UGC NET Questions (Paper – 1)

The equation (x − 3)(x + 5) = 0 is satisfied when either factor is zero. Setting x − 3 = 0 gives x = 3, and setting x + 5 = 0 gives x = −5. The sum of the roots is therefore 3 + (−5) = −2. This also agrees with the general property that for x² + bx + c = 0, the sum of the roots is −b, which is consistent if the equation is expanded.

Option A:
Option A correctly adds the two root values, 3 and −5, to obtain −2. It captures the essence of the relationship between factorized form and the numerical roots.

Option B:
Option B, 2, would arise if one mistakenly subtracted −5 from 3 or ignored the sign of one root, but that does not reflect the actual arithmetic.

Option C:
Option C, −8, is the product of the roots, not their sum. It reflects a different relation in quadratics and confuses two separate concepts.

Option D:
Option D, 8, would require both roots to be positive or both negative with different magnitudes, which is not the case for this factorization.

The groups are three consecutive letters and each new group starts two letters after the start of the previous one. The starting letters A, C, E and G correspond to positions 1, 3, 5 and 7, all increasing by 2. The next starting letter must therefore be position 9, which is I, and the group formed is IJK. This satisfies both the skip of one letter between starting positions and the three letter consecutive structure.

Option A:
Option A, HIJ, begins at H, which is position 8 and only one step after G. This ignores the observed pattern of moving the starting position by two positions. Although HIJ is a valid consecutive block, it does not respect the skip rule that defines the series.

Option B:
Option B, JKL, starts at J, which is position 10 and three positions after G. This overshoots the required increment and creates a larger gap than any previous step. Because the series maintains a consistent plus two movement, JKL cannot be correct.

Option C:
Option C, KLM, is even further ahead and increases the offset between starting letters even more. It would represent a later term in a continuation, not the immediate next group. As there is no jump of four positions in the given pattern, KLM does not fit the series.

Option D:
Option D is correct because I is exactly two positions after G, and IJK is a three letter consecutive group. It preserves the pattern of skipping one letter between the starting points and keeps the internal order unchanged. For these reasons, IJK is the logical continuation of the letter group series.

This option is correct because the sum of interior angles of a simple quadrilateral is known to be 360 degrees. A quadrilateral can be divided into two triangles, each having an angle sum of 180 degrees. Adding these gives 180 + 180 = 360. Therefore, the total interior angle sum is 360 degrees.

Option A:
A quadrilateral can be split into two triangles by drawing one diagonal. Each triangle has 180 degrees as the sum of its interior angles. Adding both triangle sums gives 360 degrees, so 360 is the correct total for the quadrilateral.

Option B:
180 degrees is the sum of interior angles of a single triangle, not a quadrilateral. Since a quadrilateral has more sides, its angle sum must be greater than that of a triangle. Thus, this option is not correct.

Option C:
270 degrees does not match any standard polygon angle-sum formula for quadrilaterals. It is less than the correct total of 360 degrees and therefore cannot be the right answer.

Option D:
540 degrees is the interior angle sum of a pentagon, not a quadrilateral. Using this value would assume five sides instead of four. Hence, this option is also incorrect.

Statements A and B correctly describe what analogies test and how to approach them by first analysing the given pair. Statement D is also true because a good analogy preserves the same type of relation in both pairs. Statement C is false since numerical analogies often depend on arithmetic or power operations, and E is false because analogies are widely used to measure reasoning ability. Therefore, A, B and D only are correct.

Option A:
Option A is incomplete as it omits D and thus does not include the key requirement that the same relationship must hold across both pairs.

Option B:
Option B is incomplete because it omits A (what analogies test) and does not give the full correct set.

Option C:
Option C is wrong because it includes C (false), even though it also contains true statements.

Option D:
Option D is correct since it retains all accurate statements and discards C and E, which are false.

Five consecutive odd integers can be written as n − 4, n − 2, n, n + 2 and n + 4, where n is the middle term. The average of such a symmetric set is always the middle term, so the average being 37 means the middle integer is 37. The largest of the five is therefore 37 + 4, which is 41. This uses both the definition of average and the structure of consecutive odd numbers.

Option A:
Option A, 39, is the second largest of the five integers, not the largest. It is 2 greater than the middle term instead of 4 and therefore does not represent the end of the sequence.

