UGC NET Questions (Paper – 1)

For a conditional statement p → q, the contrapositive ¬q → ¬p is logically equivalent to the original. Here p is “the number is divisible by 4” and q is “the number is even.” The contrapositive is “if a number is not even, then it is not divisible by 4.” Both the original statement and its contrapositive are true or false together in all possible cases, so they are logically equivalent.

Option A:
Option A reverses the direction of the implication to q → p, which is the converse. The truth of the original statement does not guarantee the truth of its converse; being even does not ensure divisibility by 4.

Option B:
Option B states “If a number is not divisible by 4, then it is not even,” which is the inverse. The inverse is not logically equivalent to the original because many numbers are not divisible by 4 but are still even.

Option C:
Option C correctly expresses the contrapositive ¬q → ¬p: if a number is not even, then it is not divisible by 4. This preserves the logical content of the original implication and is therefore equivalent to it.

Option D:
Option D, “If a number is not divisible by 4, then it is even,” contradicts the original pattern and misstates the relationship between being even and being divisible by 4.

This sequence can be modelled by aₙ = 2n⁴ + n! with n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 2·1⁴ + 1! = 3, 2·2⁴ + 2! = 34, 2·3⁴ + 3! = 168, 2·4⁴ + 4! = 536 and 2·5⁴ + 5! = 1370, which matches the given terms. For n = 6 we have a₆ = 2·6⁴ + 6! = 2·1296 + 720 = 2592 + 720 = 3312. Hence 3312 is the unique value that maintains this factorial-quartic pattern.

Option A:
Option A, 3288, is 24 less than the value produced by 2n⁴ + n! at n = 6. Choosing 3288 would mean reducing the correct outcome of the formula only at the final term, which is not supported by earlier behaviour. Therefore option A is not consistent with the established rule.

Option B:
Option B, 3300, is 12 smaller than the correct value and cannot be written as 2·1296 + 720. It is numerically close but does not respect the exact functional dependence on n. As a result, option B does not correctly continue the series.

Option C:
Option C, 3324, is 12 greater than the correct term and again cannot be obtained from 2·6⁴ + 6!. Accepting 3324 would force an unjustified increase in the last value relative to the pattern. Thus option C is not a valid continuation.

Option D:
Option D, 3312, matches exactly the number generated by the rule aₙ = 2n⁴ + n! for n = 6. It preserves the relationship that combines a strong quartic component with factorial growth evident in previous terms. Because it extends the same pattern smoothly, this option is the correct answer.

This series is based on the formula aₙ = 2n³+1 for n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 2·1³+1 = 3, 2·2³+1 = 17, 2·3³+1 = 55, 2·4³+1 = 129 and 2·5³+1 = 251. For n = 6 the expression gives 2·6³+1 = 432+1 = 433. Therefore 433 is the next term that keeps the cubic pattern intact.

Option A:
Option A, 413, is 20 less than the value provided by 2·6³+1. It does not satisfy the generating formula and would require an unexplained downward adjustment. Hence 413 cannot be considered the correct continuation of the series.

Option B:
Option B, 421, is closer but still does not match 2·6³+1, so it fails to maintain the exact relationship between n and aₙ. Choosing it would make the pattern inconsistent with the earlier terms. Thus 421 is not a valid answer.

Option C:
Option C, 433, is exactly the value given by the rule aₙ = 2n³+1 when n = 6. It carries the same structure forward without alteration and fits perfectly with all preceding data. For this reason, 433 is the correct next term.

Option D:
Option D, 448, is larger than the formula’s output and cannot be obtained from 2n³+1 for n = 6. Adopting 448 would abandon the precise cubic relationship that defines the sequence, so it is not correct.

Distance is calculated using the formula Distance = Speed × Time. The given speed is 60 km/h, but the time is 30 minutes, which is half an hour or 0.5 hours. Substituting into the formula gives Distance = 60 × 0.5 = 30 km. Hence, the train covers 30 kilometres in 30 minutes at this speed.

Option A:
Option A, 20 km, would correspond to a speed of 40 km/h for 30 minutes, not the given 60 km/h. It reflects a misalignment between speed and time.

Option B:
Option B, 25 km, does not correspond to any simple fraction of 60 using half an hour and indicates an incorrect multiplication or conversion.

Option C:
Option C, 40 km, would be the distance travelled in 40 minutes at 60 km/h, or in 2/3 of an hour, not in 30 minutes. Thus, it overestimates the distance for the time given.

Option D:
Option D correctly uses 0.5 hours as the time and multiplies by 60 km/h to get 30 km. This matches both the formula and the unit conversion from minutes to hours.

