A fruit seller had a certain number of apples, bananas, and oranges at the start of the day. The number of bananas was 10 more than the number of apples, and the total number of bananas and apples was a multiple of 11. She was able to sell 70% of the apples, 60% of bananas, and 50% of oranges during the day. If she was able to sell 55% of the fruits she had at the start of the day, then the minimum number of oranges she had at the start of the day was

Let the initial amount of Apples with the fruit seller be $$a$$, it follows that the initial amount of Bananas with the fruit seller will be $$a+10$$, and that $$a+a+10=2a+10$$ is a multiple of $$11$$. Given that $$a$$ is a natural number, the possible values for $$a$$ start from $$6$$, and can be anything after $$6$$ with a difference of $$11$$, i.e.  $$6, 17, 28, 39, 50 ...$$

Let the initial amount of Oranges with the fruit seller be $$c$$.

She sold $$70\%$$ of Apples and hence $$0.7a$$ Apples, therefore $$a$$ is a multiple of $$10$$. She also sold $$60\%$$ or $$0.6a+6$$ Bananas. The total fruits that she had was $$2a+10+c$$, of which she sold $$55\%$$ or $$1.1a+5.5+0.55c$$. Thus $$c$$ has to be a multiple of $$10$$ but not of $$20$$ (in order to add with $$5.5$$ and still give an integer). 

We are told that: $$1.1a+5.5+0.55c = 0.7a+0.6a+6+0.5c$$, which gives $$0.05c-0.2a=0.5$$

The minimum value of $$a$$ that satisfies being an integer is $$50$$, so the value of $$c$$ becomes, $$0.05c= 10.5$$ or $$c= 210$$.

Since $$210$$ is a multiple of $$10$$ and not of $$20$$, it also satisfies. Thus the correct answer is option B.