UGC NET Questions (Paper – 1)

An increase in the ratio 5:6 means the population is multiplied by 6/5 from 2018 to 2020. A further increase in the ratio 6:7 multiplies it by 7/6 from 2020 to 2022. The combined effect from 2018 to 2022 is (6/5) × (7/6) = 7/5. Therefore, the 2022 population is 1,80,000 × (7/5) = 1,80,000 × 7 ÷ 5 = 2,52,000. Thus 2,52,000 is the correct population in 2022.

Option A:
Option A, 2,16,000, corresponds to multiplying by 6/5 only once and ignoring the second stage of growth. That would be the population if the increase stopped in 2020, not in 2022. Since the problem explicitly describes two successive increases, this option underestimates the final population.

Option B:
Option B, 2,40,000, might arise from an approximate or partial application of the ratios but does not correspond to the exact product 7/5. If we back-calculate, 2,40,000 × (5/7) ≈ 1,71,428.57, which does not match the original population of 1,80,000. Hence 2,40,000 is inconsistent with the data.

Option C:
Option C, 2,70,000, would require a combined multiplier of 2,70,000 ÷ 1,80,000 = 3/2. However, the actual combined multiplier from 5:6 and 6:7 is 7/5 = 1.4, not 1.5. Therefore this option overstates the population growth and cannot be correct.

Option D:
Option D is correct because it uses the proper composition of the two growth ratios, first 6/5 and then 7/6, yielding a neat combined factor of 7/5. The arithmetic leads precisely to 2,52,000, and reversing the process returns the original 1,80,000, demonstrating full consistency.

Each term is a block of three consecutive letters arranged in reverse order, and the starting letters move backward by one. ZYX uses 26 to 24, YXW uses 25 to 23, XWV uses 24 to 22 and WVU uses 23 to 21. The next starting letter is V at position 22 moving back to U and T, so the reversed block is VUT. Thus VUT maintains both the backward shift in the starting letter and the internal reverse order.

Option A:
Option A is correct because it starts at V, exactly one position before W, and lists the next two letters U and T in descending order. This matches the pattern of overlapping reversed triples and keeps the shift of the starting letter constant. It also ensures that all letters inside the group are consecutive in reverse alphabetical order.

Option B:
Option B, UTS, reverses the internal positions differently and does not use V, which should be the starting letter according to the pattern. It also disrupts the way the groups overlap with their neighbours. Since it does not follow the same mechanism of backward stepping, UTS is not the correct continuation.

Option C:
Option C, TSV, rearranges the letters and fails to present a clean three letter consecutive block in reverse order. It also does not arise from shifting the starting letter back by one from W. Because it breaks both the structural and positional rules, TSV is inconsistent with the series.

Option D:
Option D, YZW, introduces letters that are not consecutive in the correct reversed manner following WVU. It also moves back to Y instead of correctly progressing to V. As a result, YZW does not align with the established pattern.

The starting letters move backward by one: Z, Y, X, W → next V. Each term is five consecutive letters with wrap-around when needed. From V, the five-letter block is VWXYZ.

Option A:
Option A starts at U, which would be two steps before W, not one.

Option B:
Option B wraps to A unnecessarily; from V the next five letters are VWXYZ without wrap.

Option C:
Option C is correct because it starts at V (one before W) and forms VWXYZ.

Option D:
Option D repeats an earlier term and does not continue the start-letter progression.

Comparing ADK, BEL, CFM and DGN shows that each column of letters increases by one step from term to term. The first letters go A, B, C, D and the second letters go D, E, F, G, while the third letters move K, L, M, N. To continue these simple vertical progressions, the next group must use E as the first letter, H as the second and O as the third. This gives EHO, which alone matches all three column wise sequences, so it is the correct continuation.

Option A:
Option A is correct because EHO extends each column by one step in exactly the same way as the earlier transitions. The triple A→B→C→D→E, D→E→F→G→H and K→L→M→N→O is preserved with no break. Since it respects the straightforward positional rule in every column, EHO fits perfectly as the next term.

Option B:
Option B, EIP, starts with E but moves the second and third letters too far ahead compared with the previous terms. The vertical sequences would suddenly jump rather than step smoothly. Because it disturbs the regular one step progressions, EIP cannot be the correct answer.

Option C:
Option C, FIP, begins with F instead of E, so the first column would no longer increase by one from D. It also shifts the remaining letters in a way that does not align with the existing pattern. Therefore FIP fails to match the structure of the series.

