Number System & Logic Gates

The base of the Binary Number System is 2. In this system, numbers are represented using only two digits, 0 and 1. Each digit in a binary number represents a power of 2, starting from the rightmost digit as 2^0, then 2^1, 2^2, and so on. This system is widely used in computer science and digital electronics because it aligns with the on/off states of electronic switches, making it efficient for representing and manipulating data in these systems.

The base of the hexadecimal number system is 16. In this system, numbers are represented using 16 different symbols: 0-9 for values 0-9, and A-F for values 10-15. Each digit in a hexadecimal number represents a power of 16, allowing for a compact representation of large numbers. Hexadecimal is commonly used in computer programming and digital systems, as it provides a convenient way to represent binary data.

The symbol in question represents an AND gate. An AND gate is a logic gate that outputs a true (high) signal only when all of its inputs are true. In the given options, the symbol matches the representation of an AND gate.

The AND gate generates the truth table shown because it only outputs a high signal when both of its inputs are high. In the given truth table, the output is high (1) only when both inputs are high (1), and low (0) for all other combinations of inputs. This behavior matches the functionality of an AND gate, making it the correct answer.

An OR gate is equivalent to the logical addition operation. This is because an OR gate takes two inputs and outputs a true value if at least one of the inputs is true. In logical addition, the values being added are considered true if either one or both of them are true. Therefore, the OR gate and logical addition perform the same function of combining inputs and producing a true output if any of the inputs are true.

The AND gate is equivalent to the logical multiplication operation. This is because the AND gate produces an output of true (1) only when both of its inputs are true (1). In logical multiplication, the result is true (1) only when both operands are true (1), otherwise, the result is false (0). Therefore, the AND gate and logical multiplication have the same behavior and can be considered equivalent operations.

In hexadecimal number system, the digits after 9 are represented by the letters A, B, C, D, E, and F. Therefore, the digit coming after 9 is A.

The missing output in the truth table for an X-NOR gate is 1. An X-NOR gate outputs 1 when both inputs are the same (either both 0 or both 1), and 0 when the inputs are different. Since the given truth table only has one input value (0), the missing output can be determined by applying the logic of the X-NOR gate, which results in the output being 1.

A NOT gate, also known as an inverter, is a basic logic gate that has only one input. It takes the input signal and produces the opposite output signal. Therefore, a NOT gate can have only one input.

The symbol represents a NOR gate. A NOR gate is a logic gate that produces an output of true only when both of its inputs are false. It is the opposite of an OR gate, which produces an output of true when at least one of its inputs is true. The symbol for a NOR gate consists of a circle with an arc coming out of it, indicating the negation of the OR operation.

To convert a binary number to a decimal number, we need to multiply each digit of the binary number by the corresponding power of 2 and then add the results. In this case, the binary number is (10101)2. Starting from the rightmost digit, we have 1 * 2^0 = 1, then 0 * 2^1 = 0, then 1 * 2^2 = 4, then 0 * 2^3 = 0, and finally 1 * 2^4 = 16. Adding all these results together, we get 1 + 0 + 4 + 0 + 16 = 21. Therefore, the decimal equivalent of (10101)2 is 21.

The correct answer is NAND Gate. The truth table provided matches the behavior of a NAND gate, where the output is only false when both inputs are true, and true for all other combinations of inputs.

The given truth table represents the behavior of a 3-input OR gate. In an OR gate, the output is true if at least one of the inputs is true. Since the truth table shows that the output is true for all possible combinations of the three inputs, it indicates that this truth table corresponds to a 3-input OR gate.

To convert a decimal number to octal, we divide the number by 8 and record the remainder. We repeat this process with the quotient until the quotient becomes zero. In this case, when we divide 12 by 8, we get a quotient of 1 and a remainder of 4. Therefore, the octal representation of 12 is 14.

To convert a binary number to decimal, we multiply each digit by the corresponding power of 2 and sum them up. In this case, starting from the rightmost digit, we have 1 * 2^0 + 1 * 2^1 + 0 * 2^2 + 1 * 2^3 + 1 * 2^4 = 1 + 2 + 0 + 8 + 16 = 27. Therefore, the decimal equivalent of (11011)2 is 27.

To subtract (0101)2 from (1101)2, we start by subtracting the rightmost digits. Since 1 is smaller than 0, we borrow 1 from the next digit. This results in 11 - 01 = 10. Moving to the next digits, we have 1 - 0 = 1, and 1 - 1 = 0. Finally, we have 1 - 0 = 1. Therefore, the result of subtracting (0101)2 from (1101)2 is (1000)2.

The NOT gate gives the complement of the input. It is a unary gate that takes a single input and produces the logical negation of that input as the output. In other words, if the input is true, the output will be false, and vice versa. This gate is commonly used in digital circuits to invert or negate the input signal.

