UGC NET Questions (Paper – 1)

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 series consists of consecutive odd numbers, each 2 greater than the previous one. We see 11, 13, 15 and 17 with differences of 2 throughout. Adding 2 to 17 gives 19. Therefore, 19 is the next consecutive odd number and fits the pattern.

Option A:
Option A is 18, which is an even number and would break the sequence of odd integers. It also changes the difference from 2 to 1, so it cannot be correct.

Option B:
Option B is 20, another even number, and is 3 greater than 17. This disturbs both the parity and the constant difference. Thus, it does not maintain the pattern.

Option C:
Option C is 21, which is odd but lies 4 units above 17 rather than 2. That disrupts the simple increase by 2 seen in the earlier terms. Therefore, 21 is not appropriate.

Option D:
Option D equals 19, an odd number exactly 2 more than 17. The extended series 11, 13, 15, 17, 19 keeps the same difference and parity property. Hence, this option correctly continues the sequence.

The sequence lists consecutive odd numbers starting from 1, with each term 2 greater than the previous one. We have 1, 3, 5, 7 and 9, all odd and evenly spaced. Adding 2 to 9 gives 11. Hence, 11 is the correct next term that keeps the list of odd numbers in order.

Option A:
Option A is 10, which is even and would break the pattern of odd integers. It also changes the difference from 2 to 1, so it does not match the structure.

Option B:
Option B equals 11, maintaining both the odd property and the consistent step of 2. Extending the series to 1, 3, 5, 7, 9, 11 keeps the progression uninterrupted. This makes 11 the appropriate continuation.

Option C:
Option C is 13, which is an odd number but lies 4 units above 9. Choosing 13 would skip 11 and alter the fixed difference of 2. Therefore, it cannot be the immediate next term.

Option D:
Option D is 15, which is further away and would correspond to a difference of 6 from 9. This would distort the regularity of the sequence. So, 15 is not correct in this context.

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.