In this alphabet analogy, JKLM corresponds to IIII by a pattern based on positions; using the same pattern, determine the related group for PQRS: JKLM : IIII :: PQRS : ?

Introduction / Context:
This question belongs to the alphabet series and analogy category. We are given a pair JKLM : IIII and asked to find which option PQRS should map to under the same rule. The challenge is to decode how the repeated string of I's is obtained from JKLM and then apply that logic to the second group of letters.

Given Data / Assumptions:

• The English alphabet is ordered from A = 1 to Z = 26.
• JKLM is a sequence of four consecutive letters: J, K, L and M.
• IIII is a repetition of the letter I four times.
• The same transformation that turns JKLM into IIII must be used to transform PQRS into one of the options.

Concept / Approach:
One standard way to handle such analogies is to compare the positions of letters. J is the 10th letter, K the 11th, L the 12th and M the 13th. The letter I comes just before J and has position 9. A natural pattern is that each letter in JKLM is replaced by its immediate predecessor in the alphabet, resulting in I for all four positions. If that is correct, then for PQRS we should again replace each letter by its immediate predecessor and repeat that letter four times.

Step-by-Step Solution:
Step 1: Write the positions of J, K, L and M: J = 10, K = 11, L = 12, M = 13.Step 2: Notice that each of these letters has the same predecessor I (position 9) when you step back one place from the start of the sequence.Step 3: Interpreting the pattern as "map the entire group to the single letter that comes just before the first letter in the series, repeating it four times" gives us IIII.Step 4: Now consider PQRS, which is another sequence of four consecutive letters: P = 16, Q = 17, R = 18, S = 19.Step 5: The letter that comes just before P in the alphabet is O, which has position 15.Step 6: Apply the same rule as before by mapping the entire group PQRS to this predecessor letter O, repeated four times.Step 7: This gives the string OOOO, which matches option B.

Verification / Alternative check:
We can verify consistency by checking that this rule works cleanly for the original pair. The predecessor of J is I, so replacing the group JKLM with I repeated four times is exactly what we see in IIII. Applying the same idea to PQRS using the predecessor O produces OOOO, which is present among the options. This confirms both the validity and uniqueness of the pattern.

Why Other Options Are Wrong:
Option A, PPPP, would correspond to repeating the first letter itself, not its predecessor. Option C, TTTT, would involve moving forward rather than backward in the alphabet, which does not match the original pattern. Option D, OOPP, repeats two different letters and does not show a uniform mapping. Only option B, OOOO, consistently applies the rule inferred from JKLM : IIII.

Common Pitfalls:
A common error is to try to map each letter of PQRS individually to a different letter, ignoring that the original result IIII uses the same letter four times. Another mistake is to step forward in the alphabet when the pattern requires stepping backward. Always pay close attention to whether the transformation involves predecessors or successors and whether the outcome involves a single repeated letter or varied letters.

Final Answer:
Under the same pattern that maps JKLM to IIII, PQRS must map to OOOO, so option B is correct.