Option D:
Option D, DHN, reuses some letters from the last given group and does not advance all positions simultaneously. The series clearly moves all three letters forward together in each step. As DHN does not follow this rule, it is not a valid continuation.

Statement B gives the correct falsity condition for inclusive disjunction: it fails only when both components are false. C and D are De Morgan’s laws for negation of conjunction and disjunction, both of which are valid equivalences in classical logic. A is false; a conjunction requires both p and q to be true, not just at least one. E is false because in two-valued logic, a statement and its negation cannot be true together. Therefore B, C and D only are correct.

Option A:
Option A is incomplete as it omits D and so does not state the second De Morgan equivalence, leaving the treatment of negation of disjunction unfinished.

Option B:
Option B is incorrect because it includes E, which claims that a statement and its negation can both be true, contradicting the law of non-contradiction in classical logic.

Option C:
Option C is wrong since it accepts A, which incorrectly describes the truth condition for conjunction, and therefore introduces a clear error into the combination.

Option D:
Option D is correct as it gathers the proper truth condition for disjunction and both De Morgan laws, while excluding A and E, which misrepresent conjunction and the mutual exclusivity of a statement and its negation.

The first letters A, B, C and D occupy positions 1, 2, 3 and 4, increasing by 1. The second letters M, L, K and J occupy positions 13, 12, 11 and 10, decreasing by 1. The next first letter should therefore be E (5) and the next second letter should be I (9). Combining them gives EI as the unique pair that continues both directions of movement.

Option A:
Option A is correct because it places E after D in the first-letter sequence and I before J in the second-letter sequence. It maintains the symmetric convergence of the two letters toward the middle of the alphabet. This dual movement is exactly what defines the pattern.

Option B:
Option B, EK, keeps E as the first letter but uses K as the second letter, which should already have been passed in the downward sequence. The second letters must keep moving backward without repeating earlier positions. As K is out of order, EK cannot be the correct continuation.

Option C:
Option C, FH, advances the first letter by two steps from D to F and moves the second letter too far back to H. This breaks the consistent one-step changes in both positions. Because it alters the movement rule, FH is not acceptable.

Option D:
Option D, EL, moves the first letter correctly to E but leaves the second letter at L, which reverses direction from the decreasing trend. Instead of moving to 9, it jumps back to 12, contradicting the pattern. Therefore EL does not fit the sequence.

The sequence is generated by aₙ = 3n⁴ + 4n² + 1 for n starting from 1. For n = 1, 2, 3, 4 and 5 this gives 3+4+1 = 8, 48+16+1 = 65, 243+36+1 = 280, 768+64+1 = 833 and 1875+100+1 = 1976, confirming the rule. For n = 6 we compute 3·6⁴+4·6²+1 = 3·1296+144+1 = 3888+144+1 = 4033. Thus 4033 is the only value that maintains this quartic-plus-quadratic-plus-constant pattern.

Option A:
Option A, 3985, is smaller than the formula’s output and does not satisfy aₙ = 3n⁴+4n²+1 for n = 6. It would require subtracting 48 from the correctly computed value only at the final term, which is inconsistent with the earlier behaviour. Therefore option A is incorrect.

Option B:
Option B, 4017, is closer but still not equal to 4033 and again fails to arise from substituting n = 6 into the rule. Choosing 4017 would mean modifying the constant or coefficients in a way not supported by previous terms. Hence option B is incorrect.

Option C:
Option C, 4033, matches exactly the value obtained from the polynomial expression for n = 6. It preserves the relative contributions of the quartic and quadratic terms as well as the constant increment. Because the same rule explains all terms including the next, 4033 is the correct answer.

Option D:
Option D, 4065, overshoots the correct value by 32 and does not equal 3·6⁴+4·6²+1. Using 4065 would break the precise algebraic relationship that defines the series, so option D is incorrect.

The average of three numbers is the sum of the numbers divided by 3. If the average is 25, then the total sum must be 25 × 3 = 75. The two known numbers are 20 and 30, whose sum is 50. Subtracting 50 from 75 gives 25, so the third number must be 25.

Option A:
Option A, 25, exactly matches the required third number, because 20 + 30 + 25 = 75 and 75 ÷ 3 = 25. It is therefore the correct choice.

Option B:
Option B, 20, would give a total of 20 + 20 + 30 = 70, and 70 ÷ 3 does not equal the required average of 25, so this choice fails the given condition.

