Convert the following numbers to their equivalent decimal values.
a. 11001012 b. 10011011001.101102 c. B2F16 d. 011010010111BCD e. 3748
The correct answer and explanation is:
Let’s go through each conversion step by step.
a. 1100101₂ (Binary to Decimal)
To convert the binary number 110010121100101_2 to decimal, we assign powers of 2 to each digit from right to left: 11001012=(1×26)+(1×25)+(0×24)+(0×23)+(1×22)+(0×21)+(1×20)1100101_2 = (1 \times 2^6) + (1 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)
Calculating each term: =(1×64)+(1×32)+(0×16)+(0×8)+(1×4)+(0×2)+(1×1)= (1 \times 64) + (1 \times 32) + (0 \times 16) + (0 \times 8) + (1 \times 4) + (0 \times 2) + (1 \times 1) =64+32+0+0+4+0+1=101= 64 + 32 + 0 + 0 + 4 + 0 + 1 = 101
So, 11001012=101101100101_2 = 101_{10}.
b. 10011011001.1011₂ (Binary to Decimal)
The integer part is 10011011001210011011001_2, and the fractional part is 101121011_2.
Integer part: 100110110012=(1×210)+(0×29)+(0×28)+(1×27)+(1×26)+(0×25)+(1×24)+(1×23)+(0×22)+(0×21)+(1×20)10011011001_2 = (1 \times 2^{10}) + (0 \times 2^9) + (0 \times 2^8) + (1 \times 2^7) + (1 \times 2^6) + (0 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) =1024+0+0+128+64+0+16+8+0+0+1=1241= 1024 + 0 + 0 + 128 + 64 + 0 + 16 + 8 + 0 + 0 + 1 = 1241
Fractional part: .10112=(1×2−1)+(0×2−2)+(1×2−3)+(1×2−4).1011_2 = (1 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3}) + (1 \times 2^{-4}) =0.5+0+0.125+0.0625=0.6875= 0.5 + 0 + 0.125 + 0.0625 = 0.6875
So, 10011011001.10112=1241.68751010011011001.1011_2 = 1241.6875_{10}.
c. B2F₁₆ (Hexadecimal to Decimal)
Hexadecimal uses base 16. The number B2F16B2F_{16} can be expanded as follows: B2F16=(B×162)+(2×161)+(F×160)B2F_{16} = (B \times 16^2) + (2 \times 16^1) + (F \times 16^0)
Where B=11B = 11 and F=15F = 15: =(11×256)+(2×16)+(15×1)= (11 \times 256) + (2 \times 16) + (15 \times 1) =2816+32+15=2863= 2816 + 32 + 15 = 2863
So, B2F16=286310B2F_{16} = 2863_{10}.
d. 011010010111BCD (Mixed Base to Decimal)
This looks like a mix of binary and hexadecimal digits. Let’s first consider only the binary part and convert it.
The binary part is 0110100101112011010010111_2: 0110100101112=(0×211)+(1×210)+(1×29)+(0×28)+(1×27)+(0×26)+(0×25)+(1×24)+(0×23)+(1×22)+(1×21)+(1×20)011010010111_2 = (0 \times 2^{11}) + (1 \times 2^{10}) + (1 \times 2^9) + (0 \times 2^8) + (1 \times 2^7) + (0 \times 2^6) + (0 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) =0+1024+512+0+128+0+0+16+0+4+2+1=1687= 0 + 1024 + 512 + 0 + 128 + 0 + 0 + 16 + 0 + 4 + 2 + 1 = 1687
For the hexadecimal part BCDBCD, let’s convert it to decimal. BCD16=(B×162)+(C×161)+(D×160)BCD_{16} = (B \times 16^2) + (C \times 16^1) + (D \times 16^0)
Where B=11B = 11, C=12C = 12, and D=13D = 13: =(11×256)+(12×16)+(13×1)= (11 \times 256) + (12 \times 16) + (13 \times 1) =2816+192+13=3021= 2816 + 192 + 13 = 3021
Thus, the total value is: 1687+3021=47081687 + 3021 = 4708
So, 011010010111BCD16=470810011010010111BCD_{16} = 4708_{10}.
e. 3748 (Decimal to Decimal)
The number 37483748 is already in decimal, so it remains as 3748103748_{10}.
Summary of Answers:
a. 11001012=101101100101_2 = 101_{10}
b. 10011011001.10112=1241.68751010011011001.1011_2 = 1241.6875_{10}
c. B2F16=286310B2F_{16} = 2863_{10}
d. 011010010111BCD16=470810011010010111BCD_{16} = 4708_{10}
e. 374810=3748103748_{10} = 3748_{10}