UGC NET Questions (Paper – 1)

A is given as the mother of B. B and C are siblings because B is the brother of C, so A is also the mother of C. C is the father of D, which means D is A’s grandchild through C. Hence, combining these links, A’s relationship to D is that of a grandmother.

Option A:
Option A correctly traces the family relations step by step: A is parent to C, and C is parent to D, making A one generation above D’s parent. This is precisely the definition of a grandparent, and since A is female, the correct term is grandmother.

Option B:
Option B, aunt, would be appropriate if A were a sibling of D’s parent, but the problem clearly states A is the mother of B and C, not their sister. Therefore, she stands one generation above, not at the same generation as D’s parent.

Option C:
Option C, sister, would require A and D to share the same parents or at least be in the same generation. The information places A two generations above D, so this description is incompatible.

Option D:
Option D, mother, would mean A directly gave birth to D, but the statement shows that C is the father of D, and A is C’s mother, so the relationship has an extra generational step.

The condition x > 3 tells us that x is strictly greater than 3 on the number line. Adding the same number to both sides of an inequality preserves its direction, so adding 2 to get x + 2 > 3 + 2 shows that x + 2 > 5. This relationship must hold for every x satisfying the original inequality.

Option A:
Option A follows directly from adding 2 to both sides of the original inequality. Since 3 + 2 equals 5, every value of x greater than 3 will make x + 2 greater than 5 as well.

Option B:
Option B, x − 5 > 0, would require x to be greater than 5, which is stronger than the given condition. Since x may be between 3 and 5, this statement need not always be true.

Option C:
Option C, 2x < 6, implies x < 3, which actually contradicts the assumption that x is greater than 3. Option D: Option D, x/2 < 1.5, is another form of x < 3 and thus conflicts with the original inequality rather than following from it.

Statements A, B and D are correct. A states the multiplication principle for independent choices. B defines permutations as ordered arrangements. D is true because there are n! ways to arrange n distinct objects. C is false; the formula for permutations of n distinct objects taken r at a time is nPr, and E is false because identical objects reduce the number of distinct permutations below n!. Thus A, B and D only are correct.

Option A:
Option A is correct as it includes the standard counting principle, the definition of permutations and the basic n! formula, while excluding C and E, which misapply combinatorial formulas.

Option B:
Option B is incorrect since it adds C, incorrectly using nCr in place of nPr for permutations, and so combines an error with correct statements.

Option C:
Option C is wrong because it also includes E, claiming n! permutations even with identical objects, and therefore contains two incorrect statements.

Option D:
Option D is incomplete as it drops A and leaves only B, D and E, losing the foundational multiplication principle and adding E, which contradicts the effect of identical objects.

Each term is a block of four consecutive letters, and the block slides one position to the right each time: ABCD (1–4), BCDE (2–5), CDEF (3–6), DEFG (4–7). The next block must start at position 5 and cover 5–8, which is EFGH. Thus EFGH uniquely continues the sliding four-letter window.

Option A:
Option A, FGHI, is a valid four-letter block but starts at F, thereby skipping the block that should start at E. The series does not skip intermediate starting positions. Therefore FGHI is too far ahead to be the next term.

Option B:
Option B, EFGJ, begins correctly at E but breaks the requirement that letters inside the block be consecutive. J does not follow G directly. As a result, EFGJ does not maintain the strictly consecutive nature of the blocks.

Option C:
Option C, DEFG, repeats the last given term. The pattern clearly moves forward by one starting letter at each step. Since no repetition rule is indicated, DEFG cannot be the next term.

Option D:
Option D is correct because it starts at E, the letter immediately after D, and continues with F, G and H in order. It preserves both the block length and the consecutiveness of letters. Consequently, EFGH is the proper continuation of the series.

This series is generated by the rule “multiply the previous term by 3 and add 2”. We have 1×3+2 = 5, 5×3+2 = 17, 17×3+2 = 53 and 53×3+2 = 161. Applying the same rule to 161 gives 161×3+2 = 483+2 = 485. Hence 485 is the only value that continues the sequence in strict accordance with the recursive pattern.

