Binary System Questions in English

(B) To convert a decimal number to binary,we repeatedly divide the number by $2$ and record the remainders.
$\begin{array}{l|ll} 2 & 117 & \text{Remainder} \\ \hline 2 & 58 & 1 \\ \hline 2 & 29 & 0 \\ \hline 2 & 14 & 1 \\ \hline 2 & 7 & 0 \\ \hline 2 & 3 & 1 \\ \hline 2 & 1 & 1 \\ \hline & 0 & 1 \end{array}$
Reading the remainders from bottom to top,we get $1110101$.
$\therefore$ The binary equivalent of decimal $117$ is $1110101$.

(A) To convert a decimal number to binary,we repeatedly divide the number by $2$ and record the remainders.
$\begin{array}{l|cc} 2 & 52 & \text{Remainder} \\ \hline 2 & 26 & 0 \\ \hline 2 & 13 & 0 \\ \hline 2 & 6 & 1 \\ \hline 2 & 3 & 0 \\ \hline 2 & 1 & 1 \\ \hline & 0 & 1 \end{array}$
Reading the remainders from bottom to top,we get $110100$.
$\therefore$ The binary equivalent of decimal $52$ is $110100.$

(C) To convert a binary number to its decimal equivalent,we multiply each bit by its corresponding power of $2$ based on its position (starting from $0$ on the right).
The binary number is $1110101_2$.
Calculation:
$(1 \times 2^6) + (1 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)$
$= 64 + 32 + 16 + 0 + 4 + 0 + 1$
$= 117_{10}$
Therefore,the decimal equivalent is $117_{10}$.

(C) To convert a decimal number to binary,we repeatedly divide the number by $2$ and record the remainders.
$235 \div 2 = 117$ with a remainder of $1$
$117 \div 2 = 58$ with a remainder of $1$
$58 \div 2 = 29$ with a remainder of $0$
$29 \div 2 = 14$ with a remainder of $1$
$14 \div 2 = 7$ with a remainder of $0$
$7 \div 2 = 3$ with a remainder of $1$
$3 \div 2 = 1$ with a remainder of $1$
$1 \div 2 = 0$ with a remainder of $1$
Reading the remainders from bottom to top,we get $11101011_{2}$.
Therefore,$235_{10} = 11101011_{2}$.

(A) To convert a decimal number to binary,we repeatedly divide the number by $2$ and record the remainders.
$\begin{array}{l|cc} 2 & 701 & \text{Remainder} \\ \hline 2 & 350 & 1 \\ \hline 2 & 175 & 0 \\ \hline 2 & 87 & 1 \\ \hline 2 & 43 & 1 \\ \hline 2 & 21 & 1 \\ \hline 2 & 10 & 1 \\ \hline 2 & 5 & 0 \\ \hline 2 & 2 & 1 \\ \hline 2 & 1 & 0 \\ \hline 2 & 0 & 1 \end{array}$
Reading the remainders from bottom to top,we get $(701)_{10} = 1010111101_{2}$.

(B) To convert a binary number to its decimal equivalent,we multiply each digit by $2$ raised to the power of its position (starting from $0$ on the right).
For the binary number $101001_2$:
Position $5$: $1 \times 2^5 = 32$
Position $4$: $0 \times 2^4 = 0$
Position $3$: $1 \times 2^3 = 8$
Position $2$: $0 \times 2^2 = 0$
Position $1$: $0 \times 2^1 = 0$
Position $0$: $1 \times 2^0 = 1$
Summing these values: $32 + 0 + 8 + 0 + 0 + 1 = 41$.
Therefore,the decimal equivalent is $41$.

(A) To convert a binary number to its decimal equivalent,we multiply each bit by $2$ raised to the power of its position (starting from $0$ on the right).
The binary number is $10000010011_2$.
Position values:
$1 \times 2^{10} + 0 \times 2^9 + 0 \times 2^8 + 0 \times 2^7 + 0 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$
Calculating the powers of $2$:
$2^{10} = 1024$
$2^4 = 16$
$2^1 = 2$
$2^0 = 1$
Summing these values:
$1024 + 16 + 2 + 1 = 1043$
Therefore,the decimal equivalent is $1043$.

