UGC NET Questions (Paper – 1)

This option is correct because the arithmetic mean of two numbers is obtained by adding them and dividing by 2. Here, (10 + 20) ÷ 2 = 30 ÷ 2 = 15. This value lies exactly midway between 10 and 20. Therefore, the arithmetic mean is 15.

Option A:
Taking 10 as the mean would ignore the contribution of the larger number 20. The average must be between the two numbers and balanced by both. Since 10 is at the lower end, it cannot be the mean of 10 and 20.

Option B:
A value of 12 is between 10 and 20 but does not satisfy the mean formula. Using (10 + 20) ÷ 2 does not produce 12. Therefore, this number does not represent the correct arithmetic mean.

Option C:
20 is the larger of the two numbers and would be the mean only if both numbers were 20. Because the other number is 10, the mean must lie between 10 and 20, not at the greater endpoint. Thus, 20 is incorrect.

Option D:
The average 15 comes directly from the formula for mean and is equidistant from 10 and 20. It balances the two values, which is the key idea of an arithmetic mean. Hence, this option correctly answers the question.

Statements A, B, C and E describe core properties of means. A gives the definition of arithmetic mean. B is true because the overall mean of two groups must fall between their group means, weighted by group sizes. C is correct in explaining that weighted averages give more influence to values with higher weights. E holds since adding a constant to all values shifts the mean by that constant. D is false because the mean need not be one of the data values, especially for continuous data. Hence A, B, C and E only are correct.

Option A:
Option A is incomplete because it leaves out E and therefore ignores the important transformation property showing how means shift under addition of a constant. It does not include all correct statements.

Option B:
Option B is wrong since it excludes C and includes only A, B and E, failing to mention the key feature of weighted averages, which are frequently tested in aptitude problems.

Option C:
Option C is correct because it collects all four true statements while excluding D, which wrongly asserts that the mean must be a data point. This combination matches the conceptual understanding required in NET statistics questions.

Option D:
Option D is incorrect as it omits A and does not include all four true statements together, so it cannot be the correct combination.

The average of three numbers is the sum of the numbers divided by 3. If the average is 25, then the total sum must be 25 × 3 = 75. The two known numbers are 20 and 30, whose sum is 50. Subtracting 50 from 75 gives 25, so the third number must be 25.

Option A:
Option A, 25, exactly matches the required third number, because 20 + 30 + 25 = 75 and 75 ÷ 3 = 25. It is therefore the correct choice.

Option B:
Option B, 20, would give a total of 20 + 20 + 30 = 70, and 70 ÷ 3 does not equal the required average of 25, so this choice fails the given condition.

Option C:
Option C, 30, leads to 20 + 30 + 30 = 80, giving an average of 80 ÷ 3, which is greater than 25 and inconsistent with the stated mean.

Option D:
Option D, 35, produces a total of 20 + 30 + 35 = 85, resulting in an average greater than 25, again contradicting the problem’s information.

Five consecutive odd integers can be written as n − 4, n − 2, n, n + 2 and n + 4, where n is the middle term. The average of such a symmetric set is always the middle term, so the average being 37 means the middle integer is 37. The largest of the five is therefore 37 + 4, which is 41. This uses both the definition of average and the structure of consecutive odd numbers.

Option A:
Option A, 39, is the second largest of the five integers, not the largest. It is 2 greater than the middle term instead of 4 and therefore does not represent the end of the sequence.

Option B:
Option B correctly identifies that with an average of 37, the five odd numbers are 33, 35, 37, 39 and 41. Among these, 41 is clearly the largest and fits the pattern of consecutive odd integers.

Option C:
Option C, 43, would require shifting the sequence so that the average is larger than 37. That would contradict the given mean and lead to a different central value.

Option D:
Option D, 45, is even further from the centre and again cannot be part of the exact sequence whose average is 37.

The arithmetic mean of five numbers is 18, so their total sum is 18 × 5 = 90. One of the numbers is given as 24. To find the sum of the remaining four numbers, we subtract this known value from the total: 90 − 24 = 66. Thus, the combined total of the four remaining numbers is 66.

Option A:
Option A directly applies the relationship Total = Average × Number of items and then subtracts the known element to obtain the sum of the others. This method is standard in average problems and yields 66 consistently.

Option B:
Option B, 72, would correspond to a larger total of 96 if 24 were added back, which would produce an average of 19.2 instead of 18. This contradicts the given mean.

Option C:
Option C, 80, similarly gives a total of 104 when 24 is added back, leading to an average of 20.8, which is inconsistent with the initial information.

Option D:
Option D, 84, implies a total of 108 for five numbers, resulting in an average of 21.6, again conflicting with the given average of 18.

This option is correct because three consecutive integers can be written as n−1, n and n+1. Their sum is (n−1) + n + (n+1) = 3n. Dividing by 3 gives n, which is the middle integer. Therefore, the average equals the middle integer.

