UGC NET Questions (Paper – 1)

This sequence follows the rule aₙ = n⁴ + 3n³ + n for n starting from 1. Evaluating for n = 1, 2, 3, 4 and 5 gives 1+3+1 = 5, 16+24+2 = 42, 81+81+3 = 165, 256+192+4 = 452 and 625+375+5 = 1005. For n = 6, we compute 6⁴+3·6³+6 = 1296+648+6 = 1950. Therefore 1950 is the unique next term that respects this quartic-plus-cubic-plus-linear pattern.

Option A:
Option A, 1950, matches exactly the result of n⁴+3n³+n when n = 6. It maintains the same combination of powers and coefficients that explain all the earlier terms. Because it arises directly from the established rule, this option correctly continues the sequence.

Option B:
Option B, 1938, is 12 less than the computed value and cannot be written as n⁴+3n³+n for n = 6. It implies a downward adjustment that does not appear anywhere in the earlier terms. Consequently, option B breaks the pattern and is incorrect.

Option C:
Option C, 1968, overshoots the correct polynomial value and likewise does not equal 6⁴+3·6³+6. Choosing 1968 would arbitrarily raise the sixth term relative to the rule. Therefore option C is not consistent with the functional relationship governing the series.

Option D:
Option D, 1980, departs even further from the formula’s output and has no basis in the given expression. Using 1980 would discard the tight connection between term position and term value, so option D cannot be correct.

When partners invest for the same duration, their profit shares are proportional to their invested amounts. The ratio of A’s investment to B’s investment is 20,000 : 30,000, which simplifies to 2 : 3. The total ratio sum is 2 + 3 = 5 parts. Since the total profit is ₹25,000, each part equals 25,000 ÷ 5 = 5,000. B is entitled to 3 parts, so B’s share is 3 × 5,000 = ₹15,000.

Option A:
Option A, ₹10,000, represents 2 parts of 5,000 and would actually correspond to A’s share under the 2 : 3 ratio. Assigning it to B would reverse the intended proportions of investment.

Option B:
Option B, ₹12,000, does not fit an integer multiple of 5,000 and would imply a non-integer division in the underlying ratio. It is not compatible with the simple 2 : 3 relationship.

Option C:
Option C respects the proportional rule by matching B’s three parts in the 2 : 3 split. It ensures that A’s share is ₹10,000 and B’s share is ₹15,000, together adding up to the total profit of ₹25,000.

Option D:
Option D, ₹18,000, would leave only ₹7,000 for A, which would not maintain the 2 : 3 ratio. It distorts the relative contributions of the partners.

The starting letters of the groups are B, E, H and K, which occupy positions 2, 5, 8 and 11 in the alphabet, increasing by 3 each time. The groups themselves are made of three consecutive letters in forward order. Continuing the same rule, the next starting letter is position 14, which is N, and the consecutive letters after N are O and P. Therefore NOP is the only group that fits both the plus three jump in the starting letter and the three letter consecutive structure.

Option A:
Option A is correct because it begins with N, exactly three positions after K, and continues with O and P to form a consecutive three letter block. It preserves the constant increase of three positions in the first letter across all terms. It also maintains the same internal structure where each group is made of adjacent letters in correct order. These two properties together make NOP the valid continuation.

Option B:
Option B, MNO, starts at M, which is only two positions after K, so it breaks the fixed jump of three positions for the starting letters. Although MNO is itself a set of three consecutive letters, it does not arise from the same position rule used earlier. Because it violates the numerical pattern in the first letter, it cannot be the correct next term.

Option C:
Option C, OPQ, begins at O, which is four positions after K and overshoots the required step. This makes the gap between KLM and OPQ larger than all previous jumps. Since the series shows a stable plus three increment, the excessive jump in the starting letter makes OPQ inconsistent with the pattern.

Option D:
Option D, LMN, uses L as the starting letter, but L was already part of the previous group KLM and does not follow the plus three progression from K. It would make the sequence repeat an internal letter instead of moving to the next scheduled starting point. Therefore LMN fails to continue the series logically.

Statements A, B and C express standard rules from traditional syllogistic logic. A is correct because a categorical syllogism uses three and only three terms, each appearing in specific roles. B is correct since failure to distribute the middle term at least once leads to the fallacy of undistributed middle. C is also true because with two negative premises there is no logical basis to relate the subject and predicate terms, so no conclusion follows. Statement D is false because valid syllogisms cannot have both premises negative and the conclusion negative simultaneously, so only A, B and C form the correct set.