Option C:
Option C, 30, leads to 20 + 30 + 30 = 80, giving an average of 80 ÷ 3, which is greater than 25 and inconsistent with the stated mean.

Option D:
Option D, 35, produces a total of 20 + 30 + 35 = 85, resulting in an average greater than 25, again contradicting the problem’s information.

The terms follow the rule aₙ = 2n⁴ + n³ + 2 with n starting at 1. Substituting n = 1, 2, 3, 4 and 5 yields 5, 42, 191, 578 and 1377, exactly the five numbers in the series. Evaluating the same expression for n = 6 gives 2810. Hence 2810 is the only value that fits this established polynomial pattern.

Option A:
Option A, 2786, is 24 less than the value obtained from aₙ = 2n⁴ + n³ + 2 for n = 6. To choose 2786 we would need to reduce the polynomial result only at the last term. Since earlier entries follow the rule without such a change, 2786 cannot correctly continue the series.

Option B:
Option B, 2798, is still below the required value and does not equal the output of the formula for n = 6. It represents a partial adjustment towards 2810 but still breaks the algebraic relationship. Therefore this option does not preserve the exact pattern linking the terms.

Option C:
Option C, 2810, matches precisely the value produced by 2n⁴ + n³ + 2 when n = 6. It maintains the contributions of both the fourth-power and cube terms in exactly the same way as for previous indices. Because the rule explains the given terms and extends naturally to 2810 next, this option is correct.

Option D:
Option D, 2822, is 12 larger than the polynomial value and cannot be generated by the same expression at n = 6. Accepting 2822 would require adding an unexplained increment at the final step. This would break the smooth growth dictated by the formula, so 2822 is not acceptable.

This sequence is generated by the polynomial aₙ = 3n⁴ + 2n² + 1 with n starting from 1. For n = 1, 2, 3, 4 and 5 this rule gives 6, 57, 262, 801 and 1926, exactly matching the given terms. Using the same expression for n = 6 produces 3961. Thus 3961 is the only value that maintains this polynomial pattern consistently.

Option A:
Option A, 3949, is smaller than the value produced by 3n⁴ + 2n² + 1 for n = 6. To accept 3949 we would have to reduce the correct polynomial result by 12 only at the sixth term. Such a selective adjustment is not supported by the earlier terms, so this option is incorrect.

Option B:
Option B, 3961, is exactly the number obtained when n = 6 is substituted into aₙ = 3n⁴ + 2n² + 1. It keeps the same balance between the fourth-power and square terms that explains the growth of the series. Because it extends the rule without modification, 3961 is the correct next term.

Option C:
Option C, 3973, is 12 greater than the pattern value. Accepting 3973 would require adding an arbitrary constant at the final step while leaving the rule unchanged before, which breaks the algebraic consistency. Hence this option does not correctly continue the sequence.

Option D:
Option D, 3997, deviates even more from the required value. It cannot be produced by 3n⁴ + 2n² + 1 at n = 6 and would destroy the tight functional link between index and term. Therefore 3997 is not a valid continuation of the number series.

In the first alloy, copper and zinc are 7/10 and 3/10 of the weight respectively. In the second alloy, copper and zinc are 5/7 and 2/7 of the weight respectively. Taking equal weights of each alloy, total copper = 7/10 + 5/7 = (49 + 50)/70 = 99/70 and total zinc = 3/10 + 2/7 = (21 + 20)/70 = 41/70. The common denominator 70 cancels in the ratio, so copper:zinc = 99:41.

Option A:
Option A, 89:41, looks close to the correct ratio but uses 89 instead of 99, indicating an incorrect sum of copper fractions. To obtain 89, one would have to miscalculate either 7/10 or 5/7, and using this ratio would not respect the exact contributions from both alloys.

Option B:
Option B, 41:99, simply reverses the correct ratio and would imply that zinc is present in much larger proportion than copper in the mixture. This contradicts the original alloys, both of which have more copper than zinc, so 41:99 cannot represent the final proportions.

Option C:
Option C, 99:44, keeps the correct numerator 99 but inflates the zinc component from 41 to 44. This could arise from careless arithmetic in adding 3/10 and 2/7. However, if zinc were 44 parts, the total fractions would no longer sum to the actual combined mixture, so 99:44 is inconsistent with the data.

Option D:
Option D is correct because it is derived from adding the fractional parts for copper and zinc separately and then simplifying the resulting ratio. The common denominator cancels out, leaving 99:41 as the exact composition ratio for the combined alloy when equal weights of each initial alloy are used.

