UGC NET Questions (Paper – 1)

This option is correct because the reciprocal of a non-zero number a is 1/a. For the number 4, its reciprocal is 1/4. Multiplying 4 by 1/4 gives 1, which verifies that they are multiplicative inverses. Therefore, 1/4 is the required reciprocal.

Option A:
The reciprocal of 4 must satisfy 4 × reciprocal = 1. If we take 4 × 4, we get 16, not 1. This shows that 4 is not the multiplicative inverse of 4. Hence, this option is incorrect.

Option B:
The fraction 1/2 multiplied by 4 gives 2, not 1. Therefore, 1/2 does not serve as the reciprocal of 4. It reduces the number but does not invert it. Thus, this option does not match the definition.

Option C:
The reciprocal of 4 is 1/4 because 4 × 1/4 = 1. This fits the standard rule that the reciprocal of a number a is 1/a. So 1/4 correctly represents the fractional reciprocal of 4.

Option D:
The number 2 multiplied by 4 equals 8, which is far from 1. As a result, 2 cannot be considered the reciprocal of 4. This option fails the basic test for a multiplicative inverse.

The condition “x varies inversely as the square of y” means x = k/y² for some constant k. Using x = 9 when y = 2, we get 9 = k/4, so k = 36. For y = 3, we have x = 36/3² = 36/9 = 4. Therefore, the new value of x corresponding to y = 3 is 4.

Option A:
Option A, 3, would correspond to x = 36/y² giving 3 = 36/y², so y² = 12, which does not match y = 3. Hence, 3 cannot be the value of x when y is 3 under this variation law.

Option B:
Option B is correct because it uses the constant k = 36 derived from the first pair (x, y) = (9, 2) and substitutes y = 3 appropriately. The computation 36/9 is straightforward and yields 4, which fully respects the inverse-square relationship between x and y.

Option C:
Option C, 6, could result from treating x as inversely proportional to y instead of y², leading to x = 18/3. However, that would violate the stated condition involving the square of y, and it does not satisfy x = 9 when y = 2 if we consistently use the same rule.

Option D:
Option D, 8, might arise from doubling 4 under the mistaken belief that x halves when y increases from 2 to 3, but inverse-square variation is more rapid than simple inverse variation. The correct calculation yields 4, not 8, so this option is not consistent with the given dependency.

Statements A, B and C are all correct descriptions of standard percentage and profit–loss ideas. Percentages convert quantities to a base of 100, profit percent is typically calculated on cost price, and a discount, when given, lowers the marked price. Statement D is false because an x% increase followed by an x% decrease leads to an overall decrease, not a return to the original value, and E is false since percentage questions are very common in mathematical aptitude sections of UGC NET. Therefore the combination containing A, B and C only is correct.

Option A:
Option A is incomplete because it omits C, even though the understanding that discounts reduce marked price is crucial for many profit–loss and discount problems. Without C, the description of applied percentage use is partial.

Option B:
Option B is correct since it collects all the true statements and excludes D and E, which give misleading information about successive percentage changes and exam pattern. It accurately reflects how percentages operate in typical NET aptitude questions.

Option C:
Option C is wrong because it adds D to the set, asserting that equal increase and decrease cancel out exactly, which is mathematically incorrect. Including D makes the overall combination invalid.

Option D:
Option D is also incorrect because it includes E, which wrongly denies the presence of percentage-based aptitude questions in UGC NET, and omits A, thereby leaving out the foundational definition of percentage.

Average speed is calculated as total distance divided by total time taken. Here, the distance is 120 km and the time is 3 hours. Dividing 120 by 3 gives 40 km/h. Therefore, the car’s average speed over the journey is 40 km/h.

Option A:
Option A is correct because the straightforward computation 120 ÷ 3 = 40 yields the average speed. This aligns with the definition of speed as distance per unit time. In exam settings, quickly identifying such basic calculations saves valuable time for more complex questions.

Option B:
An average speed of 30 km/h would correspond to travelling 90 km in 3 hours, not 120 km. Since the actual distance travelled is greater, this option underestimates the speed. Hence, it is incorrect.

