UGC NET Questions (Paper – 1)

The arithmetic mean is defined as the sum of all observations divided by the number of observations. It provides a central value that represents the data set as a whole. In data interpretation and aptitude questions, the arithmetic mean is often used to compare performance or compute expected values. Thus, the description in the question is precisely the standard definition of arithmetic mean.

Option A:
Median is the middle value when the data are arranged in ascending or descending order, not the sum divided by the number of observations. It is another measure of central tendency, but it is defined differently from the formula given in the stem. Therefore, median does not correspond to the description.

Option B:
Mode is the value that occurs most frequently in a data set. It does not involve summing all values or dividing by the count. As such, it does not match the phrase "sum of all observations divided by n" given in the question. Hence, mode is not correct here.

Option C:
Arithmetic mean is correct because it uses the exact procedure mentioned: adding all observations and dividing by their number. For example, the mean of 2, 4 and 6 is (2+4+6)/3 = 4. This measure plays a key role in many quantitative questions, making it essential for UGC NET aspirants to understand.

Option D:
Range is the difference between the maximum and minimum values in a data set. It measures dispersion, not central tendency, and does not involve any division by the number of observations. Therefore, range is unrelated to the description and cannot be the correct answer.

For two positive integers a and b, the product a × b equals HCF(a, b) × LCM(a, b). Here, one number is 24, the HCF is 6 and the LCM is 72. Let the other number be n. Then 24 × n = 6 × 72. The right-hand side is 432, so n = 432 ÷ 24 = 18. Hence, the other number must be 18 to maintain both the given HCF and LCM.

Option A:
Option A, 12, would give 24 × 12 = 288, and since 6 × 72 = 432, the product condition would fail. It also changes the LCM and HCF relationship.

Option B:
Option B satisfies 24 × 18 = 432, matching 6 × 72 exactly. It ensures the pair of numbers has the stated HCF and LCM by the standard product relation.

Option C:
Option C, 36, gives a product of 864, which would require a different LCM or HCF than those given.

Option D:
Option D, 48, yields an even larger product and is inconsistent with the specified values of 6 and 72.

The series is generated by the polynomial rule aₙ = n⁴ + 6n with n starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 7, 28, 99, 280 and 655, which exactly matches the given terms. Applying the same expression for n = 6 gives a₆ = 6⁴ + 6·6 = 1296 + 36 = 1332. Therefore 1332 is the only value that continues this polynomial pattern without changing the underlying rule.

Option A:
Option A gives 1332, which is exactly the value obtained from the formula n⁴ + 6n for n = 6. It maintains the same relationship between the index and term that produced all earlier numbers. Because no adjustment or exception is needed at the sixth term, this option correctly extends the series.

Option B:
Option B, 1308, is 24 less than the value predicted by n⁴ + 6n at n = 6. To accept 1308 we would have to reduce the correct polynomial result only at the last step, which contradicts the exact agreement seen for the first five terms. Hence this option does not respect the established pattern.

Option C:
Option C, 1320, is still below the correct value and cannot be obtained from n⁴ + 6n when n = 6. It represents a near but inaccurate approximation that breaks the strict algebraic rule behind the sequence. For a UGC NET numerical reasoning item, such a deviation means this option must be rejected.

Option D:
Option D, 1344, is 12 greater than the formula’s output and likewise cannot be produced by the same rule at n = 6. Choosing 1344 would introduce an unexplained increase in the last term, destroying the consistency of the pattern. Therefore this option is not a valid continuation of the number series.

Let the original price be 100 units. A 10% decrease takes it to 90. A further 10% decrease is now applied to 90, resulting in a reduction of 9, so the new price is 81. The overall decrease from 100 to 81 is 19 units, which is a 19% reduction. This shows that successive equal percentage decreases do not simply add arithmetically.

Option A:
Option A, 19%, correctly reflects the multiplicative effect of two successive 10% decreases. It uses the idea that the combined factor is 0.9 × 0.9 = 0.81, corresponding to a 19% drop.

