UGC NET Questions (Paper – 1)

For 458₉: 4×9² + 5×9¹ + 8×9⁰ = 4×81 + 5×9 + 8 = 324 + 45 + 8 = 377. Therefore, the decimal equivalent is 377.

Option A:
Option A correctly applies base-9 positional weights and sums to 377.

Option B:
Option B (365) is too small and reflects an error in weighting or addition.

Option C:
Option C (389) is too large and would require extra value not present in the digits.

Option D:
Option D (401) is far above the correct sum and cannot arise from 4,5,8 with powers 81,9,1.

For 321₄, the digits correspond to 3×4² + 2×4¹ + 1×4⁰ = 3×16 + 2×4 + 1 = 48 + 8 + 1 = 57. Therefore, the decimal representation of 321₄ is 57.

Option A:
Option A correctly computes the contributions from each digit using powers of 4. It sums the products to obtain 57, consistent with the rules of base-4 positional notation.

Option B:
Option B, 61, arises from an incorrect addition or mis-weighting of digits and does not match the computed sum.

Option C:
Option C, 45, underestimates the result and suggests one of the digits was underweighted or omitted.

Option D:
Option D, 49, is too small and fails to account for the full value contributed by the highest place.

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.

The decimal number 255 is one less than 256, which is 16². In hexadecimal, 16² corresponds to 100₁₆, so 255 must be written as 1 less than that, which is FF₁₆. Each F represents 15, and 15×16 + 15 = 240 + 15 = 255. Thus, FF₁₆ is the correct hex representation of 255₁₀.

Option A:
Option A correctly uses the relationship between powers of 16 and the decimal system. It recognises that FF₁₆ is the maximum value in two hex digits, analogous to 255 in decimal. This makes it a standard boundary value in byte-oriented systems.

Option B:
Option B, F0₁₆, equals 15×16 + 0 = 240 in decimal, which is smaller than 255. It neglects the contribution of the least significant digit required to reach 255. Hence it is an underestimation.

Option C:
Option C, FE₁₆, represents 15×16 + 14 = 254 in decimal, which is close but still one less than the required value. This corresponds to 254₁₀, not 255₁₀.

Option D:
Option D, F1₁₆, corresponds to 15×16 + 1 = 241 in decimal, which again does not equal 255. Since only FF₁₆ gives 255, this option cannot be correct.

In base-4, the first digit after the point has weight 4⁻¹ = 1/4, so 0.3₄ = 3×(1/4) = 3/4 = 0.75.

Option A:
Option A correctly uses fractional place weight 1/4 and multiplies by 3 to get 0.75.

Option B:
Option B would match 0.2₄.

Option C:
Option C would match 0.1₄.

Option D:
Option D does not equal 3/4 and is not the value of 0.3₄.

For 324₅, the digits correspond to 3×5² + 2×5¹ + 4×5⁰. This evaluates as 3×25 + 2×5 + 4×1 = 75 + 10 + 4 = 89. Hence, the decimal equivalent of 324₅ is 89.

Option A:
Option A correctly uses the powers 25, 5, and 1 for the positions and multiplies by the corresponding digits. The sum 89 matches the structure of 324₅, confirming the validity of the conversion.

Option B:
Option B, 94, would require additional value or a misinterpretation of one digit or weight, and does not equal the evaluated expression.

Option C:
Option C, 86, underestimates the correct value and would arise from a missing or misweighted term in the expansion.

Option D:
Option D, 79, is significantly lower than the correct total and thus does not represent the same base-5 number.

For 452₆: 4×6² + 5×6¹ + 2×6⁰ = 4×36 + 5×6 + 2 = 144 + 30 + 2 = 176. Therefore, the decimal equivalent is 176.

Option A:
Option A correctly applies base-6 positional weights and sums to 176.

Option B:
Option B (166) is too small and indicates an error in weighting or addition.

Option C:
Option C (182) is too large and would require extra value not present in the numeral.

Option D:
Option D (174) is slightly low and results from incorrect arithmetic.

For 352₇, compute 3×7² + 5×7¹ + 2×7⁰ = 3×49 + 5×7 + 2 = 147 + 35 + 2 = 184. Therefore, the decimal equivalent is 184.

Option A:
Option A is correct because it follows standard positional conversion from base-7 to decimal and gives 184.

Option B:
Option B (179) would result from incorrect weighting or arithmetic and does not match 147 + 35 + 2.

Option C:
Option C (173) underestimates the correct sum and indicates an error in place-value contribution.

Option D:
Option D (191) overshoots the correct total and would require larger digit-weight products than present.

The 8-bit pattern 11111111₂ can be grouped into two nibbles: 1111₂ and 1111₂. Each nibble corresponds to the hex digit F (decimal 15). Combining them gives FF₁₆. Therefore, 11111111₂ maps to FF₁₆.

Option A:
Option A correctly uses nibble-to-hex mapping: 1111₂ = F, so 11111111₂ = FF₁₆.

Option B:
Option B, FE₁₆, would correspond to 11111110₂ and thus to decimal 254, one less than the maximum.

Option C:
Option C, F0₁₆, corresponds to 11110000₂, which is decimal 240 and not the full 8-bit maximum.

Option D:
Option D, 0F₁₆, corresponds to 00001111₂, a value much smaller than 11111111₂.