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Difficulty: HardAnalogy and Classification

Four letter-group pair analogies are given below. For each analogy pair, calculate the total sum of forward alphabetical shift steps required to transform the first word into the second word (using standard position values A=1A=1 to Z=26Z=26, and wrapping around from ZZ to AA where necessary). Arrange the four items in descending order based on their total shift sum value.

  1. 1MINT:SRAZ\text{MINT} : \text{SRAZ}
  2. 2BARK:GGUQ\text{BARK} : \text{GGUQ}
  3. 3DUST:HXZX\text{DUST} : \text{HXZX}
  4. 4GLOW:JOSB\text{GLOW} : \text{JOSB}

Answer

The correct sequence of items arranged in descending order of their total forward shift sum is: MINT:SRAZ\text{MINT} : \text{SRAZ} (34), followed by BARK:GGUQ\text{BARK} : \text{GGUQ} (20), then DUST:HXZX\text{DUST} : \text{HXZX} (18), and finally GLOW:JOSB\text{GLOW} : \text{JOSB} (15).
Calculating the total forward alphabetical shift steps for each pair yields: MINT:SRAZ\text{MINT} : \text{SRAZ} has a sum of 34, BARK:GGUQ\text{BARK} : \text{GGUQ} has a sum of 20, DUST:HXZX\text{DUST} : \text{HXZX} has a sum of 18, and GLOW:JOSB\text{GLOW} : \text{JOSB} has a sum of 15. Sorting these values in descending order (34>20>18>1534 > 20 > 18 > 15) produces the correct sequence.

Step-by-Step Solution

1
Calculate the total forward alphabetical shift sum for DUST:HXZX\text{DUST} : \text{HXZX}
D(4)→H(8)=+4D(4) \rightarrow H(8) = +4, U(21)→X(24)=+3U(21) \rightarrow X(24) = +3, S(19)→Z(26)=+7S(19) \rightarrow Z(26) = +7, T(20)→X(24)=+4T(20) \rightarrow X(24) = +4. Sum = 4+3+7+4=184 + 3 + 7 + 4 = 18.
Determine the step-by-step positional shift for each letter position.
2
Calculate the total forward alphabetical shift sum for GLOW:JOSB\text{GLOW} : \text{JOSB}
G(7)→J(10)=+3G(7) \rightarrow J(10) = +3, L(12)→O(15)=+3L(12) \rightarrow O(15) = +3, O(15)→S(19)=+4O(15) \rightarrow S(19) = +4, W(23)→B(28)=+5W(23) \rightarrow B(28) = +5 (since 23+5=28≡2 mod 2623 + 5 = 28 \equiv 2 \text{ mod } 26). Sum = 3+3+4+5=153 + 3 + 4 + 5 = 15.
Account for wrap-around shift past letter ZZ.
3
Calculate the total forward alphabetical shift sum for MINT:SRAZ\text{MINT} : \text{SRAZ}
M(13)→S(19)=+6M(13) \rightarrow S(19) = +6, I(9)→R(18)=+9I(9) \rightarrow R(18) = +9, N(14)→A(27)=+13N(14) \rightarrow A(27) = +13 (since 14+13=27≡1 mod 2614 + 13 = 27 \equiv 1 \text{ mod } 26), T(20)→Z(26)=+6T(20) \rightarrow Z(26) = +6. Sum = 6+9+13+6=346 + 9 + 13 + 6 = 34.
Determine wrap-around shift from N(14)N(14) to A(1)A(1).
4
Calculate the total forward alphabetical shift sum for BARK:GGUQ\text{BARK} : \text{GGUQ}
B(2)→G(7)=+5B(2) \rightarrow G(7) = +5, A(1)→G(7)=+6A(1) \rightarrow G(7) = +6, R(18)→U(21)=+3R(18) \rightarrow U(21) = +3, K(11)→Q(17)=+6K(11) \rightarrow Q(17) = +6. Sum = 5+6+3+6=205 + 6 + 3 + 6 = 20.
Evaluate individual letter shifts.
5
Order the computed values in descending sequence
Comparing 34>20>18>1534 > 20 > 18 > 15 gives the sequence: MINT:SRAZ\text{MINT} : \text{SRAZ}, BARK:GGUQ\text{BARK} : \text{GGUQ}, DUST:HXZX\text{DUST} : \text{HXZX}, GLOW:JOSB\text{GLOW} : \text{JOSB}.
Arrange from the highest calculated shift total to the lowest.

Key Concept

Positional Alphabetical Shift and Cyclic Wrap-Around Analysis in Letter Analogies
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Topics

Analogy and Classification

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