Option B:
Option B, 20%, comes from the mistaken belief that decreases add directly (10% + 10%), but this ignores that the second decrease is on an already reduced base.

Option C:
Option C, 21%, overestimates the net effect and does not correspond to the computed final value of 81.

Option D:
Option D, 25%, greatly overstates the combined impact and has no support in the actual price sequence produced by the successive reductions.

Statements A and B are correct descriptions of bar graphs and pie charts, and D is true because meaningful comparison of graphical data demands careful reading of scales, units and legends. Statement E is also correct since many exam questions allow or even require approximations for quick decision-making. Statement C is false because ignoring scales can lead to serious misinterpretation of data. Thus, the combination that includes A, B, D and E while excluding C is correct.

Option A:
Option A is incomplete because although it lists three true statements, it omits E and does not highlight the practical exam strategy of using approximations where appropriate. This reduces the applied insight necessary for UGC NET aptitude.

Option B:
Option B is insufficient since it leaves out A, failing to acknowledge the fundamental principle that bar lengths are proportional to values, which is basic to understanding bar graphs. Without A, the description of graphical interpretation is partial.

Option C:
Option C is incorrect because it includes C, the statement that minimises the importance of scales, and therefore accepts a guideline that would lead to faulty interpretation. Even though A, B and D are true, adding C makes the combination unsound.

Option D:
Option D is correct because it keeps the four accurate statements and rejects C, which directly contradicts best practice in reading graphs. It aligns with the skills expected in data interpretation sections of UGC NET.

This option is correct because the expansion of (a + b)² gives a² + 2ab + b². We obtain this by multiplying (a + b)(a + b) and collecting like terms. The cross terms ab and ba combine to give 2ab. Therefore, the complete identity is (a + b)² = a² + 2ab + b².

Option A:
The expression a² + b² omits the middle term 2ab that arises from the cross products in the expansion. It would represent (a − b)² only if combined with a negative cross term, but as written it is incomplete. Hence, it is not the correct identity here.

Option B:
The expression a² − 2ab + b² is the expansion of (a − b)², not (a + b)². The negative sign on 2ab indicates subtraction rather than addition. Therefore, this identity does not match the given binomial square.

Option C:
The expression a² + 2ab + b² correctly represents the square of a sum. It includes the squares of both terms and twice their product. This matches the standard algebraic identity for (a + b)², so this option is correct.

Option D:
The expression 2a² + b² does not arise from any simple binomial square formula. It has an incorrect coefficient on a² and is missing the cross term. Thus, it cannot represent (a + b)².

Statement B is true because adding an odd and an even number always yields an odd result. C is correct since the product of two odd numbers is odd. D is also correct because an odd times an even number always has a factor 2 and so is even. A is false because the sum of two odd integers is even, and E is false because not every even number is divisible by 3. Hence B, C and D only are correct.

Option A:
Option A is incomplete as it leaves out D and therefore does not include the important rule about the product of an odd and even number, which is regularly tested in parity questions.

Option B:
Option B is correct because it gathers all three standard parity rules, covering sums and products of odd and even numbers, while excluding A and E, which misstate these relationships.

Option C:
Option C is incorrect since it includes A, which wrongly claims that two odd numbers add to an odd result, contradicting parity arithmetic.

Option D:
Option D is wrong as it brings in E, which incorrectly generalises divisibility by 3 to all even numbers, and so mixes a false divisibility statement with otherwise correct parity rules.

Out of 100 students, 20% failed in both subjects, so 80 students passed in at least one of Mathematics or English. Let the number who passed in both subjects be x. Using inclusion–exclusion, the number who passed in at least one is 60 + 70 − x. This must equal 80, giving 130 − x = 80 and hence x = 50. Therefore, 50 students passed in both Mathematics and English.

Option A:
Option A, 40, would correspond to 60 + 70 − 40 = 90 students passing in at least one subject, which contradicts the given figure that only 80 passed at least one.

Option B:
Option B, 45, gives 60 + 70 − 45 = 85 students passing in at least one, again inconsistent with the total implied by 20 students failing both.