Option A:
The smallest integer in the set is n−1. The average of n−1, n and n+1 is n, which is larger than the smallest. Thus, the average cannot equal the smallest integer.

Option B:
The largest integer in the set is n+1. Since the average is n, which lies between n−1 and n+1, it is less than the largest value. Therefore, it is not equal to the largest integer.

Option C:
Saying the average equals the sum of the integers would ignore the division by the number of terms. The sum is 3n, while the average is 3n ÷ 3 = n. Thus, "sum of" does not describe the average correctly.

Option D:
The middle integer n is exactly the average of the three consecutive integers. It balances the smaller and larger values symmetrically. Hence, the average is always equal to the middle integer, making this option correct.

Statements A, B and C give the standard description and formula for arithmetic and weighted means. A is correct because the simple mean assigns equal weight to each observation. B and C correctly explain that weighted means use different weights and are computed as the weighted sum divided by the total weight. Statement E is true because if all weights are equal, the weighted mean collapses to the ordinary mean. Statement D is false since a weighted mean, like the arithmetic mean, can lie between observed values and need not coincide with any one observation, so the correct set is A, B, C and E only.

Option A:
Option A is incomplete because it omits C and E and so does not show how weighted means are actually computed or how they relate back to the arithmetic mean when weights are equal. It therefore does not represent all correct statements.

Option B:
Option B is also incomplete as it does not include E, ignoring the simplifying case in which weights are equal, even though it correctly includes A, B and C. The missing link to the arithmetic mean makes it insufficient.

Option C:
Option C is wrong because, although it includes B, C and E, it omits A, which states the key contrast between arithmetic and weighted means, and still does not highlight the equal-weight case as a special instance in context.

Option D:
Option D is correct because it brings together the accurate description of both kinds of means and the proper computation rule, while explicitly excluding D, which incorrectly insists that the mean must be one of the data points. This aligns with the conceptual treatment of averages in NET questions.

Statements A, B, C and E are all correct: they define the arithmetic mean, describe how adding a constant shifts the mean, explain that the updated mean lies between the old mean and the new value, and note that weighted averages incorporate differing importance through weights. Statement D is false because the average need not coincide with any actual data point; for example, the mean of 2 and 4 is 3, which is not in the list. Therefore, the combination that includes A, B, C and E and excludes D is correct.

Option A:
Option A is incomplete since it does not mention E and therefore overlooks the important idea that some aptitude problems require weighted averages rather than simple means. Without E, the concept coverage is narrower than intended.

Option B:
Option B is also incomplete: although B, C and E are true, the omission of A means the basic definition of arithmetic mean is missing. A must be included to fully represent how averages are computed.

Option C:
Option C is correct as it brings together all the true statements about how means behave and how weighted averages work, while rejecting D, which mistakenly claims that the mean must be an actual data value. It provides a complete conceptual picture for NET-level questions.

Option D:
Option D is wrong because it includes D, the false statement about averages always equalling a data point, and omits B, which correctly summarises the effect of adding a constant. This mixture makes the option logically inconsistent.

Statements A, B and C provide standard definitions of mean, median and mode. Statement E is also correct because in skewed distributions extreme values can distort the mean, making the median more representative. Statement D is false since equality of mean, median and mode holds only for certain symmetric distributions and not for every data set. Therefore, the combination A, B, C and E only is correct.

Option A:
Option A is correct as it collects all the accurate statements while leaving out D, which overgeneralises the equality of the three measures.

Option B:
Option B is incomplete because it omits E and therefore ignores the practical guidance on skewed data.

Option C:
Option C is wrong since it includes D, the false assertion about universal equality.

Option D:
Option D is wrong because it includes D and omits A, losing a core definition while keeping a false rule.

Statements A, B and E state standard facts about central tendency. A is correct because mean can lie left or right of the median in skewed distributions or coincide in symmetric cases. B is correct for perfectly symmetric unimodal distributions such as the normal distribution. E is correct since the median position splits an ordered data set into two halves with equal counts. C is false because mean and median need not always be equal, and D is false because mode may be non-unique or even absent, so A, B and E only is the correct combination.

Option A:
Option A is correct since it includes all and only the true statements: A about relative positions of mean and median, B about equality in symmetric unimodal distributions and E about the median’s partition property. It excludes C and D, both of which overstate what must always happen.

Option B:
Option B is incorrect because it adds C, which wrongly claims that median always equals mean for any data set. Including C introduces a false universal statement about equality of mean and median.

Option C:
Option C is wrong as it includes C again, making the set A, B, C, E contain one false member. A correct combination must consist entirely of true statements, so this option cannot be chosen.

Option D:
Option D is incorrect because it includes D, which asserts that the mode must always be unique. In reality, a distribution may be bimodal, multimodal or have no mode, so adding D makes this option incorrect.

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.

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.

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.