UGC NET Questions (Paper – 1)

First and last letters are same and increase: B,C,D,E → F. Middle letters move X,Y,Z,A → next B. So next group is FBF.

Option A:
Option A is correct because outer letters become F and the middle becomes B by continuing the sequence.

Option B:
Option B breaks symmetry rule because outer letters are not identical (F and H).

Option C:
Option C skips the middle-letter progression by jumping to C instead of B.

Option D:
Option D repeats the middle letter A instead of moving to B.

A number is divisible by 6 if and only if it is divisible by both 2 and 3. The numbers 12, 18 and 24 are all even and their digit sums are divisible by 3, so each is divisible by 6. The number 35 is not even and its digit sum is 3 + 5 = 8, which is not divisible by 3. Therefore, 35 is not divisible by 6 and stands out from the rest.

Option A:
Option A, 12, is divisible by 2 and by 3, which means it is divisible by 6 and thus similar to 18 and 24 in this property. It cannot be regarded as the odd one when considering divisibility by 6.

Option B:
Option B, 18, also satisfies the test for divisibility by 6, being divisible by 2 and 3, so it remains within the main group.

Option C:
Option C, 24, is divisible by 2 and 3 as well, so it again fits the common pattern shared by the first three numbers in the list.

Option D:
Option D, 35, fails to meet either of the conditions for divisibility by 6, because it is not even and the sum of its digits is not divisible by 3. This makes it uniquely different in the given set.

The given sequence increases from 5 to 9, from 9 to 13, and from 13 to 17. In each step, the increase is 4, which means the common difference is 4. Such a sequence with constant difference is an example of an arithmetic progression. Therefore, adding 4 each time correctly describes the pattern generating the sequence.

Option A:
The option 4 is correct because 9−5, 13−9 and 17−13 all equal 4. In arithmetic progression questions, identifying the constant difference is the key step to predicting the next term. If we add 4 to 17, we get 21 as the next term, which fits the discovered pattern. Hence, 4 is the only value that maintains the same step throughout the sequence.

Option B:
Adding 5 each time would give 5, 10, 15, 20, which is not the given sequence. The actual terms 9, 13 and 17 would not be obtained by repeated addition of 5 starting from 5. This shows that 5 does not match the observed pattern. So, this option is incorrect.

Option C:
Adding 3 at each step would produce 5, 8, 11, 14 and so on, which again does not coincide with the given terms 9, 13 and 17. The difference between each pair of consecutive terms is clearly 4, not 3. Therefore, this option does not fit the sequence.

Option D:
Adding 2 each time gives the series 5, 7, 9, 11, which is different from the one given in the question. Although 2 is a possible constant difference in other series, it cannot describe the pattern here because the actual increments are larger. Hence, this option must be rejected.

The given relation can be written as x = k·y/√z for some constant k. Using x = 18 when y = 6 and z = 9, we have 18 = k·6/3, so 18 = 2k and k = 9. For y = 8 and z = 4, we get x = 9·8/√4 = 72/2 = 36. Thus, the new value of x is 36.

Option A:
Option A, 16, might be obtained by incorrectly treating the dependence on z as inverse linear (y/z) and misapplying the parameter values, but it does not follow from x = k·y/√z with k = 9. Substituting y = 8 and z = 4 into the correct formula clearly yields 36, not 16.

Option B:
Option B, 24, could arise from ignoring the square root and computing x = 9·8/3, but that would contradict the original pair (18, 6, 9) if we required a single consistent formula. The condition involves √z explicitly, so 24 is not compatible with that relationship.

Option C:
Option C, 30, does not match any straightforward misinterpretation of the formula that still respects both data points. When we use the correct k = 9 and plug in y = 8 and z = 4, we obtain 36, so 30 cannot be reconciled with the algebraic model.

Option D:
Option D is correct because it uses the same constant of proportionality derived from the first situation and correctly applies the dependence on y and √z. The computation 9·8/2 = 36 strictly follows x = k·y/√z and satisfies both given data conditions.

The series can be described recursively by a₁ = 7 and aₙ₊₁ = 4aₙ + n³ for n ≥ 1. Using this rule we obtain a₂ = 4·7+1³ = 29, a₃ = 4·29+2³ = 124, a₄ = 4·124+3³ = 523 and a₅ = 4·523+4³ = 2156, which reproduces the given terms. For n = 5 the next term is a₆ = 4·2156+5³ = 8624+125 = 8749. Therefore 8749 is the unique value that continues the same recurrence for the sixth term.

