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 which word is coded as ‘LDZMR’? 

• A. MEALS
• B. TRIMS
• C. MEATS
• D. TRAIN

Hint:

When decoding is confusing, test the given options by applying the coding rule directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a decoding problem using the same rule as the previous questions. We are given the coded word and need to find the original word by applying the inverse of the encoding rule.

Step 2: Key Formula or Approach:
The decoding rules, derived previously, are based on the coded letter's position 'c': - If 'c' is odd, the original position is \(m=c-1\). - If 'c' is even, the original position is \(m=c+1\).

Step 3: Detailed Explanation:
The coded word is LDZMR. Let's find the position of each letter: - L is the 12th letter (even). - D is the 4th letter (even). - Z is the 26th letter (even). - M is the 13th letter (odd). - R is the 18th letter (even). Now, apply the decoding rules: - L (12, even): The original position is \(m = 12+1 = 13\). The 13th letter is M. - D (4, even): The original position is \(m = 4+1 = 5\). The 5th letter is E. - Z (26, even): The original position is \(m = 26+1 = 27\), which corresponds to A. - M (13, odd): The original position is \(m = 13-1 = 12\). The 12th letter is L. - R (18, even): The original position is \(m = 18+1 = 19\). The 19th letter is S. Combining the decoded letters, we get M-E-A-L-S. The original word is MEALS.

Step 4: Final Answer:
The decoded word is MEALS, which corresponds to option (A).