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Q1:
30 people were invited to a dinner party. If 17 of the guests fed on meal A, 15 on meal B, and 20 on meal C, what is the least number of people that could feed on all three meals?
A. 2
B. 3
C. 5
D. 6
E. 8
Q2 (related to Q1, with same options):
Supposing meal A=25, meal B=21, and meal C=20, but still with 30 total people. What is the minimum number of
people who eat all three meals?
A.2 B.3 C.5 D.6 E. 8
What are the specific clue to take rather than simply picking the smallest of the numbers in the option?
Kindly elaborate on ur strategy.
30 people were invited to a dinner party. If 17 of the guests fed on meal A, 15 on meal B, and 20 on meal C, what is the least number of people that could feed on all three meals?
A. 2
B. 3
C. 5
D. 6
E. 8
Q2 (related to Q1, with same options):
Supposing meal A=25, meal B=21, and meal C=20, but still with 30 total people. What is the minimum number of
people who eat all three meals?
A.2 B.3 C.5 D.6 E. 8
What are the specific clue to take rather than simply picking the smallest of the numbers in the option?
Kindly elaborate on ur strategy.
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09 Dec 2012, 05:26
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You can simplify the problem by considering below scenario.
"30 people, invited to a dinner party. Inviter has three different types of meal pases- A, B, C. He is going to distribute these in such that no one gets multiple passes of same type and least number of get all three passes."
In this question, answer is whatever number that is surplus of 60 i.e. (30 X 2).
Q 1) People=30
A= 17
B= 15
C= 20
Answer is 0. but to chose the least option in the given choices, I would choose 2 i.e. option A.
Q 2) People =30
A= 25
B=21
C=20
least number of people, who get 3 passes=
= (25+21+20)- 60
= 6
Answer is 6. Option D.
Below is one more similar analogy that will help to understand the problems.
There are 30 boxes placed in a row. You have three types of marbles- A, B & C. You are told place such that, no box hold 2 marbles of same type and least number of boxes should hold all types of marbles.
Consider below image of puts and marbles places inside.
"30 people, invited to a dinner party. Inviter has three different types of meal pases- A, B, C. He is going to distribute these in such that no one gets multiple passes of same type and least number of get all three passes."
In this question, answer is whatever number that is surplus of 60 i.e. (30 X 2).
Q 1) People=30
A= 17
B= 15
C= 20
Answer is 0. but to chose the least option in the given choices, I would choose 2 i.e. option A.
Q 2) People =30
A= 25
B=21
C=20
least number of people, who get 3 passes=
= (25+21+20)- 60
= 6
Answer is 6. Option D.
Below is one more similar analogy that will help to understand the problems.
There are 30 boxes placed in a row. You have three types of marbles- A, B & C. You are told place such that, no box hold 2 marbles of same type and least number of boxes should hold all types of marbles.
Consider below image of puts and marbles places inside.
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File comment: Analogy of marbles for the question.
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09 Dec 2012, 06:06
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I do not understand completely the questions about these overlapping sets
if someoneclarify me please ??
I'm not sure , though, they are good question, do not seem gmat like.
if someoneclarify me please ??
I'm not sure , though, they are good question, do not seem gmat like.
available.