Option A:
Option A, 8605, is less than the recurrence output and does not equal 4·2156+5³. It would correspond to adding -19 to 4·2156 instead of the required 125, which contradicts the index-dependent cubic correction. Hence option A is incorrect.

Option B:
Option B, 8681, is closer but still fails to reach the correct value 8749 produced by the recurrence. Selecting 8681 would suggest a smaller adjustment by 57 instead of adding 125, again violating the rule aₙ₊₁ = 4aₙ+n³. Thus option B is incorrect.

Option C:
Option C, 8717, differs by 32 from the computed value and cannot be obtained from 4·2156+5³. It implies a partial use of the cubic term, which is not how the earlier terms were formed. Therefore option C is incorrect.

Option D:
Option D, 8749, matches exactly the result of applying the recurrence to a₅ with n = 5. It maintains the structure of quadrupling the previous term and then adding the cube of the index. Because this rule has generated all previous terms, 8749 is the correct next term in the series.

For a, b and c to be in continued proportion, we have a:b = b:c, which implies b² = ac. Given a:b = 2:5, we can take a = 2k and b = 5k. Then b² = ac gives (5k)² = 2k·c, so 25k² = 2kc and c = (25k²)/(2k) = 25k/2. Therefore, c:b = (25k/2):(5k) = 25:10 = 5:2.

Option A:
Option A is correct because it results from applying the definition of continued proportion and simplifying c:b. It shows that c must be larger than b when a:b is 2:5, which aligns with the idea that b is a geometric mean between a and c and hence lies between them.

Option B:
Option B, 2:5, simply restates the ratio a:b and would imply that c is smaller than b, contradicting the fact that c should lie on the opposite side of b from a in the geometric mean setting. Using 2:5 for c:b does not satisfy b² = ac with the given a:b.

Option C:
Option C, 4:5, comes from a possible misreading that c is closer to b than a is, but if c:b were 4:5, the product ac would no longer equal b². Substituting 4:5 fails the algebraic condition linking the three terms in continued proportion.

Option D:
Option D, 5:4, suggests that c is only slightly larger than b, but the actual computation shows c is 25k/2 while b is 5k, making c significantly larger. The ratio 5:4 therefore understates c and does not satisfy the required equality b² = ac.

A is correct because typical answer choices in these questions classify sufficiency patterns for the two statements. B is true as understanding what is being asked is the first step before checking sufficiency. D is also correct since the primary skill tested is judging sufficiency, not heavy computation. C is false because if either statement alone is sufficient, then both statements together are also sufficient even if redundant, and E is false because when both together suffice, the overall information is treated as sufficient. Thus A, B and D only are correct.

Option A:
Option A omits D and therefore fails to convey the main focus of these questions on sufficiency rather than calculation. Without D, the description of purpose is incomplete.

Option B:
Option B is correct because it combines the structural pattern in answer choices, the need to read the question and the intended skill focus, while excluding C and E, which misrepresent how combined sufficiency is treated.

Option C:
Option C includes E, which wrongly labels a jointly sufficient pair as insufficient, and omits A, so it fails both structurally and conceptually.

Option D:
Option D contains E as well and leaves out B, thereby misclassifying a false statement while losing an important instruction about reading the main question.

A is correct because x > 5 means any number greater than 5 is allowed. B is also correct since solving 2x + 3 < 11 gives 2x < 8 and hence x < 4. C is wrong because 1 1 and x ≤ 3, not x < 3. D is wrong because multiplying or dividing by a negative number requires reversing the inequality sign. E is correct as x ≥ −2 includes −2 and all greater numbers. Therefore, C and D only are the wrong statements. Option A: Option A lists only C as wrong and overlooks the error in D, which ignores sign reversal when multiplying by a negative number. It thus fails to capture all incorrect statements. Option B: Option B is correct because it selects exactly C and D, the statements that mis-handle combined inequalities and transformation rules, while recognising A, B and E as accurate. It aligns with how inequalities are treated in exam questions. Option C: Option C wrongly labels E as wrong, even though E correctly describes the number line representation of x ≥ −2. Including E makes this option inaccurate. Option D: Option D adds E to the set of wrong statements and so misclassifies a true description of the solution set, making the entire combination invalid.

The highest common factor is the largest number that divides both given numbers exactly. The prime factorization of 18 is 2 × 3 × 3 and of 30 is 2 × 3 × 5. The common prime factors are 2 and 3, and multiplying them gives 2 × 3 = 6. Since no larger number divides both 18 and 30 without remainder, 6 is the highest common factor.

