5.2: Applying Analogical or Rule-Based Transformations

5.2: Applying Analogical or Rule-Based Transformations

This type involves deducing a transformation rule from given examples and applying it to new letters or sequences.

Problem 1

Question: If AB → CD, then what does FG map to?

Solution Process:

• Analyze AB → CD: A=1→C=3 (+2), B=2→D=4 (+2).
• Rule: Each letter shifts forward by 2.
• Apply to FG: F=6→6+2=8=H, G=7→7+2=9=I.

Solution: HI

Detailed Explanation:

• The transformation AB → CD shifts each letter by +2 (A to C, B to D).
• For FG (F=6, G=7), apply +2: F→H (8), G→I (9).
• This tests deducing and applying a simple shift rule.

Problem 2

Question: If XY → ZA, then what does BC map to?

Solution Process:

• Analyze XY → ZA: X=24→Z=26 (+2), Y=25→A=1 (25+2=27≡1 mod 26).
• Rule: Shift forward by 2 (mod 26).
• Apply to BC: B=2→2+2=4=D, C=3→3+2=5=E.

Solution: DE

Detailed Explanation:

• XY → ZA shows a +2 shift: X (24) to Z (26), Y (25) to A (1, since 25+2=27≡1 mod 26).
• For BC (B=2, C=3), apply +2: B→D (4), C→E (5).
• This tests handling modular shifts.

Problem 3

Question: If AB → BA, then what does CD map to?

Solution Process:

• Analyze AB → BA: A (1st) swaps with B (2nd).
• Rule: Swap the two letters.
• Apply to CD: C (1st) swaps with D (2nd) → DC.

Solution: DC

Detailed Explanation:

• The transformation AB → BA reverses the order of the two letters.
• For CD (C, D), swapping gives DC.
• This tests a simple positional transformation rule.

Problem 4

Question: If PQ → RS, QR → ST, then what does TU map to?

Solution Process:

• Analyze PQ → RS: P=16→R=18 (+2), Q=17→S=19 (+2).
• Analyze QR → ST: Q=17→S=19 (+2), R=18→T=20 (+2).
• Rule: Shift each letter forward by 2.
• Apply to TU: T=20→20+2=22=V, U=21→21+2=23=W.

Solution: VW

Detailed Explanation:

• Both transformations (PQ → RS, QR → ST) show a +2 shift per letter.
• For TU (T=20, U=21), apply +2: T→V (22), U→W (23).
• This tests consistency across multiple examples.

Problem 5

Question: If ABC → CDE, then what does EFG map to?

Solution Process:

• Analyze ABC → CDE: A=1→C=3 (+2), B=2→D=4 (+2), C=3→E=5 (+2).
• Rule: Shift each letter forward by 2.
• Apply to EFG: E=5→5+2=7=G, F=6→6+2=8=H, G=7→7+2=9=I.

Solution: GHI

Detailed Explanation:

• ABC → CDE shifts each letter by +2.
• For EFG (E=5, F=6, G=7), apply +2: E→G (7), F→H (8), G→I (9).
• This tests applying a shift rule to a three-letter sequence.

Problem 6

Question: If MN → OP, then what does KL map to?

Solution Process:

• Analyze MN → OP: M=13→O=15 (+2), N=14→P=16 (+2).
• Rule: Shift forward by 2.
• Apply to KL: K=11→11+2=13=M, L=12→12+2=14=N.

Solution: MN

Detailed Explanation:

• MN → OP shows a +2 shift for each letter.
• For KL (K=11, L=12), apply +2: K→M (13), L→N (14).
• This tests a straightforward transformation.

Problem 7

Question: If AB → DD, then what does EF map to?

Solution Process:

• Analyze AB → DD: A=1→D=4 (+3), B=2→D=4 (+2).
• Rule: First letter +3, second letter +2 (or both map to the same letter, but let’s assume positional shifts).
• Apply to EF: E=5→5+3=8=H, F=6→6+2=8=H.

Solution: HH

Detailed Explanation:

• AB → DD suggests a rule where A→D (+3) and B→D (+2), or both letters transform to D. Assuming positional shifts, we use +3 for the first, +2 for the second.
• For EF (E=5, F=6): E→H (+3), F→H (+2).
• This tests a rule with different shifts per position.

Problem 8

Question: If XYZ → ZAB, then what does UVW map to?

Solution Process:

• Analyze XYZ → ZAB: X=24→Z=26 (+2), Y=25→A=1 (25+2=27≡1 mod 26), Z=26→B=2 (26+2=28≡2 mod 26).
• Rule: Shift forward by 2 (mod 26).
• Apply to UVW: U=21→21+2=23=W, V=22→22+2=24=X, W=23→23+2=25=Y.

Solution: WXY

Detailed Explanation:

• XYZ → ZAB shows a +2 shift, handling modular arithmetic for Y and Z.
• For UVW (U=21, V=22, W=23), apply +2: U→W (23), V→X (24), W→Y (25).
• This tests modular shifts in a three-letter sequence.

Problem 9

Question: If AC → EG, then what does BD map to?

Solution Process:

• Analyze AC → EG: A=1→E=5 (+4), C=3→G=7 (+4).
• Rule: Shift forward by 4.
• Apply to BD: B=2→2+4=6=F, D=4→4+4=8=H.

Solution: FH

Detailed Explanation:

• AC → EG shifts each letter by +4.
• For BD (B=2, D=4), apply +4: B→F (6), D→H (8).
• This tests a larger shift value.

Problem 10

Question: If AB → CB, BC → DC, then what does DE map to?

Solution Process:

• Analyze AB → CB: A=1→C=3 (+2), B=2→B=2 (no shift).
• Analyze BC → DC: B=2→D=4 (+2), C=3→C=3 (no shift).
• Rule: First letter +2, second letter unchanged.
• Apply to DE: D=4→4+2=6=F, E=5→E=5 (unchanged).

Solution: FE

Detailed Explanation:

• AB → CB and BC → DC show the first letter shifts +2, second letter stays the same.
• For DE (D=4, E=5), apply the rule: D→F (+2), E→E (unchanged).
• This tests a mixed transformation rule.