UGC NET Questions (Paper – 1)

The average of four numbers being 26 means that their total sum is 26 × 4 = 104. The sum of the three known numbers 20, 25 and 29 is 20 + 25 + 29 = 74. The fourth number must make the overall sum 104, so it is 104 − 74 = 30. Therefore, the missing number is 30, which is Option B.

Option A:
Option A, 25, would lead to a total of 20 + 25 + 29 + 25 = 99, and 99 ÷ 4 is 24.75, which is less than 26, so the average condition fails.

Option B:
Option B, 30, makes the sum of the four numbers 104, which divided by 4 gives exactly 26, perfectly satisfying the definition of the average.

Option C:
Option C, 32, would produce a total of 106, and 106 ÷ 4 equals 26.5, which is higher than the stated average.

Option D:
Option D, 34, makes the total 108, giving an average of 27, again inconsistent with the given value.

A fair coin has two equally likely outcomes: head (H) and tail (T). The sample space thus contains two outcomes, and only one of them is a head. Probability is calculated as favourable outcomes divided by total outcomes, so P(H) = 1/2. Therefore, the probability of getting a head in a single toss is 1/2.

Option A:
Option A is correct because it directly follows from the definition of probability for equally likely events. With two outcomes and one favourable event, the fraction must be 1/2. This is one of the most basic probabilities and forms the foundation for more complex problems involving coins and other experiments.

Option B:
A probability of 1/3 would imply that there are three equally likely outcomes, only one of which is a head. However, a standard coin has only two sides, so this interpretation is impossible in this context. Hence, 1/3 cannot be correct.

Option C:
A probability of 1/4 suggests four equally likely outcomes, which might occur in other experiments but not with a single coin toss. Since there are only two outcomes here, 1/4 misrepresents the situation. Therefore, it is not the correct answer.

Option D:
A probability of 2/3 would mean that heads are more likely than tails, which contradicts the assumption of a fair coin with equal chances. The probability must be symmetric between head and tail, making 2/3 an invalid option.

The relation 4:6 = 10:x can be expressed as 4/6 = 10/x. Cross-multiplying gives 4x = 60, so x = 60/4 = 15. Therefore, 15 is the fourth proportional to 4, 6 and 10, making the two ratios 4:6 and 10:15 equal.

Option A:
Option A, 12, would give 10:12 = 5:6, while 4:6 simplifies to 2:3. Since 2:3 ≠ 5:6, the equality of ratios is not preserved, and 12 cannot be the correct fourth proportional.

Option B:
Option B is correct because substituting x = 15 yields 10:15 = 2:3, which matches 4:6 = 2:3. This confirms that the proportionality 4:6 = 10:15 holds perfectly, satisfying the definition of fourth proportional.

Option C:
Option C, 18, leads to 10:18 = 5:9, which again does not equal 2:3 and therefore fails the test of proportion. Even though 18 is larger than 15, it does not maintain the required ratio.

Option D:
Option D, 20, produces 10:20 = 1:2, while 4:6 = 2:3, so the two ratios differ markedly. As a result, 20 is not consistent with the proportional relationship given in the question.

In each group, the letters are spaced three positions apart within the alphabet. ADG uses positions 1, 4 and 7; BEH uses 2, 5 and 8; CFI uses 3, 6 and 9; and DGJ uses 4, 7 and 10. The starting letter increases by one each time, so the next group must start at E and use letters at positions 5, 8 and 11, which are E, H and K. Thus EHK is the correct continuation.

Option A:
Option A is correct because it preserves both the constant gap of three between letters and the progression of the starting letters. E, H and K lie in exactly the same positional relationship as the letters in all previous groups. This consistent structure shows that EHK naturally follows DGJ in the series.

Option B:
Option B, EHL, uses the right first two letters but ends with L, which is position 12 rather than 11. That makes the last gap four instead of three. Since the groups are defined by a fixed difference of three between successive letters, EHL does not respect the pattern.

