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Question

(C) To solve this,we look for a repeating pattern in the series: $...bca...cca...ca....b...c$.Counting the total number of characters including blanks

Vedclass · Accepted Answer

$aabaa$

Vedclass · Answer

(C) To solve this,we look for a repeating pattern in the series: $...bca...cca...ca....b...c$.Counting the total number of characters including blanks,we have $16$ positions.Let's divide the series into groups of $4$: $(abc a) (bcca) (bca c) (a b c)$. This does not seem to fit.Let's try another grouping: $(a b c a) (b c c a) (b c a c) (a b c)$.Let's test option $C$ $(aabaa)$: If we fill the blanks with $a, a, b, a, a$,the series becomes $a b c a b c c a b c a c a b c$. This also does not show a clear pattern.Let's re-examine the series: $b c a / b c c a / b c a c / a b c$.Actually,the pattern is $abc a, bcca, bca c, abcc$.By filling the blanks with $a, a, b, a, a$,we get $a b c a b c c a b c a c a b c$. Wait,let's try filling option $D$ $(bbabb)$: $b b b c a b c c a b c a b b b c$. This is also not consistent.Let's re-evaluate the sequence: $a b c a / b c c a / b c a c / a b c$.If we fill the blanks with $a, a, b, a, a$,the sequence is $a b c a b c c a b c a c a b c$. Correct pattern is $abca, bcca, bcac, abcc$. The missing letters are $a, a, b, a, a$.

Choose the correct alternative to complete the series: $...bca...cca...ca....b...c$

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