Option C:
A speed of 50 km/h over 3 hours would give a distance of 150 km, which is higher than the given 120 km. This shows that 50 km/h is too high for the conditions described.

Option D:
A speed of 60 km/h for 3 hours would produce a distance of 180 km, which is much larger than 120 km. Therefore, 60 km/h does not match the data and cannot be the correct answer.

Let the total number of people be represented as 100%. If 20% like neither tea nor coffee, then 80% like at least one of the two beverages. By the inclusion–exclusion principle, the percentage who like at least one is the sum of tea-likers and coffee-likers minus those who like both. Thus, 80 = 70 + 40 − (both). Solving this gives (both) = 70 + 40 − 80 = 30%.

Option A:
Option A, 20%, equals the proportion who like neither, not those who like both, and so confuses the outside region with the intersection.

Option B:
Option B calculates the intersection using inclusion–exclusion and ensures that the total does not exceed 100%. It matches the only value that makes the percentages internally consistent.

Option C:
Option C, 40%, would give 70 + 40 − 40 = 70% liking at least one, which contradicts the information that 20% like neither and thus only 80% like at least one.

Option D:
Option D, 50%, creates an even larger overlap and leads to a sum exceeding the allowed total, making the data inconsistent.

This series is governed by the rule aₙ = 2n⁴ + n² + 6 with n starting from 1. For n = 1, 2, 3, 4 and 5 we obtain 9, 42, 177, 534 and 1281, which perfectly matches the given sequence. For n = 6, the formula yields a₆ = 2·6⁴ + 6² + 6 = 2·1296 + 36 + 6 = 2592 + 42 = 2634. Therefore 2634 is the only value that continues this quartic-based pattern correctly.

Option A:
Option A, 2610, is 24 less than the value given by 2n⁴ + n² + 6 at n = 6. To choose 2610 we would have to reduce the correct result at this stage, despite the formula fitting earlier terms exactly. This inconsistency makes option A incorrect.

Option B:
Option B, 2622, is 12 less than the correct answer and likewise does not equal 2·6⁴ + 6² + 6. It provides a close but still inaccurate figure that breaks the exact functional relation. Hence option B cannot be accepted as the next term.

Option C:
Option C, 2646, is 12 greater than the correct value 2634 and cannot be obtained from the rule at n = 6. Accepting 2646 would amount to inflating the polynomial output at only one position, which is not supported by the data. Thus option C is not a valid continuation of the number series.

Option D:
Option D, 2634, matches exactly the value produced by the formula for n = 6. It keeps the same contributions from the quartic term, the square term and the constant used earlier. Because this number arises directly from the established pattern, option D is the correct answer.

Statements A and C both capture the cross-multiplication property of equal ratios in a proportion. Statement B is a direct algebraic consequence of A when all terms are non-zero. Statement D is correct because multiplying or dividing each term by the same non-zero factor leaves the ratio unchanged. Statement E is false, since if a:b = 2:3, then b:a should be 3:2, not 3:4. Therefore, A, B, C and D only form the correct set, which is option D.

Option A:
Option A is incomplete as it has only A, B and C and omits D, thereby ignoring the important rule about scaling ratios, which is frequently used in simplifying or comparing them.

Option B:
Option B is incorrect because it omits B (true) and so fails to include the fraction form a/c = b/d implied by a:b = c:d.

Option C:
Option C is incomplete since it excludes A; without stating ad = bc explicitly, the key proportionality check is missing even though some other statements are true.

Option D:
Option D is correct because it gathers all the fundamental properties of proportions and ratio scaling while excluding E, which misstates the inverse ratio of 2:3.

When order does not matter, the number of ways to choose r objects from n is given by the combination formula nCr. Here n is 5 and r is 2, so the number of committees is 5C2. Using the formula, 5C2 = 5! ÷ (2! × 3!) = (5 × 4) ÷ (2 × 1) = 10. Therefore, there are 10 distinct ways to select a pair of students from the group.

Option A:
Option A, 5, corresponds to the number of students rather than the number of pairs that can be formed. It ignores the multiple distinct combinations that are possible when selecting two members.

