UGC NET Questions (Paper – 1)

The series is generated by the rule aₙ = n⁴ + 3n, where n is the position of the term starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 1⁴+3·1 = 4, 2⁴+3·2 = 22, 3⁴+3·3 = 90, 4⁴+3·4 = 268 and 5⁴+3·5 = 640, which matches the given terms exactly. For n = 6, the same formula yields 6⁴+3·6 = 1296+18 = 1314. Thus 1314 is the only value that continues the established functional pattern without any adjustment.

Option A:
Option A, 1304, is close to the expected value but does not equal 6⁴+3·6. Choosing 1304 would mean reducing the correct formula output by 10, which is not supported by the behaviour of earlier terms. Therefore this option breaks the consistent polynomial rule and cannot be correct.

Option B:
Option B, 1308, also fails to match the value obtained from n⁴+3n for n = 6. It would require introducing an arbitrary correction of −6 to the formula result only at the last step. Because the first five terms follow the rule exactly, such a change is logically inconsistent with the observed pattern.

Option C:
Option C, 1314, matches exactly the value of 6⁴+3·6 and therefore respects the rule used for all previous terms. It maintains the same dependence on the index n without any modification or exception. Since every term including the next one can be explained using this single expression, 1314 is the correct continuation of the series.

Option D:
Option D, 1320, exceeds the polynomial value by 6 and cannot be written as n⁴+3n for n = 6. Choosing 1320 would require altering the underlying algebraic relationship at the last term. Hence this option does not preserve the internal logic of the sequence and is incorrect.

In this pattern, the first letters are B(2), C(3), D(4) and E(5), increasing by 1 each time. The second letters are Y(25), X(24), W(23) and V(22), decreasing by 1 each time. The next first letter after E is F and the next second letter before V is U. Therefore FU is the only pair that continues both the forward and backward one-step movements.

Option A:
Option A, FT, gives the correct first letter F but takes T as the second letter, which is two steps before V instead of one. This breaks the strict decrement of one in the second-letter positions. As a result, FT does not follow the series rule.

Option B:
Option B, FU, is correct because it takes F as the next first letter and U as the next second letter, maintaining a one-step change in each direction. It keeps the symmetry of moving forward by one and backward by one in the two positions. This exact mirroring is what defines the series.

Option C:
Option C, GU, changes the first letter too far ahead to G rather than F, while keeping U as the second letter. This leads to a jump of two steps in the first position instead of one. The unequal step sizes mean GU does not preserve the consistent structure of the series.

Option D:
Option D, FV, keeps the second letter V unchanged, failing to move it backward at all. The second letter must continue to step down one position at each term. By freezing the second letter, FV violates a key element of the pattern.

Statement D is the only wrong statement here. A is correct because typical classification sets have three similar items and one different item. B is true since the task is to find the item that does not fit the pattern. C is correct; number properties like odd/even or prime/composite are common grouping criteria. E is also true because when multiple properties are present, the dominant common property helps determine the odd one out. D is false since familiarity to the candidate has nothing to do with correctness; the odd item is defined by the underlying property.

Option A:
Option A is incorrect as it groups B with D as wrong, even though B correctly states the central objective of classification questions.

Option B:
Option B is wrong because it treats A as wrong, but A accurately describes the common “three similar, one different” structure.

Option C:
Option C is incorrect since it labels C as wrong even though number-based classification commonly uses parity/divisibility/primality, and it still includes D; the wrong set is not “C and D”.

Option D:
Option D is correct because it singles out D (familiarity-based) as the only incorrect statement while accepting the other accurate descriptions of classification reasoning.

From the first statement, we know that every square belongs to the set of rectangles. The second statement says that no rectangles belong to the set of circles, so the sets of rectangles and circles are disjoint. Since squares are a subset of rectangles, and rectangles do not overlap with circles, squares also cannot overlap with circles. Therefore, the conclusion that no squares are circles is logically valid.

Option A:
Option A, “Some squares are circles,” contradicts the second premise. If any square were also a circle, it would simultaneously be a rectangle and a circle, which is impossible when no rectangles are circles.

Option B:
Option B, “No squares are circles,” correctly applies the idea that if rectangles and circles are disjoint sets, then any subset of rectangles (such as squares) must also be disjoint from circles. Hence, no square can be a circle.

Option C:
Option C, “All circles are squares,” reverses and overextends the relationship. We are told only that all squares are rectangles, not that all circles are contained in the set of squares.

