UGC NET Questions (Paper – 1)

The integer part 10₂ equals 2. The fractional part .1011₂ equals 1/2 + 0/4 + 1/8 + 1/16 = 0.6875. Total = 2 + 0.6875 = 2.6875.

Option A:
Option A correctly computes fractional weights and adds them to the integer part.

Option B:
Option B (2.625) misses the 1/16 term and matches 10.101₂.

Option C:
Option C (2.75) would match 10.11₂, not 10.1011₂.

Option D:
Option D (2.5625) corresponds to a different fractional-bit pattern and is too small.

The integer part 100110₂ equals 1×32 + 0×16 + 0×8 + 1×4 + 1×2 + 0×1 = 32 + 4 + 2 = 38. The fractional part .101₂ equals 1/2 + 0/4 + 1/8 = 0.5 + 0.125 = 0.625. Adding both parts yields 38.625. Thus, the decimal equivalent is 38.625.

Option A:
Option A correctly interprets the integer and fractional bits and uses appropriate weights for each bit. The sum 38 + 0.625 = 38.625 is consistent with the structure of the binary number.

Option B:
Option B, 38.75, would require the fractional part to be .11₂ (0.75), which has both first and second fractional bits set. That does not match .101₂.

Option C:
Option C, 38.5, corresponds to a fractional part of .1₂ (0.5), ignoring the contribution of the 2⁻³ bit. This underestimates the true value.

Option D:
Option D, 39.125, suggests both an incorrect integer or fractional part, as 39.125 cannot be obtained from the given combination of bits without changing the pattern.

To convert 101101.011₂ to decimal, we split it into integer and fractional parts. The integer 101101₂ equals 1×32 + 0×16 + 1×8 + 1×4 + 0×2 + 1×1 = 32 + 8 + 4 + 1 = 45. The fractional .011₂ equals 0×1/2 + 1×1/4 + 1×1/8 = 0.25 + 0.125 = 0.375. Adding these gives 45.375. Hence, 45.375 is the correct decimal value.

Option A:
Option A carefully evaluates both the integer and fractional bits to obtain 45.375. It shows correct use of base-2 positional weights, including negative powers for the fractional part. Because both parts sum to 45.375, this option accurately reflects the binary input.

Option B:
Option B, 45.625, would require the fractional part to be .101₂ (0.5 + 0.125) instead of .011₂. Therefore it misinterprets which bits are set in the fractional portion. As a result, it slightly overestimates the correct value.

Option C:
Option C, 45.5, assumes the fractional part contributes only 0.5, which corresponds to .1₂, not .011₂. This ignores one of the lower-order fractional bits and therefore under-represents the actual value of the binary fraction.

Option D:
Option D, 46.375, incorrectly increases the integer part by 1, as if 101110₂ were used instead of 101101₂. This mismatch in the integer component significantly changes the total, so it does not represent the given binary number.

To convert 7F₁₆ to decimal, we use positional weights of base 16. The digit 7 is in the 16¹ place and F represents 15 in the 16⁰ place. So the decimal value is 7×16 + 15×1 = 112 + 15 = 127. Therefore, the correct decimal equivalent of 7F₁₆ is 127.

Option A:
Option A, 119, would correspond to a different combination of hex digits because 119 is not 7×16 + 15. It implies an incorrect evaluation of the place values or the value of F. Thus, it underestimates the contribution of the least significant digit.

Option B:
Option B, 125, is close to the correct value but still results from a miscalculation of 7×16 + 15. It might arise from incorrectly adding 112 + 13 instead of 15, showing an error in interpreting F. Hence 125 does not match the precise conversion.

Option C:
Option C correctly applies the base-16 positional rule: 7×16 = 112 and F = 15, giving a total of 127. This option not only uses the right weights but also interprets F as 15, which is crucial in hexadecimal arithmetic. Therefore it is the accurate decimal representation of 7F₁₆.

Option D:
Option D, 129, overshoots the correct value and would imply the least significant digit contributed 17 instead of 15. That is not possible because a single hex digit ranges from 0 to 15. Therefore, 129 cannot be the correct conversion of 7F₁₆.

To convert 555₈ to decimal, we expand using powers of 8. The digits are 5×8² + 5×8¹ + 5×8⁰ = 5×64 + 5×8 + 5×1 = 320 + 40 + 5 = 365. Thus, the decimal representation of 555₈ is 365.

Option A:
Option A correctly computes each positional contribution and sums them. It uses 64, 8, and 1 as place values and multiplies by 5 each time, yielding a total of 365.

Option B:
Option B, 349, results from an arithmetic miscalculation or incorrect weighting of one of the positions. It does not match the standard expansion of 555₈.

Option C:
Option C, 373, is close but implies that one of the place values or digit weights was slightly misapplied, possibly adding an extra 8 or subtracting a 0 somewhere.

Option D:
Option D, 381, overshoots the correct result and similarly indicates a mis-evaluation of one or more terms in the expansion.

7B₁₆ = 7×16 + 11 = 112 + 11 = 123.

Option A:
Option A correctly interprets B as 11 and sums to 123.

Option B:
Option B is too large and not equal to 112 + 11.

Option C:
Option C is too small and indicates arithmetic error.

Option D:
Option D is too large and incorrect.

The hex number 16D₁₆ has digits 1, 6, and D. Using positional weights, we compute 1×16² + 6×16¹ + 13×16⁰ = 1×256 + 6×16 + 13×1 = 256 + 96 + 13 = 365. Therefore, 16D₁₆ equals 365 in decimal.

Option A:
Option A correctly interprets D as 13 and applies the weights 256, 16, and 1. The sum 256 + 96 + 13 yields 365, matching the result of a proper base-16 to base-10 conversion.

Option B:
Option B, 349, represents a different combination of digit values and weights and does not equal the computed sum of 1×256 + 6×16 + 13. It indicates an arithmetic or mapping error.

Option C:
Option C, 269, underestimates the value and would imply lower contributions from the hex digits than they actually have. It does not correspond to the given pattern.

Option D:
Option D, 285, neither matches the correct total nor any plausible misinterpretation that preserves digit order while using appropriate weights.

19A₁₆ = 1×16² + 9×16¹ + 10×16⁰ = 256 + 144 + 10 = 410.

Option A:
Option A correctly converts each hex digit using powers of 16 and sums to 410.

Option B:
Option B (394) is smaller and implies miscalculation of place values.

Option C:
Option C (402) is also incorrect due to arithmetic error.

Option D:
Option D (418) overshoots the correct total.

The integer part 111₂ equals 1×4 + 1×2 + 1×1 = 7. The fractional part .001₂ equals 0×1/2 + 0×1/4 + 1×1/8 = 1/8 = 0.125. Adding these components yields 7 + 0.125 = 7.125. Therefore, the decimal equivalent of 111.001₂ is 7.125.

Option A:
Option A correctly computes both the integer and fractional contributions and combines them. It uses the appropriate negative powers of 2 for the bits after the point, giving a precise decimal value of 7.125.

Option B:
Option B, 7.25, would require the fractional part to be .01₂ (1/4) rather than .001₂. This misinterprets which bit is set in the fractional portion and hence is not correct.

Option C:
Option C, 7.5, corresponds to a fractional part of .1₂, where only the 1/2 place is set. This does not align with .001₂, which has only the 1/8 position set.

Option D:
Option D, 7.375, equals 7 + 0.375, which would correspond to fractional bits .011₂. Since the given binary has only the last fractional bit as 1, this value is inconsistent.