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

3.75 = 011.11 =01.111 x 2^1 exponen - 127 = nilai anjakan exp = 1 + 127 exp = 128 exp = 10000000 fraction = 11100000000000000000000 3.75 = 01000000011100000000000000000000

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

-55 23/64 = -55.359375 -55 23/64 = 110111.010111 = 1.10111010111 x 2^5 exp - 127 = nilai anjakan exp = 5 + 127 exp = 132 = 10000100 fraction = 10111010111000000000000 -55 23/64 = 11000010010111010111000000000000

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

64000 = 1000000 64000.0 = 1000000.0 = 1.0 x 2^6 exp - 127 = nilai anjakan exp = 6 + 127 exp = 133 = 10000101 fraction = 00000000000000000000000 64000 = 01000010100000000000000000000000

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

0 1000000 00000000000000000000000 exp = 128 exp - 127 = nilai anjakan 128 - 127 = 1 128 x 2^1

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

N = (-1)s x 1.fraction x 2 (exp-127) 11000001100010000000000000000000 sign= 1 fraction = 0.03125 exp= 10000011 = 131 = 131 - 127 = 4 N =(-1)1 x 1.03215 x 2(4) = -16.5

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

01111111100000000000000000000000 sign = 0 fraction = 0 exp = 11111111 = 255 = 255 - 127 = 128 N= (-1)s x 1.fraction x 2exp-127 N = (-1)0 x 1.0 x 2(128) =2^128

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

1100000001001000000000000000000 sign = 1 fraction = 0.53215 exp = 128 = 128 -127 = 1 N= (-1)s x 1.fraction x 2exp-127 N=(-1)^1 x 1.53215 x 2^1 N= -3.0643

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

1101 0001 1010 1111 = D 1 A F = xD1AF

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

0001 1111 = 1 15 = 1 F =x1F

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 =14 13 11 2 = xEDB2#

12. Convert the following hexadecimal number x10 to binary.

x10 =0001 0000 =00010000#

13. Convert the following hexadecimal number x801 to binary.

x801 =1000 0000 0001 =100000000001#

14. Convert the following hexadecimal number xF731 to binary.

xF731 =1111 0111 0011 0001 =1111011100110001#

15. Convert the following hexadecimal number x0F1E2D to binary.

x0F1E2D = 15 1 14 2 13 = 0000 1111 0001 1110 0010 1101 =11110001111000101101#

16. Convert the following hexadecimal number xBCAD to binary.

xBCAD =11 12 10 13 =1011 1100 1010 1101 =1011110010101101#

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

F 0 =1111 0000 0000 1111 + 1 = 0001 0000 = -16 #

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

7 F F =0111 1111 1111 =(0X2^11) + (1X2^10) + (1X2^9) + (1x2^8) + (1X2^7) + (1X2^6) + (1X2^5) + (1X2^4) + (1X2^3) + (1X2^2) + (1X2^1) + (1X2^0) =2047#

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

x16 = 0001 0110 = (1 x 2^4) + (0 x 2^3) + (1 x 2^2) + (1 x 2^1) + (0 x 2^0) = 22 #

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

x8000 = 1000 0000 0000 0000 2) 0111 1111 1111 1111 = (1 x 2^14) + (1 x 2^13) + (1 x 2^12) + (1 x 2^11) + (1 x 2^10) + (1 x 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) = 32768 = -32768 #

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

=256 256/2 = 128 Remainder = 0 128/2 = 64 Remainder = 0 64/2 = 32 Remainder = 0 32/2 = 16 Remainder = 0 16/2 = 8 Remainder = 0 8/2 = 4 Remainder = 0 4/2 = 2 Remainder = 0 2/2 = 1 Remainder = 0 1/2 = 0 Remainder = 1 =0001 0000 0000 = 1 0 0 =100(base 16) #

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

=111 111/2 = 55 Remainder = 1 55/2 = 27 Remainder = 1 27/2 = 13 Remainder = 1 13/2 = 6 Remainder = 1 6/2 = 3 Remainder = 0 3/2 = 1 Remainder = 1 1/2 = 0 Remainder = 1 =0110 1111 = 6F(base 16) #

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

32 16 8 4 2 1 1 0 1 1 0 0 =0010 1100 =1101 0011 + 1 1101 0100 = 13 4 =D4

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 = x2939 #

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 = x16E36 #

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

1010 0011 1001 0111 +1010 0011 0101 1101 = 10100 0110 1111 0100 =x146F4#

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

7D96= 0111 1101 1001 0110 7412= 0111 0100 0001 0010 0111 1101 1001 0110 +0111 0100 0001 0010 = 1111 0001 1010 1000 =xF1A8 #

28. What is the hexadecimal representation of 675.625?

675.625 675=001010100011 0625=0.1010 =001010100011.1010 =001.0101000111010 x 2^1 exp-127=9 exp=136

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

01010111 + 11010111 =00101110#

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

1111111111 + 100001111 =1100001110 +0000000111 =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 = 0011 1111 0000 =1234(base 8) = 001 011 100 0011 1111 0000 +0010 1001 1100 = 0110 1000 1100 =x68C#

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

=2345 + 1234 =[(2x52) + (3x51) + (4x50)] + [(1x42) + (2x41) + (3x40)] =(50+15+4) + (6+8+3) = 69+27 = 96(base 10) #