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Sets - some tricky problems!!

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Sets - some tricky problems!!

by \'manpreet singh » Tue Jun 26, 2012 12:13 pm
Q.1) There are 6 stores in town that had a total of 20 visitors on a particular day.However, only 10 people went shopping that day; some people visited more than one store. If 6 people visited exactly two stores each, and everyone visited at least one store, what is the largest number of stores anyone could have visited?

Q.2) There are 26 students who have read a total of 56 books among them. The only books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only read Aye, and 8 students have read only Cod and Dee, what is the smallest number of books any of the remaining students could have read?

Q.3) X and Yare sets of integers. X | Y denotes the set of integers that belong to set X or set Y,but not both. If X consists of 10 integers, Y consists of 18 integers, and 5 of the integers are in both X and Y,then X | Y consists of how many integers?

These are some tricky word problems i got them wrong on my first attempt in the practice test.Share your methods of approach when you such kind of problems...
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by eagleeye » Tue Jun 26, 2012 12:51 pm
Hi \'manpreet singh:

Q.1) There are 6 stores in town that had a total of 20 visitors on a particular day.However, only 10 people went shopping that day; some people visited more than one store. If 6 people visited exactly two stores each, and everyone visited at least one store, what is the largest number of stores anyone could have visited?

This is how I would do them. Start with the maximum and keep taking out the numbers on both sides.
First I draw a grid like this. We have 10 people and 20 visits overall

I would draw it like this
20-------10

Then 6 people visited exactly two stores each, so a total of 12 stores.
So we have 12----------6.

I would then subtract them both.

20----------10
12----------6
8------------4

At this point I know that I have 4 people with 8 visits. Only one needs to visit the maximum number and the rest need to visit at least 1 store. So I have to minimize 3, and the rest will be maximized. So for those 3, we have:
3-----------3

Subtracting this from the last one

8----------4
3----------3
5----------1
We get 1 person can visit maximum of 5 stores. This is how I would do them. In reality, you can just keep doing this process like a continued subtraction, as you keep reading the conditions.

Q.2) There are 26 students who have read a total of 56 books among them. The only books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only read Aye, and 8 students have read only Cod and Dee, what is the smallest number of books any of the remaining students could have read?

Again, going the same way that I came up with in the earlier post, we have 56 books and 26 people. So we start with

56------------26

Now we know that 10 people read only Aye, so

56------------26
10-------------10
46-------------16

So at this point we have 16 people and 46 books read. Also, 8 people read only Cod and Dee. So 16 books are read by those people. Then, subtracting in the same way.

46--------------16
16--------------8
30--------------8.

So at this point I have 8 people left and 30 book readings left. Only 1 needs to read the minimum number. To minimize the 1, I will maximize the other 7. The maximum number of books anyone can read is 4 (since there are only 4 books). So total readings by these 7 = 7*4= 28.

So we continue with this. Subtracting the bottom one which we just thought of from the top one, we get
30---------------8
28---------------7
2-----------------1

So one person can read a minimum of 2 books.

Q.3) X and Yare sets of integers. X | Y denotes the set of integers that belong to set X or set Y,but not both. If X consists of 10 integers, Y consists of 18 integers, and 5 of the integers are in both X and Y,then X | Y consists of how many integers?

In this one we are asked for an overlapping sets problem where we need to find the integers unique to each set.
This is how I think about it.

Items unique to X = Total Items in X - Total Items in XY. = X - XY
The same way, Items Unique to Y = Y-XY.

So X|Y = X-XY + Y-XY = 10+18 - 2*5 = 18.

I don't know whether there is a better or more general way of doing the first two, but this is what I thought of when I read the questions.

Let me know if this helps :)
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by GMATGuruNY » Tue Jun 26, 2012 6:17 pm
There are 6 stores in town that had a total of 20 visitors on a particular day.However, only 10 people went shopping that day; some people visited more than one store. If 6 people visited exactly two stores each, and everyone visited at least one store, what is the largest number of stores anyone could have visited?
Since 6 people visited exactly two stores each, the total number of visits attributed to these 6 people = 6*2 = 12.

Since there are a total of 20 visits, the number of visits remaining = 20-12 = 8.
Since there are a total of 10 shoppers, the number of shoppers remaining = 10-6 = 4.

To MAXIMIZE the number of stores visited by one of these 4 people, we must MINIMIZE the number of stores visited by the other 3 people.
Since each of the other 3 people must visit at least 1 store, the minimum number of visits that could be delegated to these 3 people = 3*1 = 3.

The number of remaining visits = 8-3 = 5.
Since the one remaining person could use these 5 visits to shop at 5 different stores, the largest number of stores that anyone could have visited = 5.
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by GMATGuruNY » Tue Jun 26, 2012 6:33 pm
'manpreet singh wrote: X and Y are sets of integers. X | Y denotes the set of integers that belong to set X or set Y, but not both. If X consists of 10 integers, Y consists of 18 integers, and 5 of the integers are in both X and Y, then X | Y consists of how many integers?
Draw a Venn diagram and work from the inside out:
Image

Step 1: Fill in the OVERLAP
Since 5 integers are in BOTH sets, put 5 in the OVERLAP between the two circles.

Step 2: SUBTRACT the overlap
Since X=10, only X = 10-5 = 5.
Since Y=18, only Y = 18-5 = 13.

X | Y = only X + only Y = 5+13 = 18.
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by \'manpreet singh » Wed Jun 27, 2012 3:52 am
Thanks Mitch and eagle, it was helpful!!! i like the way you guys approach the problem and make it look simple.. :D :) :lol:
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