UGC NET Questions (Paper – 1)

Statements A, B and E are correct because inductive reasoning generalises from specific instances, deductive reasoning applies general rules to particular cases and many analogy questions involve inferring general pattern relationships from given examples. Statement C is false since inductive arguments support conclusions with probability, not logical guarantee, and D is false because valid deductive arguments provide necessity rather than mere probability when premises are true. Therefore, the only combination that collects exactly the true statements and excludes the false ones is A, B and E.

Option A:
Option A is incomplete because although A and B are true, it omits E, which correctly notes the inductive character of many analogy items in aptitude tests. Leaving out E fails to recognise an applied aspect of inductive reasoning important for UGC NET.

Option B:
Option B is correct as it gathers the three true statements and avoids both C and D, each of which misdescribes the strength of inductive and deductive arguments. It properly distinguishes probability in induction from necessity in valid deduction.

Option C:
Option C is wrong because it includes C, which incorrectly claims that inductive reasoning guarantees truth, and also excludes B, so it does not fully capture how deductive reasoning works in deriving specific conclusions from general premises. The presence of C makes the combination unacceptable.

Option D:
Option D is incorrect because it includes D, which wrongly asserts that deductive arguments never aim at necessity, and it omits A, ignoring the fundamental description of inductive reasoning. Thus it mixes a false statement with an incomplete set of true ones.

A, B and C are correct standard facts: absolute value measures distance, ∣x∣ < 3 implies −3 < x 4 implies x 4. D is false because absolute value cannot be negative, so ∣x∣ = −2 has no real solution. E is false because ∣x∣ = 5 has two real solutions, x = 5 and x = −5, not exactly one. Therefore, the correct set is A, B and C only.

Option A:
Option A is incomplete because it omits C, which is also a correct rule for solving strict “greater than” absolute value inequalities.

Option B:
Option B is incorrect because it includes E, which is false since ∣x∣ = 5 has two solutions, not one.

Option C:
Option C is correct because it lists exactly the true statements (A, B, C) and excludes both false statements (D, E).

Option D:
Option D is incorrect because it includes D (which is impossible) and E (which is also false), so it mixes false statements with true ones.

To find 15% of 200, we interpret 15% as 15 out of 100. The calculation is (15 ÷ 100) × 200. Simplifying, 15 × 2 = 30, so the result is 30. Thus, 15% of 200 equals 30.

Option A:
Option A, 25, corresponds to 12.5% of 200 and arises if someone confuses 1/8 with 15%. It does not follow from the correct percentage calculation.

Option B:
Option B matches the formula for a percentage of a quantity and uses the simple fraction representation of 15%. The arithmetic shows that 30 is the exact proportion of 200 described by 15%.

Option C:
Option C, 35, is greater than 30 and would correspond to 17.5% of 200, not 15%. It therefore overestimates the desired portion.

Option D:
Option D, 40, equals 20% of 200 and is again inconsistent with the requested 15% rate.

The third proportional c to numbers a and b is defined by a:b = b:c. For a = 5 and b = 20, we have 5:20 = 20:c. Writing this as a fraction gives 5/20 = 20/c, so c = (20 × 20)/5 = 400/5 = 80. Thus, 80 is the third proportional to 5 and 20.

Option A:
Option A, 40, produces the ratio 20:40 = 1:2, whereas 5:20 = 1:4. Since these ratios are not equal, 40 cannot be the third proportional under the definition a:b = b:c.

Option B:
Option B, 64, might emerge from mistakenly squaring only part of a fraction or choosing a convenient number, but 5:20 = 1:4 and 20:64 = 5:16, which do not match. Therefore, 64 does not satisfy the required proportional relationship.

Option C:
Option C is correct because with c = 80 we get 5:20 = 1:4 and 20:80 = 1:4. The equality of these ratios confirms that 80 is the valid third proportional to 5 and 20.

