UGC NET Questions (Paper – 1)

The decimal system uses ten symbols, namely 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Because a positional system's base equals the number of distinct digit symbols, the base of decimal is 10. Therefore, the number of distinct digit symbols in decimal is 10.

Option A:
Option A, 9, would omit one symbol and cannot represent all decimal digits. Since we clearly use ten distinct symbols, this count is too small.

Option B:
Option B is correct because counting from 0 to 9 gives ten unique digits used in everyday arithmetic. This property defines decimal as a base-10 system.

Option C:
Option C, 8, would correspond to an octal system, not to decimal. Octal digits range from 0 to 7 only.

Option D:
Option D, 16, describes the hexadecimal system, which uses digits 0–9 and letters A–F, not the decimal system.

Everyday counting and most written arithmetic activities use the decimal system. This system is based on ten symbols, 0 through 9, and is therefore called a base-10 or decimal system. Historically it is linked to humans having ten fingers, which made grouping in tens natural. Thus, the decimal system is the usual human counting system.

Option A:
Option A is correct because decimal notation underlies standard arithmetic taught in schools and used in currencies and measurements. Its base 10 structure is reflected in common units such as tens, hundreds and thousands. This makes it the default system for human counting.

Option B:
Option B, binary, is primarily used inside digital circuits and computer processors where two distinct states are easy to implement. It is not the system used in ordinary human counting in daily life.

Option C:
Option C, octal, uses base 8 with digits 0 to 7 and is occasionally used as a shorthand for binary. However, it is not the dominant system for general human activities.

Option D:
Option D, hexadecimal, is a base-16 system widely used in computing for compact representation of binary values, but it is not used for routine human counting tasks.

The decimal system is the standard system used in everyday counting and arithmetic. It employs ten distinct digit symbols from 0 to 9. Because the number of symbols in a positional system equals its base, the base of the decimal system is 10. All place values are powers of 10 such as 1, 10, 100 and 1000.

Option A:
Option A suggests base 2, which is used for binary and involves only digits 0 and 1. This is not consistent with decimal, which uses more symbols. Therefore base 2 cannot be correct.

Option B:
Option B proposes base 4, which would require exactly four distinct digits, often 0 to 3. Since decimal requires ten, base 4 does not match its structure.

Option C:
Option C indicates base 8, which corresponds to the octal system with digits 0–7. It fails to accommodate the digits 8 and 9 that appear in decimal notation.

Option D:
Option D is correct because decimal uses ten symbols and is defined as a base-10 positional system. Every position represents a power of 10, confirming that 10 is its base.

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.

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 expression 300 + 40 + 5 represents the sum of hundreds, tens and units. Adding them stepwise, 300 + 40 = 340 and then 340 + 5 = 345. This matches the standard place value representation of the three-digit number 345. Hence, the combined value is 345.

Option A:
Option A correctly aligns each term with its place value: 300 in the hundreds place, 40 in the tens place and 5 in the ones place. Adding them produces 345, which is the number whose digits reflect these contributions.

Option B:
Option B, 354, rearranges the tens and ones positions compared to the calculation and cannot be produced by this particular sum of components.

Option C:
Option C, 435, places the 4 in the hundreds place instead of the tens place, which would require 400 rather than 40 as one of the terms.

Option D:
Option D, 453, similarly misplaces the digits and implies 400 + 50 + 3 rather than 300 + 40 + 5.

In a positional decimal system, the rightmost digit is in the units place. Its positional weight is 10^0, which equals 1. Multiplying the digit by 10^0 leaves its value unchanged, reflecting that it counts units. Hence the exponent for the rightmost digit is 0.

Option A:
Option A is correct because any number raised to the power 0 equals 1, and the units place must have a weight of 1. This matches how we interpret the rightmost digit in any decimal number.

Option B:
Option B, exponent 2, would give a weight of 10^2 = 100, which belongs to the hundreds place, not the units place. Assigning 10^2 to the rightmost digit would distort the positional structure.

Option C:
Option C, exponent 3, yields 10^3 = 1000, corresponding to the thousands place. This is far from the unit's role and clearly inappropriate.

Option D:
Option D, exponent 1, represents the tens place, where each digit counts groups of ten. The rightmost digit is not in this position, so exponent 1 is not correct for it.

Three decimal digit positions can each take any of 10 values from 0 to 9. The total number of possible combinations is 10×10×10 = 1000. These range from 000 to 999, providing 1000 distinct integer values.

Option A:
Option A, 9, dramatically underestimates the combinations, considering only the range of a single digit rather than three positions.

Option B:
Option B, 10, is the number of symbols available for each digit but not the total combinations formed by three positions.

Option C:
Option C, 100, is the number of two-digit combinations from 00 to 99. With three digits, the combinations grow by another factor of 10.

Option D:
Option D is correct because three positions in base 10 yield 10^3 = 1000 unique patterns. This is a direct application of the multiplication principle.

Statements A, B and D are correct descriptions of how data is represented in binary form in ICT. The binary system does use only the digits 0 and 1, each digit is called a bit, and a byte is generally made up of eight bits. Statement C is wrong because the decimal system is base 10, not base 2. Therefore, the combination that includes A, B and D but excludes C is the correct answer.

Option A:
Option A is correct because it contains exactly the true statements A, B and D. It correctly identifies the digits of the binary system, the term for each binary digit and the usual size of a byte. By excluding the false statement about the decimal system being base 2, it reflects proper understanding of number systems.

Option B:
Option B is incorrect because it omits statement D while including only A and B. Although A and B are true, D is also correct regarding a byte being 8 bits. Since one correct statement is left out, this option does not capture the full set of true statements.

Option C:
Option C is incorrect because it includes statement C along with B and D. Statement C incorrectly labels the decimal system as base 2, so including it mixes true and false information. A valid combination must exclude this incorrect statement.

Option D:
Option D is incorrect because it assumes all four statements are correct. By including statement C, which misrepresents the base of the decimal system, this option fails to meet the requirement of containing only accurate statements.