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

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)

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


min students having all = 100 * 4 = 400
and 93 + 90 + 81 + 75 + 62 = 401

thus 401 - 400 = 1

ans 1

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

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

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

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

-Firstly, we must find the least value of people who do both A and B. Starting from this formula, we have: \(Total=A+B-Both+Neither=>Both= A+B-Total+Neither\). Since A, B, Total are fixed, \(Min(both)=Max(A+B-Total,0)\). Now, let's apply our transformation to our question.
-Min(mansion & car collection)\(=Max(93+90-100,0)=83\)
-Min(mansion & car collection, private jet)\(=Max(83+81-100,0)=64\)
-Min(mansion & car collection & private jet, private island)\(=Max(64+75-100,0)=39\)
-Min(mansion & car collection & private jet & private island, art collection)\(=Max(39+62-100,0)=1\)
=>The least number of the billionaires who have all five items is 1
The answer is 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.

We can work on it by taking the people without the least value
without Art collections: 38
without Private island: 25
without private Jet : 19
without car collection: 10
without mansion: 7

which calculates to 38+25+19+10+7=99. So the billionaires with all five: 1


ANS: 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

Mansion = M =93
car collection = C = 90
private jet = J = 81
private island = PI = 75
art collection = A = 62
To find out the least number, we need to find out the minimum overlap.
M+ A = 93+62-100 = 55
C+J = 90+81-100 = 71
M+A+C+J = 55+71-100 = 26
M+A+C+J+PI = 26+75-100 = 1

IMO: B

Given : 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.

To minimize the number of members who have all 5 items we will consider two items at a time and calculate the minimum number of members that can have those two items. Then take those members and proceed in similar way with each other item.

1. Members than can have a mansion and a car collection

93 have a mansion;
90 have a car collection;

From above we can see that 7 does not have mansion and 10 does not have car collection .
This information can be used to calculate the minimum and maximum number of members that can have both the things. To get minimum number of members who can have both things we have to maximize the people who do not have both things.
So max people who have do not have both a mansion and a car collection = 7 + 10
Now the remaining people have both a mansion and a car collection = 100 - 17 = 83

2. Members than can have a mansion and a car collection and private jet

As we got from 1.
Members with mansion and a car collection = 83
Members with private jet = 81

--> 17 don't have mansion and a car collection
--> 19 don't have private jet

max People who don't have these 3 items = 17+19 = 36
min number of members than can have a mansion ,a car collection, and private jet = 100-36 = 64

3.Members than can have a mansion ,a car collection ,private jet, and private island

Members with a mansion ,a car collection, and private jet = 64
Members with private island= 75

--> 36 don't have mansion ,a car collection, and private jet
--> 25 don't have private island

max People who don't have these 4 items = 36 + 25 = 61
min number of members than can have a mansion and a car collection, private jet , and private island = 100-61 = 39

4. Members than can have a mansion ,a car collection ,private jet, private island and an art collection

Members with a mansion ,a car collection,private jet, and private island = 39
Members with an art collection= 62

--> 61 don't have a mansion ,a car collection,private jet, and private island
--> 38 don't have an art collection

max People who don't have these 5 items =61 + 38 = 99
min number of members than can have a mansion and a car collection, private jet, and an art collection = 100-99 = 1

Thus, the min number of members who have all 5 items = 1

to find least number of the billionaires who have all five items, we can find most no. of billionaires with atleast one item and subtract that from 100.
i.e 100 - (100-93) - (100 - 90) - (100 - 81) - (100-75) - (100 - 62) = 1
option B

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

93 have a mansion; (100-93=7)
90 have a car collection; (1100-90=10)
81 have a private jet; (100-81=19)
75 have a private island; (100-75=25)
62 have an art collection. (100-62=38)

What is the least number of the billionaires who have all five items ?
=100 - (7+10+19+25+38) = 100-99 = 1 [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 ?

Answer: Least number of billionaires who have all five items
93 have a mansion, so 7 does not have mansion.So 7 is excluded
90 a car collection so lets say 10 does not car collective and this 10 overlaps mansion and 7 from mansion overlaps car collection
so 83 has mansion and car collection

81 have private jet, so 19 does not have private jet
so lets say this 19 overlap with mansion and car collection
so 83-19= 64 does have private jet, mansion and car collection

75 have a private land, so 25 does not have a private land
so lets say this 25 overlap with private jet, mansion and car collection
so 64-25=39 have private land, private jet, mansion and car collection

62 have art collection, so 38 does not have art collection
so lets say this 38 overlap with private land, private jet, mansion and car collection
so 39-38 =1 has art collection, private land, private jet, mansion and car collection

Answer is 1

To solve this problem we utilize the complementary set property. Let U be the universe n(U) = 100. We are given the items as follows: 93 have a mansion (A); 90 have a car collection (B); 81 have a private jet (C); 75 have a private island (D); 62 have an art collection (E). Definitely, there must be certain overlap between item classes to make set 100.

Cardinality of complementary sets
n(A') = U - n(A), --(i)

Now, consider the set A u B, we have
n(A u B) <= n(A) + n(B) --(ii)

Let S = A' u B' u C' u D' u E'

Therefore, the only region not considered in S is the region of all 5, hence S'=U-S, gives the answer.
ans = n(U) - n(S), using --(i)
ans >= n(U) - [n(A) + n(B)]
= 100 - [(100-93)+(100-90)+(100-81)+(100-75)+(100-62)]
= 1

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 ?
Since we need the least number of the billionaires who have all five items we must try to obtain the least number which overlaps with all the five items
93 have a mansion; so 7 don't have - least overlap possible = 93
90 have a car collection; so 10 don't have - least overlap possible = 90 - 7 = 83
81 have a private jet; so 19 don't have - least overlap possible = 81 - (10+7) = 64
75 have a private island; so 25 don't have - least overlap possible = 75 - (10+7+19 = 36) = 39
62 have an art collection; so 38 don't have - least overlap possible = 62 - (10+7+19+25 = 61) = 1
Hence, the least number of the billionaires who have all five items can be 1.
Answer B