Option B:
Option B correctly identifies that with an average of 37, the five odd numbers are 33, 35, 37, 39 and 41. Among these, 41 is clearly the largest and fits the pattern of consecutive odd integers.

Option C:
Option C, 43, would require shifting the sequence so that the average is larger than 37. That would contradict the given mean and lead to a different central value.

Option D:
Option D, 45, is even further from the centre and again cannot be part of the exact sequence whose average is 37.

All of these groups are four letter palindromes where the first and last letters are the same and the middle pair are also the same. From BZZB to CYYC to DXXD to EWWE, the outer letters move forward B, C, D, E while the inner letters move backward Z, Y, X, W. To continue this dual movement, the next outer letter should be F and the next inner letter should be V, one step before W. This produces FVVF, which alone maintains both the mirror structure and the opposing motions.

Option A:
Option A, FWWF, advances the outer letters correctly to F but keeps the inner letters at W instead of moving them backward. This freezes the inner sequence and breaks the clear descent Z, Y, X, W. Because it ignores the change in the middle pair, FWWF does not follow the pattern precisely.

Option B:
Option B, FXXF, pairs F with X but that would repeat an inner letter that already appeared with DXXD rather than continue the sequence. It does not move the inner letters one step back from W as required. Since it disturbs the ordered progression of middle letters, FXXF is not a correct continuation.

Option C:
Option C is correct because FVVF uses F as the next outer letter and V as the next inner letter, continuing the simple stepping patterns on both sides. The palindromic form with matching first and last letters and matching middle pair is preserved completely. This perfect consistency with the mirror structure and letter movements makes FVVF the only suitable next term.

Option D:
Option D, GVVG, shifts the outer letter too far ahead to G. It no longer respects the one step outer progression (E to F) shown in the series. Therefore it cannot be accepted as the answer.

This sequence is defined by aₙ = 3n³−n for n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 3·1³−1 = 2, 3·2³−2 = 22, 3·3³−3 = 78, 3·4³−4 = 188 and 3·5³−5 = 370. For n = 6 the formula yields 3·6³−6 = 648−6 = 642. Thus 642 is the next term that maintains this cubic-minus-linear relationship.

Option A:
Option A, 622, is 20 less than the formula’s outcome and does not equal 3·6³−6. It would require arbitrarily reducing the term, which is not supported by the pattern of the sequence. Therefore 622 cannot be the correct continuation.

Option B:
Option B, 642, matches exactly the result of applying aₙ = 3n³−n for n = 6. It keeps the dependence on the index unchanged and fits neatly after 370. Because the same rule works for all known terms and the next one, 642 is the correct answer.

Option C:
Option C, 630, also deviates from the computed value and is not of the form 3n³−n for the next integer n. Adopting 630 would make the formula inconsistent at the final step. Hence 630 is not valid.

Option D:
Option D, 657, overshoots the correct value and again is not produced by the expression for n = 6. Using 657 would spoil the perfect algebraic fit of the sequence, so it is not correct.

Within each group the jumps between letters follow a specific pattern of unequal steps. ACFJ uses one distance between the first pair of letters and a larger one between the next pairs, and the same structure appears in BDGK, CEHL and DFIM. From one group to the next, every letter moves forward by one position in the alphabet, so the columns each form a simple sequence. Shifting every letter in DFIM forward by one gives EGJN, which keeps the same internal spacing while extending each column, so it is the correct next term.

Option A:
Option A, EGIN, starts with the right two letters but then changes the internal differences between the later letters. The progression from G to I to N no longer matches the jumps used inside the earlier groups. Because it distorts the internal pattern, EGIN cannot be the next term in this series.

Option B:
Option B is correct because EGJN is obtained by adding one to each letter of DFIM and at the same time preserves the unequal but consistent gaps between positions. The first letters move from A to D to E in order and the other columns also extend cleanly. This combination of column wise advancement and preserved internal structure shows that EGJN properly continues the sequence.

Option C:
Option C, FHKO, moves the starting letter two steps ahead of D and disrupts the simple one step growth across the first column. It also alters the spacing between inner letters so that they no longer mirror the earlier terms. Hence FHKO does not align with the defined pattern.

