CS Free MCQ Bank

[ Option A ]

A weighted code is a binary code in which each bit position has a fixed weight, and the value of a digit is obtained by adding the weights of the bits that are 1.
In the 8421 Code, the weights of the four bits are 8, 4, 2, and 1, and the decimal value is calculated directly from these weights.
E.g.:

• Decimal 5 in 8421 code is 0101, and its value is calculated as  0*8+1*4+0*2+1*1=5.
• Decimal 13 in 8432 code is 1101, and its value is calculated as 1*8+1*4+0*2+1*1=13.

BCD (Binary Coded Decimal) is also based on the 8421 code, so it is a weighted code as well.

Note:
In BCD, each decimal digit from 0 to 9 is represented using 4 bits. Only 10 combinations are valid from 0000 to 1001. The remaining combinations (1010 to 1111) are invalid. Therefore, BCD is strictly limited to digits 0 to 9.

In Excess-3 Code, we first add 3 to the decimal digit and then convert it into binary. For example, to represent decimal 4, we add 3 and get 7, whose binary form is 0111.

• Excess-3 is limited to decimal digits from 0 to 9, just like BCD.

Although Excess-3 follows a fixed procedure, the bit positions themselves do not have fixed weights, so it is treated as a Non-Weighted Code.

In Gray code, there are no positional weights at all. Its main feature is that consecutive numbers differ by only one bit. 
For example, decimal 2 is 010 in binary but 011 in gray code, and decimal 3 is 011 in binary but 010 in gray code.
RULE FOR GENERATING GRAY CODE:

• First of all, convert the given decimal into binary.
• The MSB bit of Gray code is the same as the MSB of binary.
• Each next Gray Bit is obtained by XOR (⊕) of the current binary bit and the previous binary bit, i.e., G_i = B_i⊕B_i-1.

For example, obtained the gray code for the number 13.
Binary Number : 13 : 1101

So Gray code for binary 1101 is 1011.

In Gray code, consecutive numbers (next higher number) differ by only one bit. Means only one bit position changes either from 0 to 1 or from 1 to 0. All other bits remain the same. This property is called the Unit Distance property.
E.g.:
BINARY VS GRAY (3-BIT NUMBERS):

• Gray code has no digit limit. Its range depends only on the number of bits used.
• In Gray Code, only one-bit changes between consecutive numbers, unlike binary code where multiple bits may change.

We cannot calculate the decimal value by adding bit weights, which shows that Gray Code is not a weighted code.
Here both Excess-3 and Gray Code are Non-Weighted Code, but the most standard and commonly expected answer in exams is GRAY CODE, as it is clearly and purely non-weighted.

[ Option B ]

In 2’s complement form, one bit is used as the sign bit and the remaining bits are used to represent the magnitude. For an n-bit 2’s complement number, the range of values that can be represented as -2ⁿ⁻¹ to +(2ⁿ⁻¹ -1).
Now, for 8 bits, -2⁷ to + (2⁷-1) = -128 to +127
So, the smallest number that can be stored is -128, and the largest number is +127.

[ Option C ]

In a 2’s complement representation, a negative number is obtained by taking the 1’s complement of its positive binary form and then adding 1.
First convert (147)₁₀ to binary and write in 12-bit representation.
(147)₁₀=(000010010011)₂
1’s complement of 000010010011 = 111101101100
2’s complement = 111101101100 + 1 = 111101101101
Finally, (-147)₁₀=(111101101101)₂

[ Option A ]

ASCII (American Standard Code for Information Interchange) is a character encoding scheme used to represent letters, digits, and special symbols in computers. In its original and standard form, ASCII defines 128 different characters.
To represent 128 different characters, we need a minimum number of bits such that  2ⁿ=128. Since 2⁷=128, 7 bits are sufficient to store one ASCII character. That is why standard ASCII uses 7 bits per character.

