UGC NET Questions (Paper – 1)

In hexadecimal, the digit F represents decimal value 15. The binary representation of 15 uses four bits all set to 1, because 15 = 8 + 4 + 2 + 1. Therefore, the correct 4-bit binary equivalent of F is 1111₂.

Option A:
Option A, 1101₂, equals 8 + 4 + 0 + 1 = 13, which is less than 15. It does not fully represent the hex digit F.

Option B:
Option B, 1110₂, equals 8 + 4 + 2 + 0 = 14. It is still smaller than 15 and misses the 1's place value.

Option C:
Option C, 1100₂, sums to 8 + 4 = 12 and is even further from 15. It lacks both the 2's and 1's contributions.

Option D:
Option D is correct because 1111₂ sums all four available place values 8, 4, 2 and 1, giving 15. This exactly matches the value of F in hexadecimal.

For binary 1111, we sum the place values corresponding to each bit that is 1. From right to left, the bits represent 1, 2, 4 and 8. Since all four bits are 1, the sum is 8 + 4 + 2 + 1. This equals 15, so the decimal equivalent is 15.

Option A:
Option A suggests 13, which would come from the pattern 1101₂ (8 + 4 + 1). In 1111, all four bits are set, so the sum is larger than 13. Therefore 13 cannot be correct.

Option B:
Option B gives 14, which corresponds to 1110₂ (8 + 4 + 2). The given number has a 1 also in the 1's place, making its value 15, not 14. Thus this option is too small by one.

Option C:
Option C suggests 12, which would arise from 1100₂ (8 + 4). The pattern 1111 includes additional contributions from the 2's and 1's places, so 12 is clearly incorrect.

Option D:
Option D is correct because 1111₂ = 8 + 4 + 2 + 1 = 15. All place values are used, giving the highest possible value for a 4-bit binary number. This matches the required decimal equivalent.

Statements A and B correctly characterise the decimal and binary systems, D describes the method of converting a binary numeral to decimal and E points out common exam tasks involving conversions and comparisons of numbers in different bases. Statement C is false because (1010)₂ equals 10 in decimal, not 5. Hence, the combination A, B, D and E only is the correct set.

Option A:
Option A is incomplete since it omits E, thereby failing to acknowledge the typical exam requirement to convert or compare numbers across bases. Without E, the link to aptitude questions is weaker.

Option B:
Option B is correct because it groups all four true statements while excluding C, which miscalculates the value of a simple binary number. It summarises both the structure and the practical use of different base systems for NET aspirants.

Option C:
Option C is wrong as it includes C, the incorrect evaluation of (1010)₂, and omits A, leaving out the definition of decimal place value. The combination of an error with an omission invalidates the option.

Option D:
Option D is incorrect because it contains C and therefore accepts a wrong binary-to-decimal conversion, even though some of its other statements are true. This makes the overall option unsuitable as a correct answer.

The binary number 10₂ has a 1 in the 2^1 place and a 0 in the 2^0 place. This gives 1×2 + 0×1 = 2. Therefore, the decimal value corresponding to 10₂ is 2.

Option A:
Option A, 1, corresponds to 1 in binary, which is written as 1₂ or 0001₂. It does not match the pattern 10₂ where a higher place is set.

Option B:
Option B is correct because the only nonzero bit is in the 2's position, giving a value of 2. This simple conversion is often the first example when learning binary.

Option C:
Option C, 3, would correspond to 11₂, where both the 2's and 1's bits are set. Since 10₂ has only one bit set, 3 is not appropriate.

Option D:
Option D, 4, would be represented by 100₂. The given number has only two bits, so it cannot represent 4.

In a binary number, the rightmost bit represents the smallest positional value, 2^0 or 1. Because it contributes the least weight to the overall number, it is called the least significant bit (LSB). This bit is important in operations like parity checking and incrementing. Therefore, the term least significant correctly describes the rightmost bit.

Option A:
Option A names the "most significant" bit, which actually refers to the leftmost bit that has the highest positional value. It denotes the largest weight in the number and is not located on the right. Hence this cannot describe the rightmost bit.

Option B:
Option B is correct because the rightmost bit carries the smallest positional weight and so is least significant. In many algorithms, changes in the LSB affect the number by one unit, reinforcing its role as the least significant bit. This terminology is standard in digital electronics.

Option C:
Option C refers to a "control" bit, which is not a standard name for the rightmost position. Control bits can appear in various positions depending on the protocol or format; they are not inherently tied to the rightmost place.

