SSC CGL 2024 Tier-I · 19 September 2024, Shift 3 · General Intelligence and Reasoning · Question 25 of 99

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Select the option that represents the letters that when sequentially placed from left to right in the blanks below will complete the letter series. I _ _ I _ I _ _ I _ J _

(a)J I J I J I I
(b)J I I I J J I
(c)I J I I J J I
(d)I I J J J I J

Answer

Why

Correct — A. Rule: the series is I J I repeated to fill twelve slots, so J falls on every third position, starting at position 2.

The stem has twelve slots. It prints I at positions 1, 4, 6 and 9, and J at position 11.

That single J is the anchor. If J sits at 11 and recurs every third slot, the J positions are 2, 5, 8 and 11, and everything else is I. Each printed I fits, since 1, 4, 6 and 9 are none of those.

The seven blanks are at positions 2, 3, 5, 7, 8, 10 and 12. Filling them in that order gives J, I, J, I, J, I, I — option (a).

The completed series reads I J I I J I I J I I J I.

Why the others are wrong

(b)
J I I I J J I — It puts I at position 5, breaking the every-third rhythm, and drops a second J at position 10 immediately beside the printed J at 11. The block never sets two Js side by side.
(c)
I J I I J J I — Starting I at position 2 pushes the first J out to position 3, leaving a gap of five to the next J instead of three. It also doubles the J at positions 10 and 11.
(d)
I I J J J I J — It spends five Js where a twelve-slot run of I J I holds only four, and it stacks them at positions 7 and 8 and again at 11 and 12.

Concept

A letter series with blanks is pattern completion, not deduction. The printed letters are there to identify one repeating block, and the block then fills every gap.

The efficient route is to find the shortest block consistent with all the printed letters, then verify it against every printed letter before writing anything into the blanks.

The rarest printed letter makes the best anchor, because its position pins the phase of the cycle. Here a single J does that work for the whole series.

Count the twelve slots before filling anything. The option strings are seven letters long, and a miscount of the underscores shifts every letter you place after it.

Key facts

The completed series is I J I I J I I J I I J I, which is the block I J I four times.
J occupies positions 2, 5, 8 and 11, and every other position holds I.
The stem has twelve slots and seven blanks, matching the seven letters in each option.
The printed J at position 11 is the only non-I letter given, which makes it the anchor.

Study next

Letter series solved by identifying the shortest repeating blockUsing a rare printed letter to fix the phase of a cycleSeries in which blanks and printed letters are interleaved across a fixed slot count

Common traps

Filling left to right by feel instead of anchoring on the printed J
Allowing two Js to sit next to each other, which the block never does
Miscounting the underscores and matching seven option letters to the wrong slots

A number series on the same find-the-step logic is set at Reasoning Q.8, 19 Sep 2024, 16:00, where 58, 70, 83, 97 rises by 12, 13, 14 and then 15. Letters or numbers, the work is the same: name the repeating step, then check it against every term already printed.

Related PYQs

No directly related past PYQ was found.