UGC NET Questions (Paper – 1)

The first letters are C(3), G(7), K(11) and O(15), each increasing by 4. The second letters are F(6), J(10), N(14) and R(18), also increasing by 4. Therefore, the next first letter is 15 + 4 = 19 (S) and the next second letter is 18 + 4 = 22 (V). Hence SV is the only pair that continues the uniform +4 progression in both positions.

Option A:
Option A, SV, is correct because it places S (19) after O (15) by +4 and V (22) after R (18) by +4. It preserves the same arithmetic step used in every previous pair. Therefore SV is the correct continuation.

Option B:
Option B, SW, keeps S correct but changes the second letter to W (23), which is +5 from R, not +4. This breaks the constant increment rule in the second position. Hence SW is incorrect.

Option C:
Option C, SX, keeps S correct but takes X (24), which is +6 from R. This increases the step size without any basis in the series. Therefore SX does not fit the pattern.

Option D:
Option D, TX, changes the first letter to T (20), which is +5 from O instead of +4. Since the first position must also increase by exactly 4, TX cannot be the correct next term.

The first-level differences are 9, 19, 29 and 39. These differences themselves form an arithmetic sequence with common difference 10. To preserve this pattern, the next difference must be 49. Adding 49 to 102 gives 151 as the next term. Thus 151 is the only option that keeps the second-level pattern perfectly intact.

Option A:
Option A corresponds to adding 49 to 102, giving 151 and extending the differences to 9, 19, 29, 39, 49. This preserves the constant increase of 10 between successive differences. Because both the original series and the sequence of gaps remain consistent, 151 is the correct continuation.

Option B:
Option B would make the next difference 47, so the differences become 9, 19, 29, 39, 47. The jump from 39 to 47 is 8, not 10, so the second-level pattern breaks. Hence 149 does not follow the intended rule of steadily increasing differences by 10.

Option C:
Option C would require a difference of 51 from 102, giving differences 9, 19, 29, 39, 51. Here the increment from 39 to 51 is 12, which again violates the constant second-level difference of 10. Therefore 153 cannot be accepted as the next term.

Option D:
Option D implies a difference of 45 from 102, so the last increment in the difference sequence is only 6. This contradicts the observed +10 progression that generates 9, 19, 29 and 39. As a result, 147 is not a valid continuation of the series.

This series is generated by the formula (a_n = n^4 + n) for n starting from 1. For n = 1, 2, 3, 4 and 5 we get 1+1 = 2, 16+2 = 18, 81+3 = 84, 256+4 = 260 and 625+5 = 630, which match the given terms. For n = 6 the same rule yields (6^4 + 6 = 1296 + 6 = 1302). Hence 1302 is the unique next term that respects this quartic-plus-linear pattern.

Option A:
Option A, 1286, is 16 less than the formula’s output and does not equal (6^4 + 6). It would require altering the rule arbitrarily for the last term, which is not supported by the earlier data. Therefore 1286 is not correct.

Option B:
Option B, 1294, is closer but still not equal to 1302 and fails to satisfy the expression (n^4 + n) for n = 6. Choosing 1294 would weaken the exact algebraic structure of the series. Thus 1294 is not a valid continuation.

Option C:
Option C, 1310, overshoots the correct value and again cannot be written as (6^4 + 6). Adopting 1310 would break the close correspondence between the formula and the actual terms. Therefore 1310 is not acceptable.

Option D:
Option D, 1302, equals (6^4 + 6) exactly and continues the same quartic pattern without modification. It maintains the functional relationship that has generated every earlier term, so 1302 is the correct next term.

Statements A, C and E correctly describe blood relation terminology and strategy. A is true because maternal and paternal terms indicate sides of the family. C is correct since drawing family trees is a widely recommended method for organizing information. E accurately defines “son-in-law” as the husband of one’s daughter. B is false, as B’s gender is not specified by A being a brother of B, and D is false because siblings include both brothers and sisters. Thus A, C and E only are correct.

Option A:
Option A is correct because it includes the precise conceptual statements about terminology and problem-solving strategies while excluding B and D, which misrepresent gender implications and the definition of siblings. It reflects the conventions used in NET reasoning questions.

Option B:
Option B is incorrect as it adds B, wrongly asserting that B must be male when A is B’s brother, and therefore introduces a false assumption into the set of correct statements.