Option D:
Option D suggests a "status" bit, which is a general term for bits that signal conditions but not necessarily the rightmost one. Status indicators may reside anywhere in a control word or register.

Decimal 3 can be written as the sum of powers of 2, namely 2 + 1. In binary, this is represented by having a 1 in the 2's place and a 1 in the 1's place. That pattern is written as 11₂. Therefore, 11 is the correct binary representation of decimal 3.

Option A:
Option A, 1₂, equals 1 in decimal because only the 1's place is set. It does not include the 2's place contribution needed to reach 3.

Option B:
Option B, 10₂, represents 2 in decimal, with a 1 only in the 2's place and 0 in the 1's place. This is still short of 3 and therefore incorrect.

Option C:
Option C, 100₂, is equal to 4 in decimal as it sets the 4's place and zeroes elsewhere. This value is greater than 3, so it cannot be the correct representation for 3.

Option D:
Option D is correct because 11₂ has ones in both the 2's and 1's place, giving 2 + 1 = 3. This satisfies the requirement for the binary equivalent of decimal 3.

Statements A, B, D and E are correct binary facts. A describes the positional value system in base 2. B is true because an odd number has last bit 1 and an even number has last bit 0. D correctly states the conversion method from binary to decimal. E is true for multiples of 4, whose binary form ends with two zeros. C is false; adding 1 flips only a block of trailing ones and the bit after that, not necessarily all bits. Therefore, A, B, D and E only are correct.

Option A:
Option A is incomplete since it omits E, thereby missing a useful rule for recognising multiples of 4 from binary patterns, which is sometimes tested in reasoning questions.

Option B:
Option B is incorrect because it includes C, wrongly claiming that adding 1 always flips all bits, which contradicts simple examples like 0010 becoming 0011.

Option C:
Option C is wrong since it leaves out A and includes only B, D and E, losing the fundamental statement about powers of 2 that underlies the entire binary system.

Option D:
Option D is correct because it gathers positional value, parity detection, conversion method and the multiple-of-4 criterion while excluding C, which misstates the effect of incrementing a binary number.

The binary number system is defined as a positional system that uses only two symbols, 0 and 1, so its base must be 2. In any positional number system, the base equals the number of distinct digit symbols available. All place values in binary, such as 1, 2, 4, 8 and so on, are powers of 2. Therefore, the only correct base for the binary system among the options is 2.

Option A:
Option A is correct because binary has exactly two symbols and therefore a base of 2. Every binary position corresponds to 2 raised to some power, which shows that the base is 2. Without base 2, the structure of binary place values would not work correctly.

Option B:
Option B refers to base 8, which is used in the octal system, not in binary. Octal digits range from 0 to 7 and thus require eight different symbols. Therefore base 8 cannot describe a system that has only two digits, 0 and 1.

Option C:
Option C refers to base 10, which belongs to the decimal system, the usual system for everyday counting. Decimal uses ten symbols from 0 to 9, unlike binary which uses only 0 and 1. Hence 10 cannot be the base of the binary system.

Option D:
Option D refers to base 16, which is associated with the hexadecimal system. Hexadecimal uses symbols 0–9 and A–F to represent sixteen distinct values. This makes it unsuitable as a base for binary, which has only two symbols.

A base-2 system uses two symbols, typically 0 and 1, to represent numbers. This system is universally known as the binary number system. It forms the foundation of digital electronics and computer logic. Therefore, the correct alternative name for base 2 is binary.

Option A:
Option A is correct because "bi" signifies two, matching the two symbols used in the system. All computations in digital computers are carried out using this binary representation.

Option B:
Option B, octal, refers to base 8 and uses eight symbols from 0 to 7. It does not describe a base-2 system.

Option C:
Option C, decimal, describes base 10 and uses ten symbols from 0 to 9. This is the everyday counting system, not base 2.

Option D:
Option D, hexadecimal, refers to base 16 which uses sixteen symbols, including letters A–F. It is not equivalent to base 2, although it relates to binary through groups of four bits.

The binary system is defined by having two distinct symbols, typically 0 and 1. The base, or radix, of a number system equals the number of distinct digit symbols it uses. Since binary uses exactly two symbols, its base is 2. This is why powers of 2 appear throughout digital design.

Option A:
Option A correctly recognises that the binary system has two digits and hence base 2. All positional weights are powers of 2 in such a system, which aligns with logical signals that are either low or high.

Option B:
Option B, 8, refers to the octal system, which uses digits 0 through 7. It is sometimes used as a compact representation of binary, but it is not the base used directly for bit-level operations.

Option C:
Option C, 10, is the base of the decimal system commonly used in everyday arithmetic. Digital hardware, however, is built on binary logic rather than decimal.

