UGC NET Questions (Paper – 1)

Statements A, B, D and E highlight the main types of patterns used in number series questions and the techniques to detect them, so they are all correct. Arithmetic and geometric patterns are common, alternating patterns occur frequently, examining differences or ratios is a standard technique and many series use squares, cubes or factorials. Statement C is false because UGC NET series questions can use a wide variety of non-uniform rules, not just simple fixed additions. Therefore, the combination that includes A, B, D and E but excludes C is the only fully correct set.

Option A:
Option A is incorrect because although A, B and D are true, this option omits E, which correctly notes that higher powers and factorials often underlie series in competitive exams. Leaving out E makes the description incomplete and therefore not the best answer.

Option B:
Option B is correct as it brings together all the true statements about typical patterns and solution strategies while excluding the restrictive and incorrect claim in C. It reflects the variety of rules that a NET aspirant should expect in number series questions.

Option C:
Option C is wrong because it includes C, which wrongly asserts that every NET series is just fixed addition, and it omits A, so it fails to state explicitly that arithmetic and geometric patterns are central. This mixture of a false statement and an omission makes the option invalid.

Option D:
Option D is incorrect since it also contains C, the false statement about fixed addition, and does not include B, ignoring alternating or multi-step patterns. By endorsing C, this option misleads students about the range of series types in exams.

The least common multiple (LCM) of two numbers is the smallest positive integer that is divisible by both. Listing multiples, we get for 6: 6, 12, 18, 24, ... and for 8: 8, 16, 24, 32, ... . The first common multiple is 24. Therefore, the LCM of 6 and 8 is 24.

Option A:
12 is a multiple of 6, but it is not a multiple of 8 since 12 ÷ 8 is not an integer. Therefore, 12 cannot be a common multiple of both numbers. This makes it an invalid candidate for their LCM.

Option B:
18 is a multiple of 6 but not of 8, because 18 ÷ 8 is not integral. Thus, it fails the requirement of being divisible by both numbers. Hence, this option is incorrect.

Option C:
24 is correct because it is the smallest number that appears in the multiples of both 6 and 8. Additionally, 24 ÷ 6 = 4 and 24 ÷ 8 = 3, confirming it is divisible by both. This value is therefore the least common multiple sought in the question.

Option D:
36 is a multiple of 6 but not a multiple of 8 since 36 ÷ 8 does not give an integer. Consequently, it cannot be considered a common multiple of 6 and 8. Thus, 36 is not the LCM in this case.

When three fair coins are tossed, there are 2³ = 8 equally likely outcomes. The easiest way to find the probability of at least one head is to subtract the probability of getting no heads (i.e., all tails) from 1. The probability of all tails is 1 outcome (TTT) out of 8, which is 1/8. Therefore, the probability of at least one head is 1 − 1/8 = 7/8.

Option A:
Option A, 1/8, gives the probability of the complementary event (no head) rather than at least one head. It represents the chance of all tails.

Option B:
Option B, 3/8, might come from counting only single-head outcomes and ignoring cases with two or three heads, so it underestimates the event.

Option C:
Option C, 5/8, still misses some favourable outcomes and does not match the simple complement calculation.

Option D:
Option D correctly uses the complement rule and recognizes that all outcomes except the single all-tail result satisfy “at least one head,” leading to 7/8.

These are triangular numbers formed by adding consecutive natural numbers: 1, 1+2 = 3, 1+2+3 = 6, 1+2+3+4 = 10 and 1+2+3+4+5 = 15. The next triangular number is 1+2+3+4+5+6 = 21. Therefore, 21 is the correct next term.

Option A:
Option A is 16, which does not arise from summing the first 6 natural numbers. It is only one more than 15 and ignores the triangular pattern. Hence, it is not correct.

Option B:
Option B is 18, which is between 15 and 21 but not equal to the next triangular value. Including 18 would make the pattern unclear. Therefore, this option is unsuitable.

Option C:
Option C equals 21, which is the sum of the first 6 positive integers and matches the known triangular number T₆. The series 1, 3, 6, 10, 15, 21 stays consistent with this definition. Thus, 21 is the valid continuation.