Let the total volume be T litres. Initially milk = (7/10)T and water = (3/10)T. When 10 litres of well-mixed solution are removed, milk removed is (7/10) × 10 = 7 litres and water removed is 3 litres. So milk becomes (7/10)T − 7 and water becomes (3/10)T − 3. After adding 10 litres of water, water becomes (3/10)T + 7. The new ratio is [(7/10)T − 7] : [(3/10)T + 7] = 7:5, which simplifies to T = 60.

Option A:
Option A is correct because substituting T = 60 gives initial milk 42 litres and water 18 litres, matching the 7:3 ratio. Removing 10 litres of mixture takes away 7 litres of milk and 3 litres of water, leaving 35 litres of milk and 15 litres of water. After adding 10 litres of water, we have 35 litres of milk and 25 litres of water, giving 35:25 = 7:5, exactly as required.

Option B:
Option B, 50 litres, would give initial milk 35 litres and water 15 litres. Removing 10 litres of mixture would remove 7 litres of milk and 3 litres of water, leaving 28 litres of milk and 12 litres of water, which after adding 10 litres becomes 28:22 = 14:11, not 7:5. Therefore 50 litres does not satisfy the given final ratio.

Option C:
Option C, 70 litres, leads to initial milk 49 litres and water 21 litres. After removing and replacing 10 litres, the final quantities do not simplify to the ratio 7:5; the numbers no longer align with the required proportion. Thus 70 litres is inconsistent with the conditions.

Option D:
Option D, 80 litres, similarly yields initial and final quantities whose ratio becomes something different from 7:5 after the described operation. Checking the arithmetic reveals that the milk and water amounts are not in the required proportion, so 80 litres cannot be the original quantity.

Statements A, B and D describe genuine features of letter series. A is correct because positions like A=1, B=2 are routinely used to detect shifts. B is true since many series move letters by a fixed pattern that repeats. D is also correct because repeated blocks such as ABCABC suggest a repeating cycle whose length is diagnostic. C is false because there is no requirement that vowels and consonants alternate, and E is false as alternate positions can follow a second rule, so A, B and D only are correct.

Option A:
Option A is incomplete since it leaves out D. Without checking repeating blocks, the strategy for handling many cyclic letter series is not fully described, so A and B alone do not constitute the entire correct set.

Option B:
Option B is correct because it gathers the alphabet-position idea, cyclic shifts and block detection, all of which are central to solving letter series, and excludes C and E, which over-restrict possible patterns.

Option C:
Option C is wrong as it includes E, claiming that two rules cannot govern alternate positions, which is directly contradicted by many exam examples, and it omits B, losing an important cyclical feature.

Option D:
Option D is incorrect because it accepts E and drops A, thereby discarding the basic alphabet-position tool while also denying multi-rule patterns, so it cannot represent the set of correct statements.

This series is generated by the formula aₙ = 5n³+n² for n starting from 1. For n = 1, 2, 3, 4 and 5 we have 5·1³+1 = 6, 5·2³+4 = 44, 5·3³+9 = 144, 5·4³+16 = 336 and 5·5³+25 = 650. For n = 6 the same expression yields 5·6³+36 = 1080+36 = 1116. Therefore 1116 is the correct next term that preserves this cubic-plus-square relationship.

Option A:
Option A, 1076, is 40 less than the computed value and does not satisfy aₙ = 5n³+n² for n = 6. It weakens the consistency between the algebraic rule and the numeric sequence. Hence 1076 is not the correct continuation.

Option B:
Option B, 1092, is somewhat closer but still different from 1116 and cannot be obtained from the formula without changing coefficients. This would make the pattern inconsistent at the final term. Thus 1092 is not valid.

Option C:
Option C, 1146, is larger than the correct value and again fails to equal 5n³+n² for n = 6. Choosing 1146 would distort the functional relationship that holds throughout the sequence. Therefore it is not the right answer.

Option D:
Option D, 1116, exactly matches the result given by the rule when n = 6. It keeps the cubic and quadratic contributions aligned in the same way as for earlier terms and thus is the correct next term.

A is correct because bar height/length represents magnitude. C is correct because relative heights show ranking even without exact labels. D is correct because clustered bars require a legend/key to identify series. B is false because bar charts can be horizontal. E is false because the encoded value is usually height/length, not width. Thus A, C and D only are correct.

Option A:
Option A is incomplete because it omits D, which is important for interpreting grouped/clustered bar charts.

Option B:
Option B is incorrect because it adds E, which wrongly claims bar width encodes the data value.

