CSC 205 (201A) – Computer Organization Mid Term Exam

Q1. Anything that can be done with software can also be done with hardware, and anything that can be done with hardware can also be done with software.

True
False

Q2. The 4 generally accepted generations of computers are:

(1) Vacuum tubes, (2) transistors, (3) Integrated circuits, and (4) Very Large Scale Integration (VLSI)
(1) Abacus, (2) Turing Machines, (3) the ENIAC, and (4) IBM mainframes
(1) The Analytical Engine, (2) ENIAC, (3) mainframes, and (4) the IBM PC
(1) the Pascalene, (2) Difference engine, (3) Punched cards (Hollerith), and (4) Artificial Intelligence

Q3. The main components of a computer are:

A processor (CPU), a memory (RAM), and an Input/Output mechanism
The hard disk drive, the motherboard, and the processor (CPU)
Keyboard, monitor, and mouse
None of the these

Q4. A nanosecond is one billionth of a second.

True
False

Q5. If A and B are two 8-bit numbers represented in 2’s complement, adding A and B results in an overflow condition. A = 01001010, B = 11100010

True
False

Q6. Adding the 2 8-bit unsigned binary numbers shown 00101110 and 10100110 yields which of the following sums

01101100
11010100
01000001
10011001

Q7. The negative decimal (base 10) integer value of -27 is represented in 8-bit 2’s complement as:

11100100
10011011
11100101
00011011

Q8. To represent real binary numbers in floating point notation, we use a 14-bit format with 1 bit for the sign, 5 bits for the exponent, and 8 bits for the significand. Assuming an excess-16 exponent bias, and no imlied 1 bit to the left of the radix point, the binary number +11.0010101111 is represented as

00001011001010
01001011001011
01001011001010
11001011001010

Q9. Convert -123 to 8-bit 2’s complement notation

10000101
00000101
11111011
10000100

Q10. The positive decimal number 24 is represented in 8-bit 2’s complement as

111001111
00011000
00000000
11110000

Q11. The positive decimal number 185 cannot be represented as a binary number in an 8-bit 2’s complement format

True
False

Q12. Convert the decimal number 12.25 to binary (precision of 8 positions to the right of the radix point.

Cannot be represented
01010011000100
1100.01000000
None of these

Q13. To represent real binary numbers in floating point notation, we use a 14-bit format with 1 bit for the sign, 5 bits for the exponent, and 8 bits for the significand. Assuming an excess-16 exponent bias, and no implied 1 bit to the left of the radix point, the negative binary (real) number -0.0000100101

10110010010100
11110010010100
11010010010100
00110010010100

Q14. Convert the unsigned binary string to hexadecimal 010001110001100

21614
None of these
C832
238C

Q15. Convert the following octal (base 8) number to binary 3572

011101111010
100010000101
1110111110
None of these

Q16. Convert from hexadecimal to decimal (assume an unsigned integer number) 2F

Q17. Given the truth table shown below, derive the sum-of-products boolean expression (unsimplified) representing the output Q.

x y z Q

0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1

(x+y+z)(x+y’+z’)(x’+y’+z’)
(x’y’z’ )(x’yz)(xyz)
xyz + xyz + xyz
x’y’z’ + x’yz + xyz

Q18. Given the truth table shown below, derive the product-of-sums boolean expression (unsimplified) representing the output Q

x y z Q

0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

(x’y’z)(x’yz’)
(x’+y’+z)(x’+y+z’)
(x’y’+z)(x’y+z’)
(x+y+z’)(x+y’+z)

Q19. Simplify the following boolean expression F9x,y,z) = x’yz + x’yz’ + xz

yz’
x’y + xz
xy’ + x’z’
x

Q20. Using DeMorgan’s Law, find the complement for the boolean function (not simplified) F(x,y,z) = x(y’ + z)

x’y + z’
x + y’
(x + yz)’
x’ + yz’