UGC NET Questions (Paper – 1)

Successive terms decrease by 3: 17 − 20 = −3, 14 − 17 = −3 and 11 − 14 = −3. This forms an arithmetic progression with common difference −3. Subtracting 3 from the last term 11 gives 8. Therefore, 8 is the next term in the series.

Option A:
Option A keeps the same step because 11 − 3 = 8, maintaining the constant change of −3. The series 20, 17, 14, 11, 8 stays perfectly regular under this rule. Hence, this option correctly extends the pattern.

Option B:
Option B is 10, which is only 1 less than 11 and would change the difference to −1. That conflicts with the consistent decrease by 3 observed earlier. Therefore, 10 cannot be accepted.

Option C:
Option C is 9, giving a difference of −2 from 11 and again disturbing the established constant difference. The sequence would become irregular at the end. Thus, 9 is not correct.

Option D:
Option D is 7, which is 4 less than 11, producing a difference of −4. This also contradicts the steady difference of −3 that defines the progression. Therefore, 7 is not the right choice.

Statements A and B restate De Morgan’s laws, showing how negation distributes over “and” and “or”, C correctly defines logical equivalence in terms of matching truth values and D is true because De Morgan’s laws explicitly connect conjunction, disjunction and negation. Statement E is false since contradictory statements cannot both be true in any possible situation; one must be false whenever the other is true. Hence, the combination A, B, C and D only is correct.

Option A:
Option A is incomplete because, although it lists three true statements, it omits D and thus does not explicitly say that De Morgan’s laws are the ones relating “and”, “or” and “not”. Without D, the named connection between these operators is not emphasised.

Option B:
Option B is also incomplete because it leaves out A and thereby fails to include the first De Morgan equivalence concerning the negation of “p and q”. That omission makes the description of the laws partial.

Option C:
Option C is correct as it bundles all four true statements, fully describing De Morgan’s laws and logical equivalence, while rejecting E, which mischaracterises contradictions. It aligns with the logical tools used in UGC NET reasoning.

Option D:
Option D is wrong because it includes E, the false statement about contradictions, and omits C, the formal definition of logical equivalence. This combination mixes error with omission.

Let the original salaries be 7k and 9k. After adding ₹5600 to each, the new salaries become 7k + 5600 and 9k + 5600, which are in the ratio 9:11. So (7k + 5600)/(9k + 5600) = 9/11. Solving this equation gives k = 2800, so B’s original salary is 9k = 9 × 2800 = ₹25,200. Thus, ₹25,200 is the unique value that satisfies both the original and changed ratios.

Option A:
Option A, ₹19,600, equals 7 × 2800 and represents A’s original salary, not B’s. If we treat ₹19,600 as B’s original salary, we cannot express A’s salary as a smaller value in the ratio 7:9 while still matching the increment condition. Therefore it does not satisfy the relationship implied by the change from 7:9 to 9:11.

Option B:
Option B is correct because it matches the 9k term obtained from solving the proportion equation. Substituting ₹25,200 for B and ₹19,600 for A, we get new salaries ₹25,200 + ₹5600 = ₹30,800 and ₹19,600 + ₹5600 = ₹25,200, whose ratio is 25,200:30,800 = 9:11. This verifies that ₹25,200 is consistent with all given information.

Option C:
Option C, ₹28,000, would imply k = 3111.11… if we forced 9k = ₹28,000, which is not an integer and contradicts the natural-number scaling in the original ratio 7:9. Using that figure in the increment equation will not yield the new ratio 9:11. Hence, ₹28,000 cannot be B’s original salary.

Option D:
Option D, ₹30,800, is actually B’s new salary after the increment in the correct solution. Treating it as the original salary would lead to an even larger value after adding ₹5600 and a ratio that no longer matches 9:11. Therefore, ₹30,800 is not consistent with the given conditions.

The pattern in the series is that each term increases by 3. From 4 to 7 the difference is 3, from 7 to 10 it is 3 and from 10 to 13 it is again 3. To continue this arithmetic progression, we add 3 to the last term 13. That gives 13 + 3 = 16 as the next term.