(C) To convert a binary number to its decimal equivalent,we multiply each digit by $2$ raised to the power of its position (starting from $0$ on the right).
For the binary number $111011_2$:
$1 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0$
$= 32 + 16 + 8 + 0 + 2 + 1$
$= 59$
Therefore,the decimal equivalent is $59$.

(B) To add the binary numbers $1001$ and $0101$, we perform the addition column by column from right to left:
$\begin{array}{r} 0101 \\ +1001 \\ \hline 1110 \end{array}$
Step-by-step calculation:
$1 + 1 = 10$ (write $0$, carry $1$)
$0 + 0 + 1 (\text{carry}) = 1$
$1 + 0 = 1$
$0 + 1 = 1$
Thus, the sum is $1110$.

(A) To add the binary numbers $11100$ and $11010$,we perform the addition column by column from right to left:
$0 + 0 = 0$
$0 + 1 = 1$
$1 + 0 = 1$
$1 + 1 = 10$ (write $0$,carry $1$)
$1 + 1 + 1$ (carry) $= 11$
Combining these,we get $110110$.
$\begin{array}{r} 11100 \\ +11010 \\ \hline 110110 \end{array}$

(C) To add the binary numbers,we perform column-wise addition:
$11111_{2} = 31_{10}$
$10001_{2} = 17_{10}$
$1011_{2} = 11_{10}$
Sum in decimal: $31 + 17 + 11 = 59_{10}$
Now,convert $59_{10}$ to binary:
$59 \div 2 = 29$ remainder $1$
$29 \div 2 = 14$ remainder $1$
$14 \div 2 = 7$ remainder $0$
$7 \div 2 = 3$ remainder $1$
$3 \div 2 = 1$ remainder $1$
$1 \div 2 = 0$ remainder $1$
Reading remainders from bottom to top,we get $111011_{2}$.
Alternatively,adding column-wise:
$11111 + 10001 + 01011 = 111011_{2}$.

(A) To add the binary numbers $11001_{2}$, $11011_{2}$, and $11111_{2}$, we perform column-wise addition from right to left:
$11001_{2}$
$11011_{2}$
$11111_{2}$
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$1$. Column $1$ (rightmost): $1 + 1 + 1 = 3_{10}$. In binary, $3_{10} = 11_{2}$. Write $1$, carry $1$.
$2$. Column $2$: $0 + 1 + 1 + 1 (\text{carry}) = 3_{10} = 11_{2}$. Write $1$, carry $1$.
$3$. Column $3$: $0 + 0 + 1 + 1 (\text{carry}) = 2_{10} = 10_{2}$. Write $0$, carry $1$.
$4$. Column $4$: $1 + 1 + 1 + 1 (\text{carry}) = 4_{10} = 100_{2}$. Write $0$, carry $10_{2}$.
$5$. Column $5$: $1 + 1 + 1 + 2 (\text{carry}) = 5_{10} = 101_{2}$.
Combining these, we get $1010011_{2}$.

(B) To solve the addition of binary numbers,we perform column-wise addition following the rules of binary arithmetic $(0+0=0, 0+1=1, 1+1=10_{2})$:
$11_{2} = 3_{10}$
$111_{2} = 7_{10}$
$1111_{2} = 15_{10}$
$11111_{2} = 31_{10}$
Sum in decimal: $3 + 7 + 15 + 31 = 56_{10}$.
Now,convert $56_{10}$ to binary:
$56 \div 2 = 28$ remainder $0$
$28 \div 2 = 14$ remainder $0$
$14 \div 2 = 7$ remainder $0$
$7 \div 2 = 3$ remainder $1$
$3 \div 2 = 1$ remainder $1$
$1 \div 2 = 0$ remainder $1$
Reading remainders from bottom to top,we get $111000_{2}$.
Alternatively,using column addition:
$\begin{array}{r} 11 \\ 111 \\ 1111 \\ + 11111 \\ \hline 111000 \\ \hline \end{array}$

(C) To add binary numbers $111_{2}$ and $101_{2}$,we perform addition column by column from right to left:
$1$. $1 + 1 = 10_{2}$ (write $0$,carry $1$)
$2$. $1 + 0 + 1$ (carry) $= 10_{2}$ (write $0$,carry $1$)
$3$. $1 + 1 + 1$ (carry) $= 11_{2}$ (write $11$)
Thus,$111_{2} + 101_{2} = 1100_{2}$.