Option D:
Option D, 100, yields 20:100 = 1:5, which differs from 1:4, the simplified form of 5:20. While 100 is a round number, it does not preserve the proportional structure defined in the question.

Statements A and B give the usual representation of individual and combined work rates in time and work problems. Statement D is also correct because increasing the number of equally efficient workers generally reduces the time for a fixed job when total work is unchanged. Statement C is false, since a faster worker takes less time, and E is false because alternate-day work patterns depend on how many days are counted and cannot always be reduced to a simple average. Hence the correct set of statements is A, B and D only, matching option B.

Option A:
Option A is incomplete as it leaves out D, thereby ignoring the inverse relation between number of workers and time, which is a standard pattern tested in exam questions.

Option B:
Option B is correct because it collects the ideas of one-day work, combined rate and inverse proportionality while excluding C and E, which misstate how speed affects time and oversimplify alternate working-day scenarios.

Option C:
Option C is wrong since it brings in E, which incorrectly claims a simple average works in all alternate-day situations, and it omits B, so it both adds an error and loses a fundamental rule.

Option D:
Option D is incorrect because it includes C, which reverses the comparison of fast and slow workers, and it omits A, dropping the basic 1/x representation of one-day work.

B gives the standard combined-rate formula. C is also correct: independent daily rates add. A is false because working together they finish faster than either alone, so the time is less than 6 days (and certainly not greater than 10). D is false because doubling workers halves time only under ideal assumptions, not “always”. E is false because different workers can have different speeds. Thus only B and C are correct.

Option A:
Option A is incomplete because it omits C, the general rate-addition rule for more than two workers.

Option B:
Option B is correct because it includes exactly the true rate principles (B and C) and excludes A, D and E, which are false generalisations.

Option C:
Option C is wrong because it includes A, which incorrectly claims the combined time is greater than 10 days.

Option D:
Option D is incorrect because it includes D, which is not universally true, even though B and C are true.

Statements A, B, C and D correctly describe bar graphs and histograms. A states the proportional representation of values by bar lengths. B is true because bar graphs are suited to comparing discrete categories such as years or regions. C accurately distinguishes histograms as bar-like representations of continuous class intervals with no gaps. D is correct since in practice examinees sometimes estimate values when scale divisions are not fine. E is false because both vertical and horizontal bar graphs are acceptable ways to present data. Therefore, A, B, C and D only are correct.

Option A:
Option A is correct because it brings together the essential properties and uses of bar graphs and histograms and leaves out E, which wrongly forbids horizontal bars. It reflects the standard data interpretation conventions in NET examinations.

Option B:
Option B is incomplete because it omits D, failing to mention the realistic need for approximation when reading from coarse scales, which is a practical aspect of graph use. This omission makes the option less comprehensive.

Option C:
Option C is incorrect as it includes only B, C and D and omits A, so it never states the basic proportionality between bar length and value. Without A, the central representational feature of bar graphs is missing.

Option D:
Option D is wrong because it contains E, which incorrectly bans horizontal bar charts, and drops B, so it both introduces an error and loses a key comparative property of bar graphs.

Conditional probability is written using a vertical bar “|”. P(A|B) means “the probability of A given that B has already occurred,” i.e., the sample space is restricted to outcomes where B happens.

Option A:
P(A ∩ B) is the probability that both A and B occur together (intersection). It does not represent “given B occurred,” so it is not conditional probability.

Option B:
P(A ∪ B) is the probability that A or B (or both) occur (union). It does not express “given B,” so it is not the required notation.

Option C:
P(A|B) is correct because it explicitly means “probability of A given B.” This is the standard notation for conditional probability in probability theory and statistics.

Option D:
P(B|A) means “probability of B given A,” which reverses the condition. The question asks for “A given B,” so this is not correct.

The ratio 3 : 5 means that for every 3 parts of the smaller number there are 5 parts of the larger number. If 3 parts correspond to 24, then 1 part is 24 ÷ 3 = 8. The larger number has 5 such parts, so it is 5 × 8 = 40. Thus, a pair of numbers 24 and 40 has the ratio 3 : 5.

