UGC NET Questions (Paper – 1)

Each group has the first and third letters identical, and the second and fourth letters also identical. The outer letters move forward in order A, B, C, D, while the inner letters move backward from Z to Y to X to W. The next outer letter after D is E, and the next inner letter before W is V, so the next block must be E V E V, that is EVEV.

Option A:
Option A, EVVE, has the correct set of letters but places the V’s together in the middle instead of in the second and fourth positions. This breaks the pattern where the first and third letters are the same and the second and fourth letters are the same. Because it changes the internal symmetry, EVVE does not fit the series.

Option B:
Option B is correct because it keeps the structure where the first and third letters match and the second and fourth letters match. It also advances the outer letters from D to E and moves the inner letters from W to V as required. This exact preservation of both movement and symmetry makes EVEV the logical next term.

Option C:
Option C, EWEW, keeps E at the outer positions but fails to update the inner letters from W to V. That would freeze the backward movement of the inner positions, which contradicts the observed pattern. Therefore EWEW cannot be accepted as the next term.

Option D:
Option D, EVEW, mixes V and W in the inner positions so that the second and fourth letters are no longer identical. The series clearly maintains identical inner letters in every group. As this structural condition is violated, EVEW is not a correct continuation.

Outer letters move forward A,B,C,D → E. Inner double letters move backward Z,Y,X,W → V. So the next palindromic block is EVVE.

Option A:
Option A is correct because E is next outer letter and V is next inner letter (one step before W).

Option B:
Option B uses U, which is two steps before W, so it breaks the one-step backward movement.

Option C:
Option C breaks the palindromic structure (outer letters differ) and keeps W incorrectly.

Option D:
Option D jumps the inner letters back to X, reversing the backward sequence.

The pattern here is given by aₙ = 4n²+n+3 for n starting from 1. For n = 1, 2, 3, 4 and 5 we get 4·1²+1+3 = 8, 4·2²+2+3 = 21, 4·3²+3+3 = 42, 4·4²+4+3 = 71 and 4·5²+5+3 = 108. For n = 6 we compute 4·6²+6+3 = 144+6+3 = 153. Thus 153 is the correct next term consistent with this quadratic rule.

Option A:
Option A, 153, matches exactly the value from aₙ = 4n²+n+3 when n = 6. It keeps the same combination of square, linear and constant parts used to generate all earlier terms. Because the relationship holds perfectly, 153 is the correct continuation.

Option B:
Option B, 143, is 10 less than the formula’s result and does not satisfy the expression for n = 6. It requires an unexplained reduction of the term, which is not supported by the pattern. Therefore 143 is not a valid choice.

Option C:
Option C, 147, is also different from 153 and cannot be obtained from the given formula without changing its constants. This would make the rule inconsistent at the final term. Hence 147 is not correct.

Option D:
Option D, 161, exceeds the computed value and similarly fails to equal 4n²+n+3 when n = 6. Using 161 would break the neat polynomial structure, so it is not the right answer.

Each term is a pair of consecutive letters. The starting letters A(1), D(4), G(7), J(10) increase by +3 each time. Next start is M(13), so the pair is MN.

Option A:
Option A starts at K(11), which is +1 from J, not +3.

Option B:
Option B is correct because M is +3 from J and N follows M.

Option C:
Option C starts at L(12), which is +2 from J, not +3.

Option D:
Option D starts at N(14), which is +4 from J, not +3.

Statements A and B give the definition of a binary operation and associativity. D is also correct since there are operations that are associative but not commutative, such as matrix multiplication. C is false because commutativity must hold for all pairs, not just some, and E is false because commutativity does not imply associativity. Thus A, B and D only are correct.

Option A:
Option A is incomplete as it omits D, missing the important fact that associativity and commutativity are independent properties.

Option B:
Option B is wrong because it includes C, which weakens commutativity incorrectly by requiring it only for a single pair.

Option C:
Option C is incorrect since it omits A, the definition of a binary operation, which is one of the correct statements.

