UGC NET Questions (Paper – 1)

Decimal 8 can be expressed in octal as one group of eight, which is 10₈. In octal, the place values are 1, 8, 64 and so on, similar to how decimal uses 1, 10 and 100. Thus a single 8 corresponds to 1 in the eights place and 0 in the ones place, giving 10₈.

Option A:
Option A, 7₈, equals seven in decimal, which is one less than 8. It does not include the necessary value in the eights place.

Option B:
Option B, 11₈, equals 1×8 + 1 = 9 in decimal, which is greater than 8. It overshoots the correct value and therefore cannot be the right representation.

Option C:
Option C is correct because 10₈ means 1×8 + 0×1 = 8. This exactly matches the decimal value and follows the octal place value structure.

Option D:
Option D, 8, is not a valid single octal digit since octal digits range from 0 to 7. Writing 8 as a digit in an octal number would violate the base-8 rules.

To convert 110111010101 to decimal, we expand it using powers of two. The bit pattern corresponds to 2^11 + 2^10 + 2^8 + 2^7 + 2^6 + 2^4 + 2^2 + 2^0. This sum is 2048 + 1024 + 256 + 128 + 64 + 16 + 4 + 1 = 3541. Therefore, the decimal equivalent of 110111010101₂ is 3541.

Option A:
Option A, 3521, is smaller than the correct sum. If we re-evaluate the place values, the powers of two do not add up to 3521. This number reflects a slightly different pattern of bits than the one given, so it cannot be correct.

Option B:
Option B, 3540, is just one less than the actual decimal value. A correct conversion must match the exact weighted sum of all set bits. Since the correct expansion yields 3541, this near value is not acceptable.

Option C:
Option C, 3451, changes the hundreds digit significantly. This would require a very different combination of powers of two. Because the given bit pattern clearly sums to 3541, 3451 is inconsistent with the binary representation.

Option D:
Option D, 3541, exactly matches the value obtained from summing the weights of all ones. Both manual calculation and calculator verification confirm this result. Hence, this option correctly represents the decimal value of the given binary number.

A three-digit octal number has the form abc where each digit ranges from 0 to 7. The largest such number is 777 in base eight. Its decimal value is 7×8^2 + 7×8 + 7 = 7×64 + 56 + 7 = 448 + 63 = 511. Therefore, 511 is the maximum decimal value representable by a three-digit octal number.

Option A:
Option A, 255, is the maximum decimal value for an 8-bit unsigned binary pattern, not for three-digit octal. An octal number 377 corresponds to 255, which is smaller than the maximum 777. Hence, 255 does not represent the full range of three-digit octal numbers.

Option B:
Option B, 343, is just an intermediate decimal value and has no special significance as a maximum here. When converting 777 from octal, we obtain 511, which is clearly larger than 343. Thus, 343 cannot be the maximum decimal value for three-digit octal numbers.

Option C:
Option C, 512, is 1 greater than 511 and equals 8^3. It represents the total count of distinct three-digit octal combinations starting from 000. However, since numbering starts at 0, the highest representable value is 512 - 1 = 511, not 512.

Option D:
Option D, 511, results from converting 777₈ using place values 64, 8, and 1. The calculation 7×64 + 7×8 + 7 = 511 confirms it is the largest decimal value achievable. Hence, it is the correct maximum for three-digit octal numbers.

To convert 1111100111₂ to decimal, we sum the powers of two corresponding to each 1 bit. The pattern represents 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^2 + 2^1 + 2^0. This gives 512 + 256 + 128 + 64 + 32 + 4 + 2 + 1 = 999. Therefore, the decimal value of 1111100111₂ is 999.

Option A:
Option A, 995, is slightly smaller than the correct sum. If we recompute the place values, the total clearly comes to 999, not 995. Hence, this option does not preserve the exact equality required for base conversion.

Option B:
Option B, 997, is also close but still not equal to 999. A correct binary-to-decimal conversion must yield an exact match. Because the powers of two add up precisely to 999, 997 is not the correct result.

