CSAT Solved Papers/ 2021/Q65

2021 CSAT — Q65

Quant Logical & quantitative reasoning 2.5 marks Medium

In the series $\_\,b\,\_\,a\,\_\,b\,a\,\_\,b\,\_\,a\,b\,a\,b\,\_\,a\,a\,b$ ; fill in the six blanks $(\_)$ using one of the following given four choices such that the series follows a specific order.

A bababa
B baabba
C bbaabb
D ababab Answer

Worked rationale

The string has $18$ positions with blanks at positions $1,3,5,8,10,15$ . Try to make it a clean repeating block of length $6$ . Insert option (d) ababab into the six blanks in order ( $a,b,a,b,a,b$ ):

\underbrace{a\,b\,b\,a\,a\,b}_{1\text{–}6}\ \underbrace{a\,b\,b\,a\,a\,b}_{7\text{–}12}\ \underbrace{a\,b\,b\,a\,a\,b}_{13\text{–}18}

Filling positions $1,3,5,8,10,15$ with $a,b,a,b,a,b$ yields the block " $abbaab$ " repeated three times — a single specific order.

No other option reproduces a clean repeating block: e.g. (a) bababa gives $b,b,a,a,b,b,\dots$ which breaks after the first group.

Answer: (d) ababab.

Why the other options miss

A

the pattern breaks down: starts plausibly but the resulting string is not a single repeating block; the pattern collapses after the first six.
B

misreads the setup: produces an irregular string with no period- $6$ structure.
C

the wrong pattern forced: forces a doubled-letter pattern that the fixed (non-blank) letters contradict.

Specialist insight

The winning strategy is to hypothesise a repeating block and test each option by substituting it into the blank positions in order, then checking that the whole $18$ -letter string splits into three identical $6$ -blocks. Option (d) yields " $abbaab$ " thrice — the only choice consistent with every fixed letter. Don’t read the series left-to-right guessing; lock the period first ( $18 = 3\times 6$ ) and let it filter the options.

The trap, in one line

Filling the six blanks with $a,b,a,b,a,b$ makes " $abbaab$ " repeat three times $\Rightarrow$ (d).

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