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overlapping

by resilient » Tue Mar 25, 2008 8:39 am
If x and Y are sets of integers, x#y denotes the set of integers that belont to set x or set y, but not both. If x consists of 10 integers, y consists of 18 integers 6 of the integers are in both x and y, then x#y consists of how many integers?

a.6
b.16
c.22
d.30
e.174

qa is b. I originally wanted to answer 28 but then switched to 22. What is the question asking. The rest is easy





p.s. I am using the double matrix variable chart. Question is very simple just trying to understand the question. This is the problem!
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by tmmyc » Tue Mar 25, 2008 9:03 am
I would use a Venn diagram for this one.

Draw two circles that overlap in the middle:

1. In the overlapping section, place a 6.

2. In the circle representing x, we should have 10 total. Since we already counted 6 from the middle, we place a 4 in the non-overlapping area.

3. In the circle representing y, we should have 18 total. Since we already counted 6 from the middle, we place a 12 in the non-overlapping area.

x#y is the total number in the non-overlapping areas: 4+12 = 16.

Edit: Stacey's double-set matrix solution below is much more elegant than this one.
Last edited by tmmyc on Wed Mar 26, 2008 9:21 am, edited 2 times in total.
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thanks

by resilient » Tue Mar 25, 2008 10:21 am
I am only using the Venn Diagrams for 3 variables and using the double matrix chart for 2 variables. I dont see why 28 is not the answer that explain 18 y plus 10 x with 6 overlap = 28-6 =22/ 22 is the answer in my logic.

how is 16 qa. what am I missing?
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by Stuart@KaplanGMAT » Tue Mar 25, 2008 12:45 pm
Perfect question for the overlapping sets formula:

Total # of items = (# in group 1) + (# in group 2) + (# in neither group) - (# in both groups)

In this case:

Total # = 10 + 18 + 0 - 6

[in this question, there aren't any items outside of both x and y, so # in neither group = 0]

Total # = 28 - 6 = 22

Step 4 of the Kaplan method for problem solving: double check the question!

We're not asked how many total integers there are - we're asked how many terms are in either x or y, but NOT both.

In other words, we want to solve for:

Total # - both = 22 - 6 = 16: choose (b).
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by Stacey Koprince » Tue Mar 25, 2008 7:50 pm
Hey, engin

I'll show the double-set matrix method below in case you want to see how that applies to this problem; I also like Stuart's solution. The below might show you conceptually why you are making the mistake you're making, though.

Given:
................in x.....not in x......total
in y............6........................18
not in y..................0
total..........10

We want to find the ones that ARE in x and ARE in y but are NOT in both. Start filling in the remainder of the matrix.

................in x.....not in x......total
in y............6...........12..........18
not in y.......4.........0..............4
total..........10..........12..........22

So look at the cells that correspond to:
in y but NOT in x: 12
in x but NOT in y: 4
Add those to get 16

What you are doing is adding the total x (10) and the total y (18) to get 28, and then subtracting out that 6 once to get 22.

If you subtract out the 6 once, you are still counting those 6 integers once in the set; you are only subtracting out the overlap. But the question asks us to remove the 6 integers ENTIRELY from the final answer, so you'd have to subtract out the 6 TWICE, which would give you 28-6-6 = 16.

You might want to try this with some real numbers - actually write out some sets of integers for x and y that fit the parameters in the matrix above so that you can see what I mean about subtracting out the 6 overlapping integers twice instead of once.
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by sampleresume » Mon Apr 07, 2008 11:17 pm
Right, the answer is 22.
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