Option A:
Option A correctly uses the idea of common prime factors and combines 2 and 3 to produce 6. It respects both the definition of a factor and the requirement that the factor be as large as possible while still dividing both numbers exactly. This makes 6 the appropriate highest common factor for 18 and 30.

Option B:
Option B, 3, is a common factor but not the greatest one because 6 also divides both numbers and is larger than 3. Choosing 3 ignores the possibility of combining more common prime factors.

Option C:
Option C, 5, divides 30 but does not divide 18, so it fails the basic requirement that the highest common factor must be a factor of both numbers. It is therefore not a common factor at all in this pair.

Option D:
Option D, 9, divides 18 but not 30 because 30 ÷ 9 does not yield an integer. Hence, it also cannot be regarded as a common factor of the two numbers.

Observing BEHK, EHKN, HKNQ and KNQT shows that each new group is obtained by adding the same number of positions to every letter of the previous group. BEHK shifts to EHKN, then to HKNQ and then to KNQT through a fixed forward movement in all four positions. Applying the same transformation to KNQT moves K, N, Q and T each ahead by the same amount, giving NQTW. Since NQTW alone is produced by this uniform large shift, it is the correct next term.

Option A:
Option A is correct because NQTW continues the chain of starting letters B, E, H, K to N and does the same for the other three columns. The internal relations among the letters remain unchanged as each one advances together. This preserves the defining property of the series and makes NQTW the logical continuation.

Option B:
Option B, NQUW, alters the third letter differently from the way earlier groups were transformed and does not reflect the same equal shift in every position. Its letters no longer align cleanly with the vertical sequences. Because it breaks the common movement, NQUW is not valid.

Option C:
Option C, OQTW, starts with O instead of N, pushing the first letter too far ahead and changing the step size. The rest of the letters also do not follow the precise transformation seen in the series. Therefore OQTW does not satisfy the pattern.

Option D:
Option D, MPSV, changes several letters in a way that cannot be explained by a single consistent shift applied to KNQT. It invents a different structure rather than extending the existing one. As a result, MPSV cannot be accepted as the next term.

This sequence is defined recursively by a₁ = 4 and aₙ₊₁ = 4aₙ + 2ⁿ for n ≥ 1. Applying the rule gives a₂ = 4·4 + 2¹ = 18, a₃ = 4·18 + 2² = 76, a₄ = 4·76 + 2³ = 312 and a₅ = 4·312 + 2⁴ = 1264, exactly matching the given terms. For n = 5 the next term is a₆ = 4·1264 + 2⁵ = 5088. Therefore 5088 is the unique number that satisfies the recurrence and continues the series.

Option A:
Option A, 5056, is 32 less than the value given by aₙ₊₁ = 4aₙ + 2ⁿ when n = 5. Accepting 5056 would mean reducing the correct result by the size of the power term 2⁵, breaking the pattern of adding that full power at each stage. Hence option A does not follow the established rule and is incorrect.

Option B:
Option B, 5088, matches exactly the value obtained from 4a₅ + 2⁵ with a₅ = 1264. It maintains the structure of multiplying the previous term by four and then adding the appropriate power of 2. Because this mechanism generates all earlier terms and leads to 5088 next, option B correctly continues the sequence.

Option C:
Option C, 5104, is 16 greater than the recurrence value and cannot be produced by 4a₅ + 2⁵. It would require adding more than the specified power of 2 at the final step, contradicting the rule. Thus option C does not preserve the exact relationship and is not valid.

Option D:
Option D, 5120, deviates even more and again fails to equal 4·1264 + 32. Using 5120 would ignore the precise recurrence that describes the sequence. Therefore option D is not a correct continuation.

The series follows the rule aₙ = 2n⁵ − n + 3 with n starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 4, 65, 486, 2047 and 6248, which are exactly the terms in the question. Using the same formula for n = 6 produces 15549. Hence 15549 is the only value that continues this quintic pattern without altering the rule.

Option A:
Option A, 15549, is precisely the value obtained from aₙ = 2n⁵ − n + 3 when n = 6. It maintains the strong fifth-power growth together with the linear correction seen in earlier terms. Because it emerges directly from the established expression, this option correctly represents the next term.

Option B:
Option B, 15521, is 28 less than the formula’s output at n = 6. Accepting 15521 would require arbitrarily lowering the quintic value only for this term. Since all previous entries fit the rule exactly, such an adjustment is unjustified and makes option B incorrect.

