UGC NET Questions (Paper – 1)

In this series each term is a four letter group where the distance between consecutive letters inside a group is fixed. From BGLQ to CHMR to DINS to EJOT, every letter in each position moves forward by one step in the alphabet, so the vertical columns form simple increasing sequences. Applying the same shift to EJOT, E goes to F, J goes to K, O goes to P and T goes to U, giving FKPU. No other option arises from this consistent column wise movement with the same internal spacing, so FKPU is the correct continuation.

Option A:
Option A maintains the internal structure of the earlier groups because the gaps between F, K, P and U match those seen in BGLQ, CHMR, DINS and EJOT. It also continues each column sequence smoothly, for example the first letters B, C, D and E are followed naturally by F. Since it respects both the within group spacing and the across group progression, FKPU fits perfectly into the series.

Option B:
Option B, GLQV, looks similar but pushes every letter two steps ahead of EJOT instead of one, which breaks the steady one step advancement seen earlier. As a result, its letters no longer align with the vertical sequences formed by the previous groups. Because it distorts the regular shift pattern, GLQV cannot be the correct next term.

Option C:
Option C, FJPV, changes the differences between letters inside the group so that they do not match the constant spacing of the given series. It also mixes letters that do not arise from shifting EJOT uniformly forward. This double mismatch in spacing and transformation makes FJPV an invalid continuation.

Option D:
Option D, EJPU, keeps the first two letters unchanged and alters only the last two, whereas the pattern clearly moves all four letters together. Its internal distances differ from those of the original terms as well. Therefore EJPU does not satisfy the rule governing the series.

Let the original numbers be 4x and 7x. After adding 8 to each, the new numbers are 4x + 8 and 7x + 8 and their ratio is 5:8. So (4x + 8)/(7x + 8) = 5/8. Cross-multiplying gives 8(4x + 8) = 5(7x + 8), which simplifies to 32x + 64 = 35x + 40 and then to 3x = 24, so x = 8. The original numbers are 32 and 56, and the larger is 56.

Option A:
Option A, 40, does not appear as either 4x or 7x when x = 8 and would not preserve the required ratio 4:7 together with its pair. If we assume 40 as the larger number, the smaller would be less than 40, leading to a ratio different from 4:7 and a different transformed ratio after adding 8. Thus 40 is inconsistent with the given conditions.

Option B:
Option B, 48, might be guessed by partial calculation but does not satisfy the equation (4x + 8)/(7x + 8) = 5/8 when treated as the larger number. With 48 as the larger term, the corresponding smaller term from ratio 4:7 would be non-integral or lead to a different new ratio after adding 8. Hence 48 is not correct.

Option C:
Option C is correct because with x = 8 the original numbers are 4 × 8 = 32 and 7 × 8 = 56, which are in the ratio 4:7. After adding 8, we get 40 and 64, whose ratio is 40:64 = 5:8, exactly matching the condition. This confirms 56 as the unique value for the larger original number.

Option D:
Option D, 64, is the larger number after the increment rather than before it in the correct solution. Treating 64 as the original larger number would produce a much bigger value after adding 8 and the resulting ratio would no longer become 5:8. Thus 64 does not align with the stated transformation.

The series follows the formula aₙ = n³+2n² for n starting from 1. For n = 1, 2, 3, 4 and 5 we calculate 1+2 = 3, 8+8 = 16, 27+18 = 45, 64+32 = 96 and 125+50 = 175. For n = 6 the expression gives 216+72 = 288. Hence 288 is the term that correctly extends this cubic-plus-square pattern.

Option A:
Option A, 278, is 10 less than the required value and does not equal n³+2n² for n = 6. It breaks the exact numeric relationship that has generated all previous terms. Therefore 278 is not the correct continuation.

Option B:
Option B, 282, is closer but still not equal to 216+72. It would imply a small but unjustified alteration of the formula at the last step. Thus 282 does not fit the pattern.

Option C:
Option C, 296, exceeds the computed value and again does not come from n³+2n² when n = 6. Choosing 296 would destroy the perfectly consistent algebraic structure of the sequence. Hence it is not valid.

Option D:
Option D, 288, is exactly 216+72 and comes directly from the rule aₙ = n³+2n². It preserves both the cubic and quadratic contributions in the same way as before, making 288 the correct next term.