[ Option D ]

A number written with a subscript r means it is expressed in base (radix) r.
First remove the square root by squaring both sides and then convert both numbers into decimal form.
(224)_r=(13)²_r
Convert each side into decimal:
Left-hand side: (224)_r = 2r²+2r¹+4r⁰ = 2r²+2r+4
Right-hand side: (13)_r = 1r¹+3r⁰ = (r+3)
After squaring (r+3)² = r²+6r+9
Now the equation,
= 2r²+2r+4 = r²+6r+9
= 2r²+2r+4-r²-6r-9=0
= r²-4r-5=0
=(r−5)(r+1)=0
=r=5 or r=−1
The r can be either 5 or -1 but the radix cannot be negative and also the largest digit used is 4, so radix must be greater than 4.
Therefore, r=5.

[ Option A ]

To find the decimal equivalent of the hexadecimal operation (A10+B21)₁₆, we first convert each hexadecimal number into decimal form and then add the results.
(A10)₁₆ = 10*16²+1*16¹+0*16⁰=2560+16+0 = (2576)₁₀
(B21)₁₆ = 11*16²+2*16¹+1*16⁰=2816+32+1 = (2849)₁₀
Now, adding 2576+2849 = 5425.

[ Option D ]

Computers store real (floating-point) numbers using the IEEE 754 standard. In 32-bit single-precision floating-point representation, the total 32 bits are divided into three parts:

[ Option B ]

In the binary number system, the Least Significant Bit (LSB) is the rightmost bit. This bit plays an important role in determining whether a number is odd or even. If the LSB is 0, the number is even. If the LSB is 1, the number is odd.
Now consider any odd decimal number, such as 1, 3, 5, 7, or 9. When these numbers are converted into binary form, they always end with 1.

[ 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 ]

In 2’s complement representation, if the Most Significant Bit (MSB) is 1, the number represents a negative value. To find its decimal equivalent, we can use either of the following two correct methods.
In the first method, we subtract the unsigned binary value from 2ⁿ, where n is the total number of bits, and then place a negative sign before the result.
In the second, more commonly used method, we first find the 1’s complement of the binary number by inverting all bits, then add 1 to obtain the 2’s complement, and finally convert the result to decimal and assign a negative sign.
First, let us find the value of the divisor:
USING FIRST METHOD:
Given Number : (11111011)₂
This is an 8-bit number with MSB = 1, so it is negative.
256-251 = 5 : Value : -5
USING SECOND METHOD:
Given Number : (11111011)₂
1’s Complement : 00000100
2's Complement : 00000100 + 1 = (00000101)₂                  
(000000101)₂ = 5 : Value : -5
So, the question becomes, which option is divisible by -5? Remember, if a number is divisible by 5, it is also divisible by -5 and a number divisible by -5 must be a multiple of 5.

[ Option D ]

[ Option B ]

In 2’s complement, if the MSB is 0, the number is positive, and its value is the same as ordinary binary.
Here, in the given number (011010)₂ has MSB is 0. So, it is positive number.
Now, convert binary to hexadecimal. Group the bits into 4-bit nibbles. If needed pad with leading zeros.
(011010)₂ : (0001 1010)₂
(0001)₂ = (1)₁₆
(1010)₂ = (A)₁₆
So, (011010)₂ = (1A)₁₆

[ Option D ]

To find the remainder when (37)₁₆ is divided by (17)₁₆, it is easiest to first convert both hexadecimal numbers into decimal, perform the division, and then convert the remainder back into hexadecimal.
(37)₁₆ = (55)₁₀
(17)₁₆ = (23)₁₀
Now, 55/23 is 2 with a remainder of 9.
Finally, convert the remainder back into hexadecimal and the decimal 9 is written as 09 in hexadecimal.

[ Option A ]

For signed 2’s complement numbers, multiplication by 8 is done by left shifting the binary representation by 3 bits and keeping only the lower 16 bits. Applying this rule to (F87B)₁₆ gives (C3D8)₁₆.

[ Option C ]

To find the result of (10101)₂*(11101)₂ in hexadecimal form, we first convert both binary numbers into decimal, then multiply them, and finally convert the result into hexadecimal.

[ Option B ]

The main drawback of 1’s complement representation is that it has two different representations of zero, which can lead to confusion and complications in arithmetic operations.
In 1’s complement, a negative number is obtained by inverting all bits of its positive counterpart. Let us consider an 8-bit system.
Positive Zero (+0) : 00000000
Negative Zero (-0) : Take 1’s complement of +0 : 11111111
So, in 1’s complement:

• +0 = 00000000
• -0 = 11111111

Both represent the same value zero, but they have different binary patterns.