Option A:
Option A keeps the constant difference of 3 between consecutive terms, which is the defining rule of this series. It extends the pattern 4, 7, 10, 13 in a straightforward way. Because it is obtained by applying the same operation as before, 16 is the only value that fits the given progression.

Option B:
Option B, 17, would require an increase of 4 from 13, which breaks the uniform +3 difference observed in the earlier steps. It introduces a new rule that is not supported by the given data.

Option C:
Option C, 18, implies adding 5 to 13 and similarly fails to maintain the steady increment of 3. It cannot be justified from the structure of the sequence.

Option D:
Option D, 19, would mean adding 6 to 13, making the jump much larger than any of the previous differences and clearly inconsistent with the pattern.

Statements A, B, C and E are all correct: they define the arithmetic mean, describe how adding a constant shifts the mean, explain that the updated mean lies between the old mean and the new value, and note that weighted averages incorporate differing importance through weights. Statement D is false because the average need not coincide with any actual data point; for example, the mean of 2 and 4 is 3, which is not in the list. Therefore, the combination that includes A, B, C and E and excludes D is correct.

Option A:
Option A is incomplete since it does not mention E and therefore overlooks the important idea that some aptitude problems require weighted averages rather than simple means. Without E, the concept coverage is narrower than intended.

Option B:
Option B is also incomplete: although B, C and E are true, the omission of A means the basic definition of arithmetic mean is missing. A must be included to fully represent how averages are computed.

Option C:
Option C is correct as it brings together all the true statements about how means behave and how weighted averages work, while rejecting D, which mistakenly claims that the mean must be an actual data value. It provides a complete conceptual picture for NET-level questions.

Option D:
Option D is wrong because it includes D, the false statement about averages always equalling a data point, and omits B, which correctly summarises the effect of adding a constant. This mixture makes the option logically inconsistent.

Statements A, B and C correctly define tautology, contradiction and contingency by looking at truth across all valuations. Statement E is also true because the negation of a formula that is always true must be false on every valuation, hence a contradiction. Statement D is impossible in classical logic, since no statement can be both always true and always false at once. Therefore, the complete set of correct statements is A, B, C and E only, which matches option C.

Option A:
Option A is incomplete because it omits E and therefore fails to mention the important relationship between tautologies and contradictions via negation. It lists correct information but not the full set of true statements for this question.

Option B:
Option B is incorrect since it leaves out B and includes only A, C and E, ignoring the explicit definition of contradiction that is central to the topic. Without B the description of the three categories is not complete.

Option C:
Option C is correct as it brings together the accurate definitions of the three types of statements and the fact that the negation of a tautology is a contradiction, while excluding D, which suggests an incoherent situation in classical logic.

Option D:
Option D is wrong because it includes D, which falsely claims that a statement can be both tautology and contradiction and calls that “contingent”, and it omits A, so it both accepts a logical impossibility and loses a key correct statement.

To find the smallest number to be added, we first determine the remainder when 625 is divided by 7. Since 7 × 89 = 623, the remainder is 625 − 623 = 2. A number is divisible by 7 when the remainder becomes zero, so we must add a number that brings the remainder 2 up to 7. Therefore, we add 5 (because 2 + 5 = 7), making 625 + 5 = 630, which is divisible by 7.

Option A:
Option A correctly uses the idea that the required adjustment is 7 minus the remainder. With a remainder of 2, adding 5 eliminates the remainder and yields a number (630) that lies exactly on the 7-times table.

Option B:
Option B, 2, would produce 627, and 627 ÷ 7 still leaves a remainder because 7 × 89 = 623 and 7 × 90 = 630. Thus, 627 is not divisible by 7.

Option C:
Option C, 4, gives 629, which also falls between multiples of 7 and leaves a nonzero remainder, so it does not satisfy the condition.

Option D:
Option D, 6, would lead to 631, which again does not align with a multiple of 7, because the nearest multiples are 630 and 637.

A’s one-day work is 1/12 of the job and B’s one-day work is 1/20 of the job. The combined one-day work is therefore 1/12 + 1/20. Using the least common multiple of 12 and 20, which is 60, this becomes (5 + 3) / 60 = 8 / 60 = 2 / 15. The total time required is the reciprocal of 2 / 15, which is 15 / 2 or 7.5 days. Thus, together they will finish the work in 7.5 days.