Option D:
Option D is 24, which is not a triangular number at this position. While 24 has other mathematical roles, it does not fit the specific summation pattern here. So, it cannot be chosen.

The groups descend in blocks of three letters: XYZ (24–26), UVW (21–23), RST (18–20) and OPQ (15–17). Each starting letter reduces by 3 positions (24, 21, 18, 15). The next starting position should be 12, corresponding to L, and the next block is LMN (12–14). Hence LMN is the only group that continues the pattern of moving each block three letters earlier in the alphabet.

Option A:
Option A, LMN, is correct because it begins at L, which is 3 letters before O, and forms the next three-letter block in descending order. It respects the length of the groups and the step size of −3 for the starting letter. This makes LMN a perfect fit for the series.

Option B:
Option B, KLM, starts too early at K, corresponding to a step of −4 from O, not −3. This over-adjustment does not match the observed behaviour of the series. As such, KLM cannot be considered correct.

Option C:
Option C, NOP, reverses the direction by moving forward after OPQ instead of backward. It also overlaps with OPQ by reusing O and P in a way that contradicts the blockwise descent. Therefore NOP does not align with the series rule.

Option D:
Option D, MNO, starts at M, which is 2 letters before O, not 3 letters before O. It breaks the fixed interval between the initial letters of successive blocks. Because it ignores the −3 step pattern, MNO is not a valid continuation.

Statement A is correct because grids structure the information. Statement B is true since elimination narrows options. Statement D is also true because final assignments must satisfy all clues simultaneously. Statement C is false (not ruled out ≠ definitely true), and E is false (deduction is central). Therefore, A, B and D only are correct.

Option A:
Option A is incomplete because it omits D (consistency requirement).

Option B:
Option B is correct as it collects all true statements and excludes C and E.

Option C:
Option C is wrong because it includes C (false).

Option D:
Option D is incorrect because it includes E (false) and omits B.

Simple interest is calculated using the formula I = PRT/100, where P is principal, R is rate and T is time in years. Substituting P = 5000, R = 8 and T = 3 gives I = 5000 × 8 × 3 / 100 = 5000 × 24 / 100 = 1200. This shows that the total interest earned over 3 years is ₹1200. Hence, 1200 is the correct answer.

Option A:
An interest of ₹1000 would correspond to different values of rate or time, such as 5000 × 5 × 4 / 100, and does not match the given data. If we plug 5000, 8% and 3 years into the formula, we do not obtain 1000. Therefore, this option is inconsistent with the standard simple interest calculation.

Option B:
₹1100 is close to the correct value but still underestimates the interest. It might result from an incorrect intermediate step or rounding, but simple interest at whole percentages with these numbers does not require rounding. Since the exact formula gives ₹1200, this option is incorrect.

Option C:
₹1150 again does not align with the precise formula I = PRT/100. No simple substitution of P=5000, R=8 and T=3 yields 1150 as the result. Therefore, this value cannot be justified mathematically in this context.

Option D:
₹1200 is correct because the multiplication 5000 × 8 × 3 equals 120000, and dividing by 100 gives 1200. This straightforward computation demonstrates the direct application of the simple interest formula. In exam settings, carefully performing such calculations avoids avoidable errors.

Statements A, B, C and E correctly describe geometric aspects of inequalities. A states that a single linear inequality shades a half-plane. B and C distinguish between including and excluding the boundary line based on the inequality sign. E is true because shading regions on the Cartesian plane is a standard way to solve systems of inequalities. D is false because the intersection of inequalities means satisfying all of them, not just at least one. Thus A, B, C and E only are correct.

Option A:
Option A is incomplete as it excludes E and therefore misses the point that these inequalities are often solved graphically in exam problems.

Option B:
Option B is also incomplete since it omits C and thereby fails to stress the difference between strict and non-strict inequalities with respect to boundary lines.

Option C:
Option C is wrong because it omits A, which states the key half-plane interpretation.

Option D:
Option D is correct because it includes every true statement about the geometric representation and graphical solution of inequalities, while excluding D, which confuses intersection with union.