Option A:
Option A, 30, would make the ratio 24 : 30, which simplifies to 4 : 5, not 3 : 5. So it does not preserve the given ratio between the numbers.

Option B:
Option B, 36, gives a ratio of 24 : 36, which reduces to 2 : 3. This is again different from the required 3 : 5 and does not meet the condition.

Option C:
Option C, 40, is obtained by scaling the ratio correctly: each part is 8 and the larger number gets 5 parts. This respects the original ratio 3 : 5 and maintains the exact proportional relationship.

Option D:
Option D, 45, would produce 24 : 45, which does not simplify to 3 : 5 and therefore cannot be accepted as consistent with the given ratio.

Statements A, C and E are correct: surds are indeed irrational roots, rationalising eliminates surds from denominators and simple index laws such as a⁰ = 1 are frequently used to simplify expressions in aptitude questions. Statement B is false because the rule √a × √b = √(ab) holds only when a and b are non-negative, not for all real numbers. Statement D is also false as the correct product rule is aᵐ × aⁿ = aᵐ⁺ⁿ, not aᵐ⁻ⁿ. Thus, the combination consisting of A, C and E only is correct.

Option A:
Option A is incomplete since it omits E and therefore does not mention how index laws are applied in practice to simplify complex expressions in exam questions. Without E, the connection to aptitude simplification is weaker.

Option B:
Option B is wrong because it includes D, which misstates the product law for exponents, and leaves out E, so it both accepts a wrong algebraic rule and omits a true practical fact. This combination cannot be correct.

Option C:
Option C is correct as it contains exactly those statements that align with standard definitions and simplification techniques while excluding B and D, both of which misapply or overgeneralise algebraic rules. It provides the right conceptual picture for NET-level problems.

Option D:
Option D is incorrect because it includes B, which fails to note that the square-root product rule requires non-negative radicands, and it omits E, again ignoring the exam relevance of simple index laws. This mixture makes the option unsatisfactory.

In these three-letter groups, the last two letters “AT” remain constant while only the first letter changes. The first letters form the sequence R, S, T, U, which are consecutive in the alphabet. The next letter after U is V. Hence VAT is the term that continues the pattern of increasing the first letter by one while keeping “AT” fixed.

Option A:
Option A is correct because it respects both parts of the pattern simultaneously. It uses V as the next letter after U and retains the common ending “AT” without modification. This makes VAT a natural and unique extension of the given series.

Option B:
Option B, WAT, would be the next term if the series continued beyond VAT, but it skips the immediate successor required here. It jumps directly to W as the first letter, ignoring V, and therefore does not represent the very next element in the sequence. This violation of immediate succession makes WAT incorrect.

Option C:
Option C, RAA, changes both the first and last two letters, breaking the stable suffix “AT” that all terms share. It also does not follow the alphabetical order of the first letters. Since it fails to preserve the key structural feature of the series, RAA is not acceptable.

Option D:
Option D, RAT, repeats the very first term of the sequence without any forward progression. The pattern is clearly based on moving the first letter forward, not on cycling back prematurely. Therefore RAT cannot serve as the next term in this context.

The differences between consecutive terms are 9, 13, 17 and 21. These differences form an arithmetic sequence increasing by 4 each time. Continuing this rule, the next difference should be 25. Adding 25 to the last term 65 gives 90, which fits perfectly with the established pattern. Thus, 90 is the term that correctly continues the series.

Option A:
Option A gives 90, which is 25 more than 65 and retains the difference progression 9, 13, 17, 21, 25. This keeps the second-level pattern of increasing differences completely intact. Therefore, 90 is the most appropriate choice for the next term in the series.

Option B:
Option B provides 88, which corresponds to a difference of 23 from 65 and does not match the expected increase of 4 in the difference sequence. Using 23 would break the systematic growth in the differences. Hence, 88 is not consistent with the observed pattern.

