UGC NET Questions (Paper – 1)

Initially, alcohol = (5/12) × 72 = 30 litres and water = (7/12) × 72 = 42 litres. Let x litres of water be added. Alcohol remains 30 litres, water becomes 42 + x and the new ratio is 30:(42 + x) = 5:9. Cross-multiplying gives 30 × 9 = 5(42 + x), so 270 = 210 + 5x, which leads to 5x = 60 and x = 12. Thus, 12 litres of water must be added.

Option A:
Option A, 9 litres, gives water = 51 litres and alcohol = 30 litres, yielding a ratio 30:51 which simplifies to 10:17, not 5:9. Hence, 9 litres is insufficient to reach the required proportion of water to alcohol.

Option B:
Option B, 10 litres, leads to water = 52 litres and a ratio 30:52, which reduces to 15:26. This is still not 5:9, indicating that 10 litres does not adjust the mixture to the desired composition.

Option C:
Option C, 11 litres, produces water = 53 litres and a ratio 30:53 which does not simplify to 5:9. Therefore, it also fails to create the target ratio between alcohol and water.

Option D:
Option D is correct because with water increased to 54 litres, the ratio becomes 30:54, which simplifies to 5:9 when both terms are divided by 6. This exactly matches the condition specified in the problem.

In this series, the first letters are A, E, I, M, which correspond to positions 1, 5, 9 and 13, all increasing by 4. The second letters are D, H, L, P, i.e. positions 4, 8, 12 and 16, also increasing by 4. The next first letter after M (13) is Q (17) and the next second letter after P (16) is T (20). Hence QT is the only pair that continues the uniform +4 progression in both positions.

Option A:
Option A, QS, provides the correct first letter Q but gives S as the second letter, which is only 3 steps ahead of P. The series demands a constant increase of 4 positions for the second letter as well. Since S is inconsistent with the established step size, this option disrupts the pattern and is therefore incorrect.

Option B:
Option B is correct because Q and T are exactly 4 positions ahead of M and P respectively. This keeps the uniform step size intact for both components of the pair. By preserving the same arithmetic structure as in all previous terms, QT is the only option that maintains strict logical continuity in the series.

Option C:
Option C, QU, pushes the second letter forward to U, which is 5 positions after P. This overshoots the fixed increment and changes the rule mid-series. Since the first and second letters must follow the same step of +4, QU cannot be considered a valid continuation.

Option D:
Option D, QV, again uses the correct first letter Q but moves the second letter to V, which is 6 positions ahead of P. This introduces a larger increment than used before, breaking the constant-difference pattern that defines the sequence. Therefore option D does not fit the series.

This sequence follows the rule aₙ = n³+2n for n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 1³+2 = 3, 2³+4 = 12, 3³+6 = 33, 4³+8 = 72 and 5³+10 = 135, matching the given terms. For n = 6 the formula gives 6³+12 = 216+12 = 228. Therefore 228 is the only value that continues the same cubic relationship without any adjustment.

Option A:
Option A, 218, does not equal n³+2n for n = 6 and would require subtracting 10 from the value given by the formula. This distortion breaks the clear pattern connecting each term to its index. Thus 218 cannot be regarded as the correct next term.

Option B:
Option B, 222, is still below the correct cubic value of 228 and does not satisfy the rule aₙ = n³+2n. It represents an arbitrary modification rather than a logical extension of the pattern. For this reason it is not the right answer.

Option C:
Option C, 228, is precisely the value obtained from the generating rule when n = 6. It respects the same algebraic structure used for all earlier terms and keeps the series internally consistent. Hence 228 is the correct continuation.

Option D:
Option D, 236, exceeds the formula’s prediction and has no basis in the expression n³+2n for the next index. Adopting 236 would undermine the transparent cubic pattern, so it is not a valid option.

Each term is a block of four consecutive letters written in reverse order, and the starting letter moves forward by four positions each time. The starting letters D, H, L and P have positions 4, 8, 12 and 16, all increasing by 4. The next starting letter is position 20, which is T, and writing the four letter block from T in reverse gives TSRQ. This preserves both the plus four progression and the internal reverse arrangement.

