UGC NET Questions (Paper – 1)

The series is described by the quadratic formula aₙ = 3n²+2n+1 for n starting from 1. Substituting n = 1, 2, 3, 4 and 5 gives 6, 17, 34, 57 and 86, which match the given terms exactly. For n = 6 we obtain 3·6²+2·6+1 = 108+12+1 = 121. Hence the next term must be 121 to preserve the same quadratic relationship.

Option A:
Option A, 111, does not equal 3·6²+2·6+1 and therefore fails to extend the polynomial rule consistently. It would require lowering the value obtained from the formula by 10, which has no justification in the structure of the sequence. Thus 111 cannot be the correct continuation.

Option B:
Option B, 121, is exactly the value produced by the formula aₙ = 3n²+2n+1 when n = 6. It maintains the same dependence on n that explains all earlier terms. Because the pattern holds perfectly when this value is used, 121 is the correct next term.

Option C:
Option C, 115, lies between 111 and 121 but does not arise from substituting n = 6 into the quadratic expression. Selecting 115 would break the neat quadratic pattern and make the rule inconsistent. Therefore it is not the right answer.

Option D:
Option D, 129, is larger than the value predicted by the formula and again cannot be written as 3n²+2n+1 for n = 6. Using 129 would lose the tight algebraic link between n and aₙ, so it is not a valid continuation.

Average speed is defined as total distance divided by total time, not the average of the two speeds. The car travels 120 km each way, so the total distance is 240 km. The time taken in the first leg is 120 ÷ 40 = 3 hours and in the return leg is 120 ÷ 60 = 2 hours, giving a total of 5 hours. Dividing 240 by 5 yields 48 km/h as the average speed for the entire journey.

Option A:
Option A, 45 km/h, might be guessed by taking a simple average of the two speeds or by an approximate calculation, but it does not correspond to the correct total distance over total time.

Option B:
Option B uses the exact definition of average speed and calculates both distances and times explicitly. It also matches the harmonic mean formula for a round trip with equal distances, which simplifies to 2ab/(a + b).

Option C:
Option C, 50 km/h, exceeds the true average because more time is spent at the lower speed, pulling the overall average below the midpoint between 40 and 60.

Option D:
Option D, 52 km/h, is even higher and clearly inconsistent with spending a substantial part of the journey at just 40 km/h.

The rate of work of one man can be found by noting that 4 men finish the work in 12 days, so 1 man’s daily work is 1 divided by (4 × 12), which is 1/48. Similarly, 6 women finish the work in 18 days, so 1 woman’s daily work is 1 divided by (6 × 18), which is 1/108. For 2 men and 3 women together, the combined daily work is 2/48 plus 3/108. Simplifying, 2/48 is 1/24 and 3/108 is 1/36. The sum of 1/24 and 1/36 is 5/72. The total time to complete one work is 1 divided by 5/72, which equals 72/5 or 14.4 days.

Option A:
Option A, 12 days, would require the mixed group to be as effective as 4 men alone, which is not supported by the calculated rate of 5/72 per day. The combination of 2 men and 3 women is somewhat less efficient than that.

Option B:
Option B uses individual rates to compute a combined rate and then takes the reciprocal to determine total time. This standard time and work approach produces 14.4 days, which is the only option consistent with the carefully derived daily work rate.

Option C:
Option C, 15 days, is close to but not equal to 14.4 days, indicating a rounded or estimated value that does not precisely follow the computed reciprocal.

Option D:
Option D, 18 days, equates to the time taken by 6 women alone and ignores the additional contribution from the 2 men working alongside them.

The tap fills 1/12 of the tank per hour, while the leakage empties 1/18 of the tank per hour. When both operate together, the net filled amount per hour is 1/12 − 1/18. Taking the least common multiple of 12 and 18 as 36, we get the net rate as 3/36 − 2/36 = 1/36 of the tank per hour. Thus, the tank will be completely filled in 36 hours.

Option A:
Option A, 9 hours, would imply a net rate of 1/9 per hour, which is much faster than 1/36 and cannot be achieved under the given rates of filling and emptying. This ignores the slowing effect of the leakage.

Option B:
Option B, 12 hours, equals the time taken by the tap alone without any leakage. Since the leakage reduces the effective filling rate, the actual time must be longer than 12 hours.

