Which set of letters when sequentially placed at the gaps in the given letter series shall complete it?
A _ B C _ B B C _ B B _
- ABAAC
- BBBBC
- CBABA
- DABBC
Solution & Step-by-step Explanation
The given series consists of 12 letter positions:
A
BC
BBC
BB
Let us divide the series into groups of 3 letters each:
A
B∣C
B∣BC
∣BB
Let's test Option C (B, A, B, A):
1
st
blank =B→A B B
2
nd
blank =A→C A B
3
rd
blank =B→B C B
4
th
blank =A→B B A
This pattern does not show clear repetition.
Let's test Option B (B, B, B, C):
1
st
blank =B→A B B
2
nd
blank =B→C B B
3
rd
blank =B→B C B
4
th
blank =C→B B C
This gives us: ABB∣CBB∣BCB∣BBC. No clean repeating pattern.
Let's group it by 4 letters instead:
A
BC∣
BBC∣
BB
Let's test Option A (B, A, A, C):
1
st
blank =B→A B B C
2
nd
blank =A→A B B C
3
rd
blank =A→A B B
4
th
blank =C→C
This forms the repeating pattern A B B C:
A B B C∣A B B C∣A B B C
The letters filled in the blanks are B, A, A, C.
A
BC
BBC
BB
Let us divide the series into groups of 3 letters each:
A
B∣C
B∣BC
∣BB
Let's test Option C (B, A, B, A):
1
st
blank =B→A B B
2
nd
blank =A→C A B
3
rd
blank =B→B C B
4
th
blank =A→B B A
This pattern does not show clear repetition.
Let's test Option B (B, B, B, C):
1
st
blank =B→A B B
2
nd
blank =B→C B B
3
rd
blank =B→B C B
4
th
blank =C→B B C
This gives us: ABB∣CBB∣BCB∣BBC. No clean repeating pattern.
Let's group it by 4 letters instead:
A
BC∣
BBC∣
BB
Let's test Option A (B, A, A, C):
1
st
blank =B→A B B C
2
nd
blank =A→A B B C
3
rd
blank =A→A B B
4
th
blank =C→C
This forms the repeating pattern A B B C:
A B B C∣A B B C∣A B B C
The letters filled in the blanks are B, A, A, C.