UGC NET Questions (Paper – 1)

Each term is a four-letter consecutive block. Starting letters B(2), F(6), J(10), N(14) increase by +4. Next start is R(18), so the block is RSTU.

Option A:
Option A starts at Q(17), which is only +3 from N, not +4.

Option B:
Option B is correct because R is +4 from N and S,T,U follow consecutively.

Option C:
Option C starts at S(19), which is +5 from N, not +4.

Option D:
Option D starts at O(15), which is +1 from N and also disrupts the forward pattern.

The relation “x varies directly as the square of y” can be written as x = k·y² for some constant k. Using x = 50 when y = 5 gives 50 = k·25, so k = 50/25 = 2. For y = 7, we have x = 2·7² = 2·49 = 98. Thus, the new value of x corresponding to y = 7 is 98.

Option A:
Option A, 70, might result from incorrectly assuming x varies directly as y (not y²), which would give x proportional to 5 → 50 and 7 → 70. However, this ignores the square dependence stated in the problem, so 70 cannot satisfy the given variation law.

Option B:
Option B, 84, could arise from partially accounting for the square by mixing linear and quadratic reasoning, but substituting y = 7 into x = 2y² clearly yields 98, not 84. Therefore, 84 fails the direct substitution test in the correct formula.

Option C:
Option C is correct because it uses the exact variation relationship x = 2y² determined from the first pair (x, y) = (50, 5). Substituting y = 7 gives 98, which is consistent with the nature of direct variation with the square and the original data point.

Option D:
Option D, 112, would correspond to x = 2·8² if y were 8, or to another mistaken scaling, but does not arise from y = 7 in the function x = 2y². Hence, 112 is incompatible with the established proportional rule.

Each of these terms is a four letter palindrome where the first and last letters are the same and the middle pair are also the same. From ANNA to BOOB to CPPC to DQQD, the outer letters move forward A, B, C, D while the inner letters move forward N, O, P, Q. To follow the pattern, the next outer letter must be E and the next inner letter must be R, one step after Q. This yields ERRE, which alone preserves both the palindromic form and the steady increase of the inner and outer letters, making it the correct next term.

Option A:
Option A, ESSE, uses E at the ends but changes the inner letter to S, which is not the next letter after Q in the sequence. The inner series N, O, P, Q would be broken if S were used. Because it neglects the simple one step growth of the inner letters, ESSE cannot be correct here.

Option B:
Option B is correct because ERRE places E at both ends as the next outer letter and R in the middle as the next inner letter after Q. The structure mirrors that of all previous terms and continues both sequences smoothly. This perfect match with the dual progression of outer and inner letters confirms ERRE as the right continuation.

Option C:
Option C, ERTE, breaks the rule that the middle two letters must be identical by using R and T. The series has consistently kept the second and third letters the same. Since ERTE does not retain this palindromic property, it does not follow the pattern.

Option D:
Option D, FRRF, starts with F instead of E, jumping the outer letter two steps ahead rather than one. It also interferes with the simple progression of the inner letters. Therefore FRRF is inconsistent with the rule governing the series.

Each group is a block of four consecutive letters. The starting letters are E, G, I and K (positions 5, 7, 9, 11), increasing by 2 each time. The next start is M (13), giving MNOP.

Option A:
Option A is correct because it begins at M, exactly two positions after K, and continues with N, O and P consecutively.

Option B:
Option B starts at N (14), which is +3 from K, not +2, so it breaks the step.

Option C:
Option C starts at L (12), only +1 from K, so it breaks the step.

Option D:
Option D starts at O (15), +4 from K, so it breaks the step.

The daily work rates of A, B and C are 1/10, 1/15 and 1/30 respectively. Working together, their combined rate is 1/10 + 1/15 + 1/30, which simplifies to 1/5 of the work per day. In the first 2 days, they therefore complete 2 × 1/5 = 2/5 of the work. The remaining 3/5 of the work is then done by B and C alone, whose combined rate is 1/15 + 1/30 = 1/10. At this rate, they need (3/5) ÷ (1/10) = 6 more days. Adding the initial 2 days gives a total of 8 days to finish the work.

Option A:
Option A correctly adds the contributions in the two phases: an initial phase with three workers and a second phase with two workers. It respects the different group rates and sums the corresponding time intervals to reach 8 days.

Option B:
Option B, 7 days, underestimates the time and would require the second phase to progress faster than the combined rate of B and C allows.

Option C:
Option C, 6 days, is only the time needed for B and C to complete 3/5 of the work, not including the first 2 days already spent by all three.

Option D:
Option D, 9 days, overestimates the total time and suggests a slower rate than actually achieved by the workers as a group.