In this series, the first letter K remains fixed while the second and third letters move forward by two positions each time. The second letters L, N, P and R are at positions 12, 14, 16 and 18, and the third letters M, O, Q and S are at positions 13, 15, 17 and 19. Continuing this plus two progression, the next second and third letters should be T (20) and U (21), giving KTU.

Option A:
Option A, KUV, uses U and V for the second and third letters, which would be four positions ahead of R and S. This doubles the required step size and breaks the consistent plus two movement. As such, KUV does not conform to the series rule.

Option B:
Option B, KST, sets S and T in the second and third positions, which are only two and three positions ahead of R and S collectively. This creates unequal shifts and does not follow the clear parallel movement of both letters by two. Therefore KST is not an accurate continuation.

Option C:
Option C is correct because T and U are exactly two positions beyond R and S respectively, maintaining the same parallel increase. The first letter K is unchanged, reflecting the fixed initial position, and the second and third letters move in lockstep. This makes KTU the logical extension of the pattern.

Option D:
Option D, KTV, advances the second letter correctly to T but moves the third letter to V, which is three positions after S. This lack of parallelism in the movement contradicts the even shifting behaviour found in the series. Consequently, KTV cannot be chosen as the next term.

Each group is a five letter palindrome in which the first and last letters match, the second and fourth letters match and the centre letter is distinct. From APQPA to BQRQB to CRSRC to DSTSD, the outer letters move forward A, B, C, D, the inner pair moves forward P, Q, R, S and the central letter moves forward Q, R, S, T. To continue, the next outer letter should be E, the inner pair should be T and the middle letter should be U. This gives ETUTE, which maintains all three coordinated movements and the palindromic shape, making it the correct next term.

Option A:
Option A, ETUET, places T in the middle and U as the inner pair, reversing their roles compared with earlier groups. This contradicts the way the second and fourth positions move together and the middle letter moves separately. Because it changes the structure of the pattern, ETUET is not correct.

Option B:
Option B, ESVSE, keeps E at the ends but drops the inner pair back to S, which would undo the forward step from R to S already seen. The middle letter V does not follow directly from the previous T either. As it disrupts both inner and central progressions, ESVSE cannot be the answer.

Option C:
Option C, EUROE, breaks the rule that the second and fourth letters are the same by placing U and O there. The groups in the series have strictly mirrored second and fourth positions. Therefore EUROE does not follow the series structure.

Option D:
Option D is correct because ETUTE moves each of the three distinct positions forward by exactly one step from DSTSD while keeping the palindrome intact. The outer E’s, inner T’s and central U follow naturally from D, S and T respectively. This complete consistency with the multi level pattern makes ETUTE the only valid continuation.

Statements A, C and D give standard facts about quantifiers. A correctly identifies “for all n” as a universal quantifier. C is true because the negation of a universal statement is an existential statement with the predicate negated. D is also correct since negating an existential claim yields a universal statement with negated predicate. B is false because no natural number squared equals 2, and E is false because not all natural numbers are even. Therefore A, C and D only are correct.

Option A:
Option A is correct as it lists all the true statements: it recognises the universal quantification in A and the standard negation rules in C and D, while properly excluding B and E, which misstate facts about ℕ.

Option B:
Option B is incomplete since it omits D, ignoring the second important pattern for negating an existential statement. It therefore does not capture the full set of correct statements here.

Option C:
Option C is wrong as it drops A and includes only C and D, failing to mention the basic identification of a universal quantifier in statement A.

Option D:
Option D is incorrect because it adds E, which incorrectly claims that every natural number is even, so the combination contains a false universal statement.

For this series, the first letters are B(2), F(6), J(10) and N(14), each increasing by 4. The second letters are D(4), H(8), L(12) and P(16), also increasing by 4. The next first letter is 14 + 4 = 18, which is R, and the next second letter is 16 + 4 = 20, which is T. Thus RT is the pair that continues the uniform +4 progression in both positions.

Option A:
Option A is correct because both letters follow the established pattern of adding four positions. It places R after N and T after P, matching the arithmetic sequence. This preserves the symmetry between the first and second letters in the series.

Option B:
Option B, QS, has Q as the first letter, which is only 2 steps ahead of N, contradicting the +4 rule. Its second letter S is also only 2 steps beyond P. Since both gaps are too small, QS fails to maintain the consistent structure.

