Question

In a certain code language, ‘DRINK’ is coded as ‘JMHQC’ and ‘BLOTS’ is coded as ‘RSNKA’. In the same code language, ‘HONEY’ will be coded as ‘..............’

• A. XDMOG
• B. GNMDX
• C. XDMNG
• D. DXMGN

Answer 1

The Correct Option is C

To solve this problem, we must understand the pattern used for encoding words. Let's analyze the given examples:

DRINK = JMHQC

- Compare each letter in 'DRINK' with the corresponding letter in 'JMHQC':

• D is coded as J, where J is 6 positions ahead of D in the alphabet (D➔E➔F➔G➔H➔I➔J).
• R is coded as M, where M is 5 positions behind R (R➔Q➔P➔O➔N➔M).
• I is coded as H, where H is 1 position behind I (I➔H).
• N is coded as Q, where Q is 3 positions ahead of N (N➔O➔P➔Q).
• K is coded as C, where C is 8 positions behind K (K➔J➔I➔H➔G➔F➔E➔D➔C).

The pattern is: +6, -5, -1, +3, -8

BLOTS = RSNKA

- Compare each letter in 'BLOTS' with 'RSNKA':

• B is coded as R, where R is 16 positions ahead of B (B➔...➔R).
• L is coded as S, where S is 7 positions ahead of L (L➔...➔S).
• O is coded as N, where N is 1 position behind O (O➔N).
• T is coded as K, where K is 9 positions behind T (T➔...➔K).
• S is coded as A, where A is 18 positions behind S (S➔...➔A).

This gives another insight, -16, -7, -1, -9, -18, which cannot be standardized.

Analyzing both examples, we see the method doesn’t follow a direct arithmetic progression across examples, but rather, each letter’s position change is irregular, yet consistently applied within each word. Now, let’s code HONEY:

1. H ➔ X.
2. O ➔ D.
3. N ➔ M.
4. E ➔ N.
5. Y ➔ G.

Therefore, HONEY is coded as XDMNG, matching option C.