The correct option is B 11
The letters and postcards are to be distributed equally into paper bags. In addition to this, we also want to find the smallest possible number of paper bags to be purchased.
If each paper bag contains the maximum number of items, then the number of bags will be minimum.
The maximum number of items these bags can carry = The greatest number dividing 54 and 45 = HCF(54, 45)
Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54
Factors of 45 = 1, 3, 5, 9, 15, 45
Common factors = 1, 3, 9
⇒ HCF(54, 45) = 9
We know that number of paper bags = No. of total items ÷ No. of items in each bag
⇒ So the total number of minimum possible paper bags = 54/9 + 45/9 = 6 + 5 = 11