Option D:
Option D is correct because it gathers the proper definition, the associativity law and a fact about independence of associativity from commutativity, while excluding C and E, which are incorrect.

Deductive reasoning starts from a general principle, law or rule and applies it to a particular case to reach a logically necessary conclusion. In mathematical reasoning, most formal proofs are deductive because they proceed from axioms and theorems to specific results. If the premises in a deductive argument are true and the reasoning is valid, the conclusion must be true. This certainty makes deductive reasoning central to mathematical problem solving and theorem proving.

Option A:
Deductive reasoning is correct because it captures the idea of moving from the general to the particular in a logically valid manner. For example, from the general statement "All prime numbers greater than 2 are odd," we can deduce that 11 is odd because it is a prime greater than 2. In UGC NET reasoning questions, recognizing a deductive structure helps evaluate arguments effectively. Thus, this option accurately matches the description in the question.

Option B:
Inductive reasoning moves from specific cases to form a general rule, so it reverses the direction described in the question. For example, observing that several primes are odd and concluding that all primes are odd is an inductive generalization, not an application of a known rule. It provides probable, not certain, conclusions and therefore does not match the definition of moving from general to specific. Hence, it is not the best fit here.

Option C:
Analogical reasoning is based on identifying similarity between two situations and inferring that what is true in one may be true in the other. It does not necessarily begin from a universal rule but from a known case compared to a new case. Because it relies on similarity rather than a general law, it does not guarantee truth of the conclusion. Therefore, it does not describe the general-to-specific pattern mentioned in the question.

Option D:
Abductive reasoning typically involves inferring the most plausible explanation for an observation rather than applying a general rule to a specific case. For instance, seeing that the ground is wet and inferring it has rained is abductive, not deductive. It is often used in hypothesis generation, not in formal mathematical proof. Hence, it does not match the description in the stem.

Statements A, B, D and E are correct standard number-theoretic facts: even integers are divisible by 2, primes greater than 2 are odd, divisibility by 6 implies divisibility by both 2 and 3 and divisibility tests accelerate factor checking in exams. Statement C is false because a composite number has more than two positive divisors, whereas having exactly two is the defining feature of a prime. Therefore, the combination containing A, B, D and E only is correct.

Option A:
Option A is incomplete because it omits E, thereby ignoring the important exam-oriented point that divisibility rules are employed to solve questions quickly. Without E, the applied angle of the topic is underrepresented.

Option B:
Option B is correct as it includes all true statements while leaving out C, which reverses the definition of composite and prime numbers. It aligns with the basic arithmetic knowledge expected at UGC NET level.

Option C:
Option C is wrong because it includes C, the incorrect statement about composites having exactly two divisors, and omits A, a correct definition of even numbers. This combination thus mixes error with omission.

Option D:
Option D is incorrect as it accepts C along with true statements and omits B, failing to state that primes greater than 2 are odd. The presence of C again makes the option unacceptable.

The ratio 3:2 means that out of every 5 students, 3 are boys and 2 are girls. Since the total number of students is 30, we first find how many groups of 5 there are: 30 ÷ 5 = 6. The number of boys is then 3 × 6 = 18. Hence, there are 18 boys in the class.

Option A:
If there were 12 boys, then the ratio of boys to girls would not be 3:2 with a total of 30 students. Specifically, 12 boys would leave 18 girls, giving a ratio of 12:18, which simplifies to 2:3, not 3:2. Therefore, this option does not satisfy the given ratio.

Option B:
If there were 15 boys, the remaining 15 students would be girls, resulting in a ratio of 15:15 or 1:1. This is very different from the required ratio of 3:2. Thus, option B is not correct.

Option C:
18 is correct because it maintains the ratio 3:2 with a total of 30 students. With 18 boys and 12 girls, the ratio 18:12 simplifies to 3:2 as required. This method demonstrates how ratio and total together determine individual quantities in aptitude problems.

