UGC NET Questions (Paper – 1)

There are 4 aces in a standard 52-card deck. The probability that the first card is an ace is 4/52, since any of the 4 aces could appear among the 52 cards. After one ace has been drawn, 3 aces remain in a reduced deck of 51 cards, so the probability that the second card is also an ace is 3/51. Because the draws are without replacement and the events are dependent, we multiply the probabilities: (4/52) × (3/51) = 12/2652, which simplifies to 1/221. Therefore, the correct probability that both drawn cards are aces is 1/221.

Option A:
Option A (1/13) usually comes from considering only the probability that a single card is an ace (4/52 = 1/13) and ignoring that we need two consecutive aces. It does not incorporate the second dependent draw, so it does not represent the joint probability of both events occurring. Because it treats the problem as if only one ace were required, it significantly overestimates the true probability. Hence, 1/13 is not correct for “both cards are aces.”

Option B:
Option B (1/169) typically arises from incorrectly assuming that the two draws are independent with replacement and computing (1/13) × (1/13). However, in this question the cards are drawn without replacement, which means the second probability changes after the first card is drawn. Using 1/169 ignores the reduced deck on the second draw and does not reflect the correct dependent-event structure. Therefore, it underestimates the correct probability.

Option C:
Option C (2/221) looks close to the correct answer but is exactly double 1/221, and there is no step in the correct derivation that would justify multiplying the final probability by 2. This kind of error often happens when students miscount favourable outcomes or mistakenly think order does not matter when they have already accounted for it. Since the exact product (4/52) × (3/51) simplifies to 1/221, any multiple of that value, including 2/221, is incorrect.

Option D:
Option D (1/221) matches the precise result from multiplying the probability of the first ace and the second ace without replacement: (4/52) × (3/51). The numerator 12 counts the ordered ways to pick two aces in succession, and the denominator 2652 counts all ordered pairs of cards, which simplifies to 1/221. This value correctly reflects both the limited number of aces and the shrinking deck size after the first draw. Hence, it is the only option fully consistent with the rules of probability for dependent events.

From (a + b):(a − b) = 7:1, we have (a + b)/(a − b) = 7. This gives a + b = 7a − 7b. Rearranging, we get 8b = 6a, so a/b = 4/3. Therefore, the required ratio a:b is 4:3.

Option A:
Option A is correct because it is directly obtained by solving the equation derived from the ratio of sum to difference. Substituting a:b = 4:3 back into (a + b):(a − b) reproduces 7:1, verifying consistency with the original condition.

Option B:
Option B, 3:4, simply inverts the correct ratio and would imply that b is larger than a. Plugging a:b = 3:4 into (a + b):(a − b) yields a different ratio and does not give 7:1, so it fails the check.

Option C:
Option C, 5:3, might arise from mismanipulating terms or incorrectly handling coefficients when rearranging the equation. However, if a:b = 5:3, then (a + b):(a − b) simplifies to a ratio different from 7:1, so it is not acceptable.

Option D:
Option D, 7:3, uses the numerator from the given ratio as part of the answer but does not satisfy the algebraic relationship when substituted back. The resulting sum and difference ratio deviates from 7:1, demonstrating that 7:3 is incorrect.

Statements A, C and D correctly explain basic Venn diagram ideas. A is true because Venn diagrams use overlapping closed curves to depict set relationships. C is correct as union collects elements from either set or both. D is also correct since the intersection consists of elements common to both sets. B is false because the common region represents intersection, not union, and E is false as the complement contains elements not in A. Thus A, C and D only are correct.

Option A:
Option A is incomplete since it omits D, leaving out the explicit description of intersection, which is fundamental to reading Venn diagrams. It therefore does not capture all the correct statements.

Option B:
Option B is incorrect because it includes only C and D and omits A, failing to mention what a Venn diagram is, and it also leaves out one of the true descriptive statements.

Option C:
Option C is correct as it gathers all the accurate statements about the depiction of sets, their union and intersection, while excluding B and E, which misidentify the regions and the meaning of complements.

Option D:
Option D is wrong because it includes E, which reverses the meaning of complement by saying it contains all elements of A, and therefore mixes a false statement with true ones.

Statement A is correct because the proposed action must respond directly to the issue described. Statement C is also true since realistic feasibility and reasonableness are central to judging whether a course of action is strong. Statement E is correct as some questions intentionally include more than one acceptable course, and candidates must recognise all that are reasonable. Statement B is false because emotional appeal alone does not outweigh impracticality. Statement D is false since polite language cannot make an impractical or irrelevant action strong. Hence the combination A, C and E only is correct.