Option C:
Option C, RU, gives the correct first letter R but moves the second letter to U, which is 5 steps ahead of P. This breaks the equal step requirement for both letters. Because it inflates the second-letter increment, RU cannot be accepted.

Option D:
Option D, ST, advances the first letter to S, which is 5 positions ahead of N. This oversteps the observed increment and loses the alignment with the second letter’s movement. As a result, ST does not match the series rule.

To find the compound ratio p:s we multiply the fractional forms of the three given ratios: (3/5) × (10/11) × (4/7). The numerator becomes 3 × 10 × 4 = 120 and the denominator becomes 5 × 11 × 7 = 385. Simplifying 120/385 by dividing numerator and denominator by 5 gives 24/77. Therefore, the compound ratio p:s is 24:77.

Option A:
Option A, 12:35, could come from multiplying only part of the factors or simplifying incorrectly, for example by missing one of the denominators. If we back-calculate using 12:35, the intermediate ratios cannot be reconciled with 3:5, 10:11 and 4:7 simultaneously. Hence 12:35 does not reflect the full chained effect of all three ratios.

Option B:
Option B is correct because it directly results from the precise product (3/5) × (10/11) × (4/7) after proper simplification. Each given ratio is fully accounted for, and the cancellation is done using a common factor of 5, leading uniquely to 24:77. No other option preserves all three relationships in a single combined ratio.

Option C:
Option C, 24:49, keeps the correct numerator 24 but replaces 77 with 49, which is the denominator of only part of the product. This ignores the contribution of one of the original ratios when forming the compound. As a result, 24:49 cannot be the correct chain ratio from p to s.

Option D:
Option D, 30:77, uses the correct denominator but an inflated numerator, perhaps from erroneous multiplication 3 × 10 = 30 without considering the factor 4 and required simplification. When we test 30:77 against the given ratios, it fails to reproduce them, so 30:77 is not valid.

Statements A and B state the two standard De Morgan identities for sets. Statement C is correct because these laws explicitly relate complements of unions and intersections. Statement E is also true; in Venn diagrams one can see that the shaded regions for both sides of each law coincide. Statement D is false since De Morgan’s laws hold in general, not only for finite sets. Hence the correct set of statements is A, B, C and E only, corresponding to option A.

Option A:
Option A is correct because it collects the two algebraic forms of the laws, their conceptual role and their Venn-diagram visualisation, while rightly excluding D, which wrongly restricts them to finite sets.

Option B:
Option B is incomplete as it omits E and so fails to mention the important geometric interpretation in Venn diagrams, which is often tested in NET questions on sets.

Option C:
Option C is incorrect because it includes D, the false claim about failure for infinite sets, and so mixes correct principles with an incorrect limitation.

Option D:
Option D is wrong since it leaves out A and picks only B, C and E; without A, one of the two core algebraic identities of De Morgan’s laws is missing, so the description is not complete.

Starting letters D(4), G(7), J(10), M(13) increase by +3. Next start is P(16), giving PQRS.

Option A:
Option A is correct because P is +3 from M and Q,R,S follow consecutively.

Option B:
Option B starts at Q (+4 from M), not +3.

Option C:
Option C starts at R (+5 from M), not +3.

Option D:
Option D starts at O (+2 from M), not +3.

In this series, each position in the triplets increases by one letter from term to term. First letters: A, B, C, D move to E; second letters: B, C, D, E move to F; third letters: O, P, Q, R move to S. Combining these gives the next group as EFS. Thus EFS preserves the simultaneous +1 shift in all three positions.

Option A:
Option A, EFT, has the correct first two letters E and F but changes the third letter to T instead of S. The third letters should increase in order O, P, Q, R, S, not jump to T. Because it breaks the regular one-step progression in the third position, EFT is not correct.

Option B:
Option B, EFR, keeps E and F but reuses R as the third letter. The third position should move from R to S, not stay at the same letter. Repetition of R violates the strictly increasing pattern in the last position, so EFR cannot be the right continuation.

Option C:
Option C is correct because E, F and S are each exactly one step ahead of D, E and R respectively. It respects the coordinated forward shift in all three positions, which is the hallmark of this series. Therefore EFS is the only triplet that fits the established pattern.

Option D:
Option D, EGT, alters both the second and third positions by more than one step. The series does not support unequal or larger jumps in any position. As it disrupts the neat parallel movement of all three letters, EGT is not a valid next term.

