CSAT Solved Papers/ 2023/Q80

2023 CSAT — Q80

Quant Logical & quantitative reasoning 2.5 marks Hard

The letters of the word “INCOMPREHENSIBILITIES” are arranged alphabetically in reverse order. How many positions of the letter/letters will remain unchanged?

  1. A None
  2. B One
  3. C Two Answer
  4. D Three

Worked rationale

Index the 2121 letters of INCOMPREHENSIBILITIES:

I1 N2 C3 O4 M5 P6 R7 E8 H9 E10 N11 S12 I13 B14 I15 L16 I17 T18 I19 E20 S21\underset{1}{I}\,\underset{2}{N}\,\underset{3}{C}\,\underset{4}{O}\,\underset{5}{M}\,\underset{6}{P}\,\underset{7}{R}\,\underset{8}{E}\,\underset{9}{H}\,\underset{10}{E}\,\underset{11}{N}\,\underset{12}{S}\,\underset{13}{I}\,\underset{14}{B}\,\underset{15}{I}\,\underset{16}{L}\,\underset{17}{I}\,\underset{18}{T}\,\underset{19}{I}\,\underset{20}{E}\,\underset{21}{S}

Sort all 2121 letters in reverse alphabetical order (Z→\toA). Letter counts: I×5I\times 5, E×3E\times 3, N×2N\times 2, S×2S\times 2, and one each of T,R,P,O,M,L,H,C,BT, R, P, O, M, L, H, C, B. Descending order T>S>R>P>O>N>M>L>I>H>E>C>BT>S>R>P>O>N>M>L>I>H>E>C>B gives:

T1 S2 S3 R4 P5 O6 N7 N8 M9 L10 I11 I12 I13 I14 I15 H16 E17 E18 E19 C20 B21\underset{1}{T}\,\underset{2}{S}\,\underset{3}{S}\,\underset{4}{R}\,\underset{5}{P}\,\underset{6}{O}\,\underset{7}{N}\,\underset{8}{N}\,\underset{9}{M}\,\underset{10}{L}\,\underset{11}{I}\,\underset{12}{I}\,\underset{13}{I}\,\underset{14}{I}\,\underset{15}{I}\,\underset{16}{H}\,\underset{17}{E}\,\underset{18}{E}\,\underset{19}{E}\,\underset{20}{C}\,\underset{21}{B}

Compare position by position. The five II‘s now occupy positions 1111–1515; the original II‘s sat at 1,13,15,17,191, 13, 15, 17, 19. The overlap is positions 1313 and 1515 (both II before and after). Every other position differs.

So two positions remain unchanged.

Answer: (c) Two.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2023 CSAT Q80 — Logical & quantitative reasoning
Tap the drawing to open it full size for the fine detail.

Why the other options miss

  • A
    missed a case: never checks the II-block overlap, assuming a full reshuffle leaves nothing fixed.
  • B
    counted one too few: spots one II overlap (position 1313 or 1515) but misses the second.
  • D
    an arithmetic slip: mis-places the II-block (e.g. positions 1111–1515 vs 1212–1616) or miscounts a letter frequency, inventing a third fixed point.

Specialist insight

The whole item is frequency bookkeeping: the only letter appearing enough times to plausibly overlap is II (five copies). In reverse-sorted order the II-block lands at positions 1111–1515; intersect that with the original II-positions {1,13,15,17,19}\{1,13,15,17,19\} to get {13,15}\{13, 15\} — exactly two fixed points. Count each letter’s multiplicity, place the sorted blocks, and intersect the repeated-letter’s positions; isolated letters almost never stay put.

The trap, in one line

Reverse-sorted II-block sits at 1111–1515; original II's at {1,13,15,17,19}\{1,13,15,17,19\}, overlap ={13,15}=\{13,15\} ⇒\Rightarrow two unchanged == (c).

← All 2023 CSAT questions