Option A:
Option A is correct because it collects the three statements that reflect how course of action questions are supposed to be analysed and excludes B and D, both of which downplay logic and feasibility.

Option B:
Option B is incorrect as it includes B, which suggests that emotional but impractical steps might be considered strong, contradicting the logical focus of such items.

Option C:
Option C is wrong because it includes D, which is clearly false as politeness does not guarantee acceptability, and it omits A, which is essential.

Option D:
Option D is also incorrect because it retains D while omitting C, losing the explicit feasibility criterion and keeping a faulty one.

A prime number has exactly two distinct positive divisors, 1 and itself. The numbers 11, 13 and 17 each have no divisors other than 1 and themselves and are therefore prime. The number 15, however, is divisible by 1, 3, 5 and 15, so it has more than two divisors and is composite. Hence 15 is the odd one out in the list.

Option A:
Option A, 11, is a prime number because its only positive divisors are 1 and 11. It shares this property with 13 and 17 and therefore does not stand out in this group.

Option B:
Option B, 13, is also prime and fits the same classification as 11 and 17. It does not have additional factors that would make it different.

Option C:
Option C, 15, clearly deviates from the prime pattern, since it can be expressed as 3 × 5. Its composite nature makes it unique among the options and justifies treating it as the different number.

Option D:
Option D, 17, remains prime since no integer between 2 and 16 divides it without remainder. It aligns with 11 and 13 rather than differing from them.

A is correct because logical following means the conclusion must necessarily hold if the statements are true. D is also correct since the test requires ignoring outside knowledge, and E is true as some items use several statements jointly. B is wrong because factual truth in real life does not guarantee that the conclusion is logically implied by the given statements. C is wrong because conclusions that go beyond the given information are usually treated as not following, even if they appear plausible. Therefore, B and C only are the wrong statements.

Option A:
Option A recognises only B as wrong and ignores C, which also mischaracterises strong conclusions by equating them with extra information. Hence it does not capture all incorrect statements.

Option B:
Option B picks only C as wrong and overlooks B, which conflates real-world truth with logical implication. This again fails to identify the full set of wrong statements.

Option C:
Option C is correct because it groups together B and C, the two statements that mislead about how logical following is judged, while leaving A, D and E as the accurate descriptions of statement–conclusion reasoning.

Option D:
Option D incorrectly adds D to the list of wrong statements, even though D is true about ignoring outside knowledge, and so misclassifies a correct rule.

Each group here consists of four letters that are all equally spaced from each other in the alphabet. ADGJ has evenly increasing positions, and BEHK, CFIL and DGJM repeat the same internal spacing with their own starting letters. From one group to the next, every letter moves forward by one place, so the columns form simple increasing sequences. Moving each letter of DGJM forward by one produces EHKN, which is the only option that keeps the equal spacing and column wise step, making it the correct continuation.

Option A:
Option A is correct because EHKN follows directly after DGJM when each column is advanced by one letter. The distances between E, H, K and N match those inside the earlier groups exactly. It therefore extends both the vertical progressions and the internal even steps of the series.

Option B:
Option B, EHKM, uses the correct first three letters but changes the last letter so that the gap from K to M differs from the established spacing. This small alteration breaks the exact pattern that has been followed consistently. Because the internal distances are disturbed, EHKM cannot be the next term.

Option C:
Option C, FIKN, begins with F instead of E and alters multiple letters, which no longer come from a simple one step shift of DGJM. Its internal arrangement does not fit the regular ladder structure of the series. Hence FIKN is not acceptable as the correct answer.

Option D:
Option D, EGKM, mixes letters that do not align properly with the existing column sequences and ignore the equal spacing between positions. The changes in the inner parts of the group do not match the mechanical rule used earlier. Therefore EGKM does not continue the series correctly.

Each group is a block of three consecutive letters written in reverse order: ZYX (26–24), WVU (23–21), TSR (20–18) and QPO (17–15). The starting letters move backward by 3 positions: Z, W, T, Q. The next starting letter is N (14), and the reversed block from 14–12 is NML. Thus NML continues both the block size and the backward progression.

Option A:
Option A, ONM, uses the same letters but in a different arrangement. It begins with O (15), not N (14), breaking the backward step of 3 from Q. Therefore ONM does not satisfy the starting-letter rule.

