Q.Binary number of decimal number (8)_10 is (A) (1000)_2
(B) (1001)_2
(C) (111)_2
(D) (110)_2

Decimal 27 converts to binary 11011 → option (D).

Repeatedly divide 27 by 2 and read the remainders from bottom up:

• 27 ÷ 2 = 13, remainder 1
• 13 ÷ 2 = 6, remainder 1
• 6 ÷ 2 = 3, remainder 0
• 3 ÷ 2 = 1, remainder 1 …

25 = 16 + 8 + 1 = 11001 in binary.

Convert 25 to binary by repeated division by 2:

• 25 ÷ 2 = 12 remainder 1
• 12 ÷ 2 = 6 remainder 0
• 6 ÷ 2 = 3 remainder 0
• 3 ÷ 2 = 1 remainder 1
• 1 ÷ 2 = 0 remainder 1 …

8 in decimal = 1000 in binary.

Convert 8 to binary by repeated division by 2:

• 8 ÷ 2 = 4, remainder 0
• 4 ÷ 2 = 2, remainder 0
• 2 ÷ 2 = 1, remainder 0
• 1 ÷ 2 = 0, remainder 1 …

(1001)₂ = 9 in decimal.

Convert binary to decimal by multiplying each bit by its place value (powers of 2), starting from the right: …

(1100)₂ = 8 + 4 = (12)₁₀.

Convert each binary number using place values $2^3, 2^2, 2^1, 2^0 = 8, 4, 2, 1$:

• $(1100)_2 = 1\cdot8 + 1\cdot4 + 0\cdot2 + 0\cdot1 = 12$ ✓
• $(1001)_2 = 8 + 0 + 0 + 1 = 9$ …

Converting the binary number 10011 to decimal by place values gives 19.

Binary 10011 has place values (from right to left) 2⁰, 2¹, 2², 2³, 2⁴:

…