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Question

If in a certain language, TEAM = 5514 and LARK = 8914, then what is the code for BEAM?

This question was previously asked in
SSC Stenographer 2025 Question Paper (06-Aug-2025) Shift 2
The correct answer is
3514

Decode BEAM using Letter Coding Logic

This solution explains the logic used to find the numerical code for the word 'BEAM' based on the provided examples TEAM = 5514 and LARK = 8914.

Understanding the Coding Pattern

The coding pattern seems to depend on the alphabetical position of each letter and its position within the word. Let's analyze the given examples:

  • TEAM = 5514, where T=20, E=5, A=1, M=13
  • LARK = 8914, where L=12, A=1, R=18, K=11

Comparing TEAM (5514) with the target word BEAM, we notice that the letters E, A, M appear in the same positions (2nd, 3rd, 4th). In both TEAM and the likely code for BEAM (3514), these letters correspond to the digits 5, 1, and 4 respectively.

This suggests the following rule for letters other than the first one:

  • If the alphabetical position (P) is a single digit, use that digit directly. For example, E (position 5) maps to 5, and A (position 1) maps to 1.
  • If the alphabetical position (P) is a double digit, sum its digits (\(S(P)\)). For example, M (position 13) maps to \(S(13) = 1 + 3 = 4\).

Now, let's determine the rule for the first letter:

  • In TEAM, the first letter T has position P=20. It maps to the digit 5.
  • In BEAM, the first letter B has position P=2. It maps to the digit 3.

We need a rule that converts P=20 to 5 and P=2 to 3. Let's consider the sum of the digits of the position, \(S(P)\).

  • For T (P=20), \(S(20) = 2 + 0 = 2\). We need to get 5.
  • For B (P=2), \(S(2) = 2\). We need to get 3.

A possible rule that fits these two cases is:

  • If the position P is greater than 10 (\(P > 10\)), the code is \(S(P) + S(P) + 1\).
    • For T (P=20): \(S(20)=2\). Code = \(2 + 2 + 1 = 5\). This matches.
  • If the position P is less than or equal to 10 (\(P \le 10\)), the code is \(S(P) + 1\).
    • For B (P=2): \(S(2)=2\). Code = \(2 + 1 = 3\). This matches.

Applying the Rules to Find BEAM's Code

Let's apply the derived rules step-by-step to the word BEAM:

  1. B: This is the first letter. Its alphabetical position is P=2. Since \(P \le 10\), we use the rule \(Code = S(P) + 1\). The sum of digits \(S(2) = 2\). Therefore, the code for B is \(2 + 1 = 3\).
  2. E: This is the second letter. Its alphabetical position is P=5. Since P is a single digit, the code is 5.
  3. A: This is the third letter. Its alphabetical position is P=1. Since P is a single digit, the code is 1.
  4. M: This is the fourth letter. Its alphabetical position is P=13. Since P is a double digit, we sum its digits: \(S(13) = 1 + 3 = 4\). The code is 4.

Combining these digits gives the code for BEAM.

B -> 3
E -> 5
A -> 1
M -> 4

Therefore, the code for BEAM is 3514.

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Important Questions from Coding decoding

  1. The letters from A to Z are numbered from 1 to 26 respectively. If GHI = 1578 and DEF = 912, then what is ABC equal to?

  2. What is the missing term in the following? ACPQ : BESU:: MNGI: @

  3. what is the largest number among the following?

  4. What is the greatest length x such that 3 ½ m and 8 ¾ m are integral multiples of x?

  5. In a certain code, '256' means 'red colour chalk', '589' means 'green colour flower' and '254' means 'white colour chalk'. The digit in the code that indicates `white' is

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