Last visit was: 21 Jun 2024, 07:19 It is currently 21 Jun 2024, 07:19
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93944
Own Kudos [?]: 633615 [21]
Given Kudos: 82411
Intern
Joined: 30 Apr 2023
Posts: 18
Own Kudos [?]: 41 [15]
Given Kudos: 79
Location: India
General Discussion
Intern
Joined: 04 Apr 2017
Posts: 34
Own Kudos [?]: 59 [2]
Given Kudos: 309
Director
Joined: 16 Jul 2019
Posts: 522
Own Kudos [?]: 229 [2]
Given Kudos: 152
Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
1
Kudos
1
Bookmarks
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)
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5223
Own Kudos [?]: 4077 [2]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
2
Kudos
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
Manager
Joined: 28 Dec 2020
Posts: 106
Own Kudos [?]: 135 [1]
Given Kudos: 522
Re: Around the World in 80 Questions (Day 3): Out of 100 members of the [#permalink]
1
Kudos
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
Manager
Joined: 05 Nov 2012
Status:GMAT Coach
Posts: 157
Own Kudos [?]: 287 [0]
Given Kudos: 65
Location: Peru
GPA: 3.98
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

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over \$20,000 in prizes: Courses, Tests & more

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.
B
Tutor
Joined: 26 Jun 2014
Status:Mentor & Coach | GMAT Q51 | CAT 99.98
Posts: 450
Own Kudos [?]: 791 [0]
Given Kudos: 8
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
 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