Question

In a code language, \(m^{\text{th}}\) letter is coded as \((m-1)^{\text{th}}\) letter if \(m\) is odd and \((m+1)^{\text{th}}\) letter if \(m\) is even. Then the code word for ‘BUST’ is

• A. CSRU
• B. CTRU
• C. CTRS
• D. CTRR

Hint:

If alphabet position is odd, move one step backward; if even, move one step forward.

Answer 1

The Correct Option is B

Concept:
Coding rule: \[ \text{Odd position letter}\Rightarrow \text{previous letter} \] \[ \text{Even position letter}\Rightarrow \text{next letter} \]

Step 1: Write positions of letters in BUST. \[ B=2,\quad U=21,\quad S=19,\quad T=20 \]

Step 2: Apply the rule. Since \(B=2\) is even: \[ B\rightarrow C \] Since \(U=21\) is odd: \[ U\rightarrow T \] Since \(S=19\) is odd: \[ S\rightarrow R \] Since \(T=20\) is even: \[ T\rightarrow U \]

Step 3: Write the code. \[ \text{BUST}\rightarrow \text{CTRU} \]

Step 4: Final answer. \[ \boxed{\text{CTRU}} \]