UGC NET Questions (Paper – 1)

This option is correct because the mode is defined as the value that appears most often in a data set. It focuses on frequency rather than magnitude or order. In many practical situations, the mode represents the most common or popular value. Therefore, the most frequent value is called the mode.

Option A:
The mean is the arithmetic average of all values in the data set. It is obtained by summing all observations and dividing by the count. It does not specifically represent the most frequent value, so it is not the correct term here.

Option B:
The median is the middle value when the data are arranged in order. It divides the data into two equal halves but does not necessarily correspond to the most frequent observation. Hence, median is not the right answer.

Option C:
The range measures the spread of the data by subtracting the smallest value from the largest. It does not identify any particular value in terms of frequency. Therefore, range is unrelated to the notion of "most frequently occurring value."

Option D:
The mode is the observation with the highest frequency in the data. Datasets can have one mode, multiple modes or sometimes no mode if all values occur equally. Since the question focuses on the most frequent value, mode is the accurate term.

The series is described by the formula aₙ = 2n⁴ + 3n³ + n² with n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 2+3+1 = 6, 32+24+4 = 60, 162+81+9 = 252, 512+192+16 = 720 and 1250+375+25 = 1650. For n = 6 this expression yields 2·6⁴+3·6³+6² = 2592+648+36 = 3276. Thus 3276 is the unique next term that maintains this quartic-plus-cubic-plus-quadratic structure.

Option A:
Option A, 3210, is smaller than the formula’s result and does not satisfy aₙ = 2n⁴+3n³+n² for n = 6. It would require subtracting 66 from the correctly computed value at the last term only. Because the earlier terms follow the rule exactly, such a modification is unjustified and makes option A incorrect.

Option B:
Option B, 3276, matches precisely the value computed from the expression for n = 6. It preserves all polynomial components and coefficients that describe the growth of the series from the first to the fifth term. Since the same rule generates the sixth term without alteration, 3276 is the correct continuation.

Option C:
Option C, 3330, overshoots the formula’s output and cannot be obtained by substituting n = 6 into the rule. Choosing 3330 would mean arbitrarily inflating the sixth term, which breaks the tight correspondence between index and value. Therefore option C is incorrect.

Option D:
Option D, 3396, deviates even more from the correct value and has no basis in the expression 2n⁴+3n³+n² for the given index. Using 3396 would destroy the internal consistency of the polynomial pattern, so option D is incorrect.

Each term is a 4-letter consecutive block; starting letter moves backward by 1: K, J, I, H → next G. So next block is GHIJ.

Option A:
Option A is correct because it continues the −1 slide and keeps the 4-letter consecutive structure.

Option B:
Option B repeats the last term.

Option C:
Option C moves forward/back to an earlier term and breaks the progression.

Option D:
Option D jumps back by 2 (H to E), not by 1.

This option is correct because a prime number is greater than 1 and has exactly two distinct positive divisors. The number 2 is divisible only by 1 and 2, satisfying this definition. It is also the smallest number with this property. Therefore, 2 is the smallest prime number.

Option A:
The number 1 has only one positive divisor, namely itself. Because primes must have exactly two distinct positive divisors, 1 is not considered a prime number. Hence, this option is incorrect.

Option B:
The number 2 is the first integer greater than 1 that has exactly two positive divisors. It is divisible by 1 and itself and no other positive integer. This makes 2 the smallest prime number as required in the question.

Option C:
The number 3 is prime, but it is not the smallest prime. Since 2 is smaller than 3 and already satisfies the definition of a prime, 3 cannot be the correct answer. Therefore, this option is not appropriate.

Option D:
The number 0 is divisible by every non-zero integer and does not fit the standard definition of prime numbers. It is not even positive in the usual sense of positive integers. Thus, 0 cannot be the smallest prime.

The given series multiplies each term by a constant factor to obtain the next one. Specifically, 9 ÷ 3 = 3, 27 ÷ 9 = 3 and 81 ÷ 27 = 3, so the common ratio is 3. To continue the pattern, we multiply 81 by 3 to obtain the next term. This gives 81 × 3 = 243, which preserves the geometric progression.

Option A:
Option A, 162, arises if someone incorrectly multiplies by 2 instead of 3, giving 81 × 2. However, this contradicts the established common ratio of 3 used in all previous steps.

Option B:
Option B, 200, does not correspond to any simple multiple of 81 consistent with the progression and seems unrelated to the pattern.

Option C:
Option C, 225, is 15² and may be chosen due to familiarity with square numbers, but it does not follow from multiplying 81 by the common ratio 3.

