UGC NET Questions (Paper – 1)

Decimal 0.375 equals 3/8. In binary, the fraction 1/4 is 0.01₂ and 1/8 is 0.001₂, so 3/8 is 1/4 + 1/8. The binary sum of 0.01₂ and 0.001₂ gives 0.011₂. Therefore, 0.011₂ is the exact binary representation of 0.375.

Option A:
Option A is correct because it has ones in the 2^-2 and 2^-3 positions, giving 1/4 + 1/8 = 3/8. This matches the decimal fraction exactly.

Option B:
Option B, 0.001₂, represents 1/8 or 0.125 in decimal, which is too small compared to 0.375. It lacks the 1/4 component.

Option C:
Option C, 0.01₂, equals 1/4 or 0.25 in decimal, still smaller than 0.375. It accounts for only one of the required fractional parts.

Option D:
Option D, 0.111₂, equals 1/2 + 1/4 + 1/8 = 7/8 or 0.875 in decimal, which is much larger than 0.375.

Decimal 255 is the largest number representable with eight bits set to 1, that is 11111111₂. Grouping the bits in fours gives 1111 and 1111, which correspond to F and F in hexadecimal. Thus, the hex representation of 255 is FF₁₆.

Option A:
Option A, EF₁₆, corresponds to 14×16 + 15 = 239, which is less than 255. It has a different high-order digit than FF.

Option B:
Option B, FE₁₆, represents 15×16 + 14 = 254, one less than 255. It is frequently seen in contexts like signed ranges, but it is not equal to 255.

Option C:
Option C is correct because FF₁₆ equals 15×16 + 15 = 255. It reflects a full set of ones in binary and is widely used as a maximum byte value.

Option D:
Option D, EE₁₆, equals 14×16 + 14 = 238, which is significantly smaller than 255. Its binary form does not have all bits set.

If the exact mathematical result of an operation is too large or too small to be represented within the range of the chosen number system, we say overflow has occurred. In integer arithmetic, this often happens when adding two large positive numbers or two large negatives. The stored result then wraps around or is truncated, losing the true value. Thus, the condition is called overflow.

Option A:
Option A, underflow, refers to results that are too close to zero to be represented in a normalized floating-point format, not generally to exceeding the upper or lower integer range. It is a different phenomenon from overflow.

Option B:
Option B, wraparound, describes a behavior that may be observed after overflow occurs, such as high values wrapping to low values. However, it is not the formal name of the error condition itself.

Option C:
Option C is correct because overflow is the standard term for exceeding the representable range in a fixed-width number system. It signals that the numeric result is no longer reliable.

Option D:
Option D, truncation, refers to cutting off digits beyond a certain point, usually to limit precision. Although truncation can introduce error, it is not synonymous with exceeding range limits.

The standard rule for forming the 2's complement (negative) of a binary number is to first take its 1's complement by inverting all bits and then add 1. This process yields a representation that has desirable arithmetic properties, such as allowing addition to handle subtraction implicitly. Therefore, after the 1's complement, we add 1.

Option A:
Option A, subtract 1, would not produce the correct 2's complement. Subtracting 1 after inversion would generate a different pattern that does not correspond to the intended negative.

Option B:
Option B, add 2, would overshoot the value needed for 2's complement, leading to an incorrect representation except in trivial cases.

Option C:
Option C is correct because inverting the bits and adding 1 creates a pattern that when added to the original number produces zero with a carry, modeling negation. This is the basis of 2's complement arithmetic in computers.

Option D:
Option D, shift right, is related to division by 2 or logical shifts but has nothing to do with forming the negative of a 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.

Converting 53 to binary involves finding powers of 2 that sum to 53. The largest power less than 53 is 32, leaving 21. The next is 16, leaving 5. Then 4 leaves 1, and finally 1 leaves 0. This corresponds to bits for 32, 16, 4 and 1, giving 110101₂.

Option A:
Option A, 110011₂, equals 32 + 16 + 2 + 1 = 51, which is smaller than 53. It lacks the 4's contribution that is present in the correct representation.

Option B:
Option B is correct because 110101₂ represents 32 + 16 + 4 + 1 = 53. All necessary powers of 2 are present to match the decimal number.

Option C:
Option C, 110111₂, equals 32 + 16 + 4 + 2 + 1 = 55, which overshoots the desired value by 2.

Option D:
Option D, 111101₂, corresponds to 32 + 16 + 8 + 4 + 1 = 61. This includes an 8's bit that should not be present when representing 53.

In octal 157₈, the digits are in the 8^2, 8^1 and 8^0 positions. The value is 1×64 + 5×8 + 7×1. This computes as 64 + 40 + 7 = 111. Thus, the decimal equivalent of 157₈ is 111.

Option A:
Option A, 107, is smaller than 111 and would result from a different combination, such as 151₈. It does not match the actual weighted sum.

Option B:
Option B, 108, also fails to equal 64 + 40 + 7. It appears to be a guess near the correct answer but is not supported by correct calculation.

Option C:
Option C, 110, is close but still one less than the correct value of 111. This again indicates a small arithmetic error.

Option D:
Option D is correct because the precise computation of 1×64 + 5×8 + 7 gives exactly 111. This follows the general method of converting from octal to decimal.

Decimal 0.75 is equal to 3/4. In binary, 0.11₂ represents 1/2 plus 1/4, which sums to 3/4. Therefore, the exact binary representation of 0.75 is 0.11₂.

Option A:
Option A, 0.1₂, equals 1/2 or 0.5 in decimal. This is smaller than 0.75 and does not account for the additional 1/4.