Option A:
Option A gives 479, which cannot be written as 161×3+2 and would require subtracting an additional amount. This contradicts the consistently “+2” adjustment after multiplication by 3 in the earlier steps. Therefore 479 is not consistent with the rule.

Option B:
Option B gives 481, corresponding to 161×3−2, which would change the sign of the constant adjustment and break the established behaviour. Thus 481 is not a valid continuation of the series.

Option C:
Option C gives 483, equal to 161×3, but this omits the “+2” that appears in every transition. This makes the rule incomplete for the last step, so 483 cannot be accepted as correct.

Option D:
Option D gives 485, exactly equal to 161×3+2. It maintains the same transformation applied in all earlier steps, keeping the pattern transparent and stable. Consequently, 485 is the correct next term in the series.

Each group is made of three consecutive letters, but the starting letters do not progress uniformly. They move from A to C (+2), from C to F (+3) and from F to J (+4), so the gaps themselves increase by one each time. Following this rule, the next starting letter should be J + 5 = O, and the three consecutive letters from O are O, P and Q. Therefore OPQ is the correct continuation.

Option A:
Option A, MNO, begins at M, which would correspond to a gap of only +3 from J and ignores the observed +5 requirement. It would make the sequence of gaps 2, 3, 4, 3 instead of 2, 3, 4, 5. Since the pattern clearly shows increasing differences, MNO does not fit.

Option B:
Option B, NOP, starts at N, representing a gap of +4 from J and essentially repeats the previous increase. This would produce a flat sequence of gaps at the end instead of a steadily growing one. Because the rule is that each new gap increases by one, NOP cannot be the next term.

Option C:
Option C is correct because the starting letter O is exactly five positions after J, which extends the gap sequence 2, 3, 4 to 5. The letters P and Q follow in order, preserving the internal structure of each group as three consecutive letters. Hence OPQ is the only option that satisfies both the gap rule and the block structure.

Option D:
Option D, PQR, begins even further ahead at P, yielding a gap of +6 from J. This overshoots the expected step and makes the sequence of gaps jump from 4 to 6. As there is no evidence of such a sudden larger jump, PQR is not a valid continuation of the pattern.

The series can be generated by the formula (a_n = 2n^4 + 3n^2 + 1) for n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 6, 45, 190, 561 and 1326, which match the question. For n = 6 the expression gives (2×6^4 + 3×6^2 + 1 = 2592 + 108 + 1 = 2701). Hence 2701 is the correct next term.

Option A:
Option A, 2665, is 36 less than the formula’s output and does not equal (2×6^4 + 3×6^2 + 1). It would require an unjustified reduction at the final term, so 2665 is not valid.

Option B:
Option B, 2683, remains below 2701 and fails to satisfy the expression for n = 6. Choosing 2683 would disrupt the quartic-plus-square structure of the sequence. Thus 2683 is not correct.

Option C:
Option C, 2723, overshoots the computed value and again cannot be obtained from the given formula. Using 2723 would break the algebraic pattern that has held so far. Hence 2723 is not appropriate.

Option D:
Option D, 2701, exactly equals the value generated by the rule for n = 6. It maintains the same functional relationship between n and a_n for all terms, so 2701 is the correct continuation of the series.

The terms follow the rule aₙ = n³+n²−1 for n starting from 1. For n = 1, 2, 3, 4 and 5 we get 1+1−1 = 1, 8+4−1 = 11, 27+9−1 = 35, 64+16−1 = 79 and 125+25−1 = 149. For n = 6 the formula yields 216+36−1 = 251. Thus 251 is the only number that continues this cubic-plus-square-minus-one pattern consistently.

Option A:
Option A, 241, is 10 less than the value obtained from the formula for n = 6. It does not satisfy aₙ = n³+n²−1 and so fails to extend the pattern logically. Therefore 241 cannot be the correct answer.

Option B:
Option B, 251, matches exactly the result of applying the rule with n = 6. It maintains the same combination of n³ and n² seen in all the earlier terms. Because the algebraic structure remains unchanged, 251 is the correct continuation of the series.

Option C:
Option C, 245, is also different from the calculated value and would require altering the constant term in the expression. That change has no support in the given data, so 245 is not a valid next term.