Option C:
Option C is wrong because it includes D, which incorrectly narrows “siblings” to brothers only, so the combination mixes correct and incorrect statements.

Option D:
Option D is incorrect since it omits A and adds B, losing a key distinction between maternal and paternal relations and at the same time accepting a wrong inference about B’s gender.

The consecutive differences in the series are 7, 11, 15 and 19. These differences form an arithmetic progression with common difference 4. Therefore the next difference should be 23. Adding 23 to the last term 56 gives 79, which maintains the same second-level arithmetic structure and yields a smooth continuation.

Option A:
Option A gives 79, creating differences 7, 11, 15, 19, 23. Each difference increases by 4, so the pattern among the gaps remains consistent and clear. This makes 79 the only option that fully respects the established structure of the sequence.

Option B:
Option B gives 77, corresponding to a difference of 21 from 56. The resulting difference sequence 7, 11, 15, 19, 21 would end with a step of 2 instead of 4, breaking the uniform growth in the differences. Thus 77 does not fit the rule.

Option C:
Option C gives 81, which implies a difference of 25 from 56. That produces gaps 7, 11, 15, 19, 25, where the last increment is 6 and not 4. Consequently, 81 disrupts the simple arithmetic progression among the differences.

Option D:
Option D gives 83, yielding a difference of 27 from 56. The final increment between 19 and 27 becomes 8, which sharply departs from the steady step of 4. Therefore 83 is not a valid continuation of the series.

For a fixed distance, time is inversely proportional to speed. Let the distance between P and Q be d. The onward time is d/60 and the return time is d/40. Their ratio is (d/60):(d/40) = 1/60:1/40 = 40:60, which simplifies to 2:3. Hence, the ratio of onward time to return time is 2:3.

Option A:
Option A, 3:2, reverses the correct ratio. This would imply that the onward journey at higher speed takes more time than the return journey at lower speed, which contradicts the basic principle of inverse proportionality between time and speed for the same distance.

Option B:
Option B, 4:5, does not match the fractional comparison 1/60:1/40. If we attempted to equate 4:5 to 40:60, they would not simplify to the same reduced form. Therefore 4:5 is not aligned with the correct calculation.

Option C:
Option C is correct because it results from cancelling the common factor d and then simplifying the fraction 40:60. This ratio correctly reflects that the car spends less time on the onward trip at 60 km/h than on the return trip at 40 km/h by a factor of 2 to 3.

Option D:
Option D, 5:4, suggests only a mild difference between the times and in the wrong direction. It cannot be derived from the exact speeds given and leads to an incorrect relationship between the onward and return travel times.

A and B correctly give the formulas for permutations and combinations without repetition. C is true because allowing repetition means each of the r positions can be filled in n ways, giving nʳ arrangements. E is also correct since identifying whether order matters is essential to selecting the right counting approach. D is false because without repetition, r-length arrangements from n objects are counted by nPr, not rⁿ. Thus A, B, C and E only are correct.

Option A:
Option A misses E and therefore does not state the important strategic step of checking whether order is significant. Although it contains correct formulas, it is not complete as an answer.

Option B:
Option B omits A and so fails to mention the basic permutation formula, even though it includes other true statements. Leaving out A makes the set incomplete for the topic.

Option C:
Option C is correct because it collects all the true statements while excluding D, which misapplies exponent notation. It reflects the range of arrangements encountered in NET questions.

Option D:
Option D excludes B and so ignores the combination formula, and it does not correct the error in D. Consequently, it does not represent the full correct set.

The nth term of this sequence is given by aₙ = 4n³+3 for n starting from 1. For n = 1, 2, 3, 4 and 5 we get 4·1³+3 = 7, 4·2³+3 = 35, 4·3³+3 = 111, 4·4³+3 = 259 and 4·5³+3 = 503. For n = 6 the same rule yields 4·6³+3 = 864+3 = 867. Hence 867 is the only number that continues the sequence according to this cubic formula.

Option A:
Option A, 867, comes directly from applying aₙ = 4n³+3 when n = 6. It maintains exactly the same relationship between the index and the term that holds for all earlier values. Because no change in the rule is needed, 867 is fully consistent with the pattern and is correct.

Option B:
Option B, 827, is significantly smaller than 867 and does not equal 4·6³+3. It would require subtracting 40 from the formula’s result with no justification from the series. Therefore 827 is not a valid next term.