A’s marks are 72% of 750, which is (72/100) × 750 = 540. B’s marks are 84% of 750, which is (84/100) × 750 = 630. The ratio A:B is 540:630. Dividing both numbers by 90 gives 6:7. Hence, the ratio of A’s marks to B’s marks is 6:7.

Option A:
Option A is correct because it simplifies the raw scores 540 and 630 using their highest common factor. Dividing both by 90 yields exactly 6:7, which preserves the proportion while using smaller, more convenient numbers. This process directly matches standard ratio reduction techniques.

Option B:
Option B, 7:8, might be guessed from the numerical percentages 72% and 84%, but the percentages cannot be compared directly without considering the total marks. When we compute the actual scores, the ratio 540:630 does not simplify to 7:8, so this option is incorrect.

Option C:
Option C, 4:5, would require A’s marks to be four-fifths of B’s marks. If B had 630, then A would need to have 504, not 540. Since A’s actual score is higher than 504, the ratio 4:5 understates A’s relative performance and does not match the data.

Option D:
Option D, 9:10, suggests a nearly equal performance, but if we scale 540 and 630 to 9 and 10, we find that 540 × 10 ≠ 630 × 9. This shows that 9:10 is not equivalent to 540:630, so it cannot represent their actual mark ratio.

Each new term is obtained by doubling the previous term and then adding consecutive odd numbers. From 2 to 5 we apply 2×2+1, from 5 to 13 we use 5×2+3, from 13 to 31 we use 13×2+5, and from 31 to 69 we use 31×2+7. The series of added odd numbers is 1, 3, 5, 7, so the next odd number must be 9. Therefore the next term is 69×2+9 = 147.

Option A:
Option A would correspond to 69×2+5, reusing an earlier odd adjustment instead of moving to 9. This would make the added sequence 1, 3, 5, 7, 5, which does not show a clear progression. As such, 143 does not respect the steadily increasing odd-number adjustment rule.

Option B:
Option B equals 141, which would arise from 69×2+3. That uses an even smaller odd addition than in the previous step, reversing the pattern 1, 3, 5, 7. Hence 141 cannot be justified by the observed rule.

Option C:
Option C uses 69×2+9, giving 147 and continuing the series of added odd numbers as 1, 3, 5, 7, 9. This maintains both the doubling structure and the simple progression among the adjustments. Therefore 147 is the correct next term.

Option D:
Option D, 151, would require adding 13 to 69×2, jumping from an added 7 directly to 13 and skipping 9 and 11. This breaks the neat sequence of consecutive odd numbers that defines the pattern. Thus 151 is not a valid extension.

At 3:30, the minute hand is at 30 minutes, which is 30 × 6 = 180 degrees from 12 o’clock, since each minute corresponds to 6 degrees. The hour hand moves 30 degrees per hour plus 0.5 degrees per minute. By 3:30 it has moved 3 × 30 = 90 degrees for three full hours and 0.5 × 30 = 15 degrees for the extra 30 minutes, totalling 105 degrees. The angle between the two hands is |180 − 105| = 75 degrees, so 75 degrees is the correct answer.

Option A:
Option A, 45 degrees, might be chosen by wrongly assuming that the hour hand stays fixed exactly at 3 when the minute hand is at 6, but in reality the hour hand has advanced halfway towards 4 by 3:30.

Option B:
Option B, 60 degrees, would correspond to some other specific times but does not result from the correct calculation of the hands’ positions at 3:30. It underestimates the actual separation once the hour hand’s extra movement is considered.

Option C:
Option C carefully computes both hands’ positions using their respective angular speeds and then subtracts to find the included angle. This process gives 75 degrees, matching the standard result for this clock-angle case.

Option D:
Option D, 90 degrees, assumes that the hands are perpendicular at 3:30, but they are exactly perpendicular at 3:00; by 3:30 the hour hand has moved ahead, so the angle is less than 90 degrees.

