Jamia Millia Islamia MCA Letter Series Previous Year Questions (PYQs)

Solution: Work column-wise with a constant step of −8 in alphabet positions (A=1,…,Z=26): Col1: Z(26) → R(18) → ? ⇒ 26−8=18−8=10 ⇒ J Col2: ? → O(15) → G(7) ⇒ top must be 23 ⇒ W Col3: S(19) → ? → C(3) ⇒ middle must be 11 ⇒ K Thus the three letters (top-middle, middle-right, bottom-left) are W, K, J → “WKJ”.

Left letters $A,B,D,G,K$ with steps $+1,+2,+3,+4$ → next $+5$: $K$ stays as left? (already K from prev step), actually pattern gives next pair $K\to P$ and right $B,D,G,K\to P$; number $1,2,3,4\to5$. So $K5P$. $\boxed{K_5P}$

Left $C,D,E,F\to G$; numbers $1,3,5,7\to 9$; right $Z,Y,X,W\to V$. Needed $G9V$ — not in options.

First letters $A,B,D,G$ with jumps $+1,+2,+3$ → next $+4$: $K$. Second letters $B,D,F,H$ ($+2$ each) → $J$. Third letters $Z,Y,X,W$ ($-1$ each) → $V$. $\boxed{\text{KJV}}$

First letters $C,E,G,I\to K$ ($+2$); second $A,B,C,D\to E$ ($+1$); third $T,S,R,Q\to P$ ($-1$). $\boxed{\text{KEP}}$

Solution: Look at odd and even positions separately. Odd positions (1,3,5,7,9,11,13): b, a, ?, b, ?, b, ? → pattern = "bab" repeating ⇒ 5=b, 9=a, 13=b. Even positions (2,4,6,8,10,12,14): a, b, a, a, b, a, ? → pattern = "aba" repeating ⇒ 14=a. Missing terms (in order 5th, 9th, 13th, 14th) = b, a, b, a → **baba**.

Solution: Two interleaved sequences: - 1st,3rd,5th,7th,…: Z(26), X(24), V(22), T(20) ⇒ decreasing by 2 ⇒ next = R(18). - 2nd,4th,6th,8th,…: L(12), J(10), H(8), F(6) ⇒ decreasing by 2 ⇒ next = D(4). So the next two letters are **R, D**.

Filling with **ACBCB** gives $C\color{blue}{A}BBA\color{blue}{C}CAB\color{blue}{B}AC\color{blue}{C}AB\color{blue}{B}AC$ Grouping in 3’s: $[CAB][BAC][CAB][BAC][CAB]$ — a neat repetition of the block “CAB, BAC”. Other choices break this pattern.