CSAT Solved Papers/ 2023/Q37

2023 CSAT — Q37

Quant Logical & quantitative reasoning 2.5 marks Easy

If ‘ZERO’ is written as ‘CHUR’, then how is ‘PLAYER’ written?

A SOCAGT
B SODBGT
C SODBHT
D SODBHU Answer

Worked rationale

Find the shift from the anchor, letter by letter (positions $A = 1, \dots, Z = 26$ ), wrapping mod $26$ :

$Z(26) \to C(3)$ : $+3$ (wrap $26 \to 1 \to 2 \to 3$ )
$E(5) \to H(8)$ : $+3$
$R(18) \to U(21)$ : $+3$
$O(15) \to R(18)$ : $+3$

The rule is a $+3$ shift on every letter. Apply to PLAYER:

$P(16) \to S(19)$
$L(12) \to O(15)$
$A(1) \to D(4)$
$Y(25) \to B$ (wrap $25 \to 26 \to 1 \to 2 = B$ )
$E(5) \to H(8)$
$R(18) \to U(21)$

So PLAYER $\to$ SODBHU.

Answer: (d) SODBHU.

Why the other options miss

A

an arithmetic slip: shifts some letters by $+2$ instead of $+3$ ( $A \to C$ , $Y \to A$ ), off by one in the shift amount.
B

an arithmetic slip: correct through $B$ , then mis-shifts $E \to G$ ( $+2$ ) and $R \to T$ ( $+2$ ) in the tail.
C

counted one too few: gets five letters right but shifts the final $R \to T$ ( $+2$ ) instead of $R \to U$ ( $+3$ ).

Specialist insight

Confirm the shift on all four anchor letters before encoding — a Caesar cipher is only safe once the constant $+3$ is verified everywhere (one letter could hide a different rule). The two wrap-arounds are the mark-makers: $Z \to C$ and $Y \to B$ both cross the $Z \to A$ boundary. Compute positions and add $3$ mod $26$ ; the distractors are all “right shift, wrong tail” near-misses targeting a slip on the last one or two letters.

The trap, in one line

$+3$ Caesar shift (verify $Z\to C$ , $Y\to B$ wraps); PLAYER $\to$ SODBHU $=$ (d).

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