(A) To add the binary numbers $1000_{2}$, $1101_{2}$, and $1111_{2}$, we perform column-wise addition from right to left:
$1$. Rightmost column: $0 + 1 + 1 = 2_{10} = 10_{2}$. Write $0$, carry $1$.
$2$. Second column: $0 + 0 + 1 + 1 (\text{carry}) = 2_{10} = 10_{2}$. Write $0$, carry $1$.
$3$. Third column: $0 + 1 + 1 + 1 (\text{carry}) = 3_{10} = 11_{2}$. Write $1$, carry $1$.
$4$. Fourth column: $1 + 1 + 1 + 1 (\text{carry}) = 4_{10} = 100_{2}$. Write $100$.
Combining these, we get $100100_{2}$.

(B) To add the binary numbers $111_{2}$, $101_{2}$, and $011_{2}$, we perform column-wise addition from right to left:
$1$. Rightmost column: $1 + 1 + 1 = 3_{10}$. In binary, $3_{10} = 11_{2}$. Write $1$ and carry over $1$.
$2$. Middle column: $1 + 0 + 1 + (\text{carry } 1) = 3_{10} = 11_{2}$. Write $1$ and carry over $1$.
$3$. Leftmost column: $1 + 1 + 0 + (\text{carry } 1) = 3_{10} = 11_{2}$.
Thus, the sum is $1111_{2}$.

(A) To subtract the binary numbers,we can convert them to decimal,perform the subtraction,and convert back to binary.
$111000_{2} = (1 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (0 \times 2^0) = 32 + 16 + 8 = 56_{10}$.
$11001_{2} = (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 16 + 8 + 1 = 25_{10}$.
Subtracting the decimal values: $56 - 25 = 31_{10}$.
Now,convert $31_{10}$ back to binary:
$31 \div 2 = 15$ remainder $1$
$15 \div 2 = 7$ remainder $1$
$7 \div 2 = 3$ remainder $1$
$3 \div 2 = 1$ remainder $1$
$1 \div 2 = 0$ remainder $1$
Reading remainders from bottom to top,we get $11111_{2}$.
Alternatively,using binary subtraction:
$111000_{2} - 011001_{2} = 11111_{2}$.

(C) To subtract $1111_{2}$ from $10001_{2}$,we can perform binary subtraction directly:
$10001_{2} - 1111_{2}$
Step-by-step subtraction:
$1 - 1 = 0$
$0 - 1$ (borrow from next column) $\rightarrow 10 - 1 = 1$
$0 - 1$ (after borrow) $\rightarrow 10 - 1 = 1$
$0 - 1$ (after borrow) $\rightarrow 10 - 1 = 1$
$1 - 1$ (after borrow) $\rightarrow 0$
Result: $00010_{2} = 10_{2}$.
Alternatively,converting to decimal:
$10001_{2} = 2^{4} + 2^{0} = 16 + 1 = 17_{10}$
$1111_{2} = 2^{3} + 2^{2} + 2^{1} + 2^{0} = 8 + 4 + 2 + 1 = 15_{10}$
$17 - 15 = 2_{10}$
$2_{10}$ in binary is $10_{2}$.

