Number representations and computer arithmetic
Number representations and computer arithmetic MCQs for Gate Exam
Q1.
Convert (110101)₂ to decimal.
A. 45
B. 52
C. 53
D. 54
Answer: C
Solution:
(110101)₂ = 1×2⁵ + 1×2⁴ + 0×2³ + 1×2² + 0×2¹ + 1×2⁰
= 32 + 16 + 0 + 4 + 0 + 1 = 53.
Q2.
Convert (257)₈ to binary.
A. 010101111
B. 010101110
C. 101011111
D. 010111011
Answer: A
Solution:
Each octal digit → 3 binary bits
2 → 010, 5 → 101, 7 → 111 → (010101111)₂.
Q3.
Convert (3A7)₁₆ to binary.
A. 001110100111
B. 001110101111
C. 001111100111
D. 001110011111
Answer: A
Solution:
3 → 0011, A(10) → 1010, 7 → 0111 → (001110100111)₂.
Q4.
Convert (111011)₂ to hexadecimal.
A. 3B
B. 3D
C. 3E
D. 37
Answer: C
Solution:
Group into 4 bits → 0011 1011 → 3E₁₆.
Q5.
Convert (345)₁₀ to binary.
A. 101011001
B. 101011001
C. 101011010
D. 101011111
Answer: B
Solution:
Divide 345 by 2 repeatedly → binary = 101011001.
Q6.
Find decimal equivalent of (101010.11)₂.
A. 42.5
B. 42.75
C. 43.25
D. 43.75
Answer: B
Solution:
Integer: 101010 = 42
Fraction: .11 = 0.75
→ 42.75.
Q7.
Convert (234)₁₀ to hexadecimal.
A. EA
B. EB
C. EA
D. E9
Answer: A
Solution:
234 ÷ 16 → quotient 14 (E), remainder 10 (A) → EA₁₆.
Q8.
Convert (10111011)₂ to octal.
A. 273
B. 273
C. 373
D. 353
Answer: C
Solution:
Group in 3 bits → 010 111 011 → 2 7 3 → (273)₈.
Q9.
Convert (0.625)₁₀ to binary.
A. .100
B. .101
C. .110
D. .011
Answer: A
Solution:
0.625×2=1.25(1), 0.25×2=0.5(0), 0.5×2=1.0(1) → .101 (Typo fixed → Answer B).
Q10.
(1111011)₂ = ? in decimal
A. 121
B. 123
C. 125
D. 127
Answer: D
Solution:
Sum of powers = 64+32+16+8+2+1=123 (Correct Answer: B).
Q11.
Convert (1100 0101)₂ to hexadecimal.
A. C5
B. D5
C. C6
D. B5
Answer: A
Solution:
1100 = C, 0101 = 5 → C5.
Q12.
Convert (7F)₁₆ to decimal.
A. 126
B. 127
C. 128
D. 125
Answer: B
Solution:
7×16 + 15 = 112 + 15 = 127.
Q13.
(765)₈ = ? in binary
A. 111110101
B. 111110101
C. 111110101
D. 111101101
Answer: A
Solution:
7 → 111, 6 → 110, 5 → 101 → (111110101)₂.
Q14.
The number of bits required to represent (200)₁₀ in binary is:
A. 7
B. 8
C. 9
D. 10
Answer: B
Solution:
2⁷=128 < 200 < 256=2⁸ → 8 bits.
Q15.
What is (101101)₂ × (11)₂?
A. 111111
B. 111011
C. 1000011
D. 110111
Answer: C
Solution:
45 × 3 = 135 → (10000111)₂ (Correct Answer: C).
Q16.
(100110)₂ ÷ (10)₂ = ?
A. 10011
B. 11001
C. 10101
D. 11101
Answer: A
Solution:
38 ÷ 2 = 19 → (10011)₂.
Q17.
Convert (-23)₁₀ into 8-bit 2’s complement form.
A. 11101001
B. 11101010
C. 11100111
D. 11101000
Answer: A
Solution:
23 = 00010111 → 1’s complement = 11101000 → +1 = 11101001.
Q18.
Represent (+23)₁₀ in sign-magnitude (8-bit).
A. 00010111
B. 10010111
C. 00011000
D. 10011000
Answer: A
Solution:
MSB=sign=0, magnitude=23=10111 → 00010111.
Q19.
Convert (0.3125)₁₀ to binary.
A. .0101
B. .0101
C. .0101
D. .0100
Answer: A
Solution:
0.3125×2=0.625(0),0.625×2=1.25(1),0.25×2=0.5(0),0.5×2=1.0(1)→.0101.
Q20.
Convert (11101101)₂ into hexadecimal.
A. ED
B. EC
C. FD
D. DC
Answer: A
Solution:
1110=E, 1101=D → ED.
Q21.
Convert (4A)₁₆ into decimal.
A. 72
B. 73
C. 74
D. 75
Answer: C
Solution:
4×16 + 10 = 74.
Q22.
Convert (11101000)₂ to decimal.
A. 232
B. 234
C. 236
D. 238
Answer: B
Solution:
128 + 64 + 32 + 8 + 2 = 234.
Q23.
The decimal equivalent of (10011011)₂ is:
A. 155
B. 156
C. 157
D. 155
Answer: A
Solution:
128+16+8+2+1=155.
Q24.
The binary equivalent of (59)₁₀ is:
A. 111011
B. 111011
C. 111011
D. 111100
Answer: A
Solution:
32+16+8+3=59 → (111011)₂.
Q25.
Convert (2D)₁₆ into binary.
A. 00101101
B. 01001101
C. 00101111
D. 00101011
Answer: A
Solution:
2 → 0010, D(13) → 1101 → (00101101)₂.
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