Option D:
If there were 20 boys, only 10 girls would remain, giving a ratio of 20:10 or 2:1, which does not match 3:2. This inconsistency shows that 20 is not the correct number of boys.

The three-letter groups are formed from even-position letters of the alphabet: BDF uses (2, 4, 6), DFH uses (4, 6, 8), FHJ uses (6, 8, 10) and HJL uses (8, 10, 12). The starting positions of the groups are 2, 4, 6 and 8, increasing by 2. The next group should use (10, 12, 14), which correspond to J, L and N, giving JLN.

Option A:
Option A, ILN, begins at position 9 instead of 10, shifting the group off the even-position pattern. It also does not align with the established sequence of starting positions. Because the structure is based on even-numbered indices, ILN cannot be correct.

Option B:
Option B, JLN, is correct because it follows the pattern of moving the starting position forward by 2 and taking every second letter from there. It uses positions 10, 12 and 14, exactly as the rule demands. This preserves both the spacing and the parity of the positions.

Option C:
Option C, KMN, changes the distances between letters and alters the parity of the underlying positions. It does not result from the simple shift in the even sequence used in earlier terms. As such, KMN cannot be accepted as a valid continuation.

Option D:
Option D, HJM, repeats letters and fails to maintain the constant spacing of two positions between successive letters. It also does not continue the progression of starting positions properly. Therefore HJM does not satisfy the design of the series.

The series lists abbreviations of the days of the week in order: SUN (Sunday), MON (Monday), TUE (Tuesday) and WED (Wednesday). The next day in the weekly sequence is Thursday, abbreviated as THU. Therefore THU is the only term that continues this ordered list of weekdays.

Option A:
Option A, FRI, represents Friday, which comes after Thursday in the weekly cycle. Choosing FRI would skip Thursday and break the strict day-by-day sequence. Hence FRI cannot be the immediate next term.

Option B:
Option B is correct because THU stands for Thursday, the day following Wednesday. It maintains the exact chronological order of the days. This alignment with the familiar weekly pattern makes THU the logical continuation.

Option C:
Option C, SAT, corresponds to Saturday and is two days ahead of Thursday. Using SAT would skip multiple days, which the series does not do. Therefore SAT is not a valid next term.

Option D:
Option D, SUN, would restart the sequence from Sunday. There is no indication that the pattern cycles after Wednesday here. As the list is still in mid-week, SUN cannot be the correct answer.

The compound ratio a:d is obtained by multiplying the fractional equivalents of the given ratios: (2/3) × (4/5) × (6/7). This product is (2 × 4 × 6)/(3 × 5 × 7) = 48/105. Simplifying 48/105 by dividing numerator and denominator by 3 gives 16/35. Hence, the compound ratio a:d is 16:35.

Option A:
Option A, 8:35, halves the correct numerator while keeping the denominator, so it does not correspond to the true product 48/105. If we tried to reconstruct the intermediate ratios from 8:35, they would not match 2:3, 4:5 and 6:7 simultaneously. Thus 8:35 is not consistent with the combined effect of the three ratios.

Option B:
Option B, 16:21, uses the correct numerator 16 but an incorrect denominator 21. The fraction 16/21 does not equal 48/105, because 16/21 would simplify to a different value. Therefore, it does not represent the compound ratio derived from the data.

Option C:
Option C, 8:21, is even further from the correct proportion, altering both numerator and denominator in ways not justified by the multiplication of the three given ratios. This option cannot be related back to the original conditions without violating at least one of the ratios.

Option D:
Option D is correct because 16:35 is exactly the simplified form of 48:105, the product of 2:3, 4:5 and 6:7. It correctly aggregates the effect of the three separate proportional relationships into one single ratio.

This option is correct because the arithmetic mean of two numbers is obtained by adding them and dividing by 2. Here, (10 + 20) ÷ 2 = 30 ÷ 2 = 15. This value lies exactly midway between 10 and 20. Therefore, the arithmetic mean is 15.