[ 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 D ]

To find the 8-bit 2’s complement representation of -100, we follow the standard 2’s complement procedure, step by step.

• First find binary for positive 100 in 8-bit.
• Then, find 1’s complement by inverting all bits.
• Then, add 1 to obtain the 2’s complement.

[ 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 A ]

While adding binary numbers, addition always starts from the Right Most Bit (LSB) and moves toward the Left Most Bit (MSB), just like in decimal addition.
If a carry remains after adding the MSB, it is written as an extra bit on the left, increasing the length of the result.

Binary Addition: 11010 + 01111

Finally, 11010+01111=101001

[ Option B ]

[ Option C ]

The binary conversion of integer part (234).
234=128+64+32+8+2 = (234)₁₀=(11101010)₂
The binary conversion of fractional part (0.125).

(0.125)₁₀=(0.001)₂
Finally, (234.125)₁₀=(11101010.001)₂

[ Option C ]

A weighted code is a binary code in which each bit position has a fixed weight, and the value of a digit is obtained by adding the weights of the digits.

• In Decimal Number System, the weights are powers of 10 ( …,1000, 100, 10, 1). 
• In Binary Number System, the weights are powers of 2 (…, 8, 4, 2, 1).
• The BCD code is also a weighted code because it follows the 8421-weighting system, where each bit has a fixed weight of 8, 4, 2, and 1.
• The Excess-3 code is a Non-Weighted Code. In this code, decimal digits are represented by first adding 3 to the digit and then converting it to binary.

[ Option D ]

Minuend:

• The minuend is the number from which another number is subtracted.
• It is the first number in a subtraction.

Subtrahend:

• The subtrahend is the number that is subtracted from the minuend.
• It is the second number in a subtraction.

1’S COMPLEMENT SUBTRACTION METHOD:
In this, the subtrahend is first converted into its 1’s complement and then added to the minuend. If an end-around carry is generated, it is added back to the result. The final sum obtained after adding the carry gives the correct subtraction result.
Given:
Minuend = (1101)₂
Subtrahend = (1010)₂
Step 1: Take 1’s complement of the subtrahend (1010) = (0101)
Step 2: Add it to the minuend 1101+ 0101 = 10010
Step 3: Add the end-around carry

• End-around carry = 1
• Remaining bits = 0010

Therefore, 0010+1 = (0011)₂
2’S COMPLEMENT SUBTRACTION METHOD:
In this, the subtrahend is first converted into its 2’s complement and then added to the minuend. If a carry is generated during the addition, it is discarded. The remaining sum obtained after discarding the carry gives the correct subtraction result.

• The 2’s complement subtraction is preferred in computers because it eliminates the need for end-around carry handling.

Given:
Minuend = (1101)₂
Subtrahend = (1010)₂
Step 1: Find 2’s complement of the subtrahend.

• 1’s complement of (1010)₂ : 0101
• Add 1 to get 2’s complement: 0101+1=0110

Step 2: Add it to the minuend: 1101 + 0110 = 10011
Step 3: Discard the carry, after discarding carry 1 the remaining result = (0011)₂

[ Option D ]

(432267)₁₀ = (?)₈
To convert a decimal number to octal, repeatedly divide by 8 and note the remainders.

Now, read the remainders from bottom to top.
(432267)₁₀=(1514213)₈

[ Option C ]

To solve the expression (10111)₂*(1110)₂, we first convert both binary numbers into their decimal equivalents to make the multiplication easy.
(10111)₂ = 1*16+0*8+1*4+1*2+1*1 = (23)₁₀
(1110)₂ = 1*8+1*4+1*2+0*1 = (14)₁₀
Next, we multiply these two decimal numbers, 23*14 = (322)₁₀. Now, this decimal result must be converted into hexadecimal form.
322/16 = 20 and remainder is 2
20/16 = 1 and remainder is 4
1/16 = 0 and remainder is 1
So, (322)₁₀ = (142)₁₆
Finally, (10111)₂*(1110)₂ = (142)₁₆

[ 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 solve the expression (111)₂*(101)₂, we first convert both binary numbers into their decimal equivalents to make the multiplication easy.
(111)₂=1*4+1*2+1*1 = 7
(101)₂=1*4+0*2+1*1 = 5
Next, we multiply these two decimal numbers, 7*5 = (35)₁₀
Finally, convert (35)₁₀ back into binary, which gives the (100011)₂.

