Eating contest

[gmatdriller]

Please explain in details...thanks

Of the 30 people involved in an eating contest, 17 of them fed on meal A,
15 on meal B, and 20 on meal C.

What is the least number of people that could eat all three meals?

A.2 B.3 C.5 D.8 E. 10

[kanwar86]

Please explain in details...thanks

Of the 30 people involved in an eating contest, 17 of them fed on meal A,
15 on meal B, and 20 on meal C.

What is the least number of people that could eat all three meals?

A.2 B.3 C.5 D.8 E. 10

IMO, D is the correct answer

[gmatdriller]

Hello Kanwar86,
please could you provide explanations?

[chieftang]

What is the source of this question, please?

If:
10 people eat 1 meal = 10 meals eaten.
18 people eat 2 meals = 36 meals eaten.
2 people eat 3 meals = 6 meals eaten.

That's 30 people eating 52 meals.

Answer: A The least number of people (among the answer choices given)* that eat all 3 meals is 2.


*Among the answer choices not given, the least number of people that could eat 3 meals is actually 0.

[user123321]

IMO Should be 0,if it is present as one of the choices.

user123321

[gmatdriller]

Here we have different opinions: 2, 8, and 0(not stated).
I do not have the OA as it was coined from a text, but the
page for the answers was torn.

[chieftang]

Here we have different opinions: 2, 8, and 0(not stated).
I do not have the OA as it was coined from a text, but the
page for the answers was torn.

I've shown you how to arrive at 2 as the answer. Since it is the smallest answer choice given, it must be the answer.

But I don't think it's a valid question. The answer choices should contain the correct answer, IMO.

Again, what is the source of this question?

[Stuart@KaplanGMAT]

Please explain in details...thanks

Of the 30 people involved in an eating contest, 17 of them fed on meal A,
15 on meal B, and 20 on meal C.

What is the least number of people that could eat all three meals?

A.2 B.3 C.5 D.8 E. 10

Here's a good rule to remember for the GMAT:

When a question asks you to minimize one thing, you want to maximize something else; when a question asks you to maximize one thing, you want to minimize something else.

Here we're asked to minimize the number of people who eat all 3 meals; to do so, we'll maximize the number of people who eat exactly 2 meals.

We know that we have a total of 30 people. We also know that:

17 eat meal A;
15 eat meal B; and
20 eat meal C.

There are a number of different ways to maximize the doubles, but let's keep things simple. Let's say that:

Everyone who eats meal A also eats meal C.. so that's 17 doubles; and

3 of the people who eat meal B also eat meal C... that's another 3 doubles.

So, the maximum number of double meal eaters we can have is 20.

That leaves 12 people who eat meal B but nothing else. However, if we add up 20 doubles and 12 singles, that gives us 32 people. Since we only have 30 people, this is impossible!

Since we have 2 extra people, those 2 must eat all 3 meals... choose (A)!

[GmatMathPro]

Please explain in details...thanks

Of the 30 people involved in an eating contest, 17 of them fed on meal A,
15 on meal B, and 20 on meal C.

What is the least number of people that could eat all three meals?

A.2 B.3 C.5 D.8 E. 10

All of the below assumes that everyone ate at least one meal, and that there were no other meals at the contest besides A, B, and C.

First, consider the simpler case of only having 2 meals, A and B. If there are 30 people and 17 ate meal A and 15 ate meal B, that is a total of 32 people. The "extra" two people must have eaten both meals and thus were double counted when we added 15 and 17. Thus, exactly two people must have eaten both meals. The excess must be from the overlap between A and B in this case, because there are only two meals.

Now, if we consider the case of three meals, there are 17+15+20=52 meals being eaten by 30 people. In this case 22 meals must be distributed among the overlapping regions of the Venn diagram. Here we have more overlapping regions than before: AB, AC, BC, and ABC. We have some flexibility as to how we can distribute the 22 extra meals, but there are some restrictions we have to adhere to. We still need at least 2 in the overlapping region between A and B(17+15=32; 32-30=2), 7 in the overlapping region between A and C(17+20=37; 37-30=7), and 5 in the overlapping region between B and C (15+20=35; 35-30=5). Since we are trying to minimize the number in ABC, we should put these all in the "ate exactly two meals" regions of the Venn diagram as follows:



With these minimum satisfied, we've accounted for 14 of the 22 excess meals. Now you are free to distribute the remaining 8 excess meals however you wish among the regions where people ate exactly two meals, and then fill out the remaining numbers with "single" meals. Here are two examples:





In both cases, 0 people ate three meals, so the minimum number of people who ate three meals is zero.

