UGC NET Questions (Paper – 1)

Each octal digit represents 3 binary bits because base 8 equals 2³. For a 12-bit binary number, we need 12 ÷ 3 = 4 octal digits to represent all possible combinations. This grouping covers all 12 bits exactly without wasted digits or missing bits. Therefore, the minimum number of octal digits required is 4, which makes option B correct.

Option A:
Option A, 3 octal digits, cover only 9 bits, which is insufficient for 12-bit numbers.

Option B:
Option B is correct because 4 octal digits represent 12 bits exactly (4×3).

Option C:
Option C, 5 digits, cover 15 bits and are not minimal.

Option D:
Option D, 6 digits, cover 18 bits and are far from minimal.

Each octal digit is converted to a 3-bit binary group. For octal 27₈, the digit 2 becomes 010₂ and the digit 7 becomes 111₂. Combining these groups gives 010111₂. This pattern represents the correct binary equivalent of 27₈.

Option A:
Option A, 001111₂, corresponds to octal 17₈ and not 27₈. Its first group 001₂ represents 1 rather than 2.

Option B:
Option B, 011111₂, would map to octal 37₈, since 011₂ equals 3 and 111₂ equals 7. This is different from the given octal number.

Option C:
Option C, 011011₂, converts back to octal 33₈, as both groups 011₂ equal 3. It clearly does not match 27₈.

Option D:
Option D is correct because 2 maps to 010₂ and 7 maps to 111₂. Concatenating those groups produces the binary representation 010111₂ for 27₈.

Octal is a base-8 system, and 8 equals 2^3. This means three binary bits can represent exactly one octal digit. To convert binary to octal, we therefore group binary digits in sets of three from the right, converting each group to its octal equivalent.

Option A:
Option A, grouping by 2, gives only 4 different combinations per group, which is insufficient for representing all 8 octal digits. It would not align neatly with base 8.

Option B:
Option B is correct because grouping into threes uses the relationship 8 = 2^3. Each group of three bits corresponds directly to a single octal digit.

Option C:
Option C, grouping by 4, is appropriate for hexadecimal conversion since 16 = 2^4, not for octal. Using four-bit groups for octal would lead to ambiguous mapping.

Option D:
Option D, grouping by 8, would create groups representing 256 different values, which is unnecessarily large for mapping to only 8 octal symbols.

To convert binary to octal, we group bits in sets of three from right to left. For 111100101₂, we group as 111 100 101. These triplets correspond to 7, 4, and 5 in octal respectively. Placing these digits together, we obtain 745₈ as the octal representation.

Option A:
Option A accurately reads each 3-bit group: 111₂→7, 100₂→4, and 101₂→5. This direct mapping is standard for binary–octal conversion and yields 745₈ from the given binary sequence.

Option B:
Option B, 753₈, implies the middle group is 101₂ instead of 100₂, altering the middle digit from 4 to 5. This does not match the actual bits in the binary number, so it is incorrect.

Option C:
Option C, 765₈, changes both the middle and last triplets, implying 110₂ and 101₂ or other combinations, which are inconsistent with the original 111100101₂. Thus, it does not correspond to the given pattern.

Option D:
Option D, 755₈, also modifies the binary grouping, suggesting different 3-bit values than actually present. Since each octal digit must match a specific 3-bit group, this representation is invalid for the given binary number.

In hexadecimal, 9 corresponds to 1001₂ and C corresponds to 1100₂. Combining these 4-bit groups yields 1001 1100₂. Therefore, the binary equivalent of 9C₁₆ is 10011100₂.

Option A:
Option A, 10011010₂, would correspond to hex 9A₁₆, not 9C₁₆, because the last four bits represent 1010₂.

Option B:
Option B, 10010110₂, maps to hex 96₁₆, as the last group 0110₂ equals 6. It does not match C, which is 1100₂.

Option C:
Option C, 10011000₂, corresponds to hex 98₁₆, where the last group 1000₂ equals 8. This again differs from C.

Option D:
Option D is correct because 1001₂ for 9 and 1100₂ for C concatenate to 10011100₂. This is the standard grouping method for hex to binary conversion.

To convert 1000₂ to octal, we group the binary digits into sets of three from the right. Padding on the left gives 001 000. The group 001₂ equals 1 and 000₂ equals 0, so the octal number is 10₈. This means binary 1000 corresponds to octal 10.

Option A:
Option A, 7₈, corresponds to binary 111₂, not 1000₂. It represents a different decimal value and fails to match the binary input.

Option B:
Option B is correct because 1000₂ is 8 in decimal and 8 is written as 10₈ in octal. Grouping 001 000 demonstrates this equivalence clearly.

Option C:
Option C, 11₈, represents decimal 9 because it equals 1×8 + 1. Binary 1000₂ represents 8, so 11₈ is too large and does not match.

Option D:
Option D, 8, is not a valid single octal digit since octal digits range only from 0 to 7. Therefore, 8 cannot be the correct octal representation for any value.

To convert 111100₂ to hexadecimal, we group the binary digits into sets of four from the right. Padding on the left gives 0011 1100. The left group 0011₂ equals 3 and the right group 1100₂ equals 12, represented as C in hex. Therefore, the hexadecimal equivalent is 3C₁₆.

Option A:
Option A, 2C₁₆, would correspond to binary 0010 1100, which is different from 111100₂. The leading group 0010₂ equals 2, not 3, so this option does not match.

Option B:
Option B, 2F₁₆, converts to 0010 1111₂, which differs from 111100₂ both in the leading bits and the trailing pattern. Thus it cannot represent the given binary number.

Option C:
Option C is correct because grouping 111100₂ as 0011 1100₂ gives hex digits 3 and C. This matches the exact structure and value of the original binary pattern.