Option C:
Option C, 55, implies 60 + 70 − 55 = 75, which would mean 25 students failed in both subjects, not the stated 20.

Option D:
Option D, 50, exactly balances the inclusion–exclusion equation and yields 80 students passing at least one subject, matching the information derived from the 20 who failed both.

Statement C is wrong because in standard truth-functional logic the conditional “if p then q” is false precisely when p is true and q is false, not when p is false and q is true. Statements A and B correctly describe the conditions for truth of conjunction and inclusive disjunction, while D accurately explains the biconditional and E states the essential idea that compound truth values depend on component truth values. Hence, C alone is the incorrect statement, so the option that lists only C is correct.

Option A:
Option A is correct since it identifies C as the sole misstatement and implicitly accepts A, B, D and E as standard truth-table definitions. It preserves the conventional semantics of implication while rejecting the inverted condition given in C.

Option B:
Option B is incorrect because it also labels A as wrong, even though A correctly states that “p and q” is true only when both components are true. Treating A as incorrect contradicts the fundamental definition of conjunction.

Option C:
Option C is wrong because it adds D to the set of wrong statements, even though D accurately characterises “if and only if” as true when p and q share the same truth value. Including D among wrong statements undermines correct understanding of biconditional.

Option D:
Option D is also incorrect because by pairing B with C it misclassifies B, which correctly explains that inclusive “or” is false exactly when both disjuncts are false. This combination therefore rejects a correct connective rule.

This option is correct because standard deviation measures the spread of data around the mean and is defined using squared deviations. Squaring removes negative signs before averaging and taking the square root. The square root of a non-negative number is also non-negative. Therefore, standard deviation cannot be negative and is always non negative.

Option A:
Standard deviation can be zero only in the special case where all observations are equal to the mean. In most real data sets, values vary, giving a positive spread. Hence, "always zero" is incorrect.

Option B:
Although different data sets can have different standard deviations, the value is never negative. It can be zero or positive depending on the variation in the data. This means the correct description is that standard deviation is always non negative.

Option C:
"Always one" would suggest that every data set has the same fixed amount of spread, which is not true. Different data sets can have very large or very small variability. Therefore, this option is wrong.

Option D:
Saying standard deviation may be negative contradicts the mathematical definition involving squared deviations and square roots. Since squares are non negative, their average and square root cannot be negative either. Thus, this option is mathematically impossible.

To find the arithmetic mean, we add all the numbers and divide by how many numbers there are. Here, 4 + 6 + 8 + 10 = 28 and there are 4 observations. Dividing 28 by 4 gives 7. Therefore, the arithmetic mean of the given data set is 7.

Option A:
Option A is correct because it follows directly from the formula mean = sum of observations ÷ number of observations. The computation 28 ÷ 4 is straightforward and yields 7 exactly. In aptitude tests, performing such calculations accurately and quickly is crucial to managing time. Thus, 7 correctly represents the central value of the data.

Option B:
If we choose 8, we would be assuming that the average is equal to one of the larger values in the set, but the computation shows otherwise. Although 8 is close to the centre of the range, it does not equal the exact mean 7. Therefore, option B overestimates the average value.

Option C:
A mean of 6 would require that the total sum be 24, because 24 ÷ 4 = 6, but the actual sum is 28. This indicates that 6 is too low compared to the true average. Hence, this option does not match the correct calculation.

Option D:
A mean of 9 would imply a total sum of 36 for four values, which is much higher than the actual sum. Since the numbers lie between 4 and 10, an average as high as 9 is not consistent with the contributions of the smaller values. Therefore, this option is incorrect.

Every term contains five letters spaced in the same wide way across the alphabet, and each new group is obtained by shifting all letters one step forward. AFKPU becomes BGLQV, then CHMRW, then DINSX through this uniform motion in every column. To continue, we add one to each letter of DINSX: D becomes E, I becomes J, N becomes O, S becomes T and X becomes Y, giving EJOTY. Since EJOTY alone preserves both the internal spacing and the simple column wise shift, it is the correct next term.