Option C:
Option C, 18 hours, still underestimates the combined effect because the leakage continuously counters the filling tap. The computed net rate clearly shows that an even longer period is needed.

Option D:
Option D aligns with the net rate of 1/36 per hour obtained by subtracting the emptying rate from the filling rate. It is the only option that respects the mathematical relationship between the two processes.

The differences between consecutive terms are 10, 18, 26 and 34. These differences themselves form an arithmetic progression where each new difference is 8 more than the previous one. Therefore the next difference should be 42. Adding 42 to 91 gives 133 as the next term. This keeps the pattern at both the series and the difference levels consistent.

Option A:
Option A introduces a difference of 42, extending the sequence of gaps to 10, 18, 26, 34, 42. The increments between these differences are 8, 8, 8 and 8, preserving the arithmetic pattern. Hence 133 fits naturally as the next term.

Option B:
Option B would make the next difference 38, giving 10, 18, 26, 34, 38. The final increment from 34 to 38 is only 4, which breaks the consistent step of 8 seen earlier. Therefore 129 is not consistent with the rule.

Option C:
Option C corresponds to a difference of 44 from 91, creating the gaps 10, 18, 26, 34, 44. The increase from 34 to 44 is 10, not 8, so the progression among differences becomes irregular. Hence 135 cannot be the correct continuation.

Option D:
Option D gives a difference of 46, which results in an increment of 12 from the previous difference 34. This diverges even further from the constant +8 pattern and makes 137 an unsuitable next term.

A’s one-day work is 1/15 of the job and B’s one-day work is 1/20 of the job. Their combined daily work is 1/15 + 1/20. The least common multiple of 15 and 20 is 60, so this becomes 4/60 + 3/60 = 7/60 of the work per day. The total time to complete one whole work is therefore 1 ÷ (7/60) = 60/7 ≈ 8.57 days. Among the integer options given, 9 days is the closest approximation to 60/7, so it is the most appropriate answer.

Option A:
Option A, 8 days, underestimates the exact value of about 8.57 days; in 8 days they would complete only 8 × 7/60 = 56/60 of the work, leaving some part unfinished.

Option B:
Option B, 9 days, is the nearest whole number to 60/7 and slightly overestimates the exact time, which is why it is the best choice among the given options.

Option C:
Option C, 10 days, deviates further from the exact value and would correspond to a slower combined rate than actually calculated.

Option D:
Option D, 12 days, is much larger than 60/7 and ignores the productivity gain from having both A and B work together.

In the first hour, from 8:00 to 9:00 a.m., the first bus travels 40 km. After 9:00 a.m., the second bus starts and the relative speed between the two buses is 60 − 40 = 20 km/h. To close a gap of 40 km at a relative speed of 20 km/h, the second bus needs 40 ÷ 20 = 2 hours. Adding 2 hours to 9:00 a.m. gives 11:00 a.m. as the overtaking time.

Option A:
Option A, 10:00 a.m., would correspond to only one hour of catch-up time, which would close a distance of 20 km at a relative speed of 20 km/h, leaving 20 km still between the buses.

Option B:
Option B, 10:30 a.m., gives 1.5 hours of catch-up and would close 30 km of the gap, not the full 40 km, so the faster bus would still be behind.

Option C:
Option C, 11:00 a.m., correctly accounts for the full 2 hours needed at a relative speed of 20 km/h to cover the initial 40 km head start, exactly matching the required condition for overtaking.

Option D:
Option D, 11:30 a.m., overshoots the required catch-up time; by then, the second bus would have been ahead of the first bus for half an hour already.

The first-level differences are 13, 21, 29 and 37. These differences form an arithmetic progression with common difference 8. To keep this structure, the next difference should be 45. Adding 45 to the last term 107 gives 152, so 152 continues both the original sequence and the pattern among the differences.

Option A:
Option A gives 148, which corresponds to a difference of 41 from 107. This would produce differences 13, 21, 29, 37, 41, making the last increment among gaps equal to 4 instead of 8. Thus 148 does not respect the observed second-level behaviour.

Option B:
Option B gives 150, implying a difference of 43 from 107. The final step among differences would be 6, which still fails to match the constant increment of 8. Hence 150 does not align with the pattern.