Option D:
Option D, 259, is larger than the correct value and again does not come from n³+n²−1 when n = 6. Using 259 would break the very precise pattern built into the sequence, making it an incorrect option.

The consecutive differences are 7, 13, 19 and 25. These form an arithmetic progression with common difference 6. To maintain this structure, the next difference should be 31. Adding 31 to the last term 67 gives 98, so 98 is the term that continues both the original series and the progression among the gaps.

Option A:
Option A gives 94, which implies a difference of 27 from 67. This would turn the sequence of differences into 7, 13, 19, 25, 27, where the last increase is only 2 rather than 6. That breaks the consistent second-level behaviour, so 94 is not appropriate.

Option B:
Option B gives 98, corresponding to a difference of 31 from 67. This extends the differences to 7, 13, 19, 25, 31, where each increase is exactly 6. Because this respects the pattern at both levels, 98 is the correct next term.

Option C:
Option C gives 96, leading to a difference of 29 from 67. This would produce differences 7, 13, 19, 25, 29, with the last increment equal to 4 instead of 6. Therefore 96 does not align with the established rule.

Option D:
Option D gives 100, which yields a difference of 33 from 67. That would make the final increment among differences equal to 8, again breaking the uniform step of 6. Hence 100 cannot be the valid continuation of the series.

A and B define AP and GP. D is true because the sign of the common difference determines whether an AP increases, decreases, or stays constant. E is true because multiplying by −1 flips a non-zero term’s sign each step, alternating between a and −a. C is false (Fibonacci-type recurrence is generally neither AP nor GP). Hence A, B, D and E only are correct.

Option A:
Option A is incomplete as it omits E, a standard GP special case often tested.

Option B:
Option B is correct because it combines the core definitions with typical behaviour of APs and a specific GP example, and excludes C, which is false.

Option C:
Option C is wrong because it omits A, the definition of an arithmetic sequence.

Option D:
Option D is incorrect because it omits B, the definition of a geometric sequence.

From x:y = 4:7, we write x = 4k and y = 7k. From y:z = 21:10, we write y = 21m and z = 10m. Equating y gives 7k = 21m, so k = 3m. Substituting back, x = 4k = 12m and z = 10m, so x:z = 12m:10m = 6:5. Thus, 6:5 is the correct compound ratio of x to z.

Option A:
Option A, 4:5, would require x to be 4 units when z is 5 units, which is not consistent with x = 12m and z = 10m derived from solving the two given ratios simultaneously. If we force 4:5, we cannot maintain both x:y = 4:7 and y:z = 21:10 at the same time.

Option B:
Option B is correct because it results from a systematic elimination of the common term y and then simplifying the resulting x:z ratio. The use of a common multiple for y ensures that the structure of both original ratios is respected. This guaranteed alignment leads uniquely to 6:5.

Option C:
Option C, 8:5, would make x comparatively smaller than 12m when measured against z = 10m, breaking the proportional link with y. No consistent choice of scaling parameters k and m will give x:z = 8:5 without contradicting the given ratios.

Option D:
Option D, 12:7, mixes the numerator from x = 12m with an unrelated denominator. A ratio of 12:7 would not maintain the numerical relationships defined by 4:7 and 21:10 and therefore fails the internal consistency check.

The series can be modelled by a₁ = 5 and aₙ₊₁ = 5aₙ + n³ for n ≥ 1. Using this rule we obtain a₂ = 5·5 + 1³ = 26, a₃ = 5·26 + 2³ = 138, a₄ = 5·138 + 3³ = 717 and a₅ = 5·717 + 4³ = 3649, in agreement with the given numbers. For n = 5 the next term is a₆ = 5·3649 + 5³ = 18370. Hence 18370 is the only value that satisfies the same recurrence and properly extends the series.

Option A:
Option A, 18310, is 60 less than the value dictated by aₙ₊₁ = 5aₙ + n³ at n = 5. Choosing 18310 would require subtracting 60 from the correctly computed term, which has no basis in the earlier steps. Therefore this option does not conform to the established recurrence and is incorrect.

Option B:
Option B, 18340, is closer but still 30 less than the correct value 18370. It cannot be obtained from 5·3649 + 5³ and would imply weakening the cubic correction at the last stage. Since the recurrence is followed exactly before, option B is not a valid continuation.