Option B:
Option B, MNO, lists the letters in forward order rather than reverse. The pattern clearly uses descending triples. As it reverses the direction within each block, MNO cannot be the correct continuation.

Option C:
Option C is correct because it starts at N, which is exactly 3 positions before Q, and presents the triple in descending order N, M, L. This mirrors the structure of the previous groups and adheres to the same step size in the starting letters.

Option D:
Option D, KJI, begins too far back and uses a starting letter K that is 7 positions before Q. This does not match the consistent spacing seen between earlier groups. Hence KJI is not compatible with the pattern.

The positions of the letters are A(1), C(3), F(6), J(10) and O(15). The successive differences are +2, +3, +4 and +5, forming an increasing sequence of natural numbers. The next difference should be +6, so the next position is 15 + 6 = 21, which corresponds to U. Therefore U is the only letter that continues the pattern of steadily increasing gaps.

Option A:
Option A, T, lies at position 20, giving a difference of 5 from O. This simply repeats the last increment instead of increasing it to 6. Since the series clearly shows the gaps growing by 1 each time, T does not meet the rule.

Option B:
Option B, U, is correct because its position 21 equals 15 + 6, matching the expected next difference. It maintains the idea that each step forward in the series uses a gap one larger than before. This adherence to the underlying numerical pattern makes U the valid continuation.

Option C:
Option C, V, at position 22, would correspond to a +7 jump from O. That goes beyond the required next difference of +6 and distorts the incremental pattern. As a result, V cannot be accepted as the next term in the series.

Option D:
Option D, W, at position 23, deviates even more from the predicted increment by adding 8. This breaks the smooth progression of 2, 3, 4, 5, 6 in the differences. Hence W is not consistent with the logic of the sequence.

The sequence is governed by aₙ = 2n⁴ + n² − 2 for n starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 2·1⁴+1²−2 = 1, 2·2⁴+2²−2 = 34, 2·3⁴+3²−2 = 169, 2·4⁴+4²−2 = 526 and 2·5⁴+5²−2 = 1273, confirming the rule. For n = 6 we have 2·6⁴+6²−2 = 2·1296+36−2 = 2592+36−2 = 2626. Thus 2626 is the unique value that continues this quartic-based pattern.

Option A:
Option A, 2594, is 32 less than the value mandated by the formula and does not satisfy 2n⁴+n²−2 for n = 6. It would require an unexplained reduction in the final term relative to the established rule. Therefore this option is not compatible with the structure of the sequence.

Option B:
Option B, 2610, is closer but still not equal to 2626 and again fails to arise from substituting n = 6 into the expression. Choosing 2610 would modify the pattern only at the last point, which is mathematically unsound. Hence option B is not correct.

Option C:
Option C, 2626, exactly matches the computed value from aₙ = 2n⁴+n²−2 when n = 6. It preserves the balance between the quartic and quadratic components that generated the earlier terms. Because it follows naturally from the same algebraic rule, 2626 is the correct next term in the series.

Option D:
Option D, 2642, overshoots the formula’s output by 16 and does not respect the rule 2n⁴+n²−2. Selecting 2642 would break the precise connection between term position and term value. Therefore option D cannot be accepted as a valid continuation.

The rule states that each letter is replaced by its position in the alphabet and then all positions are summed. For NET, N is 14, E is 5 and T is 20. Adding these values gives 14 + 5 + 20 = 39. Since this computation matches the coding rule used in the example, 39 is the correct code for NET.

Option A:
Option A, 35, would require the sum of letter positions to be 35, but 14 + 5 + 20 clearly equals 39, not 35. This option therefore underestimates the actual total and ignores the correct arithmetic.

Option B:
Option B, 37, is also inconsistent with the required sum of positions, as there is no combination of N, E and T that would legitimately produce 37 under the stated rule. It likely arises from a miscalculation.

Option C:
Option C, 38, is again close to the correct answer but still incorrect because the precise sum 14 + 5 + 20 equals 39. In coding questions, even a small deviation from the rule’s outcome invalidates the option.

Option D:
Option D, 39, exactly matches the sum of the alphabetic positions of N, E and T. It faithfully applies the same coding procedure that produced 60 for WORD, making it the only option consistent with the defined rule.