Option A:
Option A is incomplete as it includes only A and B but omits C, which is an important rule prohibiting valid conclusions from two negative premises. Without C, the characterisation of valid and invalid syllogisms is not complete. This makes the option unsuitable for the correct answer.

Option B:
Option B is correct because it brings together all the true statements about term count, middle-term distribution and the restriction on negative premises. It also implicitly excludes the incorrect idea in D that a valid syllogism may have all three statements negative. This matches the logical rules examined in NET questions.

Option C:
Option C is incorrect because it includes D, which falsely allows a configuration with both negative premises and a negative conclusion, while omitting B. Losing the distribution condition on the middle term and accepting a wrong rule about negativity makes the option invalid.

Option D:
Option D is wrong because it pairs B with D, treating the clearly false D as if it were correct and leaving out A and C. The combination therefore mixes truth with error and fails to capture the overall structure of valid categorical syllogisms.

This option is correct because the probability of an event and its complement must sum to 1. If P(A) = p, then P(A') = 1 − p. This follows directly from the rule that A and A' are mutually exclusive and exhaustive. Therefore, the complement has probability 1 − p.

Option A:
If we use p as the probability of A', then P(A) + P(A') would be p + p = 2p, which equals 1 only when p = 1/2. This does not hold in general and contradicts probability axioms. Hence, this option is incorrect.

Option B:
The expression 1/p would usually exceed 1 for p less than 1 and is not bounded between 0 and 1. Probabilities cannot exceed 1, so this form cannot represent the complement probability.

Option C:
The expression p² is smaller than p for values between 0 and 1 and does not generally satisfy P(A) + P(A') = 1. It is just another function of p, not the complement. Thus, this option is not correct.

Option D:
The value 1 − p ensures that P(A) + P(A') equals 1, satisfying the basic complement rule. It also remains between 0 and 1 whenever p is between 0 and 1. Therefore, this formula correctly represents P(A').

Statements A and B give correct definitions of union and intersection, D correctly states that the empty set is a subset of every set and E is true because many reasoning problems about groups naturally use set models and sometimes Venn diagrams. Statement C is false since the complement of A in U consists of elements of U that are not in A, not elements that are in both A and U. Therefore, the combination A, B, D and E only is correct.

Option A:
Option A is incomplete because it omits E and so does not mention the common application of set theory to practical reasoning questions in exams. Without E, the conceptual link to aptitude problems is weakened.

Option B:
Option B is also incomplete as it leaves out A and therefore fails to give the basic description of union, which is central to understanding combined categories in reasoning questions.

Option C:
Option C is wrong because it includes C, which misdescribes complement, and omits E, ignoring the modelling role of sets. Including a false statement invalidates the option.

Option D:
Option D is correct as it keeps all true statements while excluding C, the incorrect account of complement. It aligns with the formal set-theoretic framework used for UGC NET reasoning items.

The days of the week repeat every 7 days. We find the effective shift by computing 50 mod 7. Since 50 = 7 × 7 + 1, the remainder is 1, meaning the day moves forward by 1 from Wednesday. One day after Wednesday is Thursday, so 50 days from now it will be Thursday.

Option A:
Option A, Thursday, is exactly one day after Wednesday and corresponds to a remainder of 1 in the modulo-7 calculation, so it matches the correct reasoning.

Option B:
Option B, Saturday, is three days after Wednesday and would be correct only if the remainder were 3, not 1.

Option C:
Option C, Sunday, is four days after Wednesday and would require a remainder of 4, which does not arise from 50 ÷ 7.

Option D:
Option D, Monday, is five days after Wednesday, again inconsistent with the actual remainder of 1.

The differences are 7, 15 and 31. Each difference is roughly double the previous one plus 1: 7×2+1 = 15 and 15×2+1 = 31. Continuing this rule, the next difference should be 31×2+1 = 63. Adding 63 to 58 gives 121. Therefore, 121 is the correct next term in the series.

Option A:
Option A is 110, which gives a difference of 52 from 58, not matching the predicted 63. This does not follow the doubling plus one structure. Hence, 110 is not correct.

Option B:
Option B is 118, implying a difference of 60, which again fails to align with 63. It underestimates the expected growth in the differences. Therefore, 118 cannot be accepted.

Option C:
Option C equals 121, which yields the required new difference of 63. The sequence of differences 7, 15, 31, 63 shows a clear pattern where each term is nearly double the previous one plus one. This confirms 121 as the correct answer.

Option D:
Option D is 125, giving a difference of 67. That overshoots the predicted 63 and disrupts the precise doubling-plus-one rule. Thus, 125 is not suitable here.