Option A:
Option A, 6 days, is too small and would imply a higher combined rate than 2/15 per day. It suggests that the two workers together are even more efficient than the calculations show.

Option B:
Option B accurately reflects the reciprocal of the combined daily work fraction. It respects the principle that joint work rates add and that the total time is found by inverting the sum.

Option C:
Option C, 8 days, overestimates the time and corresponds to a smaller combined rate than the actual 2/15 of the work per day.

Option D:
Option D, 10 days, is even less efficient and ignores the substantial contribution B adds when working alongside A.

Each group is a three letter palindrome where the first and last letters are the same and increase in order A, B, C, D. The middle letters move backwards from the end of the alphabet in the order Z, Y, X, W. The next outer letter should be E and the next middle letter in this backward sequence should be V. Thus EVE is the only group that satisfies both movements simultaneously.

Option A:
Option A, EUE, uses E correctly as the outer letter but chooses U instead of V for the middle position. The middle letters have been stepping down one letter at a time from Z, and U would skip V in that sequence. Because it breaks this simple reverse progression, EUE cannot be the right continuation.

Option B:
Option B is correct because it combines E as the next outer letter and V as the next middle letter after W. It preserves the palindromic structure and keeps the two opposing movements of outer letters forward and inner letters backward perfectly intact. This strict adherence to the observed pattern makes EVE the logical next term.

Option C:
Option C, EVF, is not a palindrome at all since the first and last letters do not match. The series clearly uses groups where the outer letters are identical. By breaking this defining structural feature, EVF fails to follow the pattern of the series.

Option D:
Option D, EWE, keeps the palindromic structure but repeats W as the middle letter instead of moving to V. This freezes the backward movement of the central letters, which has been consistent so far. Therefore EWE does not continue the pattern correctly.

This series follows the rule aₙ = 2n⁴ + 5n² + 3 with n starting from 1. Substituting n = 1, 2, 3, 4 and 5 yields 2+5+3 = 10, 32+20+3 = 55, 162+45+3 = 210, 512+80+3 = 595 and 1250+125+3 = 1378, which matches the given numbers. For n = 6 we compute 2·6⁴+5·6²+3 = 2592+180+3 = 2775. Therefore 2775 is the unique next term consistent with this quartic-plus-quadratic-plus-constant structure.

Option A:
Option A, 2715, is 60 less than the formula output and does not satisfy aₙ = 2n⁴+5n²+3 for n = 6. It introduces an arbitrary downward adjustment that is not indicated by the previous terms. Hence option A is incorrect.

Option B:
Option B, 2775, equals exactly the value obtained by applying the expression to n = 6. It preserves the same coefficients and constant term that were used to generate all earlier entries in the series. Because it fits seamlessly into the algebraic rule, 2775 is the correct next term.

Option C:
Option C, 2835, overshoots the correct value by 60 and again does not arise from substituting n = 6 into the rule. Choosing 2835 would inflate the sixth term without any support from the structure of the series. Therefore option C is incorrect.

Option D:
Option D, 2895, deviates even more from the computed value and has no foundation in the expression 2n⁴+5n²+3. Using 2895 would destroy the precise link between index and term, so option D is incorrect.

The given numbers are 1³, 2³, 3³ and 4³, showing a sequence of consecutive cubes. Following this rule, the next term should be 5³. Calculating 5³ gives 125. Hence, 125 is the correct continuation of the cube sequence.

Option A:
Option A matches the rule because 125 is 5³ and follows naturally after 4³ = 64. The extended series 1, 8, 27, 64, 125 represents cubes of 1, 2, 3, 4 and 5 in order. This neat structure is typical in exam-based number series questions.

Option B:
Option B is 100, which is not the cube of any integer. It lies between 4³ and 5³ but does not satisfy the pure cube property. Thus, it does not fit the pattern.

Option C:
Option C is 150, which has no direct connection with consecutive cubes in this context. Choosing 150 would break the exact power relationship between the terms. Therefore, it cannot be correct.