(B) To subtract $10111_{2}$ from $111101_{2}$,we can use the $2$'s complement method.
Step $1$: Find the $2$'s complement of the subtrahend $10111_{2}$.
First,find the $1$'s complement of $010111_{2}$ (padding with a leading zero to match the length of $6$ bits),which is $101000_{2}$.
Then,add $1$ to the $1$'s complement: $101000_{2} + 1_{2} = 101001_{2}$.
Step $2$: Add the minuend to the $2$'s complement of the subtrahend:
$\begin{array}{r} 111101_{2} \\ + 101001_{2} \\ \hline 1100110_{2} \end{array}$
Step $3$: Discard the carry bit (the leftmost $1$) to get the final result:
$100110_{2}$.
Alternatively,performing direct binary subtraction:
$\begin{array}{r} 111101_{2} \\ - 010111_{2} \\ \hline 100110_{2} \end{array}$
Thus,$111101_{2} - 10111_{2} = 100110_{2}$.

(C) To subtract $10001_{2}$ from $11111_{2}$,we perform binary subtraction:
$11111_{2} = 16 + 8 + 4 + 2 + 1 = 31_{10}$
$10001_{2} = 16 + 0 + 0 + 0 + 1 = 17_{10}$
Subtracting the decimal values:
$31 - 17 = 14_{10}$
Now,convert $14_{10}$ back to binary:
$14 \div 2 = 7$ remainder $0$
$7 \div 2 = 3$ remainder $1$
$3 \div 2 = 1$ remainder $1$
$1 \div 2 = 0$ remainder $1$
Reading remainders from bottom to top,we get $1110_{2}$.
Alternatively,using direct binary subtraction:
$\begin{array}{r} 11111 \\ -10001 \\ \hline 01110 \end{array}$
Thus,$11111_{2} - 10001_{2} = 1110_{2}$.

(A) To subtract $11110_{2}$ from $100001_{2}$,we can perform binary subtraction directly:
$100001_{2} - 11110_{2} = ?$
Step-by-step subtraction:
$1 - 0 = 1$
$0 - 1$ (borrow from next): $10 - 1 = 1$
$0 - 1$ (after borrow): $1 - 1 = 0$
$0 - 1$ (after borrow): $1 - 1 = 0$
$0 - 1$ (after borrow): $1 - 1 = 0$
Result: $000011_{2} = 11_{2}$.

(B) To multiply $1111_2$ by $11_2$ in the binary system,we perform the multiplication as follows:
$\begin{array}{r@{\quad}l} & 1111_2 \\ \times & 0011_2 \\ \hline & 1111_2 \\ 11110_2 & \\ \hline 101101_2 \end{array}$
Step $1$: Multiply $1111_2$ by the unit digit $1$ of $11_2$,which gives $1111_2$.
Step $2$: Multiply $1111_2$ by the second digit $1$ (which represents $2^1$ or $10_2$),which gives $11110_2$.
Step $3$: Add the two results: $1111_2 + 11110_2 = 101101_2$.
Thus,the correct option is $B$.

(A) To multiply $101_2$ by $11_2$ in the binary system,we perform the multiplication as follows:
$\begin{array}{r@{\quad}l} & 101_2 \\ \times & 11_2 \\ \hline & 101_2 \\ 1010_2 & \\ \hline 1111_2 \end{array}$
Step $1$: Multiply $101_2$ by the unit digit $1$ of $11_2$,which gives $101_2$.
Step $2$: Multiply $101_2$ by the second digit $1$ of $11_2$ (shifting one position to the left),which gives $1010_2$.
Step $3$: Add the results: $101_2 + 1010_2 = 1111_2$.
Thus,the correct option is $A$.

(C) To multiply $101101_2$ by $1101_2$:
$101101 \times 1 = 101101$
$101101 \times 00 = 0000000$
$101101 \times 100 = 10110100$
$101101 \times 1000 = 101101000$
Adding these results:
$101101 + 0000000 + 10110100 + 101101000 = 1001001001_2$

(A) To multiply $11001$ by $101$ in binary,we perform the multiplication as follows:
$11001 \times 1 = 11001$
$11001 \times 0 = 00000$
$11001 \times 1 = 11001$
Aligning the partial products:
$11001$
$000000$
$1100100$
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$1111101$
Thus,$11001_2 \times 101_2 = 1111101_2$.