Number Systems, Binary Arithmetic and Data Representation Practice Questions for GATE CS

Solve 79+ Number Systems, Binary Arithmetic and Data Representation practice questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Number Systems and Binary Arithmetic

Chapter Journey

Step 1: Radix Conversion

The foundation of how computers count. Mastering integer conversions across any base.

Step 2: Binary Fractions

Extending the logic to the fractional world. Handling precision and repeating fractions.

Step 3: Two's Complement

The hardware standard for negative numbers. Detecting when arithmetic exceeds limits.

The Core Idea: Radix and Positional Weight

The Anatomy of a Number System

Every system is defined by its radix ( $r$ ), which dictates the available digits ${0, 1, \dots, r - 1}$ .

d_{2}

d_{1}

d_{0}

r^{2}

r^{1}

r^{0}

(N)_{r} = i = 0 \sum n d_{i} r^{i}

Positions are counted from right to left, starting at $i = 0$ .

Number Systems, Binary Arithmetic and Data Representation: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Digital Logic MCQ

Assertion (A): In a 4-bit 2's complement system, adding $(- 1)$ and $(- 2)$ does not result in an overflow.

Reason (R): The addition produces a carry out of 1 from the most significant bit, which indicates an overflow in signed arithmetic.

A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is NOT the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: This is a construction question testing the precise definition of overflow and the common trap of confusing carry-out with overflow.

Step 1: Evaluate Assertion (A). $(- 1)_{10} = 111 1_{2}$ , $(- 2)_{10} = 111 0_{2}$ .

Step 2: Add them: $1111 + 1110 = 11101$ . Discard the carry out. The 4-bit result is $110 1_{2}$ , which is $(- 3)_{10}$ .

Step 3: Since $- 3$ is within the 4-bit range $[- 8, 7]$ , no overflow occurred. Assertion (A) is TRUE.

Step 4: Evaluate Reason (R). The addition does produce a carry out of 1. However, in signed 2's complement arithmetic, a carry out does NOT indicate overflow. Overflow is determined by the sign bits or XOR of carries. Thus, Reason (R) is FALSE.

Answer: A is true, but R is false.

Question 2 · Digital Logic MCQ

Which of the following statements correctly describes the hardware operation for subtracting an $n$ -bit number $B$ from $A$ in 2's complement arithmetic?

A.
Add $A$ and $B$ , then invert the result.
B.
Add $A$ to the 1's complement of $B$ and keep the carry out.
C.
Add $A$ to the 2's complement of $B$ and discard any carry out of the sign bit.
D.
Subtract $B$ from $A$ bit-by-bit and borrow from the sign bit.

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question testing the constraints of unified hardware logic for subtraction in 2's complement.

Step 1: Recall that subtraction $A - B$ is performed by adding $A$ to the 2's complement of $B$ .

Step 2: The 2's complement of $B$ is found by inverting the bits of $B$ and adding 1.

Step 3: After the addition is complete, any carry out of the most significant bit (sign bit) must be discarded in 2's complement arithmetic.

Step 4: Evaluate the options. Option C perfectly matches this exact procedure.

Answer: Add $A$ to the 2's complement of $B$ and discard any carry out of the sign bit.

Question 3 · Digital Logic MCQ

When converting the base-4 number $(132)_{4}$ to base 5, the process involves an intermediate conversion to base 10. How many division steps are required to convert this base-10 value into base 5?

A.
2
B.
3
C.
4
D.
5

Correct Answer:

Step-by-Step Solution

Key idea: This is a step-counting question based on the cross-base conversion method.

Step 1: Convert $(132)_{4}$ to base 10.

$(1 \times 4^{2}) + (3 \times 4^{1}) + (2 \times 4^{0}) = 16 + 12 + 2 = 30$ .

Step 2: Convert 30 to base 5 using repeated division.

Step 3: Count the division steps until the quotient is 0.

Step 1: $30 \div 5 = 6$ , remainder 0.
Step 2: $6 \div 5 = 1$ , remainder 1.
Step 3: $1 \div 5 = 0$ , remainder 1.

Step 4: The quotient is now 0, so we stop. There are exactly 3 division steps.

Answer: 3.

Question 4 · Digital Logic MCQ

When converting the base-4 number $(102)_{4}$ to base 5, the process requires routing through base 10. What is the minimum number of division-by-5 steps required to complete the base-10 to base-5 conversion?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: This is a contradiction question that requires performing the conversion and carefully counting the steps to avoid overcounting.

Step 1: Convert $(102)_{4}$ to base 10.

$(1 \times 4^{2}) + (0 \times 4^{1}) + (2 \times 4^{0}) = 16 + 0 + 2 = 18$ .

Step 2: Convert 18 to base 5 using repeated division.

Step 3: Count the division steps until the quotient is 0.

Step 1: $18 \div 5 = 3$ , remainder 3.
Step 2: $3 \div 5 = 0$ , remainder 3.

Step 4: The quotient is now 0, so we stop. There are exactly 2 division steps.

Answer: 2.

Question 5 · Digital Logic MCQ

When converting a decimal number to base $r$ using the repeated division method, what is the maximum possible value that any remainder can take?

A.
$r$
B.
$r - 1$
C.
$r + 1$
D.
$2 r$

Correct Answer:

Step-by-Step Solution

Key idea: This is a boundary condition question about the division algorithm.

Step 1: Recall the division algorithm: Dividend = Divisor $\times$ Quotient + Remainder.