In the decimal number system, the digits range from 0 to 9. After 9, the next digit is 10. This is because once we reach 9, we have used all the available single-digit numbers, so we need to add another digit to continue counting. Therefore, the correct answer is 10.

To convert a binary number to an octal number, we group the binary digits into sets of three starting from the rightmost digit. In this case, we have 101 111. Then, we convert each group of three binary digits to their equivalent octal digit. 101 in binary is equal to 5 in octal, and 111 in binary is equal to 7 in octal. Therefore, the binary number (101111)2 is equal to the octal number (57)8.

The given question asks us to add the binary numbers (1101)2 and (0111)2. When we add these two numbers, we get the sum (10100)2.

The question asks to add the hexadecimal numbers (59)16 and (A3)16. When adding hexadecimal numbers, we start from the rightmost digits and work our way to the left. Adding 9 and 3 gives us 12, which is represented as C in hexadecimal. Adding 5, A, and the carryover 1 gives us F. Therefore, the sum of (59)16 and (A3)16 is (FC)16.

To convert a decimal number to binary, we divide the decimal number by 2 repeatedly and record the remainder each time. Starting with 25, when we divide it by 2, we get a quotient of 12 and a remainder of 1. We repeat this process with the quotient, 12, and get a new quotient of 6 and a remainder of 0. We continue this process until the quotient becomes 0. The remainders, read from bottom to top, give us the binary equivalent of 25, which is 11001.

To convert a number from base 8 to base 10, each digit of the number is multiplied by 8 raised to the power of its position, starting from the rightmost digit. In this case, the number (73)8 can be calculated as (7 * 8^1) + (3 * 8^0), which simplifies to (7 * 8) + (3 * 1), resulting in 56 + 3 = 59 in base 10.

To convert a number from base 8 to base 2, we need to convert each digit of the base 8 number to its equivalent 3-digit binary representation. In this case, the base 8 number (46)8 has two digits. The first digit, 4, is equivalent to 100 in binary, and the second digit, 6, is equivalent to 110 in binary. Therefore, the conversion of (46)8 to base 2 is 100110.

To convert (A9)16 to ( )2, we need to convert each digit of the hexadecimal number to its binary equivalent. In this case, A is equal to 1010 in binary, and 9 is equal to 1001 in binary. Therefore, (A9)16 is equal to 10101001 in binary.

To convert a number from base 8 to base 16, we need to convert each digit of the base 8 number to its equivalent in base 16. In this case, the base 8 number is 356. Converting each digit, we get 3 as 3, 5 as 5, and 6 as 6. Therefore, the base 16 equivalent of (356)8 is 0EE.

The given binary number (11011101100)2 can be converted to hexadecimal by grouping the binary digits into groups of four starting from the rightmost digit. In this case, the groups would be 1101, 1101, and 1100. Each group can then be converted to its hexadecimal equivalent, which gives us D, D, and 4 respectively. Therefore, the hexadecimal representation of (11011101100)2 is 6EC.

To add two numbers in base 8, we start by adding the rightmost digits together. In this case, 4 + 2 equals 6. Then, we move to the next digit, which is 6 + 6. Since 6 + 6 is equal to 14 in base 10, we write down 4 and carry over 1 to the next digit. Finally, we add the carried over 1 to the leftmost digit, which gives us 1 + 3 = 4. Therefore, the correct answer is 1246.

The given circuit consists of two inputs connected to two parallel switches, and these switches are connected in series to an output. This configuration is characteristic of an AND gate. An AND gate produces an output only when both of its inputs are high (logic 1), and in this circuit, the output will be high only if both switches are closed, indicating that the inputs are both high. Therefore, the given circuit is equivalent to an AND gate.

The given answer, 001010101110, is the binary representation of the octal number 1256. To convert an octal number to binary, each digit of the octal number is replaced by its binary equivalent. In this case, the octal digits 1, 2, 5, and 6 are replaced by their binary equivalents 001, 010, 101, and 110 respectively. Therefore, the binary representation of the octal number 1256 is 001010101110.

The given question asks to convert the hexadecimal number (27A9)16 to octal. To convert from hexadecimal to octal, we can first convert the hexadecimal number to binary and then from binary to octal. In this case, the hexadecimal number (27A9)16 is equivalent to (0010 0111 1010 1001)2 in binary. Then, grouping the binary number into sets of 3 digits from right to left, we get (023651)8 in octal.

To convert a decimal number to binary, we need to repeatedly multiply the decimal portion by 2 and take the integer part as the next binary digit. In this case, when we multiply 0.625 by 2, we get 1.25. The integer part, 1, becomes the first binary digit. We then take the decimal part, 0.25, and multiply it by 2 again, giving us 0.5. The integer part, 0, becomes the second binary digit. Finally, we multiply 0.5 by 2, resulting in 1. The integer part, 1, becomes the third binary digit. Therefore, the binary representation of 0.625 is 0.101.