[ Option B ]

In 2’s complement arithmetic, overflow occurs when:

• Two positive numbers are added and the result becomes negative.
• Two negative numbers are added and the result becomes positive.

For a 4-bit 2’s complement system, the representable range is: -8 to +7
In 4-bit 2’s complement representation, the Most Significant Bit (MSB) indicates the sign:

• If MSB is 0 then it is positive number.
• If MSB  is 1 then it is negative number.

The option (A) is 1111 + 1101

• 1111 = -1
• 1101 = -3

Why is 1111 equal to -1 and 1101 equal to -3 in 4-bit 2’s complement?
Since 1111 and 1101 has MSB = 1, it represents a negative number.
Given number : 1111
Find its magnitude using 2’s complement.

• Take 1’s complement : 1111 : 0000
• Add 1 : 0000+1 = 0001 (1 in decimal)
• Finally, attach the negative sign -1

Given number : 1101
Find its magnitude using 2’s complement.

• Take 1’s complement : 1101 : 0010
• Add 1 : 0010+1 = 0011 (3 in decimal)
• Finally, attach the negative sign -3

Sum = 1111 + 1101 = (-1+-3) = -4. The sum -4 is within range. So, no overflow occur.
The same procedure is applied for all.

[ Option D ]

Given,
(12x)₃=(123)_x
The left number is in base 3.
The right number is in base x.
Convert (12x)₃ to decimal. In base 3, the positional weights are 3², 3¹ ,3⁰.
(12x)₃ = 1*3²+2*3¹+x*3⁰
(12x)₃ = 9+6+x = 15+x
Convert (123)_x to decimal. In base x, the positional weights are x², x¹, x⁰.
(123)_x = 1*x²+2*x¹+3*x⁰
(123)_x = x²+2x+3
Now the equation is
15+x = x²+2x+3
= x²+x-12=0
= (x+4)(x-3)=0
So, x=-4 or x=3
Now,

• The x=-4 is not possible because base cannot be negative.
• The x=3 is not valid because in (123)_x, digit 3 is used, which is not allowed in base 3.

So, no valid value of x satisfies the given expression. Therefore, none of these is marked as correct answer.

[ Option B ]

For an n-bit 2’s complement number, the range of representable integers is determined using the formula : -2^n-1 to (2^n-1-1)
For an 8-bit number : -2⁷ to (2⁷-1) = -128 to 127
Hence, the smallest integer that can be represented is -128.

[ Option A ]

To find the 8-bit 2’s complement representation of -93, we follow the basic 2’s complement steps.
First, write 93 in binary using 8 bits.
(93)₁₀=(01011101)₂
Next, find the 1’s complement by inverting all bits and finally add 1 to 1’s complement.
1’s complement of 01011101 is 10100010.
Finally, 10100010 + 1 = (10100011)₂

[ Option C ]

In a 10-bit computer system that uses 2’s complement representation, a negative number is represented by taking the 2’s complement of its positive value.

• Write the positive number.
• Then take 1’s complement.
• Finally add 1 to 1’s complement.

Write +35 in binary : (35)₁₀=(100011)₂
Convert it to 10-bit binary : (35) = (0000100011)
1’s complement : 0000100011 : 1111011100
Add 1 : 1111011100+1 = 1111011101
Finally, (-35)₁₀ = (1111011101)₂

[ 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 ]

[ Option A ]

Here 47 is subtracted from 38. So, 38-47 = -9
As we know, negative numbers are stored in 2’s complement form in a computer system. Therefore, we need to represent -9 in 8-bit 2’s complement form.
Step 1: Write the binary equivalent of 9 in 8 bits:
(9)₁₀ = (00001001)₂
Step 2: Find the 2’s complement of 00001001:
1’s complement : 11110110
2’s complement : 11110110+1 = 11110111
Hence, the 8-bit 2’s complement representation of -9 is 11110111.