Option D:
Option D, EGJO, keeps most letters close but sends the last letter too far ahead, changing the final gap inside the group. This breaks the consistent relation between the third and fourth letters found in ACFJ, BDGK, CEHL and DFIM. Therefore EGJO is not a valid continuation.

Statements A, B, C and E correctly describe the representational power of Venn diagrams. They visually show relationships among sets, capture intersections as overlapping regions, show disjointness through non-overlapping circles and can handle reasoning situations with three classes. Statement D is false because Venn diagrams are actually a standard tool for checking the validity of categorical syllogisms by mapping premises and examining the conclusion. Therefore the combination that lists A, B, C and E only is correct.

Option A:
Option A is incomplete because it omits E, thereby ignoring the useful fact that Venn diagrams can be extended to three sets and are often used in exam problems involving three categories. Without E, the picture of their application is narrower.

Option B:
Option B is correct since it collects all true statements and excludes D, which wrongly denies the use of Venn diagrams in syllogistic reasoning. It captures both the set-theoretic and exam-oriented aspects of Venn diagrams.

Option C:
Option C is wrong as it leaves out A, so it does not explicitly say that Venn diagrams represent relationships among sets or classes, a fundamental point about their purpose. Though B, C and E are true, the combination is incomplete.

Option D:
Option D is incorrect because it omits B, which explains how common elements are shown, and thus fails to mention how intersections appear visually. It also does not rectify D, the false claim about their use in testing syllogisms.

Using the principle of inclusion–exclusion, the number of students who like at least one of Mathematics or English is 18 + 16 − 8 = 26. The total number of students in the class is 30. Therefore, those who like neither subject are 30 − 26 = 4. This calculation separates students who fall outside both preference sets.

Option A:
Option A, 2, would imply that 28 students like at least one of the two subjects, but the inclusion–exclusion result clearly shows only 26 students in that category. Hence, 2 underestimates the count of those liking neither.

Option B:
Option B, 3, similarly assumes 27 students like at least one subject, conflicting with the properly derived figure of 26. It reflects a minor miscalculation in subtracting from the total.

Option C:
Option C correctly applies inclusion–exclusion to avoid double counting those who like both subjects, then subtracts from the total class strength. This yields 4 students who like neither Mathematics nor English.

Option D:
Option D, 6, would require at least one count to be different because it implies only 24 students like at least one subject, which contradicts the overlap-adjusted total of 26.

The work rate of A is 1/12 and that of B is 1/18 of the work per day. Together, their combined rate is 1/12 + 1/18 = (3 + 2)/36 = 5/36 of the work per day. The total time to complete one whole work at this rate is the reciprocal, which is 36/5 days, equal to 7.2 days. Hence, they can complete the work together in 7.2 days.

Option A:
If we assume 6 days, the combined work done would be 6 × (5/36) = 30/36 = 5/6 of the work, which is less than 1. This means the work would remain incomplete after 6 days, so 6 days is too little time and thus incorrect.

Option B:
If we choose 7 days, the total work done would be 7 × (5/36) = 35/36, still short of the full work. This value underestimates the time required, so 7 days is not an accurate answer for the exact completion.

Option C:
Assuming 8 days, the work done would be 8 × (5/36) = 40/36, which exceeds 1, meaning more than the total work is accounted for. This overestimates the time, indicating that 8 days is too long.

Option D:
7.2 days is correct because 36/5 is precisely the reciprocal of the combined rate 5/36. Writing it as 7.2 makes the decimal representation easier to interpret in numeric aptitude problems. This value ensures that the computed work equals exactly 1, satisfying the problem conditions.

A is correct because arranging data in a table helps track multiple attributes. B is true as Latin squares require each symbol exactly once per row and column. D is also correct since elimination based on clues is a standard approach to such puzzles. C is false because a standard Sudoku requires each number 1–9 exactly once in each row, column and each 3×3 sub-grid, and E is false as these puzzles frequently appear in reasoning sections. Thus A, B and D only are correct.

Option A:
Option A properly groups the three accurate statements about how grid puzzles are structured and solved, while rejecting C and E, both of which misdescribe Sudoku rules and exam usage. It matches the reasoning component of Unit 5.

Option B:
Option B includes C, which wrongly allows repetition within sub-grids, and therefore cannot be correct despite containing other true statements.

