Letter Classification Test - 1

A(BC)D
J(KL)M
P(QRS)T
W(XY)Z

First three combinations are obtained by writing first two letters in succession and then writing the third one by skipping two letters. On the other hand, in option (4), the third letter of the combination is written after skipping three letters. So, option (4) is the odd one out.

In all the other options, except option 2, one letter is skipped between the two given letters in each pair.

In the options,
1. AEF = A(+4)E(+1)F
2. GKM = G(+4)K(+2)M
3. RVW = R(+4)V(+1)W
4. UYZ = U(+4)Y(+1)Z

Here, in each group, three consecutive letters of the English alphabet are taken and third letter is placed in the middle. The group STU does not follow the same rule, hence the odd one.

In all the options, except option (3), one letter is skipped after each letter.

In all other options, except (2), there is a skip of 3 letters between the first and the second letters and the last one is the next letter to the second one.

A + 4 = E, E + 4 = I
F + 4 = J, J + 4 = N
P + 4 = T, T + 3 = W
K + 4 = O, O + 4 = S

B(2): 2² = 4 = D
E(5): 5² = 25 = Y
D(4): 4² = 16 = P
C(3): 3² = 9 = I ≠ J

Going by options:
1. NO = N + 1 = O
2. MP = M + 3 = P
3. UX = U + 3 = X
4. ZC = Z + 3 = C

(1) JK (L) MN (Gap of 1)
(2) ST (UV)WX (Gap of 2)
(3) XY (Z) AB (Gap of 1)
(4) EF (G) HI (Gap of 1)

Option 1: D + 2 = F, F + 2 = H, H + 2 = J
Option 2: K + 2 = M, M + 2 = O, O + 1 = P

Option 3: R + 2 = T, T + 2 = V, V + 2 = X
Option 4: Q + 2 = S, S + 2 = U, U + 2 = W

Option 5: S + 2 = U, U + 2 = W, W + 2 = Y

KMOP is different from others.