UGC NET Questions (Paper – 1)

Outer letters move forward A,B,C,D → E. Inner double letters move backward Z,Y,X,W → V. So the next palindromic block is EVVE.

Option A:
Option A is correct because E is next outer letter and V is next inner letter (one step before W).

Option B:
Option B uses U, which is two steps before W, so it breaks the one-step backward movement.

Option C:
Option C breaks the palindromic structure (outer letters differ) and keeps W incorrectly.

Option D:
Option D jumps the inner letters back to X, reversing the backward sequence.

Each group is a three letter palindrome where the first and last letters are the same and increase in order A, B, C, D. The middle letters move backwards from the end of the alphabet in the order Z, Y, X, W. The next outer letter should be E and the next middle letter in this backward sequence should be V. Thus EVE is the only group that satisfies both movements simultaneously.

Option A:
Option A, EUE, uses E correctly as the outer letter but chooses U instead of V for the middle position. The middle letters have been stepping down one letter at a time from Z, and U would skip V in that sequence. Because it breaks this simple reverse progression, EUE cannot be the right continuation.

Option B:
Option B is correct because it combines E as the next outer letter and V as the next middle letter after W. It preserves the palindromic structure and keeps the two opposing movements of outer letters forward and inner letters backward perfectly intact. This strict adherence to the observed pattern makes EVE the logical next term.

Option C:
Option C, EVF, is not a palindrome at all since the first and last letters do not match. The series clearly uses groups where the outer letters are identical. By breaking this defining structural feature, EVF fails to follow the pattern of the series.

Option D:
Option D, EWE, keeps the palindromic structure but repeats W as the middle letter instead of moving to V. This freezes the backward movement of the central letters, which has been consistent so far. Therefore EWE does not continue the pattern correctly.

The first letters A, B, C and D move one step forward each time. The middle letters Z, Y, X and W move one step backward from the end of the alphabet. The last letters B, C, D and E move one step forward from the beginning. The next first letter after D is E, the next backward middle letter after W is V, and the next forward last letter after E is F, giving EVF.

Option A:
Option A, EVG, uses E and V correctly for the first and middle positions but moves the last letter two steps ahead from E to G. The last letters in the series only advance by one position each time. Because it breaks this one step increment in the last position, EVG does not fully respect the pattern.

Option B:
Option B is correct because it combines E, V and F, each obtained by a single step move from the respective letters of DWE. E is one step after D, V is one step before W, and F is one step after E. This continues all three directional movements consistently and therefore EVF fits the series exactly.

Option C:
Option C, EUF, alters the middle letter incorrectly by taking U, which is two steps before W. The middle letters should move only one position at a time in reverse order. Since U breaks the uniform backward movement, EUF cannot be the next term.

Option D:
Option D, EWF, keeps the last letter F correct but fails to move the middle letter from W to V. It freezes the middle letter instead of shifting it backward, which goes against the established transformation. Consequently, EWF is not a valid continuation of the pattern.