There are 6 stores in town that had a total of 20 visitors

since 6 people have visited 2 stores,the total of visiting of this ppl is 6*2=12
20-12=8 (the rest of visits) 10ppl-6 ppl=4 ppl

since at least 1 store is visited by each of ppl, and since to maximize one option we need to minimize the other ones , then 1+1+1 +x=8
x=5

1 Person ---> 1 store = 1 visit. (as the question says that some people visited more than 1 store, meaning that there was at least 1 person who visited 1 store)
6 People ---> 2 store = 12 visit.
Remaining 3 People, to maximize the number of store visited by an individual, we minimize the store visited by other 2.
So,
2 people ---> 1 store = 2 visit.
Total visit remaining = 20 -1-12-2 = 5 visit. (which is done by the last person). Hope it helps.

Given:
Total visits = 20
Total people = 10

6 people visited 2 stores.
Hence total visits accounter for = 6*2 = 12

Now, we are left with 8 visits and 4 people.
To maximize the visits by one person, we need to minimize by others.

So, 3 persons visit 1 store each
Left visits = 5, Left persons = 1
Hence the maximum number of visits = 5

Option C

Though i understand and agree with the reasoning above, please can anyone let me know where did i go wrong with the below approach:

10 people went shopping, out of which 6 visited two stores each. Therefore, the number of visits=10+6=16. Since the total number of visits are 20, 4 additional visits are to be undertaken. Also, since each one has already visited minimum one store and because the number of visits has to be maximised, if one of the six guys who has already visited two stores each visits the remaining four stores, he would have visited 6 stores?

Thanks.

The error is highlighted above. We are given that those 6 people visit EXACTLY two stores, so for the remaining stores, we need to choose a person other than these 6.

total visits = 20 , number of people = 10 let the people be. restriction criteria is 6 visitors has exactly 2 visits and each individual visited at least 1

people .......A:B:C:D:E:F:G:H:I:J
Visits 2:2:2:2:2:2:1:1:1:1
thus we have 4 visits left after satisfying the restrictive criteria above and only visitors G,H,I,J can make those extra visit because A,B,C,D,E,,F are restricted to 2 each.

thus any of visitors G.H.I.J can make the remaining 4 visits to total 5 visits

C

1. There are 20 visits by 10 people.
2. 6 people made exactly 2 visits i.,e a total of 12 visits
3. There are 4 people who made the remaining 8 visits.
4. Since we want to find the maximum possible visits by one person, we minimize the visits of the other 3 which has to be 3 visits , of 1 each
5. So the maximum number of visits possible is 8-3=5

We must recognize that when the problem says there are 20 visitors, it doesn’t mean there are 20 people (since it later says there are only 10 people). It really means, if a person visited one store, then he or she is considered as one visitor. However, if that person visited two stores, then he or she is considered as two visitors since each of the two stores he or she visited will consider him or her as one visitor. Similarly, if he or she visited three stores, then he or she is considered as three visitors and so on.

Since 6 people visited exactly two stores each, then they are considered to be 12 visitors. Thus we have 20 - 12 = 8 visitors left for the remaining 10 - 6 = 4 people. In order to maximize the number of stores visited for one person, we must minimize the number of stores visited for the other 3 individuals. Thus, if those 3 people each visited exactly 1 store, then they are considered to be 3 visitors. Thus, we have 8 - 3 = 5 visitors left for the last person, which means he or she must have visited 5 stores.

Answer: C

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