Option C:
Option C is wrong because it omits A, the fundamental definition of bar charts where bar height/length is proportional to value.

Option D:
Option D is correct because it combines the basic encoding rule, comparative interpretation and the need for a legend in clustered charts, while excluding B and E, which are incorrect.

The pattern can be described by aₙ = 2n! + n² with n starting from 1. For n = 1, 2, 3, 4 and 5, this yields 2·1! + 1² = 3, 2·2! + 2² = 8, 2·3! + 3² = 21, 2·4! + 4² = 64 and 2·5! + 5² = 265, matching the given terms. For n = 6, we obtain a₆ = 2·6! + 6² = 2·720 + 36 = 1476. Thus 1476 is the only value that fits this factorial-based rule and continues the series.

Option A:
Option A, 1452, is 24 less than the correct value given by 2n! + n² at n = 6. To pick 1452 we would need to subtract 24 from the computed result only at this position, which would disrupt the pattern of combining factorial and square terms. Therefore option A is incorrect.

Option B:
Option B, 1476, exactly matches the number produced by 2·6! + 6². It preserves the factorial component and the quadratic addition in the same way as for earlier values of n. Because it extends the rule without any modification, this option correctly continues the series.

Option C:
Option C, 1464, is 12 smaller than the correct value and cannot be expressed as 2·720 + 36. It is only an approximate candidate and does not reflect the precise functional relationship. As such, option C is not a valid next term.

Option D:
Option D, 1488, is 12 greater than the required value and again fails to satisfy the formula at n = 6. Choosing 1488 would overstate the term in comparison to what the rule dictates, breaking the pattern. Hence option D is not acceptable.

This sequence is defined by a₁ = 3 and aₙ₊₁ = 4aₙ + n² + 1 for n ≥ 1. Applying the rule yields a₂ = 4·3 + 1² + 1 = 14, a₃ = 4·14 + 2² + 1 = 61, a₄ = 4·61 + 3² + 1 = 254 and a₅ = 4·254 + 4² + 1 = 1033, matching the given terms. For n = 5 the next term is a₆ = 4·1033 + 5² + 1 = 4158. Therefore 4158 is the unique number that continues the same recurrence correctly.

Option A:
Option A, 4126, is 32 less than the recurrence output for n = 5. To obtain 4126 we would need to subtract 32 from the formula result at the last step, which is not consistent with the earlier pattern. Hence this option does not follow the rule and is incorrect.

Option B:
Option B, 4142, is closer but still 16 less than the correct value 4158. It cannot be expressed as 4a₅ + 5² + 1 and would imply undercounting the contribution of the square term. As the recurrence is satisfied exactly before, option B cannot be the correct continuation.

Option C:
Option C, 4148, is slightly below the required value and also fails to satisfy the recurrence exactly. Selecting 4148 would again require modifying the output of the rule only at this step. Such a change destroys the strict structure of the sequence, so this option is not valid.

Option D:
Option D, 4158, equals exactly the result produced by aₙ₊₁ = 4aₙ + n² + 1 when n = 5. It maintains the process of quadrupling the previous term and adding a quadratic plus constant correction. Because this description explains each transition and leads naturally to 4158, option D is correct.

If A alone finishes the work in 10 days, A’s one-day work is 1/10 of the job. If B alone finishes in 15 days, B’s one-day work is 1/15. The ratio of work done by A to work done by B in one day is (1/10):(1/15) = 15:10, which simplifies to 3:2. Thus, the required ratio is 3:2.

Option A:
Option A, 2:3, reverses the correct comparison and implies B does more work per day than A, suggesting B is faster. This contradicts the completion times, because A finishes in fewer days than B and must therefore have the higher daily rate.

Option B:
Option B, 4:3, does not match the exact fraction (1/10)/(1/15) = 15/10 = 3/2. While 4:3 is close numerically to 3:2, the calculation from the given times unambiguously leads to 3:2, not 4:3. Hence 4:3 is incorrect.

Option C:
Option C is correct because it faithfully reflects the inverse relationship between time and rate for a fixed amount of work. Shorter time (10 days) corresponds to a larger rate, and the ratio of these rates simplifies precisely to 3:2.

Option D:
Option D, 5:3, exaggerates the difference between the two rates beyond what is implied by 10 and 15 days. Using 5:3 would not produce the correct individual times if we converted back from the rate ratio, so it cannot be accepted.