Option C:
Option C, 15581, is slightly larger than the required value and does not equal 2n⁵ − n + 3 when n = 6. It would introduce a positive discrepancy at the last step that is not present earlier. Therefore this option breaks the algebraic consistency of the sequence.

Option D:
Option D, 15613, deviates even more from the computed value and cannot be produced by the same formula. Using 15613 would destroy the exact match between the rule and the observed terms, so it is not a valid continuation.

The series follows the rule aₙ = n⁴ + 4n² + 7 with n beginning at 1. Evaluating this expression for n = 1, 2, 3, 4 and 5 gives 12, 39, 124, 327 and 732, which are exactly the given terms. For n = 6 we get a₆ = 6⁴ + 4·6² + 7 = 1296 + 4·36 + 7 = 1296 + 144 + 7 = 1447. Thus 1447 is the unique number that fits this quartic-plus-quadratic pattern.

Option A:
Option A, 1447, is precisely the value produced by n⁴ + 4n² + 7 when n = 6. It maintains the same combination of a dominant fourth-power term and quadratic correction plus constant that explains earlier terms. Since no deviation from the rule is required, this option correctly continues the series.

Option B:
Option B, 1423, is 24 less than the formula’s value at n = 6 and cannot be generated by the same expression. To obtain 1423 we would have to subtract 24 only at the last term, which is inconsistent with the pattern. Therefore option B is not correct.

Option C:
Option C, 1435, is 12 below the required value 1447 and again fails to equal 6⁴ + 4·6² + 7. It offers an approximate but incorrect continuation that ignores the exact algebraic relationship. Hence this option must be rejected.

Option D:
Option D, 1459, is 12 greater than the computed term and would demand increasing the expression’s value only at the sixth position. Such a change would destroy the perfect fit between formula and data, so option D does not represent the correct next term.

Statements A, B and C give the standard description and formula for arithmetic and weighted means. A is correct because the simple mean assigns equal weight to each observation. B and C correctly explain that weighted means use different weights and are computed as the weighted sum divided by the total weight. Statement E is true because if all weights are equal, the weighted mean collapses to the ordinary mean. Statement D is false since a weighted mean, like the arithmetic mean, can lie between observed values and need not coincide with any one observation, so the correct set is A, B, C and E only.

Option A:
Option A is incomplete because it omits C and E and so does not show how weighted means are actually computed or how they relate back to the arithmetic mean when weights are equal. It therefore does not represent all correct statements.

Option B:
Option B is also incomplete as it does not include E, ignoring the simplifying case in which weights are equal, even though it correctly includes A, B and C. The missing link to the arithmetic mean makes it insufficient.

Option C:
Option C is wrong because, although it includes B, C and E, it omits A, which states the key contrast between arithmetic and weighted means, and still does not highlight the equal-weight case as a special instance in context.

Option D:
Option D is correct because it brings together the accurate description of both kinds of means and the proper computation rule, while explicitly excluding D, which incorrectly insists that the mean must be one of the data points. This aligns with the conceptual treatment of averages in NET questions.

Every group is a five letter palindrome where the first and last letters match, and the second and fourth letters also match. Moving from one term to the next, all letters shift one step forward in the alphabet: ABCBA becomes BCDCB, then CDEDC, then DEFED. Continuing this rule, D becomes E, E becomes F and F becomes G for the central letter, so the new group must be EFGFE. This preserves both the palindromic structure and the uniform forward movement.

Option A:
Option A is correct because it uses E as the new outer letter, F as the new second and fourth letters and G as the new centre. These letters are each obtained by shifting the corresponding letters of DEFED forward by one position. The result is a perfect five letter palindrome that mirrors the behaviour of the earlier terms.

Option B:
Option B, EFHFE, introduces H as the central letter while the second and fourth letters remain F. That would mean the centre has moved two steps from E to H instead of one, breaking the consistent +1 progression. Since the pattern does not show unequal jumps, EFHFE cannot be the correct continuation.

Option C:
Option C, EFGGF, loses the palindromic form because the first and last letters are not the same. The series clearly maintains symmetry in every term. By changing the structure of the word, EFGGF fails to fit the given pattern.

Option D:
Option D, EFFGE, also spoils the required symmetry and does not match the simple column wise shift from DEFED. The positions of letters are altered in a way that is not supported by any earlier step. Therefore EFFGE cannot be accepted as the next term.