Around the World in 80 Questions (Day 3): Out of 100 members of the

The number of people who don’t have the items =7+10+19+25+38=98
Therefore minimum number of billionaires with all items =100-98=2.
The answer would be C.

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The least number of billionaires who have all five items would be 1.
This is because it's possible that each item is owned by a different billionaire, with one billionaire owning all five.
So, the answer is B. 1

the difference between last two 75 have a private island;
62 have an art collection is 7
which is the possibility of at least members having all of the 5 listed items

option D

The Answer is E i.e 62.
Since there are a total of 100 members and 5 items, The minimum possible number of members having all 5 items would be 62.

Out of 100 members of the billionaires' club "Scrooge McDuck":
93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection
What is the least number of the billionaires who have all five items ?

I do this one base on the concept: to ensure minimum people get too many of them we spread out the goodies.
Thus no billionaire has no item (Neither = 0)

93 have a mansion (M) => 7 dont have mansion => Give these 7 all other four items

90 have a car collection (C) => 90 - 7 = 83 left => 83 will have both M & C => 100 - 83 - 7 = 10 who have M but dont have C
=> we have: 83 have MC; 17 dont have at least 1 M or C => continue to give these 17 all other 3 items

81 have a private jet (J) => 81 - 17 = 64 left => pick 64 from 83MC
=> 64 have all three M,C & J (64MCJ) and (83-64) = 19 have M,C but not J
Then we have 17 + 19 = 36 dont have at least one items (M or C or J) => continue to give these 36 all other 2 items

75 have a private island (I) => 75 - 36 = 39 left => continue to pick 39 from 64MCJ
=> we have 39MCJI (39ppl have all 4 items) and (64 - 39) = 25 billionaires who dont have item I
Thus till now, we have 36+25 = 61 billionaires who dont have at least one above items (M or C or J or I)
=> continue to give these 61 the fifth item

62 have an art collection (A) => 62 - 61 = 1 left
Hence only 1 billionaire must receive all 5 items

p/s: Im sorry for so many words, hope someone have better solution. Please see attached image for easier understanding. Manythanks

Sum=401;
Distribute 401 chocolates among
1,2,3,4...100 people.

Let each person get 1 ...
we are left with 301,
repeat this step 2 times:
We are left with 1 chocolate.
Distribute to any one. We are left with 1
Hence ans is 401-400=1

B)

93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

The least no having all will be the minimum point of intersection of the all items

Since the minimum overlapping group cam be 62 hence least no having all the items will be 62

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Given: Out of 100 members of the billionaires' club "Scrooge McDuck":

93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

Asked: What is the least number of the billionaires who have all five items ?

To minimise number of billionaires who have all five items, we have to maximise number who don't have a particular item.

38 don't have an art collection
25 don't have a private island.
38 + 25 = 63 are excluded
Remaining 100 -25 - 38 = 100 - 63 = 37 can have all items

Let all 63 have private jet, and 81-63 = 18 have private jet and can have all items
37 -18 = 19 more are excluded
Total 63 + 19 = 82 are excluded

Let all 82 have car collection, 90-82 = 8 can have all items.
18 - 8 = 10 more are excluded
Total 82 + 10 = 92 are excluded.

Let all 92 have mansion; 93-92 = 1 can have all items.
8 -1 = 7 are excluded
Total 92 + 7 = 99 are excluded.

The least number of the billionaires who have all five items = 1

IMO B

Out of 100 members of the billionaires' club "Scrooge McDuck":

93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

What is the least number of the billionaires who have all five items ?

A. 0
B. 1
C. 2
D. 7
E. 62

Solution:
Total no. of members = 100
93 have a mansion= 7 don't have.
90 have a car collection= 10 don't have.
81 have a private jet= 19 don't have.
75 have a private island= 25 don't have.
62 have an art collection= 38 don't have.