Option A:
Option A, RQPO, rearranges letters including P and O but does not start with the correct letter T. It also breaks the strict descending order that characterises each block. Since it fails to follow the formula for the starting position and the internal structure, RQPO is not consistent with the pattern.

Option B:
Option B, SRQP, begins with S instead of T, which is only three positions ahead of P, not four. This changes the size of the jump in starting letters and disrupts the regularity of the sequence. Additionally, the letter order within the block does not correspond to a straightforward reverse of four consecutive letters starting at T.

Option C:
Option C, QPON, reverses the internal order of letters found around P but again starts with Q, not T, and does not reflect the plus four movement. It also returns to letters that are two positions behind P, which contradicts the forward movement of starting positions. Because it fails on both structural and positional grounds, QPON is not the correct continuation.

Option D:
Option D is correct because T is four positions after P and TSRQ lists T, S, R and Q as consecutive letters in descending order. It exactly mirrors the format of earlier blocks, which all reverse a run of four consecutive letters. This consistent adherence to both the starting letter progression and the reverse order makes TSRQ the valid next term.

The series lists English words for natural numbers in increasing order: ONE (1), TWO (2), THREE (3) and FOUR (4). The next number in this sequence is FIVE (5). Therefore the word FIVE is the only term that preserves the numeric progression described in word form.

Option A:
Option A, SIX, represents the number 6, which would be two steps ahead of FOUR if FIVE is skipped. The series shows no evidence of skipping intermediate numbers. As the pattern is strictly consecutive, SIX cannot be the immediate next term.

Option B:
Option B, ONE, would restart the sequence from the beginning. There is no cyclical structure indicated; the list moves forward rather than looping. Repeating ONE therefore breaks the clear numeric ordering.

Option C:
Option C, SEVEN, represents 7 and skips both FIVE and SIX. This contradicts the strictly increasing by 1 pattern in the given sequence. Hence SEVEN cannot logically follow FOUR.

Option D:
Option D is correct because FIVE is the next integer after FOUR in the natural counting order. It continues the sequence smoothly without omissions or repetitions. As the series is clearly numerical in nature, FIVE must be the next term.

Starting letters A(1), D(4), H(8), M(13) have gaps +3, +4, +5. Next gap is +6 → S(19). Pair is ST.

Option A:
Option A starts at Q(17), which is +4 from M, not +6.

Option B:
Option B is correct because S is +6 from M and T follows S.

Option C:
Option C starts at R(18), which is +5 from M, not +6.

Option D:
Option D starts at O(15), which is +2 from M, not +6.

Statements A, B and C are correct: the behaviour of inequalities under multiplication by positive and negative numbers is standard, and solving them parallels solving equations with additional attention to sign changes. Statement D is false because solutions of inequalities are often intervals or ranges of values, not single numbers, and E is false because inequality reasoning has many practical applications and is tested in aptitude sections. Therefore the option that includes A, B and C and excludes D and E is correct.

Option A:
Option A is incomplete as it omits C and thus fails to highlight the close procedural connection between solving equations and inequalities, which is a key conceptual insight in aptitude contexts.

Option B:
Option B is also incomplete because it leaves out A, ignoring the important fact that multiplication by a positive number leaves the inequality direction unchanged. Without A, some fundamental transformation rule is missing.

Option C:
Option C is correct because it gathers all three true statements about manipulation and solution processes for inequalities, while rejecting D and E, which misrepresent the nature and usefulness of inequality solutions. It matches the conceptual level expected in UGC NET.

Option D:
Option D is wrong since it includes D, the false claim that solution sets are always single numbers, and thereby contradicts the idea of intervals. Including this misstatement makes the combination logically incorrect.

The letters in each group are equally spaced by five positions. AFK is (1,6,11), BGL is (2,7,12), CHM is (3,8,13), DIN is (4,9,14). Next is (5,10,15) = EJO.

Option A:
Option A is correct because E, J and O preserve the constant five-letter spacing and the start-letter progression.

Option B:
Option B ends with Q (17) instead of O (15), breaking the spacing.

Option C:
Option C does not maintain equal gaps of five.

