CSAT Solved Papers/ 2021/Q40

2021 CSAT — Q40

Quant Logical & quantitative reasoning 2.5 marks Medium

In the English alphabet, the first $4$ letters are written in opposite order; and the next $4$ letters are written in opposite order and so on; and at the end $Y$ and $Z$ are interchanged. Which will be the fourth letter to the right of the $13$ th letter?

A N
B T Answer
C H
D I

Worked rationale

Reverse the alphabet in consecutive blocks of $4$ , and swap the final pair $Y,Z$ :

\underbrace{DCBA}_{1\text{–}4}\ \underbrace{HGFE}_{5\text{–}8}\ \underbrace{LKJI}_{9\text{–}12}\ \underbrace{PONM}_{13\text{–}16}\ \underbrace{TSRQ}_{17\text{–}20}\ \underbrace{XWVU}_{21\text{–}24}\ \underbrace{ZY}_{25\text{–}26}

Now read off positions. The $13$ th letter is $P$ . The fourth letter to the right of it is position $13+4 = 17$ , which is $T$ .

Answer: (b) T.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2021 CSAT Q40 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

stops short: counts to position $13+1=14$ (or miscounts to $N$ at $15$ ), not four to the right.
C

counts the wrong way: goes to the left of the $13$ th letter instead of to the right.
D

mis-blocks the reversal: reverses within blocks inconsistently or mislabels the block boundaries, shifting the index.

Specialist insight

Write the transformed alphabet once, in labelled blocks of $4$ , before answering anything — the reversal $DCBA\,HGFE\,LKJI\dots$ is mechanical and the $Y,Z$ swap only touches the tail. After that the question is pure position arithmetic: ” $4$ to the right of the $13$ th” $=$ position $17$ . The two traps are direction (right vs left) and the fencepost (“fourth to the right” $= +4$ , landing on $17$ , not counting the start). Build the string, then index — never index in your head.

The trap, in one line

Transformed string puts $T$ at position $17 = 13+4 \Rightarrow$ (b).

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