Option C:
Option C, EGI, has unequal gaps between its letters and does not align with the arithmetic spacing seen earlier. The distances between E, G and I differ from those in ADG, BEH, CFI and DGJ. Consequently, EGI cannot be regarded as a valid continuation.

Option D:
Option D, FHK, starts at F instead of E and therefore does not extend the simple sequence of starting letters from A to D. It also fails to keep the precise spacing structure, so FHK is inconsistent with the series rule.

The differences between successive terms are 2, 3, 4 and 5, increasing by 1 each time. The next difference should therefore be 6. Adding 6 to 18 gives 24. Thus, 24 is the correct next term that extends the pattern of gradually increasing differences.

Option A:
Option A is 22, which yields a difference of 4 from 18, repeating an earlier step instead of moving to 6. This breaks the consistent growth of the differences. Hence, 22 is not correct.

Option B:
Option B is 23, corresponding to a difference of 5 from 18. Although 5 appears earlier, the structure demands 6 next, not another 5. Therefore, 23 is unsuitable.

Option C:
Option C is 25, giving a difference of 7 from 18. That jumps beyond the expected increase of 6 and disrupts the smooth progression of 2, 3, 4, 5, 6. Hence, 25 does not fit.

Option D:
Option D equals 24, producing the required difference of 6. The resulting difference sequence 2, 3, 4, 5, 6 is perfectly coherent. This makes 24 the correct continuation.

The sequence lists consecutive odd numbers starting from 1, with each term 2 greater than the previous one. We have 1, 3, 5, 7 and 9, all odd and evenly spaced. Adding 2 to 9 gives 11. Hence, 11 is the correct next term that keeps the list of odd numbers in order.

Option A:
Option A is 10, which is even and would break the pattern of odd integers. It also changes the difference from 2 to 1, so it does not match the structure.

Option B:
Option B equals 11, maintaining both the odd property and the consistent step of 2. Extending the series to 1, 3, 5, 7, 9, 11 keeps the progression uninterrupted. This makes 11 the appropriate continuation.

Option C:
Option C is 13, which is an odd number but lies 4 units above 9. Choosing 13 would skip 11 and alter the fixed difference of 2. Therefore, it cannot be the immediate next term.

Option D:
Option D is 15, which is further away and would correspond to a difference of 6 from 9. This would distort the regularity of the sequence. So, 15 is not correct in this context.

Each term consists of three consecutive letters taken in blocks: ABC (1–3), DEF (4–6), GHI (7–9) and JKL (10–12). The starting positions of the blocks are 1, 4, 7 and 10, increasing by 3. The next block must therefore start at position 13, giving the letters M, N and O. Hence MNO is the only group that follows the established block-wise progression.

Option A:
Option A is correct because it continues the pattern of non-overlapping three-letter blocks in alphabetical order. After JKL, the natural next block is MNO, covering positions 13–15. It respects both the length of each group and the sequential order of letters. Therefore MNO is the exact logical continuation.

Option B:
Option B, NOP, begins at N instead of M, which would start a block at position 14 rather than 13. This breaks the rule that the start of each block is 3 letters ahead of the previous one. Since it shifts the entire block by one extra letter, NOP does not fit the series.

Option C:
Option C, LMN, overlaps with JKL because L already appears as the last letter of JKL. The series is constructed from disjoint groups of three letters without repetition across block boundaries. Because LMN reuses L and disturbs the block structure, it is not acceptable.

Option D:
Option D, PQR, skips several letters after L and jumps to P, breaking the continuous alphabetical order. This disrupts the regular stepping of three consecutive letters. As a result, PQR fails to maintain the defining structure of the sequence.

Statements A and B give the two De Morgan identities for complements of unions and intersections. Statement C correctly notes that these laws link complements with union and intersection operations. Statement D is also true because the universal set’s complement is the empty set relative to that universe. Statement E is false since similar De Morgan laws hold for logical connectives “and” and “or”. Thus A, B, C and D only are correct.

Option A:
Option A is correct because it gathers all the true statements and excludes E, which wrongly denies the logical counterpart of De Morgan’s laws.