Option B:
Option B, 8, does not match the combinatorial calculation for 5C2 and appears to be an arbitrary number without basis in the formula.

Option C:
Option C, 9, is close to 10 but still does not equal the true combination count as derived from factorial operations, showing a small but important error.

Option D:
Option D correctly employs the combination formula and simplifies it accurately. It recognizes that for each choice of two students, the order is irrelevant, leading to exactly 10 possible committees.

If A fills the tank in 12 hours, A’s one-hour work is 1/12; for B it is 1/18, and for C it is 1/36. The ratio A:B:C in one hour is (1/12):(1/18):(1/36). Taking the LCM of 12, 18 and 36 as 36, we get 3:2:1 as the simplified ratio. Thus, A, B and C complete the work in the ratio 3:2:1 per hour.

Option A:
Option A, 6:4:3, represents 6:4:3 which, when normalised, does not match 1/12:1/18:1/36. Dividing these by 3 gives 2:4/3:1, which is not proportional to 3:2:1, so it misstates the relative speeds of the taps.

Option B:
Option B, 3:2:2, suggests that B and C work at the same rate, which contradicts the given times. Since C takes 36 hours, which is double B’s 18 hours, C must work at half of B’s rate, not the same rate. Therefore, 3:2:2 is inconsistent with the time data.

Option C:
Option C, 2:3:6, would imply that C is working fastest, with the largest share of work per hour, while in reality C is the slowest tap. This directly contradicts the inverse relationship between time and rate, so 2:3:6 cannot be correct.

Option D:
Option D is correct because converting the one-hour work fractions to a common denominator and simplifying gives 3:2:1. This ratio accurately reflects that A is fastest, B is intermediate and C is slowest, in exact accordance with their filling times.

In this sequence each term is given by aₙ = n! + n² for n starting from 1. Checking this rule, we get 1!+1² = 2, 2!+2² = 6, 3!+3² = 15, 4!+4² = 40 and 5!+5² = 145, which matches the given terms exactly. For n = 6 the expression yields 6!+6² = 720+36 = 756. Consequently, 756 is the only term that continues the factorial-plus-square pattern.

Option A:
Option A, 720, corresponds to 6! alone and omits the n² part of the expression. It does not follow the rule aₙ = n!+n² that has generated all previous terms. Therefore option A ignores part of the pattern and cannot be accepted as correct.

Option B:
Option B, 744, adds only 24 to 720 instead of the required 36, so it does not equal 6!+6². Selecting 744 would require changing the square component from n² to a smaller value, which is inconsistent with the earlier terms. Hence option B is not a valid continuation.

Option C:
Option C, 756, matches exactly the value of 6!+6² and maintains both components of the rule: the factorial growth and the quadratic correction. It fits the behaviour observed for n = 1 to 5, where each term is one factorial plus a square. Because it preserves this combined structure, 756 is the correct next term in the series.

Option D:
Option D, 780, adds 60 to 720 and cannot be written as n!+n² for n = 6. It would imply using a different function of n for the second part of the expression. This breaks the pattern evident from the given terms, so option D is incorrect.

Statements A and B correctly describe simple random and stratified sampling. Statement C is also true as cluster sampling initially chooses whole clusters for inclusion. Statement E is correct because systematic sampling indeed selects every nth unit after a random start. Statement D is false since convenience sampling is non-probability based and can easily introduce bias. Thus, A, B, C and E only are correct.

Option A:
Option A is incomplete as it omits E (systematic sampling).

Option B:
Option B is wrong because it includes D (false).

Option C:
Option C is correct because it unites all of the true descriptions and excludes D.

Option D:
Option D is incorrect since it retains D (false) and omits A.

Profit is the difference between selling price and cost price, so here profit = 920 − 800 = ₹120. The profit percentage is calculated on cost price as (Profit ÷ Cost price) × 100, giving (120 ÷ 800) × 100 = 15%. Thus, the trader earns a profit of 15% on the article. This matches option A.