Option C:
Option C mistakenly contains E, which denies the presence of such puzzles in exams, and omits A, which explains the row–column organisation. This makes the option both erroneous and incomplete.

Option D:
Option D includes E and so inherits the error that these puzzles never appear in aptitude tests, even though it includes some correct ideas. As a result, it is an invalid combination.

Each term is a set of four consecutive letters, and the starting letter of each group increases by one: C, D, E, F. After FGHI, the next group must start with G, the next letter after F, and continue with H, I and J. Thus GHIJ is the only group that continues the sliding window pattern correctly.

Option A:
Option A is correct because it starts at G and includes the next three consecutive letters H, I and J. This maintains the group length, keeps the letters consecutive, and advances the starting position by exactly one step. These are the key features seen consistently in the earlier terms of the series.

Option B:
Option B, HIJK, begins at H, which would be the group after GHIJ rather than the immediate successor to FGHI. It therefore shifts the starting letter too far forward by skipping the block beginning at G. Because the pattern never skips a position, HIJK cannot be the correct answer.

Option C:
Option C, IJKL, moves even further ahead and ignores both the immediate G and H starting positions. It would represent a later block in the theoretical continuation, not the next one. Since the question asks for the very next term, IJKL is not appropriate.

Option D:
Option D, JKLM, similarly represents a block starting even later in the alphabet. It discards the incremental nature of the starting letters that has been established. For that reason, JKLM does not provide the correct continuation of the given series.

This sequence follows the rule aₙ = 2n⁴ + 5n² + 1 with n beginning at 1. For n = 1, 2, 3, 4 and 5 the expression gives 8, 53, 208, 593 and 1376, which are exactly the provided terms. Applying the same formula to n = 6 results in 2773. Therefore 2773 is the unique next term that maintains this quartic pattern with a quadratic adjustment.

Option A:
Option A, 2749, is 24 less than the value produced by aₙ = 2n⁴ + 5n² + 1 at n = 6. To obtain 2749 we would have to reduce the correct polynomial result arbitrarily for the sixth term. This is not supported by any change in the earlier sequence, so option A is incorrect.

Option B:
Option B, 2757, is still lower than the required value and does not match the formula’s output for n = 6. While numerically close, it fails to respect the exact algebraic structure of the sequence. Hence option B does not correctly continue the series.

Option C:
Option C, 2765, lies closer but remains 8 below the correct value 2773. Accepting 2765 would again mean departing from the precise polynomial expression only at the final term. Because the rule explains each previous entry perfectly, option C cannot be considered correct.

Option D:
Option D, 2773, equals exactly the value obtained when n = 6 is substituted into 2n⁴ + 5n² + 1. It preserves both the dominant quartic term and the quadratic contribution along with the constant term. As it fits seamlessly into the same pattern, 2773 is the correct next term in the series.

Let the original numbers be 2x and 5x. After adding 6 to each, they become 2x + 6 and 5x + 6, and the new ratio is 1:2. Thus, (2x + 6)/(5x + 6) = 1/2. Cross-multiplying gives 2(2x + 6) = 5x + 6, which simplifies to 4x + 12 = 5x + 6 and then x = 6. Therefore, the original numbers are 12 and 30, and the smaller is 12.

Option A:
Option A is correct because it corresponds to 2x with x = 6 in the ratio 2:5. Substituting 12 and 30 back into the problem, we get (12 + 6):(30 + 6) = 18:36 = 1:2, which matches the given transformed ratio. This confirms that 12 is the smaller of the original pair.

Option B:
Option B, 18, would require the larger number to be 45 to maintain the original ratio 2:5. The pair 18 and 45 has sum 63, and after adding 6 to each we get 24 and 51, whose ratio is 24:51 = 8:17, not 1:2. Hence 18 does not satisfy the transformation condition.

Option C:
Option C, 20, when paired with 50 for the 2:5 ratio, leads to new numbers 26 and 56 after the equal addition of 6. Their ratio 26:56 simplifies to 13:28, which again is not 1:2. Thus, 20 cannot be the smaller original number.

Option D:
Option D, 24, would pair with 60 to give the 2:5 ratio. After adding 6 we obtain 30 and 66, and the ratio 30:66 reduces to 5:11, not 1:2. Therefore, 24 fails to meet the required ratio change after the addition.