Option D:
Option D, “Some rectangles are circles,” is directly inconsistent with the second statement, which asserts that no rectangles are circles.

Each term after the first is obtained by doubling the previous term and adding 1. We have 3 × 2 + 1 = 7, 7 × 2 + 1 = 15, 15 × 2 + 1 = 31 and 31 × 2 + 1 = 63. Applying the same rule again gives 63 × 2 + 1 = 127. This keeps the recursive structure consistent and explains every transition in the series.

Option A:
Option A gives 127, which fits the rule "double the previous term and add one" exactly. The extended sequence 3, 7, 15, 31, 63, 127 maintains a single, simple recursive pattern. Therefore, 127 is the correct next term for this number series.

Option B:
Option B provides 125, which would correspond to subtracting 1 after doubling or using a different constant, neither of which is supported by earlier terms. This breaks the consistent "+1" part of the rule. Hence, 125 cannot be considered correct.

Option C:
Option C suggests 129, which would require adding 3 after doubling 63, changing the constant in the recursive expression. No previous step uses "+3," so this option does not follow the observed pattern. Therefore, 129 is not appropriate.

Option D:
Option D presents 130, which is 67 more than 63 and does not come from doubling and adding a fixed small constant. This makes the relationship unclear and inconsistent with the earlier rule. Thus, 130 is not the right answer.

The first letters A, B, C and D move one step forward each time. The middle letters Z, Y, X and W move one step backward from the end of the alphabet. The last letters B, C, D and E move one step forward from the beginning. The next first letter after D is E, the next backward middle letter after W is V, and the next forward last letter after E is F, giving EVF.

Option A:
Option A, EVG, uses E and V correctly for the first and middle positions but moves the last letter two steps ahead from E to G. The last letters in the series only advance by one position each time. Because it breaks this one step increment in the last position, EVG does not fully respect the pattern.

Option B:
Option B is correct because it combines E, V and F, each obtained by a single step move from the respective letters of DWE. E is one step after D, V is one step before W, and F is one step after E. This continues all three directional movements consistently and therefore EVF fits the series exactly.

Option C:
Option C, EUF, alters the middle letter incorrectly by taking U, which is two steps before W. The middle letters should move only one position at a time in reverse order. Since U breaks the uniform backward movement, EUF cannot be the next term.

Option D:
Option D, EWF, keeps the last letter F correct but fails to move the middle letter from W to V. It freezes the middle letter instead of shifting it backward, which goes against the established transformation. Consequently, EWF is not a valid continuation of the pattern.

Let the numbers be 5x and 8x. After subtracting 20 from each, they become 5x − 20 and 8x − 20, with ratio 3:5. So (5x − 20)/(8x − 20) = 3/5. Cross multiplying gives 5(5x − 20) = 3(8x − 20), which simplifies to 25x − 100 = 24x − 60 and hence x = 40. The original numbers are therefore 200 and 320, so the larger is 320.

Option A:
Option A, 200, is the smaller original number when x = 40. If we took 200 as the larger, it would contradict the initial ratio 5:8, in which 8x must exceed 5x. Substituting 200 as the larger in the equations would also fail to reproduce the ratio 3:5 after subtraction.

Option B:
Option B, 240, does not correspond to either 5x or 8x for the solved value x = 40. If we attempted to solve using 240 as one of the numbers, the resulting x would not satisfy the transformed ratio condition. Hence 240 is inconsistent with the problem’s constraints.

Option C:
Option C, 280, again fails to match the form 5x or 8x with x = 40 and does not produce the stated ratio 3:5 after subtracting 20 from both numbers. Any attempt to fit 280 into the ratio structure leads to contradictions with the proportional relationship.

Option D:
Option D is correct because 320 equals 8 × 40, matching the larger term in the original ratio 5:8. With numbers 200 and 320, subtracting 20 gives 180 and 300, and 180:300 simplifies to 3:5. This verifies that 320 is the correct larger original number.

This series follows the rule aₙ = 2n⁴ + 3n² + 5 with n starting from 1. For n = 1, 2, 3, 4 and 5 the expression yields 10, 49, 194, 565 and 1330, exactly reproducing the given terms. When n = 6, we obtain a₆ = 2·6⁴ + 3·6² + 5 = 2·1296 + 3·36 + 5 = 2592 + 108 + 5 = 2705. Thus 2705 is the only value that fits this polynomial rule and correctly continues the series.