Option A:
Option A, EJOSY, changes the last two letters so that the distances inside the group no longer match those of the earlier terms. The relationship among O, S and Y differs from the regular jumps seen in AFKPU, BGLQV, CHMRW and DINSX. Because the spacing is inconsistent, EJOSY cannot correctly continue the series.

Option B:
Option B, FKPUZ, begins with F instead of E, so the starting letter jumps two steps ahead rather than one after D. It also alters the internal distances towards the end of the group. This disruption of both starting position and spacing shows that FKPUZ is not in line with the pattern.

Option C:
Option C, EKNQY, mixes some correct letters with others that do not arise from a one step shift of DINSX. The middle letters do not follow the same vertical sequences present in earlier groups. Therefore EKNQY fails to maintain the strict mechanical rule behind the series.

Option D:
Option D is correct because EJOTY results from applying the identical transformation that links each pair of consecutive terms in the series. It keeps the same distance between each pair of adjacent letters and fits neatly after DINSX in every column. This full consistency with the observed construction confirms EJOTY as the intended continuation.

The area of a square is proportional to the square of its side, so if A₁:A₂ = 9:16, then (s₁²):(s₂²) = 9:16. Taking square roots of both sides gives s₁:s₂ = √9:√16 = 3:4. Thus, the side lengths must be in the ratio 3:4, which directly follows from the relationship between area and side for similar squares. No other option maintains this exact square–square root consistency.

Option A:
Option A is correct because it correctly applies the square root to the area ratio 9:16, converting it into a side ratio. Since area scales as the square of the side, this reverse operation is essential. The resulting ratio 3:4 is the only one consistent with both the given area ratio and the geometry of squares.

Option B:
Option B, 4:3, simply reverses the correct order and would imply that the square with area 9 units has the larger side, which contradicts the fact that 16 is greater than 9. If the second square has greater area, its side must be longer, not shorter, so 4:3 cannot be the correct side ratio.

Option C:
Option C, 9:16, repeats the area ratio rather than converting it to a side ratio and so ignores the square relationship. This would only be appropriate if areas themselves were being compared, not side lengths. Because the question explicitly asks for the ratio of sides, 9:16 is not valid here.

Option D:
Option D, 81:256, is the square of the given area ratio and would correspond to comparing higher powers of side lengths, not the sides themselves. It exaggerates the difference between the two squares and has no direct geometric justification in this context.

The word NET has three distinct letters: N, E and T. The number of permutations of n distinct objects taken all at a time is n!, so here it is 3!. Calculating 3! gives 3 × 2 × 1 = 6. Therefore, there are 6 possible distinct arrangements of the letters in NET.

Option A:
Option A uses the factorial formula for permutations correctly, recognizing that there are no repeated letters. By computing 3! as 6, it counts all possible orderings such as NET, NTE, ENT, ETN, TEN and TNE.

Option B:
Option B, 3, would be the number of letters, not the number of arrangements. It undercounts the permutations by ignoring the different orders in which letters can appear.

Option C:
Option C, 9, may come from incorrectly squaring the number of letters (3²), which does not match the combinatorial rule for arrangements of distinct objects.

Option D:
Option D, 12, overestimates the possible arrangements and could arise from misapplying permutation formulas that involve repetition or additional positions.

Let the original number be 100 for simplicity. Increasing it by 25% gives 100 × 1.25 = 125. Decreasing this new value by 20% means multiplying 125 by 0.8, giving 125 × 0.8 = 100. This returns exactly to the original value, so the net percentage change is 0%. Hence, Option A, 0%, correctly represents no overall increase or decrease.

Option A:
Option A, 0%, directly expresses that there is no net change in the value after both operations. It matches the calculation using multiplicative factors 1.25 and 0.8, whose product is exactly 1.

Option B:
Option B, 2% increase, would require the final value to be 102 when starting from 100, but the actual final value here is 100, so this option overstates the effect.

Option C:
Option C, 4% increase, similarly suggests a final value of 104, which is not supported by the computations of the successive changes.

Option D:
Option D, 0.5% decrease, implies a small net drop, but the actual multiplicative effect of 1.25 and 0.8 is neutral.