This series alternates between adding 5 and multiplying by 2. Starting from 4 we have 4 + 5 = 9, 9 × 2 = 18, 18 + 5 = 23, 23 × 2 = 46 and 46 + 5 = 51. The next step should therefore apply the multiplication operation. Hence, the next term is 51 × 2 = 102, which follows the established alternation.

Option A:
Option A gives 100, which does not result from either adding 5 or multiplying by 2 to 51. It introduces a new and unsupported operation into the sequence. Therefore, 100 is not compatible with the observed pattern.

Option B:
Option B yields 102, exactly equal to 51 × 2 and in perfect agreement with the alternating "+5, then ×2" rule. The series 4, 9, 18, 23, 46, 51, 102 remains structured and predictable under this rule. Hence, 102 is the correct next term.

Option C:
Option C suggests 104, which cannot be derived from either +5 or ×2 applied to 51. It would force an arbitrary adjustment not present in any earlier step. Thus, 104 is not the correct continuation.

Option D:
Option D presents 106, which is even further from the expected result of 51 × 2. This value does not fit the consistent alternating operations used to build the series. Therefore, 106 must be rejected.

Let the present ages of the father and son be 7x and 3x respectively. After 10 years, their ages will be 7x + 10 and 3x + 10. The future ratio is (7x + 10) : (3x + 10) = 13 : 7. Cross-multiplying gives 7(7x + 10) = 13(3x + 10), i.e., 49x + 70 = 39x + 130. This simplifies to 10x = 60, so x = 6. Therefore, the father’s present age is 7x = 7 × 6 = 42 years.

Option A:
Option A, 35 years, would make the son 15 years old (ratio 7 : 3), and after 10 years their ages would be 45 and 25, giving a ratio of 9 : 5, not 13 : 7.

Option B:
Option B, 42 years, makes the son 18 years old. After 10 years they would be 52 and 28, and 52 : 28 simplifies to 13 : 7, satisfying both the given present ratio and the future ratio.

Option C:
Option C, 49 years, would imply a son aged 21, and after 10 years their ages become 59 and 31, which do not reduce to 13 : 7.

Option D:
Option D, 56 years, would give the son 24 years at present and 66 and 34 after 10 years, with ratio 33 : 17, again not 13 : 7.

The differences between the terms are 3, 6, 12, 24 and 48. Each difference is double the previous one, forming a geometric pattern at the second level. The next difference should therefore be 96. Adding 96 to the last term 100 gives 100 + 96 = 196. This continuation respects the doubling rule in the differences.

Option A:
Option A gives 184, whose difference from 100 is 84, not the expected 96. This breaks the doubling pattern governing the increments. Hence, 184 is not consistent with the series structure.

Option B:
Option B offers 188, giving a difference of 88 from 100, again not equal to 96 and incompatible with the second-level rule. Therefore, 188 cannot be accepted as correct.

Option C:
Option C suggests 192, corresponding to a difference of 92 from 100, which still fails to match the precise value 96. This makes the pattern irregular. Thus, 192 is not appropriate.

Option D:
Option D yields 196, which is 96 more than 100 and extends the difference sequence 3, 6, 12, 24, 48, 96 in a fully consistent way. This keeps the series logically sound. Therefore, 196 is the correct next term.

The ratio 2:3:5 has a total of 2 + 3 + 5 = 10 equal parts. B’s share is 3 of these 10 parts. Therefore B receives (3/10) × ₹6600 = ₹1980. This calculation uses the standard method of dividing a sum in a given ratio, and the figure ₹1980 uniquely matches the portion associated with B’s 3 parts.

Option A:
Option A is correct because it follows directly from the fraction 3/10 applied to the total amount. Checking the other shares, A would receive (2/10) × 6600 = ₹1320 and C would get (5/10) × 6600 = ₹3300, and the three results add up to the original ₹6600. This confirms that ₹1980 is the precise share for B.

Option B:
Option B, ₹1320, corresponds to the share for A, which has only 2 parts in the ratio 2:3:5. It underestimates B’s share, since B’s part count is higher than A’s. Accepting ₹1320 for B would contradict the given ratio because B must receive more than A.

Option C:
Option C, ₹2640, would equal 4 of the 10 parts if we imagined such a division, but no person in the ratio 2:3:5 has exactly 4 parts. Using this value would require altering the original ratio and would no longer reflect 2:3:5. Hence, ₹2640 cannot be B’s share.

Option D:
Option D, ₹3300, is equal to half the total amount and actually matches the share for C, who has 5 parts in the ratio. Assigning this to B would flip the intended hierarchy of the shares and make B’s entitlement much larger than specified. Therefore, ₹3300 is not B’s correct portion.