Option A:
Option A is correct because the calculation based on the formula yields exactly 15%. A 15% increase on ₹800 is 0.15 × 800 = 120, leading to a selling price of 920, which fits the data given. Recognising this pattern helps students quickly check their answers in profit and loss questions.

Option B:
A profit of 10% on ₹800 would be ₹80, resulting in a selling price of ₹880, which is lower than the actual selling price. This does not match the information in the problem. Hence, 10% is not the correct profit percentage.

Option C:
A profit of 12% would be 0.12 × 800 = ₹96, giving a selling price of ₹896. This is still less than ₹920 and therefore inconsistent with the stated transaction. As a result, 12% is not the right answer.

Option D:
A profit of 20% would imply 0.20 × 800 = ₹160 as profit and a selling price of ₹960, which is higher than the given selling price. Thus, 20% overestimates the gain and cannot be correct.

The series shows multiples of nine: 9×1, 9×2, 9×3 and 9×4. Following this, the next term should be 9×5. Computing 9×5 gives 45, so 45 is the correct extension of the sequence.

Option A:
Option A is 40, which is not divisible by 9 and does not follow the multiplication rule. Including 40 would disrupt the pattern of multiples of nine. Therefore, it is not correct.

Option B:
Option B is 44, which again is not a multiple of 9 and lies just below 45. It has no direct connection with the established series. Hence, it does not fit the rule.

Option C:
Option C is 48, which equals 9×5 + 3 and is not a pure multiple of nine here. The pattern has not used any additional constant, so 48 cannot be accepted.

Option D:
Option D equals 45 and follows naturally as 9×5. The full sequence 9, 18, 27, 36, 45 keeps the same multiplication by 9 relationship. This makes 45 the valid next term.

Statements B and E are the wrong statements here. A is correct because assumptions are unstated ideas that the statement presupposes. C is true since many assumptions concern whether the suggested action can realistically be carried out. D is also correct because candidates should judge assumptions based mainly on what the statement reasonably implies. B is wrong because not every inference is an assumption; some inferences go beyond what is taken for granted. E is wrong because assumptions that contradict the statement are weak, not strong. So the wrong statements are B and E only.

Option A:
Option A is incorrect because it names only B as wrong and ignores E, even though E clearly mischaracterises how strong assumptions are judged in reasoning questions. This makes the option incomplete.

Option B:
Option B is correct as it identifies precisely B and E as the problematic statements. It recognises that both overgeneralising inferences and treating contradictions as strong assumptions are errors in reasoning.

Option C:
Option C is wrong because it also labels A as wrong, even though A is the standard definition of an assumption used in exam questions and reasoning textbooks.

Option D:
Option D is incorrect since it lists only A, C and D as wrong. All three actually describe accepted guidelines for evaluating assumptions, so this option misclassifies core principles as incorrect.

Let 3a = 4b = 6c = k for some positive constant k. Then a = k/3, b = k/4 and c = k/6. To eliminate denominators, multiply each term by 12, the LCM of 3, 4 and 6. This gives a:b:c = 4k:3k:2k, which simplifies to 4:3:2. Thus, the required ratio a:b:c is 4:3:2.

Option A:
Option A, 2:3:4, reverses the correct ordering and suggests that c is the largest term. But since 6c = k and 3a = k, we have c = k/6 and a = k/3, so a must be larger than c. Therefore, 2:3:4 is inconsistent with the given equalities.

Option B:
Option B, 3:4:2, places b as the largest value, which conflicts with 3a = 4b = 6c. From 3a = 4b we deduce a = (4/3)b, so a is larger than b. This means a should have the largest coefficient, which is not the case in 3:4:2.

Option C:
Option C, 4:2:3, misorders b and c relative to a. If we normalise using 3a = 4b = 6c, c must be the smallest because its divisor (6) is the largest. In 4:2:3, c’s coefficient is 3, not the smallest, so the pattern does not respect the initial conditions.

Option D:
Option D is correct because the derivation from 3a = 4b = 6c leads directly to a = k/3, b = k/4 and c = k/6, and scaling by 12 produces 4:3:2. This ordering matches the relative sizes implied by the denominators 3, 4 and 6.