Option B:
Option B is incomplete as it omits D (complement of universal set).

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

Option D:
Option D is wrong because it includes E (false) and omits A (one of the two key identities).

The groups are consecutive three-letter blocks taken in order around the alphabet: PQR (16–18), STU (19–21), VWX (22–24) and YZA (25, 26, 1). After YZA, the next three consecutive letters are B, C and D, starting from the letter after A. Therefore BCD is the immediate next block, and the series continues cyclically.

Option A:
Option A, ABC, restarts at A but overlaps with the last letter of YZA and does not begin with the next letter after A. The pattern continues forward from A to B, not back to A again. Because ABC does not follow the “next-after-A” continuation, it is not consistent with the series.

Option B:
Option B, BCD, is correct because it starts from B, the next letter after A, and forms a three-letter consecutive block. This preserves both the block length and the cyclic progression across the alphabet. The group BCD therefore fits perfectly into the established rotation.

Option C:
Option C, CDE, skips directly to C as the starting letter, ignoring B, which should come immediately after A in the cycle. This breaks the sequence of consecutive letters. Thus CDE does not align with the ordered block structure.

Option D:
Option D, AZB, does not represent the next three consecutive letters after A. It changes the internal order instead of continuing to B, C and D. Consequently, AZB fails to reflect the stepwise nature of the series.

Statements A, B, C and E faithfully describe how blood relation problems are framed and solved: they use family links, family trees or diagrams, include relations by marriage and often span several generations with qualifiers like paternal or maternal. Statement D is false because in many questions genders must be inferred from context or from terms like “son”, “daughter”, “niece” or “nephew”, and are not always explicitly labelled. Thus, the combination A, B, C and E only is correct.

Option A:
Option A is incomplete as it leaves out E and therefore underplays the complexity added by multiple generations and qualifiers, which are common in higher-level relation questions.

Option B:
Option B is also incomplete because it omits C, ignoring the fact that in-law relations are not blood ties but are nonetheless part of such problems. Without C, the picture of relation types is partial.

Option C:
Option C is correct because it includes all the true statements about content and strategy while excluding D, which incorrectly asserts that gender is always explicitly given. It matches the structure of blood relation items used in NET.

Option D:
Option D is wrong as it contains D, the false claim about always explicit gender, and omits A, thereby failing to mention basic family links. This combination mixes an error with an omission.

Let the original quantities of milk and water be 3x and 2x litres. After adding 10 litres of milk, the quantities become 3x + 10 and 2x, and their ratio is 8:5. Thus, (3x + 10)/(2x) = 8/5. Cross multiplying gives 5(3x + 10) = 16x, so 15x + 50 = 16x and x = 50. The original quantity of milk is therefore 3x = 150 litres.

Option A:
Option A is correct because with x = 50 we have initial milk 150 litres and water 100 litres, giving 3:2. After adding 10 litres of milk, the new quantities are 160 litres of milk and 100 litres of water, giving 160:100 = 8:5, which matches the stated new ratio. This confirms 150 litres as the correct original milk quantity.

Option B:
Option B, 120 litres, would imply x = 40 and water 80 litres initially. Adding 10 litres of milk would give 130 litres of milk and 80 litres of water, yielding a ratio of 130:80 = 13:8, not 8:5. Therefore 120 litres does not satisfy the condition.

Option C:
Option C, 125 litres, does not correspond to a clean multiple of 3 and would make the water quantity non-integral if we tried to preserve the 3:2 ratio. Even if we forced approximate values, the resulting ratio after adding 10 litres of milk would not simplify to 8:5.

Option D:
Option D, 160 litres, would require a smaller water quantity than 100 litres to maintain the initial ratio 3:2. After the addition of milk, the ratio would become even more extreme than 8:5, and a detailed calculation shows it does not fit the given transformation.

A ratio is defined as the comparison of two quantities of the same kind by division. For example, if there are 3 boys and 5 girls, the ratio of boys to girls is 3:5, which represents 3 divided by 5. Ratios are fundamental in many aptitude topics such as proportion, mixtures and scaling. Hence, the phrase "comparison of two quantities by division" is the textbook definition of ratio.

