CS Free MCQ Bank

[ Option D ]

BCD (Binary-Coded Decimal) represents decimal digits 0 to 9 using 4-bit binary codes.

🔶 BCD Valid Range: 0000 to 1001

🔶 1011 = 11 (which is not a valid decimal digit in BCD)

[ Option D ]

[ Option A ]

[ Option B ]

In one’s complement representation, a negative number is obtained by inverting all the bits of its positive binary form. For (−8)₁₀, we first write +8 in 5-bit binary, which is (01000)₂. Now, by inverting each bit (all 0s become 1s and all 1s become 0s), we get (10111)₂. Hence, the one’s complement representation of (−8)₁₀ is (10111)₂.

[ Option B ]

Binary subtraction works similarly to decimal subtraction, with borrowing done in base 2. When we subtract (1001)₂ from (1100)₂, we first convert them to decimal to understand the operation, 12 minus 9 equals 3. Converting 3 back to binary gives (0011)₂. Therefore, the result obtained from subtracting (1001)₂ from (1100)₂ is (0011)₂.

[ Option B ]

[ Option B ]

[ Option C ]

To solve this base arithmetic problem, we first understand that numbers in different bases can be converted to decimal (base 10) to simplify calculations. By converting each term to decimal, we can form an algebraic equation in terms of the unknown base x and then solve for it using standard algebraic methods.

Alternatively, since the bases involved are small integers, a trial-and-error method can be used to test reasonable values of x until the sum matches the decimal equivalent of the right-hand side.

Method 1: Algebraic / Decimal Conversion Method:
Convert each number to decimal:

• (146)_x = 1⋅x²+4⋅x+6 = x² + 4x + 6
• (313)_x−2 = 3⋅(x−2)²+1⋅(x−2)+3 = 3x² − 11x + 13
• (246)₈=2⋅8² + 4⋅8 + 6 = 166

Form the equation in decimal:

(x² + 4x + 6) + (3x² − 11x + 13) = 166
4x² − 7x + 19 = 166
4x² − 7x − 147 = 0
When solving this quadratic equation, we found value of x is 7.
Method 2: Trial-and-Error
Understand the Base Involved:

• The base x must be greater than or equal to 7 because the digit '6' appears in (146)_x, so x>6.
• x−2 must be greater than or equal to 4 since the digit '3' appears in (313)_x−2. Hence, x−2>3 : x>5.

Check possible integer values for x:
From the given option 5,6,7,9, try only x=7,9 (because x must be >6), by converting each term to decimal and verify if the sum equals the decimal value of (246)₈ (which is 166).
 

[ Option B ]

To compute the 8’s complement of a number, we use the (r−1)’s complement method, where r is the base. First, we find the (r−1)’s complement by subtracting each digit of the number from (r−1), and then we add 1 to the result to get the r’s complement.

For example, consider the octal number (7650)₈. Here, the base r=8, so r−1=7. To find the 7’s complement of 7650, we subtract each digit from 7 individually.

• 7 : 7 - 7 = 0
• 6 : 7 - 6 = 1
• 5 : 7 - 5 = 2
• 0 : 7 - 0 = 7

Thus, the 7’s complement of (7650)₈ is 0127. Next, to obtain the 8’s complement, we add 1 to the 7’s complement, 0127+1= 0130.

Note: The addition is performed according to the base. Here, since it is base 8 (octal), each digit is added modulo 8, and any carry is handled within octal arithmetic.

[ Option D ]

[ Option B ]

The rightmost bit of a binary number is known as the Least Significant Bit (LSB). It represents the smallest value in binary positional notation, corresponding to 2⁰=1.

The opposite is the Most Significant Bit (MSB), which is the leftmost bit and represents the largest value place in the binary number, corresponding to 2^n-1.

[ Option A ]