) = -1 x 1.5625 x 2^1 = -3.125 #

8. Convert the following unsigned binary number 1101000110101111 to hexadecimal.

1101 0001 1010 1111 = 13 1 10 15 = D 1 A F = D1AF #

9. Convert the following unsigned binary number 0011111 to hexadecimal.

001 1111 = 0001 1111 = 1 F = 1F #

10. Convert the following unsigned binary number 1 to hexadecimal.

= 0001 = 1 #

11. Convert the following unsigned binary number 1110110110110010 to hexadecimal.

1110 1101 1011 0010 = 1110 1101 1011 0010 = 14 13 11 2 = E D B 2 = EDB2 #

12. Convert the following hexadecimal number x10 to binary.

10 = 1 0 = 0001 0000 = 00010000 #

13. Convert the following hexadecimal number x801 to binary.

801 = 8 0 1 = 1000 0000 0001 = 100000000001 #

14. Convert the following hexadecimal number xF731 to binary.

F731 = F 7 3 1 = 1111 0111 0011 0001 = 1111011100110001 #

15. Convert the following hexadecimal number x0F1E2D to binary.

0F1E2D = 0 F 1 E 2 D = 0000 1111 0001 1110 0010 1101 = 000011110001111000101101 #

16. Convert the following hexadecimal number xBCAD to binary.

BCAD = B C A D = 1011 1100 1010 1101 = 1011110010101101 #

17. Convert the following hexadecimal representation of 2’s complement binary number xF0 to decimal number.

F0 = F 0 = 1111 0000 → negation = 0000 1111 + 1 = 0001 0000 #

18. Convert the following hexadecimal representation of 2’s complement binary number x7FF to decimal number.

7FF = 7 F F = 0111 1111 1111 → (0 x 2)^11 + (1 x 2)^10 + (1x 2)^9 + (1 x 2)^8 + (1 x 2)^7 + (1 x 2)^6 + (1 x 2)^5 + (1 x 2)^4 + (1 x 2)^3 + (1 x 2)^2 + (1 x 2)^1 + (1 x 2)^0 = 2047 #

19. Convert the following hexadecimal representation of 2’s complement binary number x16 to decimal number.

16 = 0001 0110 = 2 2 = 22 #

20. Convert the following hexadecimal representation of 2’s complement binary number x8000 to decimal number.

8000 = 8 0 0 0 = 1000 0000 0000 0000 → negation = 0111 1111 1111 1111 + 1 1000 0000 0000 0000 → (1 x 2)^15 + (0 x 2)^14 + (0 x 2)^13 + (0 x 2)^12 + (0 x 2)^11 + (0 x 2)^10 + (0 x 2)^9 + (0 x 2)^8 + (0 x 2)^7 + (0 x 2)^6 + (0 x 2)^5 + (0 x 2)^4 + (0 x 2)^3 + (0 x 2)^2 + (0 x 2)^1 + (0 x 2)^0 = - 32768 #

21. Convert the following decimal number 256 to hexadecimal representation of 2’s complement number.

256 → 256/2 = 128 Reminder = 0 128/2 = 64 Reminder = 0 64/2 = 32 Reminder = 0 32/2 = 16 Reminder = 0 16/2 = 8 Reminder = 0 2/2 = 1 Reminder = 0 1/2 = 0 Reminder = 1 → 0001 0000 0000 = 1 0 0 = 100 #

22. Convert the following decimal number 111 to hexadecimal representation of 2’s complement number.

111 → 111/2 = 128 Reminder = 1 55/2 = 64 Reminder = 1 27/2 = 13 Reminder = 1 13/2 = 6 Reminder = 1 6/2 = 3 Reminder = 0 3/2 = 1 Reminder = 1 1/2 = 0 Reminder = 1 → 0110 1111 = 6 F #

23.Convert the following decimal number -44 to hexadecimal representation of 2’s complement number.

-44 44 = 0010 1100 → negation = 1101 0011 + 1 = 1101 0100 = 4 D #

24. Perform the following addition x025B + x26DE. The corresponding 16-bit binary number is in 2’s complement notation. Provide your answer in hexadecimal.

025B + 26DE → 0 2 5 B + 2 6 D E = 2 9 3 9 #

25. Perform the following addition x7D96 + xF0A0. The corresponding 16-bit binary number is in 2’s complement notation. Provide your answer in hexadecimal.

7D96 + F0A0 → 7 D 9 6 + F 0 A 0 = 16 E 3 6 #

26. Perform the following addition xA397 + xA35D. The corresponding 16-bit binary number is in 2’s complement notation. Provide your answer in hexadecimal.

→ A 3 9 7 = 1010 0011 1001 0111 → A 3 5 D = 1010 0011 0101 1101 → 1010 0011 1001 0111 + 1010 0011 0101 1101 = 10100 0110 1111 0100 = 146F4 #

27. Perform the following addition x7D96 + x7412. The corresponding 16-bit binary number is in 2’s complement notation. Provide your answer in hexadecimal.

→ 7 D 9 6 = 0111 1101 1001 0110 → 7 4 1 2 = 0111 0100 0001 0010 → 0111 1101 1001 0110 + 0111 0100 0001 0010 = 1111 0001 1010 1000 = F1A8 #

28. What is the hexadecimal representation of 675.625?

675 = 0010 1010 0011 0.625 = 0.1010 → 0010 1010 0011.1010 1.0101000111010 → exp – 127 = 9 exp = 9 + 127 = 136 = 10001000 → fraction = 01010001110100000000000 → 0100 0100 0010 1000 1110 1000 0000 0000 = 9428E800 #

29. Answer the following arithmetic operation in binary 01010111 + 11010111

→ 01010111 + 11010111 = 00101110 = 00101110 #

30. Answer the following arithmetic operation in binary 1111 + 100001111 + 0111

→ 111111111 (-1) + 100001111 (-241) 1100001110 + 0000000111 (7) 1100010101 (-235) = 1100010101 #

31. Find out the answer for the following arithmetic operation x3F0 + 12348. State your answer in any number system given in the question.

3F0 + 1234 → 3F0 = 0011 1111 0000 → 1234 = 0010 1001 1100 → 0011 1111 0000 + 0010 1001 1100 0110 1000 1100 = 68C #

32. Find out the answer for the following arithmetic operation 2345 + 1234. State your answer in any number system given in the question.

234 + 123 → [(2 x 5)^2 + (3 x 5)^1 + (4 x 5)^0] + [(1 x 4)^2 + (2 x 4)^1 + (3 x 4)^0] = (50 + 15 + 4) + (6 + 8 + 3) = 69 + 27 = 96 #