Option A:
Ratio is correct because it explicitly involves dividing one quantity by another to express their relative size. This form can be written as a/b, a:b or "a to b," all representing a comparative measure. In UGC NET mathematical aptitude questions, understanding ratios is essential for solving problems on partnerships, mixtures and map scales. Therefore, this option precisely fits the description.

Option B:
A fraction is a representation of part of a whole, usually written as a/b, but it does not necessarily compare two separate quantities of the same kind. While ratios can be expressed in fractional form, the context of comparison of two distinct quantities is more specific to ratios. Thus, fraction is not the most accurate term for the definition given.

Option C:
Percentage expresses a quantity per hundred, such as 25% meaning 25 out of 100. It is a special way of writing fractions or ratios with a fixed denominator of 100. The stem, however, speaks more generally about comparison by division without specifying any base, so percentage is too narrow and not the exact term required.

Option D:
Proportion is a statement that two ratios are equal, such as 2:3 = 4:6. It builds upon the idea of ratio but involves an additional equality relation. Since the question is asking just for the basic comparison by division, proportion goes one step beyond and is therefore not correct.

This option is correct because converting a fraction to a percentage involves multiplying by 100. For 1/4, we compute (1/4) × 100 = 25. Writing this as a percentage gives 25%. Therefore, 1/4 is equivalent to 25%.

Option A:
25% means 25 parts out of 100, which simplifies to the fraction 25/100. Reducing 25/100 by dividing numerator and denominator by 25 gives 1/4. This confirms that 1/4 and 25% represent the same value.

Option B:
20% corresponds to 20/100, which simplifies to 1/5. Since 1/5 is different from 1/4, this percentage does not match the given fraction. Hence, this option is incorrect.

Option C:
40% corresponds to 40/100, which reduces to 2/5. This fraction again differs from 1/4. Therefore, 40% cannot be the correct conversion.

Option D:
50% represents 50/100, which simplifies to 1/2. This is larger than 1/4 and does not equal it. Thus, this option is also not correct.

For compound interest, the amount is given by A = P(1 + R/100)ᵗ. Here P = 10,000, R = 5% and t = 2 years. This gives A = 10,000 × (1.05)². The square of 1.05 is 1.1025, so the amount becomes 10,000 × 1.1025 = ₹11,025. This includes both the original principal and the interest earned over two years.

Option A:
Option A, ₹10,500, represents the amount after one year of compound or simple interest at 5%, not two years. It misses the second period of compounding.

Option B:
Option B, ₹11,000, may come from incorrectly adding 5% of 10,000 twice and ignoring the interest on interest, which is the essence of compounding.

Option C:
Option C, ₹11,250, overestimates the effect of compounding and does not match the precise factor (1.05)². It might arise from using an incorrect growth factor.

Option D:
Option D uses the exact compounding formula and incorporates the second year’s interest on the first year’s interest. It therefore gives the correct accumulated amount of ₹11,025.

This option is correct because the logical negation of a universal statement "All S are P" is an existential statement "Some S are not P." Here, "All students are punctual" negates to "Some students are not punctual." Thus, the correct word to fill the blank is "Some."

Option A:
Saying "No students are not punctual" would mean that every student is punctual, which actually restates the original universal claim in a different form. It does not express the negation. Therefore, this option is incorrect.

Option B:
The word "Most" suggests that a large majority of students are not punctual, which is stronger than merely saying at least one is not punctual. Negation in logic only needs one counterexample, so "most" is not the precise logical opposite of "all."

Option C:
The correct negation of a universal statement is existential, and "some" expresses the existence of at least one counterexample. Saying "Some students are not punctual" states that there is at least one student who fails the property, which contradicts the claim that all are punctual. Thus, this option correctly reflects the logical negation.

Option D:
Saying "Exactly two" students are not punctual again specifies a particular number of exceptions. Logical negation, however, only requires at least one exception and does not fix the number. Hence, this option is not the standard negation.