Statements A, B and C describe how to handle clocks that do not keep correct time. A is correct because a gaining clock runs ahead of real time. B is true since losing 5 minutes per hour over 12 hours accumulates a loss of 60 minutes, i.e. one hour. C is also correct because we treat the error per hour as a rate and compare it with a complete 60-minute error to see when the clock will be exactly one hour fast or slow and thus again coincide with the correct time on the dial. D is false because gain or loss cannot be ignored in problems about correct readings, and E is false because there is no guarantee that the error aligns perfectly at midnight. Thus A, B and C only are correct.

Option A:
Option A is incomplete as it gives only A and B and omits C, which explains the method of using the error rate to determine when the clock again shows the correct reading. Without C, the understanding of such problems is not complete.

Option B:
Option B is correct since it includes all three true statements and excludes D and E, which disregard or misdescribe the impact of gain or loss. It directly corresponds to the typical reasoning steps required in NET-level clock questions.

Option C:
Option C is wrong because it introduces D, which incorrectly claims that gain or loss can always be ignored, even though angle-based questions sometimes require perfect time. Including D mixes an error into the option.

Option D:
Option D is also incorrect as it omits A and instead lists only B and C, without clearly distinguishing faulty clock behaviour from standard angle problems, and still does not address the false ideas in D and E explicitly.

A is correct because subset means every element of A belongs to B. D is true as the empty set, having no elements, vacuously satisfies the subset condition for any set. E is also correct because mutual inclusion implies equality of sets. B is false since a proper subset is strictly smaller than the set, and C is false because the universal set contains all elements, so other sets are subsets of U, not the other way round. Thus A, D and E only are correct.

Option A:
Option A lists A and D but omits E, failing to express how mutual subset relations guarantee equality. Without E, the account of subset relations is incomplete.

Option B:
Option B includes A and E but omits D, ignoring the important property of the empty set as a subset of every set. This makes the combination less comprehensive.

Option C:
Option C is correct because it combines all three true statements and excludes B and C, which misrepresent proper subset and universal set roles. It fully reflects the basic set relations used in reasoning questions.

Option D:
Option D includes only D and E, omitting A, and therefore does not explicitly state what it means for A to be a subset of B. This leaves out a fundamental definition.

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 C give standard definitions of injective, surjective and bijective functions. E is also correct in the finite equal-cardinality case: an injective map from A to B must be surjective. D is false because surjectivity alone does not guarantee injectivity; many onto but not one-to-one functions exist. Therefore A, B, C and E only are correct.

Option A:
Option A is correct as it aggregates all the true statements about these function types, including the finite-set result, while leaving out D, which wrongly treats surjectivity as forcing injectivity.

Option B:
Option B is incorrect because it omits E, and E is a true theorem for finite sets of equal size (injective ⇒ surjective).

Option C:
Option C is incorrect because it omits A (a true definition of injective), so it does not list all correct statements.

Option D:
Option D is incorrect because it omits B (a true definition of surjective), so it does not list all correct statements.

Statements A and B are true because the weekday sequence repeats every seven days and a non-leap year of 365 days shifts the starting weekday by one day in the next year. Statement E is also correct since many calendar questions ask candidates to determine the weekday for a given date. Statement C is false because leap years have 366 days, and D is false because century years must be divisible by 400, not merely by 4, to be leap years. Hence, the combination A, B and E only is correct.

Option A:
Option A is incomplete as it omits E and does not explicitly connect the calendar structure to typical exam tasks that require finding weekdays. Without E, the applied dimension of the topic is missing.

Option B:
Option B is correct because it groups the three true statements about the cyclic nature, year shift and exam usage of calendar facts, while rejecting C and D, which misrepresent leap-year and day-count rules. It thus matches the logical content of NET-level calendar questions.

Option C:
Option C is wrong because it adds C, which incorrectly assigns 365 days to leap years, and still omits D, failing to discuss the additional condition for century years. Including C makes the option factually incorrect.

Option D:
Option D is incorrect since it includes D, the statement that ignores the special rule for centuries, and therefore endorses a mistaken description of the Gregorian calendar. This misclassification invalidates the option.

Each term is a block of four consecutive letters, but the starting letters move backward by two each time. The starting letters M, K, I and G are at positions 13, 11, 9 and 7, all decreasing by 2. Continuing this pattern, the next starting letter is position 5, which is E, and the four letter block from E is EFGH. Hence EFGH continues both the backward step and the consecutive group structure.

Option A:
Option A, FGHI, starts at F, which is one position after G instead of two positions before it. This contradicts the clear pattern of stepping back by two in the starting letters. Although FGHI is a consecutive group, it does not come from the correct starting point and therefore is not the right answer.

Option B:
Option B is correct because it uses E as the starting letter, exactly two positions before G, and forms the consecutive sequence E, F, G and H. This keeps the block length constant and preserves the minus two jump in the starting positions. It fits the way the previous terms move backward through the alphabet.

Option C:
Option C, DEFG, begins at D, which would represent a step of minus three from G rather than minus two. This changes the size of the backward movement without any indication that the pattern is altering. As the series exhibits a consistent decrement of two, DEFG cannot be accepted.