Option D:
Option D is 216, which equals 6³ and skips the cube of 5. While 216 is a cube, the series is moving stepwise through consecutive integers, so skipping 5³ is not logical. Hence, 216 is not the correct next term here.

When two fair dice are thrown, there are 6 × 6 = 36 equally likely outcomes. A sum greater than 9 corresponds to sums of 10, 11 and 12. The sum 10 can occur in 3 ways, 11 in 2 ways and 12 in 1 way, giving a total of 6 favourable outcomes. The probability is therefore 6/36, which simplifies to 1/6. This fraction represents the chances of obtaining a sum greater than 9 on a single throw of two dice.

Option A:
Option A, 1/18, undercounts the favourable outcomes and would require only 2 favourable cases out of 36, which is not the case here. It might arise from only considering one of the sums.

Option B:
Option B, 1/9, simplifies to 4/36 and still does not match the 6 favourable cases actually present in the sample space. It underestimates the event’s likelihood.

Option C:
Option C, 5/36, suggests only 5 favourable outcomes, but a systematic list shows there are 6 outcome pairs producing sums 10, 11 and 12.

Option D:
Option D, 1/6, correctly reflects the 6 favourable outcomes out of 36 total and is the simplified probability consistent with the complete sample space.

The movement forms a right-angled triangle with legs of 4 km east and 3 km north. The straight-line distance from the starting point to the final point is the hypotenuse of this triangle. By the Pythagoras theorem, the hypotenuse is √(4² + 3²) = √(16 + 9) = √25 = 5 km. So the person is 5 km away from the starting point.

Option A:
Option A uses the classic 3–4–5 right-angled triangle pattern, which fits perfectly with the given distances. Applying Pythagoras correctly yields 5 km, making this the exact displacement from the origin.

Option B:
Option B, 6 km, overestimates the distance and would not satisfy the squared relationship 4² + 3² = 5². It suggests a hypotenuse of √36 when the correct sum of squares is 25.

Option C:
Option C, 7 km, exaggerates the distance even more and is not supported by any computed combination of the given legs. It is much larger than the true hypotenuse.

Option D:
Option D, 4 km, simply repeats one leg of the triangle and ignores the northward movement, so it cannot represent the actual straight-line displacement.

For a fixed distance d, time is given by t = d/speed, so it is inversely proportional to speed. The first train’s time is d/90 and the second’s time is d/60. Their time ratio is (d/90):(d/60) = 1/90:1/60 = 60:90, which simplifies to 2:3. Thus, the first train takes 2 parts of time for every 3 parts taken by the second.

Option A:
Option A, 3:2, reverses the correct ratio and would mean the faster train takes more time, which contradicts the fundamental relationship between speed and time for the same distance. A higher speed must correspond to a shorter time, not a longer one.

Option B:
Option B is correct because it arises from properly comparing the reciprocals of the speeds. When the distance is constant, the time ratio is the inverse of the speed ratio. Since the speed ratio is 90:60 = 3:2, the time ratio must be 2:3, matching the computed result.

Option C:
Option C, 1:2, would imply that the first train takes half the time of the second, meaning its speed would have to be double the second’s speed. However, 90 km/h is only 1.5 times 60 km/h, not twice, so 1:2 overstates the difference.

Option D:
Option D, 5:6, bears no direct relationship to 90 and 60 when we follow the inverse proportionality rule. It cannot be obtained by simplifying 60:90 or any equivalent fraction, so it does not represent the correct time ratio.

A is correct because time = distance/speed for fixed distance, so time varies inversely with speed. C is true since distance = speed × time at constant speed. D is true because halving speed doubles time for the same distance. B is false because doubling speed at the same time doubles distance, not halves it. E is false; starting time is not forced by speed alone. Therefore A, C and D only are correct.

Option A:
Option A is correct as it captures both inverse and direct proportion relationships along with a key consequence for halving speed.

Option B:
Option B is incomplete because it excludes D, missing an explicit and important inverse-proportion consequence.

Option C:
Option C is wrong since it includes B, which misstates the effect of doubling speed at fixed time.

Option D:
Option D is incorrect because it adds E (false) and omits A (true), so it cannot be correct.