UGC NET Questions (Paper – 1)

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.

In the first alloy, copper and zinc are 7/10 and 3/10 of the weight respectively. In the second alloy, copper and zinc are 5/7 and 2/7 of the weight respectively. Taking equal weights of each alloy, total copper = 7/10 + 5/7 = (49 + 50)/70 = 99/70 and total zinc = 3/10 + 2/7 = (21 + 20)/70 = 41/70. The common denominator 70 cancels in the ratio, so copper:zinc = 99:41.

Option A:
Option A, 89:41, looks close to the correct ratio but uses 89 instead of 99, indicating an incorrect sum of copper fractions. To obtain 89, one would have to miscalculate either 7/10 or 5/7, and using this ratio would not respect the exact contributions from both alloys.

Option B:
Option B, 41:99, simply reverses the correct ratio and would imply that zinc is present in much larger proportion than copper in the mixture. This contradicts the original alloys, both of which have more copper than zinc, so 41:99 cannot represent the final proportions.

Option C:
Option C, 99:44, keeps the correct numerator 99 but inflates the zinc component from 41 to 44. This could arise from careless arithmetic in adding 3/10 and 2/7. However, if zinc were 44 parts, the total fractions would no longer sum to the actual combined mixture, so 99:44 is inconsistent with the data.

Option D:
Option D is correct because it is derived from adding the fractional parts for copper and zinc separately and then simplifying the resulting ratio. The common denominator cancels out, leaving 99:41 as the exact composition ratio for the combined alloy when equal weights of each initial alloy are used.

This option is correct because when equal quantities are mixed, the concentration of the mixture is the simple average of the two concentrations. The average of 10% and 20% is (10 + 20)/2 = 15%. Since both quantities are equal, no other weighting is needed. Therefore, the resulting concentration is 15%.

Option A:
If the mixture concentration remained at 10%, the presence of the stronger 20% solution would have no effect. This contradicts the idea of mixing. Therefore, 10% cannot be the concentration after mixing equal amounts.

Option B:
When equal volumes are combined, each solution contributes equally to the final concentration. Averaging 10% and 20% gives 15%, which lies between the two original values. This matches our expectation for a mixture of equal parts, so 15% is correct.

Option C:
A concentration of 20% would ignore the dilution effect of the weaker 10% solution. The final concentration cannot be as high as the stronger solution when they are mixed equally. Thus, 20% is not correct.

Option D:
30% exceeds both of the original concentrations. Mixing cannot produce a concentration stronger than the strongest component without adding extra solute. Therefore, 30% is impossible in this context.

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, C and E are correct as they capture the essence of mixture problems: combining components, summing component quantities, using the rule of alligation as a shortcut and representing scenarios algebraically or graphically. Statement D is false because mixture questions are naturally modelled by weighted averages where proportions act as weights. Thus, the combination A, B, C and E only is correct.

Option A:
Option A is incomplete as it leaves out E, not mentioning that visual or diagrammatic approaches are often used to make mixture relationships clearer in exams. Without E, the modelling variety is understated.

Option B:
Option B is also incomplete because it omits A and thereby fails to state directly that mixtures can involve various types of components like solutions or grains, which is basic contextual information.

Option C:
Option C is correct because it collects all the true statements about mixture structures, alligation shortcuts and representation methods, while excluding D, which incorrectly denies the connection with weighted averages. This matches the treatment in UGC NET aptitude.

Option D:
Option D is wrong as it includes D, which wrongly claims that weighted averages cannot be used for mixtures, and omits B, which accurately describes how component quantities combine. The presence of D invalidates the set.

Statement A correctly describes the overall concentration as a weighted average of component concentrations. Statement B is a special case for equal quantities, where the weights are equal and the average is arithmetic. Statement C states the classical alligation rule connecting mixing ratio to the differences from the mean. Statement D is false because adding pure water does not change the solute amount, and E is false since diluting with water lowers concentration but does not decrease the amount of solute. Hence A, B and C only are correct, which corresponds to option A.

Option A:
Option A is correct because it includes the general weighted-average idea, its equal-quantity special case and the alligation rule, while excluding D and E, which misunderstand how solute amount changes under dilution.

Option B:
Option B is incorrect since it includes D, which wrongly asserts that adding pure water increases solute amount, mixing an error with accurate statements.

Option C:
Option C is wrong because it includes E (false: adding water alone does not reduce solute amount) and omits B (true equal-quantity averaging), so it does not match the true set.

Option D:
Option D is incorrect as it contains D (false) and omits A (the key weighted-average principle), making it both incorrect and incomplete.

Statement A is correct since with equal maxima the candidate corresponding to the larger ratio part has the higher score. Statement C is true because when group sizes differ, the overall average must weight each group average by its size. D is also correct since percentages place different exams on a common scale and make comparisons meaningful. E is correct because ratios and percentages adjust for different totals and show relative performance better than raw marks alone. B is false because simply averaging group means without considering sizes can give wrong overall values, so A, C, D and E only are correct.

Option A:
Option A is incomplete; although it lists A and C, it omits D and E and thus fails to acknowledge the role of percentages and relative measures in comparing performance. As a result, it does not give the full conceptual picture required.

Option B:
Option B is also incomplete because it leaves out E and includes only A, C and D, so it does not stress the comparative advantage of ratios and percentages over raw scores in unequal situations.

Option C:
Option C is wrong because it includes only C, D and E and omits A, losing the simple but important fact that the greater ratio component corresponds to higher marks under equal maxima. The combination is therefore not fully correct.

Option D:
Option D is correct because it brings together the ideas of ratio comparison, weighted averaging and the usefulness of percentages and relative measures, while excluding B, which misapplies the averaging rule. It mirrors the reasoning needed for NET data interpretation problems on exam marks.