Subtracting all those who don't have something from 100 i.e. 100 - (7+10+19+25+38) = 100 - 99 = 1. ANSWER B

Out of 100 members of the billionaires' club "Scrooge McDuck":

93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

The least number of the collection would be the number of all five collections
62

Total members: 100

93 have a mansion
Mansion + Others = 93 + 7

90 have a car collection
Maximum members who have Car but not Mansion = 7
90 - 7 = 83
Maximum members who have Car and Mansion = 83
Maximum members who have Mansion but no Car = 10

81 have a private jet
Maximum members who have Car + Mansion but not Jet = 17 (7+10)
81 - 17 = 64
Maximum members who have Car + Mansion + Jet = 64
Maximum members who have Car + Mansion but no Jet = 19

75 have a private island
Maximum members who have Car + Mansion + Jet but not Island = 36 (17+19)
75 - 36 = 39
Maximum members who have Car + Mansion + Jet + Island = 39
Maximum members who have Car + Mansion + Jet but no Island = 25

62 have an art collection
Maximum members who have Car + Mansion + Jet + Island but not Art = 61 (36+25)
62 - 61= 1
Maximum members who have Car + Mansion + Jet + Island + Art = 1
Maximum members who have Car + Mansion + Jet + Island but no Art = 38

We calculate the maximum members in every case to minimise the number of members in other case.

Correct answer: 1 member

dont have mansion 100 -93 =7
dont have car collection 100-90=10
dont have private jet 100-81=19
dont have private island 100-75=25
highest no who can have all of them = 62
least no = 62-(7+10+19+25)=1 answer

Out of 100 members of the billionaires' club "Scrooge McDuck":

93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

What is the least number of the billionaires who have all five items ?
___________________________________________________
Generally, if we think that 62 people have art collection, the most number of people who have all 5 items is 62. In this case, least number of people who have all 5 items could be zero. I think this question has more logic behind than this.
Experts should help with this matter )

The correct answer is A. 0.

The least number of billionaires who have all five items is 0. This is because 93 have a mansion, but only 81 have a private jet. This means that there are at least 12 billionaires who have a mansion but not a private jet. Since all of the other items require a private jet, it follows that there can be no billionaires who have all five items.

The other answers are incorrect because they are all greater than 0. Answer choice B, 1, is incorrect because there could be 1 billionaire who has all five items, but there could also be 0. Answer choice C, 2, is incorrect because there could be 2 billionaires who have all five items, but there could also be 0 or 1. Answer choice D, 7, is incorrect because there could be 7 billionaires who have all five items, but there could also be 0, 1, 2, or 6. Answer choice E, 62, is incorrect because 62 billionaires have an art collection, but not all of them have the other four items.

Out of 100 members:
93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

My approach:
The least number of people who own one item: 62
Therefore that is the least number of the billionaires who have all five items.
There are 38 billionaires who don't have an art collection.

I will go with E.
Waiting for the OA!

We have to reduce the overlap between the groups as much as possible. We can take two groups first and then subsequently reduce the overlap of that number with the next group.

Start with those who have a mansion and those who have a car collection.



So in a venn diagram,
a+b = 93
b+c = 90
a+b+c = 100

There can be some people who fall outside both of these groups as well but to reduce the overlap as much as possible, we have to assume this number is zero. So solving for b from both these equations, b = 83.

Next, look at those who have both mansion and car collection (i.e. 83) and private jet.
Following the same logic and using the same venn diagram,
a+b = 83
b+c = 81
a+b+c = 100

hence, b = 64

Similarly, look at those who have all three (i.e. 64) and private island
a+b = 64
b+c = 75
a+b+c = 100

hence, b = 39

Now, look at those who have all four (i.e. 39) and art collection
a+b = 39
b+c = 62
a+b+c = 100

hence, b = 1

So answer is option B

Out of 100 members of the billionaires' club "Scrooge McDuck":

93 have a mansion;
90 have a car collection;
81 have a private jet;
75 have a private island;
62 have an art collection.

No idea but seems 7