Option D:
Option D, 3F₁₆, corresponds to 0011 1111₂, which has an extra 1 at the least significant position. That would represent 63 in decimal instead of the 60 represented by 111100₂.

To convert 101011₂ to hexadecimal, we group bits in sets of four from the right, padding with zeros on the left if needed. This gives 0010 1011. The left group 0010₂ is 2 and the right group 1011₂ is 11, represented by B in hexadecimal. Thus, the hexadecimal equivalent is 2B₁₆.

Option A:
Option A, 2A₁₆, would correspond to 0010 1010₂, where the right group is 1010₂ = 10. This differs from the given binary number, whose last four bits are 1011₂.

Option B:
Option B is correct because the grouping 0010 1011₂ maps directly to the hex digits 2 and B. This representation preserves the value of 101011₂ accurately.

Option C:
Option C, 1B₁₆, would correspond to 0001 1011₂, which has a different leading group and thus a smaller value.

Option D:
Option D, 1D₁₆, translates to 0001 1101₂, where the right group 1101₂ differs from the original 1011₂. This makes it an incorrect conversion.

Each hexadecimal digit represents 16 distinct values, which equals 2^4. Therefore, four binary bits map exactly to one hex digit. When converting binary to hexadecimal, we group the binary digits into sets of four starting from the right so that each group converts cleanly to a single hex digit.

Option A:
Option A is correct because grouping in fours aligns with the 2^4 relationship between binary and hexadecimal. This method ensures a direct and lossless mapping.

Option B:
Option B, grouping by 2, would only cover 2^2 = 4 values per group, which does not match the 16 possibilities for a hex digit. It would require extra steps to combine groups.

Option C:
Option C, grouping by 3, corresponds to octal conversion where each group of three bits represents one octal digit. It is not suitable for hexadecimal.

Option D:
Option D, grouping by 8, gives 2^8 = 256 values per group, which is much larger than needed for a single hex digit and does not provide a clean one-to-one mapping.

Each hexadecimal digit represents exactly 4 binary bits because base 16 equals 2⁴. To encode a 20-bit binary number, we divide 20 by 4 and obtain 5 hex digits. No fewer digits can cover 20 bits, while exactly 5 hex digits provide 5×4 = 20 bits of capacity. Therefore, the minimum number of hexadecimal digits required is 5, so option D is correct.

Option A:
Option A, 4 hex digits, cover only 16 bits, so they are insufficient for 20 bits.

Option B:
Option B, 6 hex digits, cover 24 bits; this is not minimum.

Option C:
Option C, 8 hex digits, cover 32 bits; far from minimum.

Option D:
Option D correctly matches 20 bits to 5 hex digits using 4 bits per hex digit.

To convert binary to hexadecimal, we group the binary digits into sets of four from right to left and convert each group to its hex digit. The binary 110101110010₂ groups as 11 0101 1100 10, which we pad on the left to 0011 0101 1100 10; correctly grouping from right gives 1101 0111 0010. These nibbles correspond to D,7,2 respectively, forming D72₁₆. Therefore, D72₁₆ is the correct hexadecimal representation of 110101110010₂.

Option A:
Option A correctly interprets 1101₂ as D, 0111₂ as 7, and 0010₂ as 2, giving D72₁₆. The grouping into nibbles and mapping into hexadecimal digits follows the standard binary–hex conversion rule. Hence, this option accurately captures the value of the given binary number in base 16.

Option B:
Option B, D62₁₆, implies the middle nibble is 0110₂ instead of 0111₂, which changes the value. This would correspond to a different binary pattern than 110101110010₂. Therefore, D62₁₆ does not match the given binary sequence and is incorrect.

Option C:
Option C, D7A₁₆, takes the last nibble as 1010₂ (A) rather than 0010₂ (2), implying different trailing bits. This would require the binary number to end in 1010, not 0010 as provided. Thus, D7A₁₆ is inconsistent with the original binary digits.

Option D:
Option D, C72₁₆, implies the first nibble is 1100₂ (C) instead of 1101₂ (D), which changes the most significant bits. This modifies the overall magnitude of the number and no longer matches 110101110010₂. Hence, C72₁₆ is not the correct conversion of the given binary number.

Each octal digit converts to a 3-bit binary pattern. For 73₈, the digit 7 maps to 111₂ and the digit 3 maps to 011₂. Combining these gives 111011₂. This pattern is the binary equivalent of the octal number 73.

Option A:
Option A is correct because 7→111₂ and 3→011₂, so 73₈ becomes 111011₂. This maintains the positional values exactly across bases.

Option B:
Option B, 111101₂, would represent octal 75₈, since 101₂ corresponds to 5. It therefore does not match the given octal number.

Option C:
Option C, 110111₂, maps back to octal 67₈, with groups 110₂ and 111₂ representing 6 and 7. This differs from 73₈.

Option D:
Option D, 101111₂, would correspond to octal 57₈. The grouping of bits does not align with 73₈.

To convert 101110₂ to octal, we group bits into sets of three from the right, giving 101 110. The group 101₂ equals 5 and the group 110₂ equals 6. Combining these yields octal 56₈. Thus, the correct octal equivalent is 56.

Option A:
Option A, 54₈, would correspond to binary 101100₂, where the last group is 100 instead of 110. This does not match the original binary pattern.

Option B:
Option B, 55₈, converts to 101101₂ and differs in the last bit. The original binary ends with 110, not 101, so 55₈ is incorrect.

Option C:
Option C is correct because grouping 101110₂ as 101 and 110 gives octal digits 5 and 6. The combined octal number 56 precisely represents the original binary number.

Option D:
Option D, 57₈, would correspond to binary 101111₂, which has an extra 1 at the end. This changes the original value and makes the option invalid.