Option A:
Option A, 2681, is 24 less than the value produced by 2n⁴ + 3n² + 5 at n = 6. Taking 2681 would require lowering the correct outcome only at the sixth term, which is inconsistent with the exact agreement at earlier positions. Hence this option does not preserve the algebraic pattern.

Option B:
Option B, 2705, exactly matches the result of substituting n = 6 into the rule 2n⁴ + 3n² + 5. It respects both the fourth-power component and the quadratic correction that govern the growth of the sequence. Because it extends the same functional relationship without alteration, this option is the correct answer.

Option C:
Option C, 2693, is slightly below the correct value and cannot be expressed as 2·6⁴ + 3·6² + 5. Selecting 2693 would amount to undercounting the contribution of the polynomial terms at the final step. Therefore it does not provide a legitimate continuation of the series.

Option D:
Option D, 2717, is 12 greater than the required value and would similarly require modifying the rule only for n = 6. Such a change is not justified by the previous terms, where the formula holds exactly. Consequently, option D is not a valid extension of the given pattern.

This sequence is described by the rule aₙ = n⁵ + 3n³ + n² with n starting from 1. For n = 1, 2, 3, 4 and 5 the expression yields 5, 60, 333, 1232 and 3525, exactly matching the given numbers. For n = 6, we compute a₆ = 6⁵ + 3·6³ + 6² = 7776 + 3·216 + 36 = 7776 + 648 + 36 = 8460. Therefore 8460 is the only value that keeps this polynomial pattern intact.

Option A:
Option A, 8436, is 24 less than the value predicted by n⁵ + 3n³ + n² at n = 6. To accept 8436 we would have to subtract 24 from the correct result only at this term, despite perfect agreement elsewhere. This change would break the algebraic rule, so option A is incorrect.

Option B:
Option B, 8460, exactly equals the output of the formula for n = 6. It continues the same combination of a dominant fifth-power term with cubic and quadratic corrections that shape the earlier terms. Because it arises directly from the established expression, this option correctly extends the series.

Option C:
Option C, 8448, is 12 less than the correct value and cannot be written as 6⁵ + 3·6³ + 6². It reflects a small downward shift that is not supported by any structural change in the rule. Hence this option does not provide a valid continuation of the pattern.

Option D:
Option D, 8472, is 12 greater than the required value and would similarly require inflating the polynomial result at the sixth term. This violates the consistency of the relationship between n and aₙ, so option D must be rejected.

Statements A and B define HCF and LCM correctly, D notes that prime factorisation is a standard method to compute them and E highlights their application to repeated event and synchronisation problems, all of which are true. Statement C is false because in general a × b = HCF(a, b) × LCM(a, b), not just HCF(a, b). Therefore, the combination A, B, D and E only is the set of correct statements.

Option A:
Option A is incomplete as it omits E and thus fails to mention how HCF and LCM play a role in real exam problems involving repeated cycles and meeting times. Without E, the practical context of these concepts is under-emphasised.

Option B:
Option B is correct since it includes all the true statements and rightly excludes C, which confuses the relationship between product, HCF and LCM. It reflects both the definitions and the application of these notions in aptitude questions.

Option C:
Option C is wrong because it contains C, the incorrect equality, even though it also includes some true statements. The presence of C makes the overall combination mathematically inaccurate.

Option D:
Option D is incorrect as it includes C and omits B, thereby endorsing a false formula while ignoring the correct definition of LCM as the smallest common multiple. This mixture cannot be accepted as correct.

The series follows the rule “multiply the previous term by 3 and subtract 1”. We have 3×3−1 = 8, 8×3−1 = 23, 23×3−1 = 68 and 68×3−1 = 203. Applying the same rule again gives 203×3−1 = 609−1 = 608. Thus 608 is the unique value that continues the series in exact agreement with the recursive transformation.

Option A:
Option A gives 608, exactly equal to 203×3−1. This preserves the pattern that every term (after the first) is obtained by tripling the previous term and then subtracting 1. Because it keeps the recurrence intact without modification, 608 is the correct next term.

Option B:
Option B gives 610, which would correspond to 203×3+1 and change the subtraction step into an addition. This contradicts the repeated “−1” adjustment seen at every earlier stage, so 610 is not consistent with the series.

Option C:
Option C gives 612, which would require 203×3+3 or another altered operation. That breaks the simple and regular rule and introduces unnecessary complexity. Therefore 612 cannot be accepted.

Option D:
Option D gives 606, implying 203×3−3. This again modifies the constant adjustment from −1 to −3, which is not supported by the previous transitions. Hence 606 is not a valid continuation.