The given circuit consists of two inputs connected to two parallel branches, each containing a single switch. In an OR gate, the output is true if either or both of the inputs are true. In this circuit, if either switch is closed, the output will be true, making it equivalent to an OR gate.

An X-OR gate is a logic gate that produces a high output when only one of its inputs is high. If both inputs are low or both inputs are high, the output of the X-OR gate will be low. Therefore, the correct answer is "Give a high output when only one of its inputs are high".

The given answer, 00110, represents the number (25)10 in 1's complement form. In 1's complement, to obtain the negative representation of a positive number, we invert all the bits in the binary representation. In this case, the binary representation of 25 is 11001, and when we invert all the bits, we get 00110. Therefore, 00110 is the 1's complement representation of (25)10.

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The given logic circuit configuration produces an X-OR gate. An X-OR gate outputs a high (1) signal if the number of high inputs is odd, and a low (0) signal if the number of high inputs is even. In this configuration, the inputs are not specified, but based on the options given, it can be inferred that the circuit is designed to function as an X-OR gate.

In the hexadecimal number system, the digits after F are represented by numbers 0 to 9 and then A to F. So, the digit coming after F is 10.

The given answer, 01101, represents the number 19 in 2's complement form. In 2's complement, the leftmost bit represents the sign of the number, where 0 indicates a positive number and 1 indicates a negative number. Since the leftmost bit is 0 in this answer, it indicates a positive number. The remaining bits, 1101, represent the magnitude of the number. Converting 1101 to decimal gives us 13. Therefore, the answer 01101 represents the positive number 13, which is equivalent to (19)10.

The given solution shows the step-by-step process of performing BCD addition of the numbers (45)10 and (93)10. Each row represents a step in the addition process. The final answer is obtained by adding the corresponding digits from each row, carrying over any excess to the next row. The solution starts with the least significant digits and progresses towards the most significant digits. The final answer, 000100111000, represents the BCD sum of the two numbers.

The gray code is a binary numeral system where two consecutive values differ by only one bit. To find the gray code of a decimal number, we convert the decimal number to binary and then apply the gray code algorithm. The decimal number 12 is equivalent to 1100 in binary. Applying the gray code algorithm, we start with the most significant bit and copy it as it is (1). Then, for each subsequent bit, we perform an XOR operation between the current bit and the previous bit. The gray code for 1100 is 1010.

The given circuit consists of two inputs connected to the inputs of two NAND gates, and the outputs of those NAND gates are connected to the inputs of another NAND gate. This configuration is known as a NAND gate, which is the opposite of an AND gate. In other words, the output of the NAND gate will be the complement of the output of an AND gate. Therefore, the correct answer is AND Gate.

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The given circuit is equivalent to an X-NOR gate. This can be determined by analyzing the inputs and outputs of the circuit. An X-NOR gate produces a high output only when both inputs are the same, otherwise it produces a low output. In the given circuit, the inputs are connected to two NOT gates, which invert the inputs. The inverted inputs are then connected to an AND gate, which produces a high output only when both inputs are high. Finally, the output of the AND gate is connected to a NOT gate, which inverts the output. Therefore, the circuit behaves like an X-NOR gate.

Connecting the inputs of a NAND and NOR gate together will produce a NOT gate. This is because the output of a NAND gate is the inverse of the AND gate, while the output of a NOR gate is the inverse of the OR gate. By connecting the inputs of both gates together, the outputs will be the same, resulting in a NOT gate where the output is the inverse of the input.

The gray code (1001011)2 can be converted to binary by applying the gray code to binary conversion algorithm. Starting from the leftmost bit, we keep the first bit of the gray code as is. For each subsequent bit, we XOR it with the previous bit of the gray code. So, in this case, we keep the first bit as 1. Then, we XOR the second bit with the first bit, resulting in 1 XOR 0 = 1. Continuing this process, we XOR each bit with the previous resulting bit. The final binary representation is 1110010.

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The cyclic code of (7)10 is 1011 because it is the only option that satisfies the criteria of being a cyclic code. A cyclic code is a type of error-correcting code where if a codeword is cyclically shifted, it remains a valid codeword. In this case, if we cyclically shift 1011, we get 1101, which is still a valid codeword.

The given answer, 10110101, is the result of performing XS-3 addition of the decimal numbers 47 and 35. XS-3 is a binary-coded decimal (BCD) representation where each digit is represented by a 4-bit binary number. The addition process involves adding the corresponding bits from right to left, carrying over if necessary. In this case, the addition is done correctly, resulting in the binary number 10110101.

The given circuit is an OR gate. This can be determined by observing that the output is high (1) when either input A or input B (or both) is high (1), and the output is low (0) only when both inputs are low (0). This behavior is characteristic of an OR gate, where the output is true (1) if at least one of the inputs is true (1).