Option C:
Option C gives 152, which implies a difference of 45 from 107. The sequence of differences becomes 13, 21, 29, 37, 45 and the increments remain 8 throughout. This perfect consistency at the second level makes 152 the correct next term.

Option D:
Option D gives 154, yielding a difference of 47 from 107. That would make the last increment among differences equal to 10, breaking the simple arithmetic progression. Therefore 154 is not a valid continuation.

A’s one day work is 1/10 of the job and B’s one day work is 1/15 of the job. When they work together, the daily work done is the sum of their individual rates: 1/10 + 1/15. Taking the LCM of 10 and 15 gives 30, so the combined rate is (3/30 + 2/30) = 5/30 = 1/6. Thus, together they complete 1/6 of the work each day and need 6 days to finish the entire work.

Option A:
Option A, 4 days, would imply a combined rate of 1/4 per day, which is higher than what A and B can achieve together. The actual sum of their rates does not support completion in 4 days.

Option B:
Option B, 5 days, would mean they do 1/5 of the work per day, requiring a combined rate larger than 1/6. However, the calculated total rate from their individual capacities is only 1/6.

Option C:
Option C is consistent with the reciprocal method: adding 1/10 and 1/15 gives 1/6, whose reciprocal is 6 days. This uses the standard approach for time and work problems and fits the given data exactly.

Option D:
Option D, 8 days, underestimates their combined productivity and corresponds to a rate of 1/8 per day, which is lower than the actual 1/6 per day.

A ratio is defined as the comparison of two quantities of the same kind by division. For example, if there are 3 boys and 5 girls, the ratio of boys to girls is 3:5, which represents 3 divided by 5. Ratios are fundamental in many aptitude topics such as proportion, mixtures and scaling. Hence, the phrase "comparison of two quantities by division" is the textbook definition of ratio.

Option A:
Ratio is correct because it explicitly involves dividing one quantity by another to express their relative size. This form can be written as a/b, a:b or "a to b," all representing a comparative measure. In UGC NET mathematical aptitude questions, understanding ratios is essential for solving problems on partnerships, mixtures and map scales. Therefore, this option precisely fits the description.

Option B:
A fraction is a representation of part of a whole, usually written as a/b, but it does not necessarily compare two separate quantities of the same kind. While ratios can be expressed in fractional form, the context of comparison of two distinct quantities is more specific to ratios. Thus, fraction is not the most accurate term for the definition given.

Option C:
Percentage expresses a quantity per hundred, such as 25% meaning 25 out of 100. It is a special way of writing fractions or ratios with a fixed denominator of 100. The stem, however, speaks more generally about comparison by division without specifying any base, so percentage is too narrow and not the exact term required.

Option D:
Proportion is a statement that two ratios are equal, such as 2:3 = 4:6. It builds upon the idea of ratio but involves an additional equality relation. Since the question is asking just for the basic comparison by division, proportion goes one step beyond and is therefore not correct.

The sequence is generated by aₙ = n⁴ + n³ + 3n² + 2 for n starting from 1. For n = 1, 2, 3, 4 and 5 we calculate 1+1+3+2 = 7, 16+8+12+2 = 38, 81+27+27+2 = 137, 256+64+48+2 = 370 and 625+125+75+2 = 827, matching the given terms exactly. For n = 6 this formula yields 6⁴+6³+3·6²+2 = 1296+216+108+2 = 1622. Thus 1622 is the only value that continues this specific quartic-plus-cubic-plus-quadratic pattern.

Option A:
Option A, 1622, matches perfectly the value obtained from n⁴+n³+3n²+2 when n = 6. It preserves all components of the expression and continues the same growth trend seen in the previous terms. Because no modification of the coefficients or constant is required, 1622 is the correct continuation of the series.

Option B:
Option B, 1606, is 16 less than the formula’s output and does not satisfy the rule for n = 6. Choosing 1606 would require reducing the term without any structural change in the function, which contradicts the exact fit observed earlier. Hence option B is incorrect.

Option C:
Option C, 1640, overshoots the correct value by 18 and again cannot be written as n⁴+n³+3n²+2 for n = 6. It represents an arbitrary increase that is not indicated by the previous behaviour of the series. Therefore option C is incorrect.

Option D:
Option D, 1664, deviates even more from the computed value and has no basis in the established expression. Using 1664 would break the precise mapping between index and term that defines this number series, so option D is incorrect.