Option D:
Option D, 16, corresponds to the hexadecimal system, which is often used to represent binary values more compactly, but it is not the fundamental base of binary computation.

The leftmost bit in a binary number carries the largest positional weight because it corresponds to the highest power of 2 in the representation. This makes it the most significant bit (MSB). Its value greatly influences the magnitude of the entire number. Therefore, it is correctly referred to as the most significant bit.

Option A:
Option A labels it "low order," which is usually associated with bits near the right side that carry smaller weights. Using low-order terminology for the leftmost bit contradicts standard conventions in digital design.

Option B:
Option B suggests a parity bit, which is a special bit added to detect errors. A parity bit may be placed at either end of a word and is not inherently the leftmost data bit. Thus, parity does not define the leftmost bit of any binary number.

Option C:
Option C is correct because the leftmost bit has the greatest influence on the numeric value and is therefore called the most significant. In signed representations, this bit may also indicate the sign of the number, reinforcing its significance.

Option D:
Option D uses the term "overflow" bit, which typically refers to a flag indicating arithmetic overflow in a status register. It is not a positional name for any particular bit within a numeric representation.

To represent numbers from 0 to 7 inclusive, we need 8 distinct binary patterns. Each additional bit doubles the number of representable patterns, so 1 bit gives 2 values, 2 bits give 4 values and 3 bits give 8 values. Therefore, 3 bits are enough to represent 0 through 7.

Option A:
Option A, 2 bits, can represent only 2^2 = 4 distinct values, namely 0 to 3. This is insufficient for covering the full range 0 to 7.

Option B:
Option B, 4 bits, can represent 16 distinct values, which is more than required. While 4 bits would work, it is not the minimum needed, so it does not answer the question correctly.

Option C:
Option C is correct because 3 bits give exactly 2^3 = 8 different patterns, matching the eight numbers in the range 0 to 7. This makes 3 the minimum bit length required.

Option D:
Option D, 5 bits, provides 32 combinations, which is unnecessarily large for representing only eight values. It does not satisfy the requirement of minimum number of bits.

The expression 2^5 means 2 multiplied by itself five times, which gives 32. In a binary number, the bit in the 2^5 position therefore has a weight of 32. When that bit is set to 1, it contributes 32 to the decimal value of the number.

Option A:
Option A is correct because calculating 2×2×2×2×2 results in 32. This is a fundamental power of two frequently used in computing.

Option B:
Option B, 16, corresponds to 2^4, not 2^5. It therefore has a smaller weight than required.

Option C:
Option C, 64, is 2^6, which is one exponent higher than the requested position. It represents a larger place value than 2^5.

Option D:
Option D, 8, is equal to 2^3 and corresponds to a different bit position in a binary number. It cannot represent the value of the 2^5 place.

The binary number 0001 has a 1 only in the 2^0 place, which has a weight of 1. All higher-order bits are zero, so they do not contribute to the value. Therefore, the decimal equivalent is simply 1.

Option A:
Option A is correct because it reflects a single 1 in the least significant bit position, giving a value of 1. This is the smallest positive nonzero integer representable in binary.

Option B:
Option B, 0, would require all bits to be zero. Since there is a 1 present, the number is not zero.

Option C:
Option C, 2, corresponds to 0010₂, where the 2^1 bit is set instead of the 2^0 bit. This pattern does not match 0001₂.

Option D:
Option D, 4, would be represented by 0100₂. In 0001₂, the 4's position is zero, so 4 cannot be the correct value.

To convert decimal 9 to binary, we look for powers of 2 that sum to 9. Here 9 = 8 + 1, which corresponds to 2^3 and 2^0. This gives a binary pattern with ones in the 8's and 1's places and zeros in the 4's and 2's places, namely 1001₂. Hence 1001 is the correct binary representation.

Option A:
Option A, 1000₂, equals 8 in decimal because it has a 1 only in the 8's place. It does not include the additional 1 needed to sum to 9. Therefore it is too small.

Option B:
Option B, 1010₂, corresponds to 8 + 2 = 10 in decimal. This overshoots the required value of 9 and cannot be the correct representation.

Option C:
Option C is correct because it places ones in the 8's and 1's positions, giving 8 + 1 = 9. Zeros in the 4's and 2's positions indicate there is no contribution from those powers of 2. This matches the decimal value exactly.

Option D:
Option D, 1100₂, equals 8 + 4 = 12, which is greater than 9. The presence of a 1 in the 4's place adds extra value that does not belong to the decimal number 9.