Option C:
Option C, 999, coincides exactly with the sum of all relevant powers of two. Both manual expansion and calculator verification confirm this equality. Thus, it is the correct decimal equivalent of the given binary number.

Option D:
Option D, 1001, exceeds the correct value by 2. It would correspond to a different combination of bits in the binary representation. Therefore, 1001 is not consistent with the pattern 1111100111₂.

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.

In hexadecimal, each digit represents a power of sixteen. The number 3FA means 3×16^2 + 15×16^1 + 10×16^0. This equals 3×256 + 15×16 + 10 = 768 + 240 + 10. The sum 768 + 240 + 10 is 1018, so 3FA corresponds to 1018 in decimal.

Option A:
Option A, 1000, would correspond to a different hexadecimal pattern. Converting 3FA explicitly shows the result 1018, not 1000. Therefore, while 1000 is a familiar decimal number, it is not the correct image of 3FA in base ten.

Option B:
Option B, 1010, again does not match the computed value. If 3FA were 1010, the place value calculation would need to yield that sum. Since 768 + 240 + 10 clearly equals 1018, 1010 cannot be the correct decimal equivalent.

Option C:
Option C, 1018, directly matches the detailed place value calculation for 3FA. It correctly uses the weights 16^2, 16^1, and 16^0 and the digit values 3, 15, and 10. Because the numerical sum is consistent, this is the valid decimal conversion.

Option D:
Option D, 1024, is equal to 2^10 and often appears in computing contexts. However, it corresponds to 400 in hexadecimal, not 3FA. Thus, 1024 is unrelated to the specific digits in 3FA and is not the correct answer.

In a weighted code, each digit position carries a predetermined weight, usually a power of the radix. Positional number systems like decimal and binary are weighted because the value of a digit depends on both its magnitude and its position. Therefore, such systems are described as weighted codes.

Option A:
Option A correctly identifies that fixed positional weights, such as 1, 10, 100 in decimal or 1, 2, 4, 8 in binary, define a weighted code. The overall value is found by summing the products of digits and their weights, which is characteristic of these systems.

Option B:
Option B, non-weighted, refers to codes like Gray code or some error-detecting codes where positions do not correspond to simple powers of a base. These systems do not use fixed positional weights, so this term is not appropriate here.

Option C:
Option C, redundant, describes codes that use more bit patterns than strictly necessary to represent a set of values. While some weighted codes can be redundant, redundancy is not the defining feature in this context.

Option D:
Option D, residue, refers to codes representing numbers using modular arithmetic, such as residue number systems. These systems behave differently from standard positional weighted codes.

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.

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.

The numeral 121 in base b expands as 1×b^2 + 2×b + 1. Setting this equal to 16 gives b^2 + 2b + 1 = 16. Simplifying, we get b^2 + 2b - 15 = 0. Solving this quadratic yields b = 3 or b = -5, and only the positive base 3 is valid, so b = 3.

Option A:
Option A, base 2, would make 121 equal to 1×2^2 + 2×2 + 1 = 9, not 16.

Option B:
Option B, base 4, gives 1×4^2 + 2×4 + 1 = 25, not 16.

Option C:
Option C, base 3, yields 1×3^2 + 2×3 + 1 = 16, matching the condition.

Option D:
Option D, base 5, yields 36, far from 16.

In 345, the digit 5 is in the units place, 4 in the tens place and 3 in the hundreds place. The tens place corresponds to 10^1, so the value of 4 there is 4×10^1. Therefore, the exponent for the tens place is 1.

Option A:
Option A, exponent 0, refers to the units place where the digit 5 resides. Assigning 10^0 to the tens place would ignore the role of tens in the number.

Option B:
Option B, exponent 2, gives 10^2 = 100, which corresponds to the hundreds place occupied by digit 3. It does not describe the tens place.

Option C:
Option C, exponent 3, would mean a place value of 1000, which is not present in the three-digit number 345. This exponent is not related to the digit 4 in the tens position.

Option D:
Option D is correct because the tens place always carries the weight of 10^1 in decimal positional notation. Thus, 4 in that position contributes 40 to the total value.

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.

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.