Option C:
Option C, 18370, is exactly the result of applying the rule to a₅ with n = 5. It preserves the structure of multiplying the previous term by five and then adding the cube of the index. Because this process accounts for all earlier terms and produces 18370 next, this option is correct.

Option D:
Option D, 18430, is 60 greater than the computed value and again fails to satisfy 5a₅ + 5³. Accepting 18430 would mean amplifying the adjustment term at the last step without any justification. Thus option D breaks the recurrence and is not acceptable.

The first-level differences are 4, 8, 16, 32 and 64. Each difference is double the previous one, representing a geometric progression at the second level. The next difference should therefore be 128. Adding 128 to the last term 126 gives 254, so 254 is the value that preserves the doubling structure among the differences.

Option A:
Option A gives 250, corresponding to a difference of 124 from 126. This breaks the exact doubling rule, since 124 is not twice 64. Therefore 250 does not respect the second-level geometric behaviour.

Option B:
Option B gives 252, implying a difference of 126, which again does not equal 2×64. This leads to an inconsistent change in the pattern of differences. Hence 252 cannot be accepted.

Option C:
Option C gives 256, corresponding to a difference of 130 from 126. That would overshoot the expected difference of 128 and disrupt the doubling rule for the gaps. Thus 256 is not a valid continuation.

Option D:
Option D gives 254, which is 128 more than 126. The differences become 4, 8, 16, 32, 64, 128, perfectly doubling each time. This strong and simple structure confirms 254 as the correct next term.

Each group is made of five letters that are equally spaced from each other in the alphabet, forming a regular ladder pattern. ADGJM, BEHKN, CFILO and DGJMP all have the same internal step size between successive letters and each new group is obtained by moving all letters one position forward. Applying this simple shift to DGJMP gives E, H, K, N and Q. The resulting group EHKNQ preserves the internal spacing and extends the vertical sequences of every column, so it is the correct continuation.

Option A:
Option A, EHJMQ, uses some of the right letters but places them in positions that break the constant spacing between adjacent letters. The sequence from H to J to M to Q does not match the regular step pattern of the series. Because the internal distances are wrong, EHJMQ cannot be the answer.

Option B:
Option B, EIKNQ, changes the second letter so that the equal increments between letters inside the group are lost. It does not arise from a uniform one step shift of DGJMP. As a result, EIKNQ fails to fit the exact construction rule.

Option C:
Option C, FHKNR, starts with F instead of E and moves the letters in a way that does not preserve the ladder structure or the column wise progressions. The clear transformation from one group to the next is no longer visible. Therefore FHKNR is not a valid continuation.

Option D:
Option D is correct because EHKNQ is formed by adding one to each letter of DGJMP and it keeps the same step between neighbouring letters as in all previous groups. The vertical sequences remain intact in every column. This strict adherence to the series rule makes EHKNQ the only suitable next term.

The first-level differences are 7, 13, 21 and 31. The second-level differences are 6, 8 and 10, which increase by 2 each time. Continuing this pattern, the next second-level difference should be 12, so the next first-level difference becomes 31 + 12 = 43. Adding 43 to the last term 76 gives 119, making 119 the only value that preserves the consistent second-level arithmetic structure.

Option A:
Option A gives 119, which fits the calculated first-level difference of 43 and keeps the second-level differences as 6, 8, 10, 12. This maintains a neat progression where the growth of the gaps is itself arithmetic. Because both layers of the pattern remain intact, 119 is the correct continuation of the series.

Option B:
Option B gives 115, which would imply a difference of 39 from 76. That would make the second-level differences 6, 8, 10 and 8, breaking the steady increase by 2 and introducing an inconsistency. Hence 115 does not follow the underlying logic of the sequence.

Option C:
Option C gives 123, corresponding to a difference of 47 from 76. This would yield second-level differences 6, 8, 10 and 16, which jumps too sharply and does not respect the +2 progression. Therefore 123 cannot be accepted as the next term.

Option D:
Option D gives 127, leading to a difference of 51 from 76. The resulting second-level differences become 6, 8, 10 and 20, distorting the smooth pattern of increase. Thus 127 is not compatible with the observed structure of the series.