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Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
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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)
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Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
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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
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Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
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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
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Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
Bunuel wrote:
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


 


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If we add the items, the 100 billionaires possess 401 elements. To minimize the nthe number of billionaires who have the 5 items, we should maximize the number of billionaires who have 4 items.
if we consider that all of them had 4 items, 100x4, the number of items would fall short by 1.
but idf 99 of them had 4 items, 99x4= 396. 401 - 396 = 5, then the remaining billionare would have 5 items.
Answer 1 Billionaire
B
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Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
Expert Reply
Bunuel wrote:
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
 


This question was provided by GMAT Club
for the Around the World in 80 Questions

Win over $20,000 in prizes: Courses, Tests & more

 



Least number of people having a mansion and car collection both = (93 + 90) - 100 = 83 - Let us call this category X
Least number of people who are in category X and have a jet = (83 + 81) - 100 = 64 - Let us call this category Y
Least number of people who are in category Y and have an island = (64 + 75) - 100 = 39 - Let us call this category Z
Least number of people who are in category Z and have an art collection = (39 + 62) - 100 = 1 - This category has all 5 items

Answer B
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Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
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