Option C:
Option C, 843, is closer but still fails to match the computed value, and it is not the output of 4n³+3 for any integer following n = 5. Selecting it would disrupt the clean cubic structure of the sequence. Thus 843 cannot be the correct answer.

Option D:
Option D, 897, overshoots the value given by the formula and similarly cannot be derived from 4n³+3 for n = 6. Using 897 would destroy the exact fit between the rule and the data, so it is not appropriate.

The terms satisfy the rule (a_n = 3n^4 + n) for n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 3+1 = 4, 48+2 = 50, 243+3 = 246, 768+4 = 772 and 1875+5 = 1880. For n = 6 the expression gives (3×6^4 + 6 = 3×1296 + 6 = 3888 + 6 = 3894). Hence 3894 is the term that keeps this quartic-plus-linear pattern intact.

Option A:
Option A, 3846, is 48 less than the computed value and does not equal (3×6^4 + 6). It would require unjustified modification of the rule at the final term. Therefore 3846 is not a correct continuation.

Option B:
Option B, 3862, is closer but still fails to match 3894 and does not arise from the formula for n = 6. Choosing 3862 would weaken the exact algebraic structure that defines the sequence. Hence 3862 is not valid.

Option C:
Option C, 3894, comes directly from evaluating (a_n = 3n^4 + n) at n = 6. It preserves both the quartic and linear components exactly as before, ensuring internal consistency. For this reason, 3894 is the correct next term.

Option D:
Option D, 3950, overshoots the correct value and again cannot be written as (3n^4 + n) for the next index. Using 3950 would distort the pattern, so it is not the right answer.

In this pattern, the first letters go backward in the alphabet: Z(26), Y(25), X(24), W(23), decreasing by 1. The second letters go forward: A(1), B(2), C(3), D(4), increasing by 1. The next first letter is V(22), and the next second letter is E(5), forming VE. Therefore VE is the unique pair that preserves both movements.

Option A:
Option A, UF, uses U as the first letter, which is 2 steps before W instead of 1. It also advances the second letter to F, which is 2 steps beyond D. These larger jumps break the consistent step of 1 in both directions, making UF incorrect.

Option B:
Option B, WD, simply repeats the last term of the series. The sequence is clearly moving and does not indicate any repetition pattern. Thus WD cannot serve as the next term.

Option C:
Option C is correct because V and E are exactly one step beyond W backward and D forward respectively. It keeps the first letter moving towards the beginning of the alphabet and the second letter moving towards the end. This symmetry ensures VE follows the established rule.

Option D:
Option D, VG, uses the correct first letter V but sends the second letter to G, two steps ahead of E. That distorts the +1 progression in the second position. As a result, VG does not properly continue the series.

Each group consists of four letters that are two positions apart inside the alphabet, for example ACEG uses positions 1, 3, 5 and 7. From one group to the next, every letter shifts one step forward: ACEG becomes BDFH, then CEGI, then DFHJ. Applying the same column wise +1 shift again, D becomes E, F becomes G, H becomes I and J becomes K, giving EGIK. Thus EGIK uniquely preserves the regular spacing and advancement.

Option A:
Option A, EFGH, destroys the spacing pattern because the letters are consecutive instead of being two steps apart. It does not arise from adding one to each letter of DFHJ. Since the series is defined by both even spacing and uniform shifting, EFGH does not fit the rule.

Option B:
Option B, EGIJ, starts correctly with E and G but changes the last two letters to I and J. The final letter J is only one step ahead of I instead of two steps ahead of G, breaking the uniform internal differences. Because it fails to keep both the +2 spacing and the consistent shift from DFHJ, EGIJ cannot be the correct answer.

Option C:
Option C, EGHJ, again loses the pattern of equal spacing. The letters E, G, H and J have mixed intervals, and this group is not produced by a simple +1 shift from DFHJ. Even though some letters look close to the expected ones, the underlying structure is not preserved.

Option D:
Option D is correct because E, G, I and K are each exactly one step ahead of D, F, H and J respectively. It keeps the difference of two positions between successive letters inside the group and maintains the same transformation that generated all earlier terms. Therefore EGIK is the logical next term in the series.

The starting letters Q, N, K and H occupy positions 17, 14, 11 and 8, decreasing by 3 each time. Each group is a three letter block of consecutive letters in forward order. Continuing this pattern, the next starting letter is position 5, which is E, and the following letters are F and G, giving EFG. Therefore EFG fits both the minus three jump and the internal structure of the groups.

