Number Representation - Practice MCQs for CCAT

Correct Answer: C - 1101

Sign-magnitude: MSB=1 (negative), remaining=5(101). -5 = 1101.

Correct Answer: B - 1001

1's complement: invert all bits. 0110 -> 1001.

Correct Answer: B - 1010

2's complement = 1's complement + 1 = 1001 + 1 = 1010.

Correct Answer: D - 1001

BCD: each decimal digit in 4 bits. 9 = 1001.

Correct Answer: A - Only one bit changes between adjacent values

Gray code: adjacent values differ by only 1 bit.

Correct Answer: C - 7 or 8

ASCII uses 7 bits (standard) or 8 bits (extended).

Correct Answer: A - 1000

Excess-3: add 3 to digit then convert. 5+3=8=1000.

Correct Answer: D - Sign, exponent, mantissa

Floating point: sign bit + exponent + mantissa.

Correct Answer: A - 32 bits

IEEE 754 single precision = 32 bits.

Correct Answer: B - Error detection

Parity bit detects single-bit errors in data.

Correct Answer: C - 10

1010 = 1×8 + 0×4 + 1×2 + 0×1 = 8 + 0 + 2 + 0 = 10

Correct Answer: A - 11111011

5 = 00000101. 1's complement = 11111010. Add 1 for 2's complement = 11111011.

Correct Answer: B - -128 to +127

For n bits, 2's complement range is -2^(n-1) to +2^(n-1)-1. For 8 bits: -128 to +127.

Correct Answer: A - The sign of the number

In sign-magnitude, the Most Significant Bit (MSB) represents sign: 0 for positive, 1 for negative.

Correct Answer: C - Only one representation of zero

2's complement has only one representation of zero, while 1's complement has +0 and -0.

Correct Answer: D - F0

11110000: 1111 = F, 0000 = 0. So 11110000 = F0 in hexadecimal.

Correct Answer: C - 10100101

A = 1010, 5 = 0101. So A5 = 10100101 in binary.

Correct Answer: C - When result needs more bits than available

Overflow occurs when the result of an arithmetic operation exceeds the range that can be represented with the available bits.

Correct Answer: B - Binary Coded Decimal - 4 bits per decimal digit

BCD uses 4 binary bits to represent each decimal digit (0-9), making decimal calculations easier.

Correct Answer: B - 01000111

In BCD: 4 = 0100, 7 = 0111. So 47 = 01000111.

Correct Answer: D - 72

Group binary digits in 3s from right: 111 010 = 7 2 = 72 in octal.

Correct Answer: B - Very large or very small real numbers

Floating point allows representation of very large or very small real numbers using sign, exponent, and mantissa.

Correct Answer: A - 8

IEEE 754 single precision (32-bit): 1 sign bit, 8 exponent bits, 23 mantissa bits.

Correct Answer: C - 0101

1's complement is obtained by inverting all bits. 1010 becomes 0101.

Correct Answer: D - Universal character encoding supporting many languages

Unicode is a universal character encoding standard that supports characters from virtually all writing systems.

Correct Answer: D - 01001101

The 1's complement is obtained by flipping all bits: 1→0, 0→1. So 10110010 becomes 01001101.

Correct Answer: B - 11110100

2's complement = 1's complement + 1. 1's complement of 00001100 = 11110011. Adding 1: 11110011 + 1 = 11110100.

Correct Answer: B - The sign of the number

In signed magnitude representation, the Most Significant Bit (MSB) represents the sign: 0 for positive and 1 for negative. The remaining bits represent the magnitude.

Correct Answer: B - -128 to +127

In n-bit 2's complement, the range is -2^(n-1) to 2^(n-1)-1. For 8 bits: -2⁷ to 2⁷-1 = -128 to +127.

Correct Answer: D - 0100 0111

In BCD, each decimal digit is represented by its 4-bit binary equivalent. 4 = 0100, 7 = 0111. So 47 in BCD is 0100 0111.