Statements A and B correctly capture fundamental logarithm values, D states the power rule and E refers to the change of base formula where one logarithm can be expressed as a constant multiple of another base’s logarithm. All of these are true. Statement C is false because the product rule for logarithms is logₐ(xy) = logₐ(x) + logₐ(y), not the difference. Therefore, the combination that includes A, B, D and E while excluding C is the only correct option.

Option A:
Option A is incomplete because although A, B and D are true, it leaves out E, which is an important fact about changing bases and is useful in more complex aptitude questions involving logarithms. The omission makes the set incomplete.

Option B:
Option B is also incomplete as it omits D, thereby ignoring the power rule, which is frequently used to simplify expressions like logₐ(xᵏ) in exam problems. Without D, the description of key log properties is not full.

Option C:
Option C is wrong because it excludes A and thereby omits the basic fact that logₐ(1) = 0, while still including the true D and E. Although some statements are correct, the absence of A means the option does not gather all true statements.

Option D:
Option D is correct because it retains all valid properties and discards the incorrect product rule in C, aligning with the standard set of logarithm rules that a NET candidate is expected to know.

This series alternates between adding 4 and multiplying by 3. Starting from 1, we have 1 + 4 = 5, 5 × 3 = 15, 15 + 4 = 19, 19 × 3 = 57 and 57 + 4 = 61. Following the same pattern, the next operation must be multiplication by 3. Thus, the next term is 61 × 3 = 183, preserving the alternating structure.

Option A:
Option A gives 181, which would require adding 120 to 61 or applying a different operation that is not indicated by any previous step. This has no relation to the alternating rule. Hence, 181 is not consistent with the pattern.

Option B:
Option B offers 182, which again cannot be produced by either adding 4 or multiplying by 3 to 61. It breaks the carefully maintained alternation of operations. Therefore, 182 is not appropriate.

Option C:
Option C yields 183, exactly equal to 61 × 3 and in full agreement with the "+4, then ×3" pattern. The sequence 1, 5, 15, 19, 57, 61, 183 remains logically coherent. Hence, 183 is the correct next term.

Option D:
Option D suggests 185, which requires adding 124 to 61 or some other arbitrary step, neither of which matches the earlier structure. This makes 185 an invalid continuation.

Each group consists of letters that are two positions apart inside the alphabet. Comparing successive groups shows that each letter in the group shifts forward by one position at every step. BDFH moves to CEGI, which moves to DFHJ, then to EGIK. Shifting E, G, I and K one position forward gives F, H, J and L, forming FHJL.

Option A:
Option A, FGIK, changes some letters but does not represent the result of adding one to each letter of EGIK. In particular, G should become H and I should become J, which FGIK does not implement. Because it fails to apply a uniform one step shift to all four letters, FGIK does not follow the pattern.

Option B:
Option B, FGIJ, keeps the first two letters plausible but breaks the regular spacing between the letters and the consistent one position shift from EGIK. The transformations from G to G and from K to J are not aligned with the plus one rule. Therefore FGIJ is not a logical continuation.

Option C:
Option C is correct because F, H, J and L are obtained by adding one to E, G, I and K respectively. It maintains both the two position spacing within each group and the equal shift of one between successive groups. This perfect parallel movement in all positions makes FHJL the only term that fits the series.

Option D:
Option D, GHJM, alters the spacing and does not arise from applying the same transformation to each letter of EGIK. Some letters move more than one step while others move inconsistently. As a result, GHJM fails to capture the structure of the pattern.

Each pair is formed by two consecutive letters, and the starting letter of each pair moves forward by one each time: AB, BC, CD, DE. After DE, the next consecutive pair should begin with E and include F, giving EF. Thus EF continues the sliding two-letter window.

Option A:
Option A, FG, is also a pair of consecutive letters but starts at F. It skips the pair that should begin at E, which the pattern requires. Because the series moves the starting letter by only one position at a time, FG is too far ahead.

Option B:
Option B is correct because EF starts at E, the letter immediately following D, and includes F. It preserves the same length and consecutive-letter property as the earlier terms. This perfect alignment with the rule makes EF the valid next term.

Option C:
Option C, DE, merely repeats the last pair in the series. The pattern is strictly forward-moving and does not suggest repetition. Hence DE cannot be considered the next item.

Option D:
Option D, CD, goes backward, returning to an earlier pair. This contradicts the continuous forward shift of the starting letter. As a result, CD does not logically follow DE in this sequence.