In this series, each new group is obtained by shifting every letter of the previous group one step forward in the alphabet. ACDF becomes BDEG, then CEFH, then DFGI by this uniform +1 movement in all positions. Applying the same rule once more, D becomes E, F becomes G, G becomes H and I becomes J. Therefore EGHJ is the only group that correctly continues the pattern.

Option A:
Option A is correct because it preserves the exact transformation seen between all consecutive terms. Every letter is increased by one step from the previous group DFGI, giving E, G, H and J in order. This keeps the structure and spacing of the letters unchanged, so EGHJ fits perfectly as the next term.

Option B:
Option B, EGHI, keeps the first three letters correctly but leaves the last letter as I instead of advancing it to J. This breaks the rule that every position must move forward by one step. Because the last letter does not follow the established shift, EGHI cannot be accepted as the next term.

Option C:
Option C, EFGJ, changes the second letter incorrectly to F, which does not come from shifting F in the previous group. It also disturbs the internal spacing between letters, which has been maintained in all earlier terms. Since it fails to apply the uniform +1 movement to every position, EFGJ does not fit the series.

Option D:
Option D, FGHI, looks like a familiar consecutive pattern but is not derived from DFGI by adding one to each letter. It replaces D with F, which is a jump of two positions, and creates a different spacing between letters. As it does not follow the consistent transformation rule, FGHI is not the correct continuation.

When two fair coins are tossed, the sample space consists of four equally likely outcomes: HH, HT, TH and TT. Exactly one head occurs in the outcomes HT and TH, giving 2 favourable cases. Therefore, the probability of exactly one head is 2/4, which simplifies to 1/2. Thus, 1/2 is the correct probability.

Option A:
A probability of 1/4 would correspond to just one favourable outcome out of four, but in this experiment there are two sequences that show exactly one head. Hence, 1/4 undercounts the favourable cases and is incorrect.

Option B:
1/2 is correct because the two favourable outcomes HT and TH out of four total outcomes give 2/4, which reduces to 1/2. This demonstrates a straightforward application of the probability formula for equally likely outcomes. Understanding such simple models helps in solving more advanced probability questions.

Option C:
3/4 would mean that three out of four outcomes have exactly one head, which is not true here. In reality, only two outcomes satisfy this condition, while HH and TT do not. Therefore, 3/4 exaggerates the likelihood and cannot be right.

Option D:
1/3 suggests an assumption of three equally likely outcomes, only one of which is favourable, which does not match the actual sample space of four outcomes. Thus, 1/3 is not compatible with the correct analysis of this experiment.

In commercial mathematics, if the selling price (SP) of an item is greater than its cost price (CP), the trader gains money on the transaction. This gain is called profit and is calculated as SP − CP. Profit indicates that the trader has sold the item for more than it cost to acquire. Therefore, the situation described in the question corresponds exactly to making a profit.

Option A:
Loss occurs when the cost price exceeds the selling price, meaning the trader receives less money than was spent on purchasing the item. This is the opposite situation to what is described in the question, where selling price is higher than cost price. Hence, loss cannot be the correct answer here.

Option B:
Profit is correct because it is the name given to the positive difference SP − CP when SP > CP. For example, if an article costs ₹500 and is sold for ₹600, the trader makes a profit of ₹100. Recognising this condition quickly is useful in many UGC NET aptitude problems involving multiple stages of buying and selling.

Option C:
A discount is a reduction given on the marked price of an item, usually to attract customers, and is calculated before determining the final selling price. While discounts can affect profit, the term itself does not directly describe the situation SP > CP. Therefore, discount does not match the definition in the stem.

Option D:
Commission is a fee or percentage of sales paid to an agent or salesperson for services rendered. It is not the difference between selling price and cost price. Since the question focuses on the trader's gain from SP exceeding CP, commission is not the appropriate term.

Each new group is formed by adding +1 to every letter: CAM→DBN→ECO→FDP. Next is FDP shifted to GEQ.

Option A:
Option A changes the second letter incorrectly and does not apply +1 uniformly.

Option B:
Option B uses D as the middle letter, which is not +1 from D in FDP (it should become E).

Option C:
Option C ends with R, which is +2 from P, not +1.

Option D:
Option D is correct because F→G, D→E, P→Q.