In a certain code language, ‘BUZZY’ is coded as ‘10’ and ‘KING’ is coded as ‘8’. How will ‘ABROAD’ be coded in that language?

Correct — C. Rule: code = number of letters × 2.

BUZZY has 5 letters, and 5 × 2 = 10
KING has 4 letters, and 4 × 2 = 8

ABROAD has 6 letters, so 6 × 2 = 12, which is option (c).

The count is of letters as written, so the A that appears twice in ABROAD is counted twice.

Why the others are wrong

• (a)
  7 — 7 is one more than ABROAD's six letters. The two given codes are twice the letter count, not one more than it, and 7 is odd where doubling always gives an even number.
• (b)
  10 — 10 is what you get by counting only the distinct letters of ABROAD (A, B, R, O, D). The rule counts every letter, including the repeated A.
• (d)
  9 — 9 is odd, and letters × 2 can only produce even codes, as the given 10 and 8 both are.

Not every coding question is about the alphabet. When the code is a single small number, test the word's structure first: letter count, vowel count, consonant count, or the sum of alphabet positions.

Here 5 letters to 10 and 4 letters to 8 settle it at once — the code is twice the letter count.

Always confirm against both given words. Letter count and vowel count often agree by accident on one word and separate on the next, and a rule tested once is a guess.

Key facts

• BUZZY has 5 letters and is coded 10, and KING has 4 letters and is coded 8.
• ABROAD has 6 letters, of which 5 are distinct.
• Alphabet positions in BUZZY sum to 100 and in KING to 41, which fit no single rule giving 10 and 8.

Common traps

• Counting distinct letters and answering 10
• Adding alphabet positions because BUZZY sums to 100, which looks like 10 until KING gives 41
• Assuming every coding item needs a letter shift

No directly related past PYQ was found.