Option D:
Option D maintains the constant ratio of 3 between consecutive terms. By applying 81 × 3, we get 243, which is the only option that respects the geometric structure of the series.

The rule is that a number n maps to n² + 4. For n = 2, we get 2² + 4 = 4 + 4 = 8, and for n = 3, we get 3² + 4 = 9 + 4 = 13, matching the given examples. Applying the same rule to n = 5 yields 5² + 4 = 25 + 4 = 29. Hence, 5 maps to 29 under this square-plus-four rule.

Option A:
Option A, 21, could arise from a different linear rule like 4n + 13, but it does not agree with the n² + 4 pattern shown by the examples.

Option B:
Option B, 29, is exactly 5² + 4 and therefore fits the defined function rule for this analogy.

Option C:
Option C, 31, would correspond to 5² + 6, which changes the constant term and breaks the consistency of the mapping.

Option D:
Option D, 35, is even further from 29 and does not follow from squaring and then adding a fixed constant.

Each term is a 5-letter consecutive block sliding forward by one starting letter: P, Q, R, S. Next start is T, giving TUVWX.

Option A:
Option A is correct because it starts at T (after S) and continues with U,V,W,X.

Option B:
Option B (UVWXY) would be the term after TUVWX, so it is one step ahead.

Option C:
Option C repeats the last term and stops the slide.

Option D:
Option D goes backward to an earlier term.

To solve 3x − 5 = 2x + 7, we first bring like terms together. Subtracting 2x from both sides gives x − 5 = 7. Then we add 5 to both sides to isolate x, obtaining x = 12. This value satisfies the equation when substituted back, confirming that 12 is the correct solution.

Option A:
Option A, 6, does not satisfy the equation because substituting x = 6 gives 3 × 6 − 5 = 13 and 2 × 6 + 7 = 19, which are not equal. This indicates that 6 cannot be the solution.

Option B:
Option B, 10, also fails on substitution, as 3 × 10 − 5 equals 25 while 2 × 10 + 7 equals 27, showing a mismatch between the two sides.

Option C:
Option C results in equal values on both sides when substituted, with 3 × 12 − 5 giving 31 and 2 × 12 + 7 also giving 31. This equality verifies that x = 12 satisfies the original equation.

Option D:
Option D, 15, leads to 3 × 15 − 5 = 40 and 2 × 15 + 7 = 37, again producing unequal sides and ruling out 15 as a solution.

Each term is a block of four consecutive letters and the block slides one position to the right in the alphabet. The starting letters P, Q, R and S progress forward by one step in each term. After STUV, the next block should begin with T and include the next three letters U, V and W. Hence TUVW is the unique group that continues the sliding four letter window pattern.

Option A:
Option A, UVWX, is a proper four letter block but its starting letter U is two steps ahead of S rather than one. It corresponds to the block that would come after TUVW, not immediately after STUV. Because the series advances the starting point by only one letter at a time, UVWX cannot be the immediate next term.

Option B:
Option B is correct because TUVW starts with T, the next letter after S, and continues with U, V and W in order. It keeps both the length of the group and the sliding by one letter consistent with all earlier transitions. This perfect match with the sliding window rule makes TUVW the correct continuation.

Option C:
Option C, STUV, simply repeats the last given term instead of moving the starting letter forward. The pattern does not show any repetition or cycling of groups. Using STUV again would freeze the progression, so it cannot be the right answer.

Option D:
Option D, RSTU, moves backward in the sequence to a previous block. The series clearly moves forward in the alphabet, and there is no indication of reversal. Thus RSTU does not fit as the next term.

Statement A correctly defines variance as the mean of squared deviations from the mean (in the population sense). Statement B is true because standard deviation is the positive square root of variance. Statement D is also correct since a larger standard deviation reflects more spread in the data. Statement C is false because if all observations are equal, both variance and standard deviation are zero. Statement E is false for the same reason. Therefore, A, B and D only are correct.

Option A:
Option A is incomplete; while A and B are true, it omits D and thus fails to connect standard deviation to dispersion.

Option B:
Option B is correct because it includes all three accurate statements and excludes C and E, both of which incorrectly deny the possibility of zero spread.

Option C:
Option C is wrong because it includes C (false), even though B and D are true.

Option D:
Option D is wrong because it includes E (false) and omits B (definition link), so it cannot be correct.

Statements A, B, D and E are all correct descriptions of the role of permutations and combinations. Permutations count ordered arrangements, combinations count unordered selections, seating arrangement problems rely heavily on order-based counting, and both concepts are frequently tested in UGC NET aptitude. Statement C is false because, for r greater than 1, nPr is larger than nCr since each combination can be arranged in several ordered ways. Thus the only option that includes all true statements while excluding the false one is the combination A, B, D and E.