Option D:
Option D starts at F instead of E, breaking the start-letter sequence.

The given relation “x varies directly as y and inversely as z” can be written as x = k·y/z for some constant k. Using x = 16 when y = 8 and z = 3, we have 16 = k·8/3, so k = 16 × 3 / 8 = 6. For y = 9 and z = 2, x = 6·9/2 = 54/2 = 27. Hence, the new value of x is 27.

Option A:
Option A, 18, would correspond to x = 6·y/z = 6·9/? with some other z value, but it is not obtained when z = 2. If we plug x = 18 into x = k·y/z with k = 6, we get 18 = 6·9/z, implying z = 3, which contradicts the specified z = 2. So 18 does not satisfy the given condition.

Option B:
Option B, 24, arises if we mistakenly treat the change in y as proportional but ignore the change in z. With k = 6, x would be 24 only when y/z = 4, which is not the case for y = 9 and z = 2 where y/z = 4.5. Thus, 24 is not consistent with the correct variation formula.

Option C:
Option C is correct because it uses the same constant of proportionality k = 6 found from the first condition and correctly substitutes y = 9 and z = 2. The resulting value 27 is obtained by straightforward application of x = k·y/z and maintains both the direct dependence on y and inverse dependence on z.

Option D:
Option D, 30, would require y/z to be 5 when k = 6, since 30 = 6·5. But the actual ratio y/z is 9/2 = 4.5 in the new situation. Therefore, 30 does not follow from the prescribed relationship between x, y and z and is not an acceptable solution.

Statements A and B correctly describe the length of non-leap and leap years and the divisibility rule used to identify leap years. Statement C is true because the weekly cycle repeats every 7 days, and E is also correct as calendar questions usually count odd days from a known reference day. Statement D is false because odd days are calculated modulo 7, not modulo 5. Therefore, the combination A, B, C and E only is correct.

Option A:
Option A is incomplete because it omits E and therefore misses the practical step of accumulating odd days from a reference, which is crucial for solving day-of-week problems. Without E, the method is not fully described.

Option B:
Option B is wrong as it includes D, which uses modulo 5 instead of modulo 7, and so miscalculates odd days. This incorrect rule makes the entire option unacceptable.

Option C:
Option C is incomplete since it leaves out A and B, thereby not explicitly stating the year-length facts and leap-year rule, even though it contains two true statements.

Option D:
Option D is correct because it gathers all the true statements and excludes D, the incorrect modulus rule, giving a complete and accurate picture of odd-day calculations in calendar questions.

Each term is a 5-letter block; starting letters move by +2: A, C, E, G → next I. So the next block is IJKLM.

Option A:
Option A starts at H (+1 from G), not +2.

Option B:
Option B is correct because it starts at I (+2 from G) and continues consecutively.

Option C:
Option C starts at J (+3 from G), not +2.

Option D:
Option D repeats an earlier term and does not continue the slide.

This series alternates between adding 3 and multiplying by 2. Starting from 2, we have 2 + 3 = 5, 5 × 2 = 10, 10 + 3 = 13, 13 × 2 = 26 and 26 + 3 = 29. Following the same alternation, the next operation should be multiplication by 2. Therefore, the next term is 29 × 2 = 58. This continues the alternating rule without any break.

Option A:
Option A yields 58, which honours the exact up–down pattern of "+3, then ×2" seen throughout the series. The extended sequence 2, 5, 10, 13, 26, 29, 58 shows the rule being applied consistently. Hence, 58 is the correct next term.

Option B:
Option B gives 56, which would require adding 27 to 29 or applying a different operation unrelated to the earlier steps. It does not arise from either "+3" or "×2" as used in the pattern. Therefore, 56 is not consistent with the series.

Option C:
Option C suggests 60, requiring an addition of 31 or some new rule not indicated in previous terms. This breaks the elegant alternation between addition and multiplication. Thus, 60 cannot be considered the correct continuation.

Option D:
Option D offers 62, which again does not result from multiplying 29 by 2 or adding 3 to it. It forces a new operation into the series and disrupts the established structure. Hence, 62 is not the right answer.

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.