To find how many students like at least one of the two subjects, we use the inclusion–exclusion principle. We add those who like each subject and subtract those counted twice in the overlap: 22 + 18 − 10 = 30 students. These 30 students like Mathematics, Science or both.

Option A:
Option A, 30, is the direct result of applying inclusion–exclusion and correctly represents the number who like at least one of the two subjects.

Option B:
Option B, 32, would correspond to assuming an overlap of 8 instead of 10, which contradicts the given overlap and misuses the data.

Option C:
Option C, 40, assumes that every student likes at least one of the subjects, which is not stated and need not be true.

Option D:
Option D, 50, exceeds the total number of students in the group, so it is impossible.

The pattern here is aₙ = 2n⁵ − 3n² + 4 with n beginning at 1. For n = 1, 2, 3, 4 and 5 this expression gives 3, 56, 463, 2004 and 6179, exactly reproducing the given sequence. When n = 6 we get a₆ = 2·6⁵ − 3·6² + 4 = 2·7776 − 108 + 4 = 15552 − 104 = 15448. Thus 15448 is the unique value consistent with this quintic pattern.

Option A:
Option A, 15448, is exactly the outcome of substituting n = 6 into 2n⁵ − 3n² + 4. It maintains the same dominance of the fifth-power term with a negative quadratic adjustment and a constant shift. Because this structure accounts for all earlier terms and extends naturally to 15448, this option correctly continues the series.

Option B:
Option B, 15424, is 24 less than the rule’s value and cannot be generated by the same polynomial at n = 6. Accepting 15424 would require reducing the correct result arbitrarily at the final term. This breaks the algebraic coherence of the sequence, so option B is incorrect.

Option C:
Option C, 15436, is 12 less than 15448 and again does not equal 2·6⁵ − 3·6² + 4. It represents an approximate but inaccurate candidate that does not come from the established rule. Therefore option C is not a valid next term.

Option D:
Option D, 15460, is 12 greater than the computed value and would demand increasing the polynomial output only at this step. Since earlier terms exactly match the expression, this modification is unjustified. Hence option D does not correctly extend the pattern.

Statements A, B, C and E correctly describe the behaviour of clock hands and the nature of related aptitude questions: the minute and hour hands have known revolution times, their relative speed determines meeting or opposite positions and questions hinge on both angular calculations and relative motion. Statement D is false because the hour also affects the angle; for example, 3:00 and 4:00 have different angles even though the minutes are zero. Hence, the combination A, B, C and E only is correct.

Option A:
Option A is correct because it includes all the true statements while excluding D, which oversimplifies angular dependence by ignoring the hour. It reflects the standard method used to model clock questions in NET exams.

Option B:
Option B is incomplete since it omits E and so does not mention that these problems blend geometry and speed concepts. Without E, the description of the skill set tested is less comprehensive.

Option C:
Option C is wrong because it includes D, which incorrectly denies the effect of the hour hand on the angle, and it also omits A, failing to mention the minute hand’s revolution time. This mix of omission and error makes the option invalid.

Option D:
Option D is incorrect as it leaves out B, and so does not state that the hour hand takes 12 hours to complete a revolution, which is important for computing relative motion. The combination is thus incomplete even though some statements in it are true.

The terms are four letter consecutive blocks arranged in a sliding window. The starting letters B, C, D and E move forward by one position in each step, and every block contains the next three letters after the starting one. After EFGH, the next starting letter is F, and the block from F is FGHI. This group therefore continues both the sliding and the consecutive pattern.

Option A:
Option A, GHIJ, begins at G and thus skips the block starting with F that should immediately follow EFGH. While GHIJ is a valid consecutive block, it represents a later step in the sliding sequence. As the question asks for the next term, not a later one, GHIJ is not correct.

Option B:
Option B, EFGH, repeats the last given term without moving the starting letter forward. The sequence clearly advances the starting letter by one in each new group. Using EFGH again would freeze that movement, so it does not fit the pattern.

Option C:
Option C, HIJK, starts too far ahead and does not match the one step progression of the starting letter. It is consistent with the idea of consecutive letters, but not with the specific sliding rule observed. Therefore HIJK cannot be the immediate next term.

Option D:
Option D is correct because FGHI begins exactly one letter after E and continues with three consecutive letters. It preserves the group length and the simple sliding behaviour that characterises the series. For these reasons, FGHI is the appropriate continuation.