Option A:
Option A is incomplete because it omits E, failing to note explicitly that both permutations and combinations are actively used in NET mathematical reasoning questions. Without E, the statement about exam relevance is missing.

Option B:
Option B is correct because it gathers A, B, D and E, which together present the conceptual distinction between order and selection and their examination usage, while rightly rejecting C, which reverses the relative sizes of permutations and combinations.

Option C:
Option C is incorrect since it leaves out A and therefore does not mention that permutations are arrangements where order matters, a central concept for the topic. Though it contains some true statements, it is not complete.

Option D:
Option D is wrong because it includes C, the false claim that nPr is always less than nCr, and thereby mixes an incorrect statement with true ones. The presence of C makes the entire combination unacceptable.

Statements A, B and C give the standard definitions of converse, contrapositive and inverse. Statement D is also correct because in classical logic a conditional and its contrapositive share the same truth table. Statement E is false; the converse need not be logically equivalent to the original conditional, as many counterexamples show. Hence the full and only correct collection is A, B, C and D, which appears in option B.

Option A:
Option A is incomplete because it omits D and thus fails to mention that a conditional and its contrapositive are equivalent in truth value. While A, B and C are true, they do not exhaust the set of correct statements.

Option B:
Option B is correct because it includes A, B, C and D, which are all true, and it excludes E, which is false because converse is not always equivalent to the original conditional.

Option C:
Option C is wrong because it includes E, which claims that the converse is always equivalent to the original statement. Since converse can change truth conditions, including E makes this combination incorrect.

Option D:
Option D is incorrect as it leaves out A. Although B, C and D are true, omitting A loses the correct description of converse, so this option does not list all correct statements.

Statements A and B are correct because headings define the meaning of each cell and percentages may reveal relative importance even when absolute numbers differ. Statement D is true in noting that trends over time are often more insightful than isolated figures, and E correctly states that approximation is acceptable when explicitly permitted, supporting speed in timed tests. Statement C is false because the largest absolute number may not correspond to the largest percentage if totals differ across rows. Therefore, the combination A, B, D and E only is correct.

Option A:
Option A is incomplete since it omits E and thus does not include the practical exam strategy of using approximations where allowed, which is important in NET-style data interpretation questions.

Option B:
Option B is also incomplete because although A, B and E are true, it excludes D and so fails to mention the value of analysing trends across years or categories, a key skill in interpreting tables.

Option C:
Option C is wrong because it does not include A and thereby ignores the fundamental role of headings, and it still omits the explicit rejection of C. Although B, D and E are true, leaving out A reduces completeness.

Option D:
Option D is correct as it unites all four true statements and excludes C, which oversimplifies the relationship between absolute numbers and percentages. It matches the way candidates are expected to reason with tabular data in the exam.

Statements A and B give the standard truth conditions for conjunction and inclusive disjunction. Statement D is correct because p → q is equivalent to ¬p ∨ q in classical logic. Statement E is also true since connectives allow us to build complex propositions from atomic ones. Statement C is false because ¬p is true when p is false, not when p is true. Thus A, B, D and E only form the correct set.

Option A:
Option A is incomplete as it omits E and therefore does not mention the important role of connectives in constructing compound statements, even though A, B and D are true.

Option B:
Option B is correct since it includes all true statements and excludes C, which reverses the meaning of negation. It matches the truth-table definitions that NET candidates must know.

Option C:
Option C is incorrect because it contains C, the wrong description of negation, despite including some true statements, so the overall set cannot be accepted.

Option D:
Option D is also wrong as it retains C and omits A, losing the clear definition of conjunction while endorsing an incorrect statement about negation.

This option is correct because the basic formula for uniform motion is distance = speed × time. Speed tells how much distance is covered per unit time. Multiplying it by the time gives the total distance travelled. Therefore, distance equals speed multiplied by time.

Option A:
Speed is defined as distance per unit time. When we multiply this rate by the actual time travelled, the time units cancel appropriately, leaving total distance. This matches the formula for uniform motion and makes this option correct.

Option B:
Acceleration measures how quickly speed changes over time, not directly how far an object travels. It does not multiply with time to give distance in the simple uniform motion formula. Hence, this option is not correct here.

Option C:
Momentum is the product of mass and velocity and belongs to mechanics, not basic kinematics equations of distance. It does not appear in the elementary distance-speed-time relation. Therefore, momentum is irrelevant to the question.

Option D:
Power measures the rate at which work is done or energy is transferred. It is not used to compute distance in uniform motion problems. Thus, this option does not fit the relationship described.