Option A:
Option A is correct because E lies exactly three positions before H and EFG continues with F and G as consecutive letters. It maintains the constant decrement in the starting letters and keeps the groups consistently formed. This alignment with both the numerical and structural aspects of the pattern makes EFG the correct answer.

Option B:
Option B, GHI, moves forward from H instead of backward, reversing the direction of the starting sequence. The series clearly shows the starting letters descending through the alphabet. As GHI would be a forward step, it does not conform to the observed rule.

Option C:
Option C, IJK, begins at I, which is only one position after H and also goes in the wrong direction. It neither preserves the backward movement nor maintains the required step size of three. Consequently, IJK cannot be the next term.

Option D:
Option D, ABC, starts at A, which is position 1, representing a jump of seven positions backward from H instead of three. This change in step size is not supported by any previous transition. As a result, ABC does not fit the pattern.

In partnership problems, the profit is divided in the ratio of capital multiplied by time. Taking capitals as 3k, 5k and 7k for A, B and C and times as 2, 4 and 6 years, their effective investments are 3k × 2 = 6k, 5k × 4 = 20k and 7k × 6 = 42k. The ratio 6:20:42 simplifies to 3:10:21. The sum of these parts is 34, so B’s share is (10/34) of ₹1,36,000, which equals ₹40,000.

Option A:
Option A is correct because the calculation (10/34) × 1,36,000 gives exactly ₹40,000, and using 3/34 and 21/34 for A and C yields the remaining profit. The three shares add up neatly to ₹1,36,000, confirming that the ratio of effective investments has been applied correctly. Hence, ₹40,000 accurately represents B’s entitlement.

Option B:
Option B, ₹38,000, would not correspond to the fraction 10/34 of the total profit. If we assign ₹38,000 to B, the remaining amount for A and C would not then be divisible in the ratio 3:21 without leaving a remainder or altering the initial capital-time structure. Therefore, ₹38,000 is inconsistent with the proportional rule.

Option C:
Option C, ₹48,000, would represent a larger share than warranted by the 10 parts out of 34. Using ₹48,000 for B implies that B’s effective contribution outweighs that of C more than the 10:21 ratio allows. This contradicts the original investment-time data, so ₹48,000 cannot be correct.

Option D:
Option D, ₹52,000, is even higher and would require B’s share to be close to or greater than C’s share, which is not supported by the ratio 3:10:21. The necessary rebalancing of the remaining profit would distort the intended proportions. As a result, ₹52,000 does not match the proportional distribution.

Each term is 4 consecutive letters in reverse. Starting letters Z(26), V(22), R(18), N(14) decrease by −4. Next start is J(10), so the reversed block is JIHG.

Option A:
Option A is correct because J is −4 from N and J,I,H,G are consecutive in reverse order.

Option B:
Option B starts at K (11), which is −3 from N, not −4.

Option C:
Option C starts at M (13), which is −1 from N, not −4.

Option D:
Option D moves forward to P and breaks the backward step rule.

Each group is made of four consecutive letters written in reverse order. The starting letters X, U, R and O are moving backwards through the alphabet by a fixed number of steps from one term to the next. Continuing this pattern, the next starting letter must be L and the remaining letters must step down consecutively from there. That gives LKJI, which matches both the constant backward step in starting letters and the reverse ordering inside each group, so it is the correct next term.

Option A:
Option A is correct because it begins at L, which is the next point in the backward chain after X, U, R and O, and then lists the previous three letters K, J and I in descending order. The internal structure is exactly the same as in the earlier terms. This alignment with both the starting letter sequence and the reverse block rule makes LKJI the only suitable continuation.

Option B:
Option B, MJIH, starts from M instead of L and so moves the starting letter too far back. It no longer matches the fixed step used between X, U, R and O. The letters also do not form the straightforward descending run that characterises the series, hence MJIH is not correct.

Option C:
Option C, NMLK, reverses letters but starts at N, which breaks the specific starting letter progression. It also produces a block that overlaps awkwardly with ONML rather than continuing smoothly after it. Because it does not respect the consistent backward move of the first letter, NMLK cannot be accepted.

Option D:
Option D, KJIH, uses letters that are in descending order but begins at K rather than the expected L. This would shorten the step from O to the next starting letter and contradict the constant pattern. Therefore KJIH does not fit the established rule.