Statements A and B give the standard definitions of injective and surjective functions. Statement C is also correct because a bijection is precisely a function that is both one–one and onto. Statement D is false as there are surjective functions that are not injective when the codomain is smaller than the domain. Statement E is false because the definition of a function requires exactly one output in B for each input in A. Hence, A, B and C only are correct, which matches option C.

Option A:
Option A is incomplete since it omits C and thereby fails to mention bijection, which combines the properties described in A and B and is often tested in NET.

Option B:
Option B is incorrect as it adds D, which wrongly claims that surjectivity implies injectivity, so it contains an incorrect statement along with true ones.

Option C:
Option C is correct because it precisely gathers the three accurate definitions and excludes D and E, which misrepresent either implications between properties or the basic definition of a function.

Option D:
Option D is wrong since it includes E, which permits multiple outputs per input, contradicting the concept of a function, and omits A, so it removes a key correct statement and accepts a false one.

The given terms are 1², 2², 3², 4² and 5², showing a sequence of consecutive perfect squares. Following this pattern, the next square should be 6². Computing 6² gives 36, so 36 must be the next term. This preserves the structure of consecutive squares.

Option A:
Option A is 30, which is not a perfect square of any integer. It does not follow from squaring the next whole number after 5. Thus, it fails to continue the pattern correctly.

Option B:
Option B is 32, which lies between 5² and 6² but is not itself a square of an integer. Because the rule involves exact squares in order, 32 cannot be accepted.

Option C:
Option C equals 36, which is 6² and follows naturally after 5² = 25. The extended series 1, 4, 9, 16, 25, 36 clearly shows consecutive squares of 1, 2, 3, 4, 5 and 6. Therefore, this choice correctly matches the pattern.

Option D:
Option D is 34, which again is not a perfect square and has no special relation to the square sequence here. Including 34 would break the logical connection between terms. Hence, it is not the correct answer.

Let the original numbers be 7x and 4x. After subtracting 5 from each, they become 7x − 5 and 4x − 5, and their ratio is 3:1. Thus, (7x − 5)/(4x − 5) = 3/1. This gives 7x − 5 = 3(4x − 5) = 12x − 15, which simplifies to 10 = 5x and hence x = 2. Therefore, the original numbers are 14 and 8, and the larger is 14.

Option A:
Option A, 8, is the smaller of the two numbers in the correct solution. While it satisfies the original ratio when combined with 14, the question asks for the larger number. Choosing 8 would misinterpret which quantity the problem seeks.

Option B:
Option B is correct because 14 equals 7x with x = 2 and is the larger term in the ratio 7:4. Substituting 14 and 8 into the transformed condition, we get (14 − 5):(8 − 5) = 9:3 = 3:1, which exactly matches the new ratio, confirming that 14 is the correct larger original number.

Option C:
Option C, 16, would require the smaller number to be 16 × (4/7) ≈ 9.14 to preserve the ratio 7:4. This is not an integer and, even if approximated, would not yield a new ratio of 3:1 after subtracting 5. Thus, 16 cannot be the larger number that fits all conditions.

Option D:
Option D, 18, similarly leads to a non-integer partner of 18 × (4/7) ≈ 10.29 under the original ratio and fails to produce the required new ratio of 3:1 after the subtraction. Hence, 18 is inconsistent with the given information.

A statement that is always true under every possible assignment of truth values to its components is called a tautology. The disjunction “p or not p” covers all possibilities: if p is true, the first part is true; if p is false, not p is true. Thus, in all cases the compound statement evaluates to true. This makes “p or not p” a classic example of a tautology.

Option A:
Option A, “p and not p,” is instead a contradiction because it asserts that p is both true and false simultaneously. There is no assignment of truth values that can make both p and not p true together, so this statement is never true.

Option B:
Option B, “p or not p,” correctly spans all truth possibilities for p. Since one of p or not p must be true, their disjunction is always true, demonstrating tautological status.

Option C:
Option C, “p → not p,” is not always true. If p is true, the implication would require not p to be true as well, which is impossible, so the statement can be false.

Option D:
Option D, “not p → p,” also fails to be always true; there are assignments where it becomes false, so it is not a tautology.