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.

Hi umeshpatil

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.


Can you explain this part ?
I didnt get this how u got 6?

[quote="gmatbull"]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

Ans. We calculate people having only meal A , only meal B , only meal C and add them to 30. Then from the options the number giving us the least integer value is 2. Therefore the answer is (A).

I think zero.

If we feed more than 23 in third group then only overlapping will occur in all three.

PiyushK
what is your method? i"ve never seen anything like this?

Hi Galiya,

This method is same as described above by umeshpatil, I am fixing the position of 30 people and distributing resources in descending order like 20-17-15, like first meal C from left to right 10 people left, then Meal A from right to left 13 people left on the right side, then distribution of meal B to first those who got only one meal to minimize chances of overlapping of three meals to a single person.

Thus in the diagram you can see we are short of 8+ meals to get a person who can have all three meals.

Therefore, the least number of people is zero, from answer choices we can select 2 bcz that is the least option available.

I hope explanation clarifies the method.

I think this question is a brain twister, and it can help you to develop a thought process, which you can further use in various other solutions.
Yes it is not a GMAT type question, but I liked it.