This option is correct because a linear equation in two variables, such as ax + by + c = 0, represents a straight line when graphed on the Cartesian plane. Every solution pair (x, y) lies on this line. Therefore, the geometric representation is a straight line.

Option A:
A single point would correspond to a specific solution pair rather than all possible solutions. Linear equations generally have infinitely many solutions forming a continuous set. Thus, a point is not the correct geometric representation.

Option B:
A straight line contains all ordered pairs that satisfy the given linear equation. This matches the definition of a linear relation in two dimensions. Hence, this option correctly describes the graph of a linear equation.

Option C:
A circle is described by equations like x² + y² = r², which are not linear because they include squared terms. Therefore, a circle is not the graph of a linear equation.

Option D:
A parabola corresponds to quadratic equations, such as y = ax² + bx + c. These involve squared terms and are not linear, so they are not the correct graph for a linear equation.

Statements A, B and C give the standard truth conditions for conjunction, inclusive disjunction and negation in classical logic. Statement E is also correct because truth tables are precisely the tool used to define and study these connectives. Statement D is false, since a conjunction requires both conjuncts true and not just one. Thus the full set of correct statements is A, B, C and E only, which matches option B.

Option A:
Option A is incorrect because it omits E, even though using truth tables to define connectives is a central logical technique. It therefore lists only part of the true information and does not capture the complete correct set.

Option B:
Option B is correct since it includes all and only the true statements: the right truth conditions for ∧, ∨ and ¬ together with the role of truth tables in defining them, while excluding D, which misstates when a conjunction is true.

Option C:
Option C is wrong because it includes D, the false claim that a conjunction is true if at least one conjunct is true, and it omits B, so it both accepts an incorrect statement and fails to include a correct one.

Option D:
Option D is incorrect because although B, C and E are true, it drops A, which gives the fundamental condition for conjunction, so it does not contain the entire set of correct statements for this question.

B is correct: A (“All S are P”) and O (“Some S are not P”) are contradictories. C is correct because contradictories cannot both be true and cannot both be false. E is correct: I and O are subcontraries (they cannot both be false but may both be true). A is false (it describes contraries, not contradictories). D is false because A and E are contraries, not subcontraries. Thus B, C and E only are correct.

Option A:
Option A is incomplete because it omits E, a key relation on the square involving particular propositions.

Option B:
Option B is wrong because it adds D, which misclassifies A and E as subcontraries.

Option C:
Option C is correct because it includes exactly the true relationships and excludes A and D, which are incorrect.

Option D:
Option D is incorrect because it includes A, which misdefines contradictories.

To solve the inequality 2x − 5 > 7, we first add 5 to both sides, giving 2x > 12. Then we divide both sides by 2, which is positive, so the inequality direction remains the same, yielding x > 6. Thus, the solution set consists of all real numbers greater than 6, and this is the equivalent inequality in terms of x.

Option A:
Option A captures the full result of correctly isolating x. It incorporates both the addition and division steps and expresses the final constraint directly as x > 6.

Option B:
Option B, x > 5, would follow only if the inequality were 2x − 5 > 5, not 2x − 5 > 7. It represents a weaker condition than required.

Option C:
Option C, x > 4, ignores more of the numerical shift and significantly broadens the solution set beyond what the original inequality allows.

Option D:
Option D, x > 3, is even looser and does not align with the operations needed to isolate x in the given inequality.

The number of permutations of n distinct objects taken r at a time is given by nPr = n!/(n−r)!. Here n = 5 and r = 3, so 5P3 = 5!/(5−3)! = 5!/2! = (5×4×3×2×1)/(2×1) = 5×4×3 = 60. Thus, there are 60 different ordered arrangements of 3 objects chosen from 5.

Option A:
A value of 10 corresponds to the combination 5C2 or 5C3, which counts unordered selections, not ordered arrangements. Since permutations consider order, 10 is too small and does not account for all possible orderings of 3 objects. Therefore, this option is incorrect.

Option B:
60 is correct because it counts each distinct ordering separately. For example, selecting A, B and C yields six permutations: ABC, ACB, BAC, BCA, CAB and CBA, and similar counts apply for other triplets. Multiplying the number of choices for each position, 5×4×3, naturally arrives at 60.

Option C:
A value of 20 does not match any standard permutation or combination formula for the given n and r. It likely arises from a miscalculation or misunderstanding of order versus selection. Hence, this option cannot be supported.

Option D:
120 equals 5!, which counts permutations of all 5 objects taken at a time, not just 3 out of 5. The question restricts us to arrangements of length 3, so 120 overcounts and is not correct here.