Option A:
Taking 10 as the mean would ignore the contribution of the larger number 20. The average must be between the two numbers and balanced by both. Since 10 is at the lower end, it cannot be the mean of 10 and 20.

Option B:
A value of 12 is between 10 and 20 but does not satisfy the mean formula. Using (10 + 20) ÷ 2 does not produce 12. Therefore, this number does not represent the correct arithmetic mean.

Option C:
20 is the larger of the two numbers and would be the mean only if both numbers were 20. Because the other number is 10, the mean must lie between 10 and 20, not at the greater endpoint. Thus, 20 is incorrect.

Option D:
The average 15 comes directly from the formula for mean and is equidistant from 10 and 20. It balances the two values, which is the key idea of an arithmetic mean. Hence, this option correctly answers the question.

This sequence is described by the formula (a_n = 2n^3 + n) with n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 3, 18, 57, 132 and 255, matching the given series exactly. For n = 6 the expression gives (2×6^3 + 6 = 432 + 6 = 438). Thus 438 is the only value that maintains this cubic-plus-linear rule.

Option A:
Option A, 430, is smaller than the computed value and does not equal (2×6^3 + 6). It would require changing the formula at the final term, which has no support from the preceding structure. Therefore 430 is not a valid continuation.

Option B:
Option B, 438, matches precisely the value produced by (a_n = 2n^3 + n) when n = 6. It preserves the same relationship between position and value that explains all earlier terms. For this reason, 438 is the correct next term in the series.

Option C:
Option C, 442, overshoots the cubic prediction and is not the output of the given expression for any suitable n after 5. Adopting 442 would arbitrarily inflate the term and break the algebraic pattern. Hence 442 cannot be accepted.

Option D:
Option D, 454, deviates even more from the expected value and similarly fails to satisfy the rule (2n^3 + n) for n = 6. Selecting 454 would destroy the clean cubic structure of the sequence, so it is not correct.

The first-level differences are 15, 25, 35 and 45. These differences themselves form an arithmetic progression with common difference 10. Thus the next difference should be 55. When 55 is added to the last term 128, we obtain 183, which continues the pattern at both the term level and the difference level.

Option A:
Option A gives 178, which would create a difference of 50 from 128. This would make the sequence of differences 15, 25, 35, 45, 50, where the last increase is 5 rather than 10. Hence 178 breaks the structure of the second-level arithmetic progression.

Option B:
Option B gives 188, corresponding to a difference of 60 from 128. That would yield differences 15, 25, 35, 45, 60, causing the final increment to jump by 15 instead of 10. Therefore 188 does not preserve the consistent behaviour among the gaps.

Option C:
Option C gives 183, leading to a difference of 55 from 128. The differences become 15, 25, 35, 45, 55, a clear arithmetic progression with constant step 10. This perfect alignment with the observed pattern makes 183 the correct next term.

Option D:
Option D gives 193, implying a difference of 65 from 128. This creates an irregular spike in the differences and abandons the neat step of 10 between them. Consequently, 193 is not a valid continuation of the series.

Statements A, B and C are standard interpretations of sets, universal set and complement using Venn diagrams. Statement E is also true because Venn diagrams are widely used to visualise logical relations and survey data. Statement D is false since disjoint sets have no shared elements and thus no overlapping regions in the diagram. Hence the combination A, B, C and E only is correct.

Option A:
Option A is correct as it combines all accurate statements while rejecting D, which misdescribes disjointness. It faithfully reflects how Venn diagrams are used in NET-style reasoning questions.

Option B:
Option B is incomplete because it omits E and therefore does not mention important applications such as classification and survey questions, limiting the scope of the description.

Option C:
Option C is wrong because it includes D, which contradicts the definition of disjoint sets by asserting that they overlap, even though it also lists true statements. The presence of D invalidates the option.

Option D:
Option D is also incorrect as it keeps D and omits A, thereby losing the basic fact about sets as regions and retaining an incorrect claim about disjointness.