Statements A, C, D and E describe key features of three-set Venn diagrams. A is correct because three sets generate 2³ = 8 possible membership combinations. C is true as the triple intersection appears where all three circles overlap. D correctly states that the union collects elements belonging to at least one of the sets. E is also correct in survey contexts where the region outside all circles stands for those having none of the properties. B is false since “A only” must exclude elements that also belong to B or C, so the correct combination is A, C, D and E only.

Option A:
Option A is correct because it gathers all the true statements and excludes B, which misdefines the “A only” region by including members of B and C. It captures the structure and interpretation of three-set Venn diagrams used in survey and counting questions.

Option B:
Option B is incomplete as it omits E, ignoring the important role of the outside region for respondents with none of the attributes. Without E, the account of all regions relevant to survey interpretation is not comprehensive.

Option C:
Option C is incorrect because it includes B along with A, C and D, thereby accepting the wrong description of “A only”. Even one incorrect statement within the combination makes the whole option invalid.

Option D:
Option D is wrong as it features B and therefore carries forward the same misunderstanding of exclusive regions, while also omitting A, which states the basic partitioning into eight regions. This mix of omission and error makes the option unsuitable.

Each term after the first is obtained by doubling the previous term and subtracting 2. We see 3 × 2 − 2 = 4, 4 × 2 − 2 = 6, 6 × 2 − 2 = 10, 10 × 2 − 2 = 18 and 18 × 2 − 2 = 34. Applying the same rule to 34 yields 34 × 2 − 2 = 68 − 2 = 66. This keeps the recursive structure intact.

Option A:
Option A gives 62, which would correspond to subtracting 6 after doubling 34 and introduces a new constant that does not appear in earlier steps. Thus, 62 is not consistent with the recursive rule.

Option B:
Option B offers 64, equal to 34 × 2 − 4, which again changes the subtractive constant and conflicts with the uniform "minus 2" operation. Therefore, 64 cannot be accepted as correct.

Option C:
Option C yields 66, exactly equal to 34 × 2 − 2, and fits the observed pattern at every stage. The series 3, 4, 6, 10, 18, 34, 66 remains governed by a single clear transformation. Hence, 66 is the correct next term.

Option D:
Option D suggests 68, which is simply 34 × 2 and omits the subtraction of 2 found throughout the series. This breaks the rule that defines the sequence. Therefore, 68 is not appropriate.

Statement A matches the definition of percentage as a ratio per hundred. C is correct because a 25% profit implies selling price equals cost price plus 25% of cost, giving 125% of cost. D is true since equal ratios constitute a proportion. E is also correct; if a:b = 2:3, reversing gives b:a = 3:2. B is false, because a 20% increase followed by a 20% decrease results in a net 4% decrease. Thus A, C, D and E only form the correct set.

Option A:
Option A is incomplete; while A and C are true, omitting D and E ignores important proportional and inverse ratio relationships frequently examined in NET aptitude questions.

Option B:
Option B is wrong as it includes B, which misstates the effect of successive percentage changes, and so mixes a false statement with correct ones.

Option C:
Option C is correct because it collects the accurate definitions and interpretations of percentage, profit, proportion and inverse ratio, while excluding B, which leads to a typical exam trap.

Option D:
Option D is incorrect because it includes B, allowing a false claim about successive percentage operations into the option, even though C, D and E are correct.

For a fixed distance, speed and time are inversely proportional, meaning that as speed increases, the time required decreases. When the train’s speed rises from 60 km/h to 80 km/h, it covers the same distance in less time. This relationship is fundamental in time and distance problems. Therefore, the time taken to travel the same distance will decrease.

Option A:
An increase in time would occur only if the speed decreased, not when it increased. Because speed is the rate at which distance is covered, moving faster cannot logically result in taking more time for the same distance. Thus, "increase" contradicts the basic formula time = distance ÷ speed.

Option B:
Decrease is correct because when speed rises while distance remains constant, the denominator in the time formula increases, making the overall time smaller. For example, travelling 160 km at 60 km/h takes about 2 hours 40 minutes, whereas at 80 km/h it takes only 2 hours. This clearly shows a reduction in time, validating option B.

Option C:
If time remained the same, it would imply that changing speed has no effect on duration, which violates the fundamental relationship between distance, speed and time. Since the distance is fixed, a change in speed must affect time. Therefore, this option is not logically possible.

Option D:
Time becoming double would happen only if the speed were halved for the same distance. Here, the speed is increased, not decreased, so the time cannot become double. This directly contradicts the inverse proportional relationship between speed and time.