In a certain code language, ‘TIZZ’ is written as ‘VKBB’ and ‘QOPH’ is written as ‘SQRJ’. How will ‘CKAU’ be written in that language?

Correct — C. Line each coded word up under its original and take the differences.

TIZZ → VKBB: T(20) → V(22), I(9) → K(11), Z(26) → B(2), Z(26) → B(2). Every letter is +2, and Z wraps round to B.

QOPH → SQRJ: Q(17) → S(19), O(15) → Q(17), P(16) → R(18), H(8) → J(10). +2 again.

Rule: every letter moves two places forward.

C(3) + 2 = E
K(11) + 2 = M
A(1) + 2 = C
U(21) + 2 = W

CKAU is coded EMCW, option (c).

Why the others are wrong

• (a)
  ENCV — E and C are right, but N is three places past K and V is only one past U. Both sample words move every letter by the same 2.
• (b)
  DNBV — Here C, A and U each move +1 and K moves +3. A substitution code of this kind uses one step for the whole word.
• (d)
  DMBW — M and W are right; D and B are one place past C and A rather than two. No code shifts half its letters by 1 and half by 2.

In a shift code every letter of the message moves the same number of places along the alphabet. The two given words exist so you can measure that number and then confirm it.

Work in positions, not letters. Write 20 under T and 22 under V, and the +2 is visible at once.

Repeated letters are a free check. TIZZ ends in a double Z and VKBB ends in a double B, which is exactly what a letter-for-letter substitution must produce.

Key facts

• A=1 to Z=26, and a step past Z wraps round, so Z + 2 = B.
• TIZZ to VKBB and QOPH to SQRJ both use a forward shift of 2.
• A letter-for-letter code sends repeated letters to repeated letters.

Common traps

• Shifting some letters and copying others, which is how ENCV and DMBW arise.
• Losing the wrap at Z, which leaves the ZZ in TIZZ with no coded form.
• Measuring the shift from one sample word and never checking the second.

No directly related past PYQ was found.