In the following question, a pair of letters is given, followed by its corresponding product of alphabetical positions (A = 1, B = 2, ..., Z = 26). A second pair of letters is given without its product. Identify the correct product for the second pair that maintains the same relationship as the first. AxE : 1×5 :: DxK : ?
- (a)3×9
- (b)4×11
- (c)5×10
- (d)4×12
Answer
Why
Correct — B.
Rule: each letter is replaced by its alphabetical position.
A = 1 and E = 5, giving 1×5.
D = 4, and K comes one after J = 10, so K = 11.
DxK therefore gives 4×11 → option (b).
Why the others are wrong
- (a)3×9 — 3×9 stands for C and I. D is the 4th letter, not the 3rd, and K is the 11th, not the 9th.
- (c)5×10 — 5×10 stands for E and J. D is 4, not 5, and K is 11, not 10.
- (d)4×12 — 4×12 gets D right but counts K as 12. K is the 11th letter, and L is the 12th.
Concept
This is letter-to-number coding in its plainest form: replace each letter with its place in the alphabet, A = 1 to Z = 26, and keep the × sign.
The work is knowing positions fast. Anchors help: E = 5, J = 10, O = 15, T = 20, Y = 25. From J = 10, K is one step on, 11.
The stem calls 1×5 a 'product', but every option is written as an unworked multiplication. Compare the factors, not the results: 4×11 = 44 is never needed.
Key facts
- D is the 4th letter of the alphabet and K the 11th.
- Anchor positions: E = 5, J = 10, O = 15, T = 20, Y = 25.
- A = 1 and E = 5, so AxE maps to 1×5.
Study next
Common traps
- Counting K as 12 and choosing 4×12
15 Sep 2025, 16:00, Reasoning Q.13 asks the same thing at the far end of the alphabet: HxL : 8×12 :: TxZ : ? is keyed 20×26.
24 Sep 2024, 09:00, Reasoning Q.12 writes whole words as positions: PUPIL is 162116912, and PURSE is keyed 162118195.
Related PYQs
No directly related past PYQ was found.