Option C:
Option C suggests 92, implying a difference of 27 from 65, which overshoots the expected next difference of 25. This disrupts the neat arithmetic progression in the differences. Therefore, 92 cannot be regarded as the correct continuation.

Option D:
Option D proposes 94, giving a difference of 29 from 65, moving even further away from the expected 25. Such a jump fails to respect the structure of the series and makes the pattern irregular. Thus, 94 is not a valid answer.

The differences between the terms are 3, 5, 7, 11 and 13. These differences are prime numbers in increasing order. The next prime after 13 is 17. Adding 17 to the last term 41 gives 58. This continues the prime-based difference pattern consistently.

Option A:
Option A gives 56, which yields a difference of 15 from 41 and is not a prime, so it conflicts with the rule that differences are prime numbers. Hence, 56 is not consistent with the underlying structure.

Option B:
Option B yields 58, whose difference from 41 is 17, the next prime expected in the pattern. The extended differences 3, 5, 7, 11, 13, 17 form a coherent prime-based second-level sequence. Therefore, 58 is the correct next term.

Option C:
Option C suggests 60, giving a difference of 19, which is a prime but would skip 17 and break the specific order in which primes are being used here. This leads to an inconsistency in how the differences evolve. Thus, 60 is not the right continuation.

Option D:
Option D offers 62, which results in a difference of 21 from 41, not a prime but a composite number. This choice would discard the key property that defines the pattern. Therefore, 62 cannot be accepted as correct.

Each term of this series can be expressed as n!+1 for n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 1!+1 = 2, 2!+1 = 3, 3!+1 = 7, 4!+1 = 25 and 5!+1 = 121. For n = 6 the same factorial rule gives 6!+1 = 720+1 = 721. Therefore 721 is the only value that keeps the factorial-based pattern intact.

Option A:
Option A, 701, is much smaller than 6!+1 and does not correspond to n!+1 for any integer n following 5. It breaks the rule that clearly generates the given terms. Thus 701 cannot be the correct next term in the sequence.

Option B:
Option B, 709, is closer numerically but still fails to equal 6!+1. It is not of the form n!+1 for the next natural number and would therefore require changing the underlying idea of the series. Hence 709 is not appropriate.

Option C:
Option C, 721, exactly matches the value obtained when we compute 6!+1. It preserves the simple and elegant rule that each term is one more than the factorial of its index. For this reason, 721 is the correct continuation of the series.

Option D:
Option D, 736, exceeds the correct factorial-based value and likewise cannot be written as n!+1 for the required n. Choosing 736 would abandon the factorial structure, so it is not a valid answer.

D is wrong because coding–decoding and analogy questions are an integral part of reasoning and are explicitly included under mathematical reasoning and aptitude in UGC NET Unit 5. These items require recognising and applying patterns, which is a key aspect of mathematical thinking. Statements A, B, C and E are all correct descriptions: coding relies on replacement by rules, solving requires discovering that rule, analogies use one pair to transfer a relationship to another, and letter codes often depend on alphabet positions or shifts. Hence, D alone is the incorrect statement.

Option A:
Option A is correct because it singles out D as the statement that misrepresents the syllabus and nature of reasoning questions, while leaving the accurate descriptions of coding and analogies untouched. It reflects the exam’s emphasis on pattern-based reasoning.

Option B:
Option B is incorrect because it treats C, which correctly explains the essence of analogy questions, as if it were also wrong. Analogies indeed work by mapping a known relationship onto a new pair.

Option C:
Option C is wrong since it marks B as incorrect along with D, even though B rightly points out that identifying the substitution pattern is crucial in coding–decoding. Ignoring this fact undermines the understanding of how such items are solved.

Option D:
Option D is also incorrect because it adds A to the set of wrong statements, even though A correctly states that coding–decoding involves replacing a word or number according to some rule. This option therefore misclassifies a core feature of coding questions.