Problem Solving

At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. The 3 restaurants are the family buffet, the dining room, and the snack bar. Exactly 19 employees are trained to work in the family buffet, 18 are trained to work in the dining room, and 12 are trained to work in the snack bar. If 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants?
2
3
4
5
6


Is the right answer 2 ?

leonswati wrote:At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. The 3 restaurants are the family buffet, the dining room, and the snack bar. Exactly 19 employees are trained to work in the family buffet, 18 are trained to work in the dining room, and 12 are trained to work in the snack bar. If 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants?
2
3
4
5
6


Is the right answer 2 ?

The answer should be 3.

Start by adding the three "single" groups: 19 + 18 + 12 = 49.
Now we only have 39 employees, some the difference (10 employees) are counted twice / three times as part of different groups, and need to be reduced.

4 employees are trained in 2 restaurants, so our original 49 includes these 4 names twice, and we discount these 4. We are left with 6, and these represent people who have been counted 3 times - once for each group. We want to leave one instant of these employees, so these 6 employees represent two instances of 3 employees that work in all 3 restaurants.

Hi Geva...

The answer is 3 as u said. Thanks for ur explanation also..But can u help me solve it using the formula for three intersecting sets..

We are left with 6, and these represent people who have been counted 3 times - once for each group. We want to leave one instant of these employees, so these 6 employees represent two instances of 3 employees that work in all 3 restaurants.

Hi, could someone please explain why/how we get from 6 to 3? I'm not quite clear on the explanation above.

leonswati wrote:Hi Geva...

The answer is 3 as u said. Thanks for ur explanation also..But can u help me solve it using the formula for three intersecting sets..

I'd just like to go on record and say that I *hate* formulas. Formulas switch off your brain. Every time you use a formula without thinking why and how this formula applies, an angel cries.

So the formula for a 3 sets A,B,C is

A + B + C - AB - AC - BC - 2*ABC = Total.

Where AB, AC, BC are the "doubles", and ABC are the "triples". Note - this formula is used when the question uses "exactly" to describe doubles and triples - i.e. that the groups are distinct from each other. Also note that this formula assumes that all members of the total are a member of at least one of the three groups - that there are no people that are "none". If the problem deviates from these terms on a single point, this formula will be irrelevant and using it will lead you to the wrong answer - but you'll use it anyway, because it's a FORMULA, and your brain is switched off. Do you hear that angel sob?

Plug in the numbers:

19 + 18 + 12 - 4 -2*ABC = 39

49 - 4 - 2ABC = 39

45-39 = 2ABC

6=2ABC

3=ABC.

cbaum wrote:

We are left with 6, and these represent people who have been counted 3 times - once for each group. We want to leave one instant of these employees, so these 6 employees represent two instances of 3 employees that work in all 3 restaurants.

Hi, could someone please explain why/how we get from 6 to 3? I'm not quite clear on the explanation above.

Now, that is a good question.

Think of each number of people in a group as a list of names in an excel spreadhsheet. We start by adding three lists: 19 + 18 + 12 = 49. We now have a spreadsheet with 49 names, but some of these names are duplicates:

The 4 that are members of 2 groups (doesn't matter which ones) appear on the list twice - for example, john was added twice, once as part of the 19 people in the buffet, another time as part of the 12 people in the snack bar.

And there is an unknown quantity of names that appear three times - once for each of the three restaurants they work in. Sally is a busybee that works at all three restaurants, so when we added the original lists together, we created three "sallys" who are actually the same person.

You know that there are only 39 employees, and your job is to clear up this mess - leave the list with a final 39 names, every name counted only once. What do you do?

The 4 that were counted twice (John), you only want to count once - so you subtract one instance of their name: subtract 4. Great, now we only have 49-4=45 names on the list - we still have 6 names extra.

The others that were counted THREE times (Sally), you still want to only count once - so you have to strike out TWO instances of each of the triples. So these 6 extra actually count for only 3 people, each appearing twice: two Sallys, two Thelmas, two Louises that we want to take out and leave the last instance of their names in the list of 39.