Option B:
Option B, 0.01₂, equals 1/4 or 0.25 in decimal, which is far from 0.75. It lacks the 1/2 component entirely.

Option C:
Option C, 0.101₂, equals 1/2 + 0 + 1/8 = 5/8 or 0.625 in decimal. This is closer but still not equal to 0.75.

Option D:
Option D is correct because 0.11₂ combines 1/2 and 1/4 to give exactly 3/4. This is a standard example of a terminating binary fraction.

Signed magnitude representation uses one bit, typically the most significant bit, to indicate the sign of the number. The remaining bits represent the magnitude and are the same for both positive and negative versions. Thus, negative numbers differ from their positive counterparts only in the sign bit. The magnitude bits remain unchanged.

Option A:
Option A, magnitude, is incorrect because the magnitude bits for +5 and -5 are identical in signed magnitude representation. Only the sign indicator changes.

Option B:
Option B, parity, refers to an error-checking mechanism where an extra bit indicates whether the number of ones is even or odd. It is not the primary difference between positive and negative values in this scheme.

Option C:
Option C is correct because the sign bit explicitly encodes whether the number is positive or negative while leaving the magnitude intact. This is why the representation is called signed magnitude.

Option D:
Option D, overflow, describes a condition when a calculation exceeds the representable range. It is typically recorded in a separate status flag rather than being the distinguishing bit between positive and negative values.

For 7B₁₆, the digit 7 is in the 16's place and B represents 11 in decimal. The value is 7×16 + 11×1. Calculating 7×16 gives 112 and adding 11 yields 123. Thus, the decimal equivalent of 7B is 123.

Option A:
Option A, 115, would result from a different combination of hex digits such as 73₁₆. It does not agree with the correct weighted sum for 7B.

Option B:
Option B, 119, is smaller than 123 and would require less contribution from either the 16's place or the units place. The digits 7 and B fix those contributions.

Option C:
Option C, 121, is close but still two less than 123, indicating a miscalculation. Accurate computation shows the total is 123, not 121.

Option D:
Option D is correct because 7×16 + 11 = 112 + 11 = 123. This is a direct application of hex place values to obtain the decimal result.

With a fixed number of digit positions, a higher base allows each position to hold more distinct values. This expands the total number of combinations, thereby increasing the range of representable values. For example, three digits in base 2 yield 8 values, while three digits in base 10 yield 1000 values. Hence, increasing the base increases the representable range.

Option A:
Option A is correct because the total number of distinct patterns is base^digits, so raising the base while holding digits constant raises this power. This mathematical relationship directly implies an increased range.

Option B:
Option B, decreases, is the opposite of what actually happens. Decreasing the base reduces the number of choices per position and thus shrinks the range.

Option C:
Option C, keeps unchanged, would only be true if either the base or the number of digits did not change. Changing the base without altering digits affects the range.

Option D:
Option D, first increases then decreases, does not apply here because the formula base^digits grows monotonically as the base increases for a fixed positive number of digits.

Dividing 125 by 16 gives a quotient of 7 and a remainder of 13. In hexadecimal, 13 is represented by D. Therefore, 125 in decimal is written as 7D₁₆. This reflects 7×16 + 13 = 112 + 13 = 125.

Option A:
Option A, 7C₁₆, represents 7×16 + 12 = 124, which is one less than 125. The remainder is too small.

Option B:
Option B is correct because 7×16 = 112 and adding the remainder 13, mapped to D, gives 125. This matches the standard hex conversion process.

Option C:
Option C, 7E₁₆, equals 7×16 + 14 = 126, one more than 125. It overshoots the required value.

Option D:
Option D, 7F₁₆, corresponds to 7×16 + 15 = 127, which is even further from 125.

Binary 1011011₂ can be converted by summing its positional values. From right to left, the bits correspond to 1, 2, 4, 8, 16, 32 and 64. The pattern 1011011 has ones in positions 64, 16, 8, 2 and 1. This gives 64 + 16 + 8 + 2 + 1 = 91, so the decimal equivalent is 91.

Option A:
Option A is correct because it matches the sum of the active binary weights 64, 16, 8, 2 and 1. These add up exactly to 91, faithfully representing the binary pattern.

Option B:
Option B, 89, is two less than the correct value and would require a different combination of bits. For example, switching off the 2's place would yield 89, but in 1011011 that bit is set.

Option C:
Option C, 93, is greater than 91 and would require additional active bits or different positions. The given pattern does not support that sum.

Option D:
Option D, 95, also exceeds the correct total. It would need extra contributions, such as a 4's bit, which is not set in 1011011₂.

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.

Decimal 101 can be expressed as 64 + 32 + 4 + 1, corresponding to 2^6, 2^5, 2^2 and 2^0. This means that in binary the bits for these positions are 1 while others are 0. Writing them in order gives 1100101₂. Thus, 1100101 is the correct binary representation of 101.

Option A:
Option A, 1100011₂, equals 64 + 32 + 2 + 1 = 99 in decimal, which is smaller than 101 and lacks the 4's place contribution.

Option B:
Option B, 1100100₂, equals 64 + 32 + 4 = 100 in decimal and is one less than 101. It misses the 1's bit.

Option C:
Option C is correct because 1100101₂ equals 64 + 32 + 4 + 1 = 101, matching the required decimal value exactly.

Option D:
Option D, 1100111₂, equals 64 + 32 + 4 + 2 + 1 = 103, which overshoots the target. It has an extra 2's bit set.