Step 2: By definition, the remainder must always be strictly less than the divisor. If the remainder were equal to or greater than the divisor, you could divide again.

Step 3: Since the divisor in this method is the target base $r$ , the remainder must be less than $r$ .

Step 4: Since remainders are non-negative integers, the maximum possible value is $r - 1$ .

Answer: $r - 1$ .

Question 6 · Digital Logic MCQ

Which of the following statements is TRUE regarding the conversion of a number from base 3 to base 7?

A.
It can be performed directly by grouping bits into chunks of 3.
B.
It must be routed through base 10 as an intermediate step.
C.
It can be performed directly by grouping bits into chunks of 7.
D.
It requires conversion to base 2 before converting to base 7.

Correct Answer:

Step-by-Step Solution

Key idea: This is a statement evaluation question testing the 'Golden Rule' of cross-base conversion.

Step 1: Recall the rule for direct conversion. Direct bit-grouping only works when both bases are powers of the same number (e.g., 2, 4, 8, 16).

Step 2: Check the bases in the question: base 3 and base 7. Neither is a power of 2, and they do not share a common power base.

Step 3: Apply the Golden Rule. For arbitrary bases that are not powers of each other, you must route the conversion through base 10.

Step 4: Evaluate options. Options A and C suggest direct grouping, which is invalid here. Option D suggests routing through base 2, which is only useful if the bases were powers of 2. Option B correctly states the Golden Rule.

Answer: It must be routed through base 10 as an intermediate step.

Question 7 · Digital Logic MCQ

When converting a base-10 integer to base 8 using the repeated division method, what is the maximum possible remainder that can be recorded in any single division step?

A.
6
B.
7
C.
8
D.
9

Correct Answer:

Step-by-Step Solution

Key idea: This is an observation question about the boundary conditions of the division algorithm.

Step 1: Recall the division algorithm: Dividend = Divisor $\times$ Quotient + Remainder.

Step 2: The remainder must always be strictly less than the divisor. If the remainder were equal to or greater than the divisor, the quotient would increase.

Step 3: Here, the divisor is the target base, which is 8.

Step 4: Therefore, the remainder must be strictly less than 8. The maximum integer less than 8 is 7.

Answer: 7.

Question 8 · Digital Logic MCQ

Which of the following statements correctly describes the 'Golden Rule' for converting a number from base 6 to base 9?

A.
Direct bit-grouping can be used by grouping digits into chunks of 3.
B.
The conversion must be routed through base 10 as an intermediate step.
C.
The conversion can be done directly by dividing the base-6 number by 9.
D.
The conversion must be routed through base 2 first, then to base 9.

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question testing the constraints of the 'Golden Rule' of cross-base conversion.

Step 1: Recall the Golden Rule. Direct bit-grouping only works when both bases are powers of the same number (e.g., 2, 4, 8, 16).

Step 2: Check the bases in the question: base 6 and base 9. Neither is a power of 2, and they do not share a common power base.

Step 3: Apply the rule. For arbitrary bases that are not powers of each other, you must route the conversion through base 10.

Step 4: Evaluate options. Options A, C, and D suggest invalid direct methods. Option B correctly states the mandatory intermediate step.

Answer: The conversion must be routed through base 10 as an intermediate step.

Question 9 · Digital Logic MCQ

What is the maximum decimal value that can be represented by a 4-bit binary fraction of the form $(0. b_{1} b_{2} b_{3} b_{4})_{2}$ ?

A.
0.9375
B.
0.8750
C.
0.9999
D.
0.5000

Correct Answer:

Step-by-Step Solution

Key idea: This is an observation question about the boundary of fractional representation using memorized weights.

Step 1: The maximum value occurs when all bits $b_{1}, b_{2}, b_{3}, b_{4}$ are set to 1.

Step 2: The value is the sum of the first four negative powers of 2: $2^{- 1} + 2^{- 2} + 2^{- 3} + 2^{- 4}$ .

Step 3: Using memorized weights: $0.5 + 0.25 + 0.125 + 0.0625$ .

Step 4: Summing these gives exactly $0.9375$ .

Answer: 0.9375.

Question 10 · Digital Logic MCQ

When converting the decimal fraction $(0.625)_{10}$ to binary using the repeated multiplication method, what is the minimum number of multiplication steps required until the fractional part becomes exactly zero?

A.
2
B.
3
C.
4
D.
5

Correct Answer:

Step-by-Step Solution

Key idea: This is a contradiction question where the process seems like it might repeat infinitely, but terminates cleanly if counted correctly without overcounting the final zero state.

Step 1: Multiply $0.625 \times 2 = 1.25$ . Integer part is 1, new fraction is $0.25$ . (Step 1)

Step 2: Multiply $0.25 \times 2 = 0.50$ . Integer part is 0, new fraction is $0.50$ . (Step 2)

Step 3: Multiply $0.50 \times 2 = 1.00$ . Integer part is 1, new fraction is $0.00$ . (Step 3)

Step 4: The fractional part is now exactly zero, so the process terminates. There are exactly 3 multiplication steps.

Answer: 3.

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Number Systems, Binary Arithmetic and Data Representation Practice Questions for GATE CS

Chapter Roadmap: Number Systems and Binary Arithmetic

Chapter Journey

The Core Idea: Radix and Positional Weight

The Anatomy of a Number System

Number Systems, Binary Arithmetic and Data Representation: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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