Option D:
Option D, HIJK, moves forward from G instead of backward and breaks the monotonic decrease in the starting letters. It ignores the structure of moving left through the alphabet and so does not satisfy the series rule.

Each term follows a clear rule: the first letter increases by 1 (A, B, C, D, …), the second letter decreases by 1 from the end (Z, Y, X, W, …) and the third letter increases by 1 from C onward (C, D, E, F, …). After DWF, the first letter should become E, the second V and the third G. Therefore EVG is the only option that satisfies all three simultaneous movements.

Option A:
Option A, EVF, gives the correct first letter E and second letter V but keeps the third letter at F. The third letter must also advance by one position compared to the previous term’s third letter F. Since EVF does not reflect this change, it fails to meet the full pattern.

Option B:
Option B is correct because its letters E, V and G correspond precisely to the next steps in their respective progressions. It advances the first and third letters by one and moves the second letter back by one, exactly as earlier terms did. This complete alignment makes EVG the accurate continuation of the series.

Option C:
Option C, EWG, changes the second letter from W instead of V, skipping the required step in the second position. It also disrupts the relationship between the second and third letters across terms. Because it ignores the careful backward stepping of the middle letter, EWG is not consistent.

Option D:
Option D, FVG, advances the first letter by two positions from D to F while using the correct second and third letters. This breaks the +1 rule for the first position. Thus, even though part of the term looks plausible, the entire group fails the strict pattern.

The terms follow the rule aₙ = 2n⁴ + 2n³ + n with n starting from 1. For n = 1, 2, 3, 4 and 5 this formula produces 5, 50, 219, 644 and 1505, matching the given series perfectly. Evaluating the same expression for n = 6 yields 3030. Hence 3030 is the only number that preserves this quartic-plus-cubic pattern.

Option A:
Option A, 3006, is 24 less than the value obtained from aₙ = 2n⁴ + 2n³ + n at n = 6. To accept 3006 we would need to reduce the polynomial result arbitrarily at the sixth term. This change is not reflected anywhere else in the progression, so option A is incorrect.

Option B:
Option B, 3020, is 10 less than the correct value and also fails to equal the formula’s output for n = 6. Although numerically close, it breaks the precise algebraic relationship between n and aₙ. Therefore option B does not correctly continue the series.

Option C:
Option C, 3030, matches exactly the value produced by the expression when n = 6. It maintains the same contributions from the fourth-power and cubic terms along with the linear part as in earlier terms. Because this rule explains the entire sequence and extends naturally to 3030, option C is correct.

Option D:
Option D, 3046, is 16 greater than the expected value and cannot be derived from the formula at n = 6. Choosing 3046 would introduce an unjustified increase at the last step, disrupting the pattern. Thus option D is not a valid continuation.

This option is correct because the formula for simple interest is I = PRT/100. When P and R are fixed, the interest I varies directly with T. Doubling the time doubles the interest, and halving the time halves the interest. Therefore, simple interest is directly proportional to the time period.

Option A:
The time period appears as a direct factor in the simple interest formula. If P and R stay constant, any change in T results in a proportional change in I. This shows a clear direct proportionality between interest and time, so this option correctly identifies the varying quantity.

Option B:
Compound amount includes both principal and interest in compound interest problems and is not part of the simple interest formula in this way. Simple interest is not directly proportional to compound amount. Thus, this option is not appropriate.

Option C:
Discount allowed refers to the reduction from the marked price in sales, which belongs to commercial arithmetic. It does not appear in the simple interest formula and is unrelated to the direct proportionality mentioned. Therefore, discount is not the correct answer.

Option D:
Selling price is determined by cost, profit or loss and discount, not directly by simple interest. While interest can affect selling price in some credit contexts, the simple interest formula itself does not express any direct relationship with selling price. Hence, this option is incorrect.

The series alternates between doubling and adding 2. We have 7×2 = 14, 14+2 = 16, 16×2 = 32, 32+2 = 34 and 34×2 = 68. Following this pattern, the next step should be "+2". Adding 2 to 68 gives 70. Thus, 70 is the correct next term in the alternation.

Option A:
Option A is 66, which would require subtracting 2 from 68 rather than adding. No earlier transition uses subtraction, so this choice violates the pattern. Therefore, 66 is not correct.

Option B:
Option B equals 70 and results from 68+2, preserving the alternation of ×2, +2 operations. The extended series 7, 14, 16, 32, 34, 68, 70 maintains the same structure. Hence, 70 is the valid continuation.

Option C:
Option C is 72, implying an increase of 4 from 68, which would change the "+2" part of the rule. There is no evidence of adding 4 in any previous step, so 72 cannot be accepted.

Option D:
Option D is 76, giving a jump of 8, which would introduce both a new magnitude and operation. This disrupts the clean alternating rule and therefore is not appropriate.