Correct Answer: C - 1010 (10)

BCD uses 4 bits to represent decimal digits 0-9. The 4-bit patterns 1010 (10) through 1111 (15) are invalid in BCD as they don't correspond to any single decimal digit.

Correct Answer: A - 11111011

+5 in binary = 00000101. 1's complement = 11111010. 2's complement = 11111010 + 1 = 11111011. So -5 in 8-bit 2's complement is 11111011.

Correct Answer: D - Only one representation for zero

2's complement has only one representation for zero (00000000), while 1's complement has two (+0 = 00000000, -0 = 11111111). This simplifies arithmetic operations and comparison.

Correct Answer: C - 255

In unsigned representation, all 8 bits contribute to the magnitude. 11111111 = 2⁸ - 1 = 255.

Correct Answer: B - When the result exceeds the representable range

Overflow occurs when the result of an arithmetic operation exceeds the range that can be represented with the given number of bits. In 2's complement, it occurs when adding two numbers of the same sign produces a result of the opposite sign.

Correct Answer: D - When carry into MSB differs from carry out of MSB

Overflow in 2's complement is detected when the carry into the MSB (sign bit) differs from the carry out of the MSB. This indicates that the sign of the result is incorrect.

Correct Answer: D - 0111

Excess-3 (XS-3) code adds 3 to the decimal digit before converting to binary. 4 + 3 = 7 = 0111. So Excess-3 representation of 4 is 0111.

Correct Answer: C - 010

Gray code for 3: Binary of 3 = 011. Gray code conversion: MSB stays same (0), each subsequent bit = XOR of adjacent binary bits: 0⊕1=1, 1⊕1=0. Gray code = 010.

Correct Answer: A - Only one bit changes between consecutive numbers

In Gray code, only one bit changes between consecutive values. This property is crucial in applications like rotary encoders and Karnaugh maps to avoid glitches during transitions.

Correct Answer: A - 652

The 9's complement of a decimal number is obtained by subtracting each digit from 9. 9-3=6, 9-4=5, 9-7=2. So 9's complement of 347 is 652.

Correct Answer: A - 653

10's complement = 9's complement + 1. 9's complement of 347 = 652. Adding 1: 652 + 1 = 653. So the 10's complement is 653.

Correct Answer: B - 8

IEEE 754 single precision uses 32 bits total: 1 bit for sign, 8 bits for exponent (biased by 127), and 23 bits for mantissa (fraction).

Correct Answer: B - To extend the bit width while preserving the signed value

Sign extension increases the number of bits in a signed number while preserving its value. The sign bit (MSB) is replicated to fill the new higher-order bits. For example, 4-bit 1011 (-5) sign-extends to 8-bit 11111011 (-5).

Correct Answer: A - 1's complement

In 1's complement arithmetic, if a carry is generated out of the MSB during addition, it must be added back to the LSB (end-around carry) to get the correct result. This is not needed in 2's complement.

Correct Answer: C - BA

Group the binary number into 4-bit nibbles from right: 1011 1010. Convert each: 1011 = B (11), 1010 = A (10). So 10111010 in hex = BA.

Correct Answer: A - 7

Standard ASCII uses 7 bits to represent 128 characters (0-127), including uppercase/lowercase letters, digits, punctuation, and control characters. Extended ASCII uses 8 bits for 256 characters.

Correct Answer: D - Overflow has occurred

When two positive numbers (MSB = 0) are added and the result has MSB = 1 (appears negative), overflow has occurred. The true result is too large to be represented in the given number of bits.

Correct Answer: A - -15

MSB is 1, so the number is negative. To find magnitude: take 2's complement of 11110001. 1's complement = 00001110, add 1 = 00001111 = 15. So the value is -15.

Correct Answer: D - 127

IEEE 754 single precision uses a bias of 127 for the 8-bit exponent field. The actual exponent = stored exponent - 127. This allows representation of both positive and negative exponents.

Correct Answer: C - 1001 0110

In packed BCD, two decimal digits are stored in one byte (8 bits). 9 = 1001, 6 = 0110. So packed BCD of 96 = 1001 0110.