[gmatdriller]

GmatMathPro and Stuart did a great job in their explanations to the problem above,
Thanks to you and all others for your analysis.

[gmatdriller]

A question to both GmatMathPro and STuart; I crave your indulgence to please
explain what to do in the event of such question...

It seems one could simply select the least value among the options
without doing any work...

What is your take on this?

[GmatMathPro]

A question to both GmatMathPro and STuart; I crave your indulgence to please
explain what to do in the event of such question...

It seems one could simply select the least value among the options
without doing any work...

What is your take on this?

No, don't do this. It all depends on what numbers are given. For example, in this problem, suppose instead of A=17, B=15, and C=20, it was A=25, B=21 and C=20, but still with 30 total people. The minimum number of people who eat all three meals in this case will be greater than zero. See if you can figure out what it would be. I can post a detailed explanation later if you'd like.

[gunjan1208]

GMAT Math Pro,

I tried solving it with the numbers you have simulated:--

Finally I reach that we need to explain 36 people and in the given scenario we end up explaining 16+11+15=42 people which are 6 extra.

Will that mean that 6 people ate all 3 foods or we need to keep 2 inbetween the common portion for all three?

Thanks.

[GmatMathPro]

GMAT Math Pro,

I tried solving it with the numbers you have simulated:--

Finally I reach that we need to explain 36 people and in the given scenario we end up explaining 16+11+15=42 people which are 6 extra.

Will that mean that 6 people ate all 3 foods or we need to keep 2 inbetween the common portion for all three?

Thanks.

At least 6 people have to go in the area common to all three. Let's see what happens if we try to construct a scenario with any fewer than 6 people:

As you said, the minimum number of people who eat two meals is as follows:

AB=16
AC=15
BC=11

Focusing on managing the overlap: If we say that 5 people ate all 3 meals, then 16-5=11 people ate only AB, 15-5=10 people ate only AC, and 11-5=6 people ate only BC. This is reflected in the following image:




But now we've got serious problems. The number of people in A is 26, the total number of people in B is 22, the total number of people in C is 21, and the total number of people is 32. As the number in the middle gets lower, these discrepancies gets worse and worse. Try as you might, you simply won't be able to satisfy all of the things you need to satisfy simultaneously with a number less than 6.

Note, however, that 6 people is merely the minimum. We could have MORE than 6 people in the middle. We could have as many as 18 (do you see why?).

So, this is all instructive for building scenarios that fit all the criteria. If we are ONLY interested in the minimum value of the middle number, however, we can do the following:

Pick any two meals, let's say A and B, and add them together. A+B=46 and then subtract the total: 46-30=16. So we know at least 16 people have to go in the overlap between A and B. But we also need 20 people to go in C. But 20+16=36, which is 6 more than the total number of people. In other words, there's no way all 16 of these people can be outside of C, with 20 totally different people constituting C. There must be some overlap. And the minimum overlap can be calculated just like the others: 36-30=6. Note that it works out the same way no matter which two you start with: A+C=45, 45-30=15. Then add this with B: 15+21=36. 36-30=6. Starting with B and C: B+C=41. 41-30=11. Then add this with A: 11+25=36. 36=30=6.

Putting it all together: A,B,C is the size of the groups. T is the total. A+B-T= excess between A and B. Then add this excess to C and again subtract the total: (A+B-T)+C-T = A+B+C-2T. So the quick way to find the minimum is add all the groups and subtract twice the total. Of course, if any of these values come back negative, then the minimum is zero.

[gunjan1208]

Please explain in details...thanks

Of the 30 people involved in an eating contest, 17 of them fed on meal A,
15 on meal B, and 20 on meal C.

What is the least number of people that could eat all three meals?

A.2 B.3 C.5 D.8 E. 10

That would mean that the answer to the original question can be 17+15+20-2*30=52-60=-8 Which indicates that the minimum should be zero. (The fall of eight would be fitted the food eaten by two but not three). Should i take it as a rule?