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If X and Y are sets of integers, X@Y denotes the set of inte

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If 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 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

Problem Solving
Question: 18
Category: Arithmetic Properties of numbers
Page: 64
Difficulty: 600


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SOLUTION

If 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 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

The number of integers that belong to set X ONLY is 10-6=4;
The number of integers that belong to set Y ONLY is 18-6=12;

The number of integers that belong to set X or set Y, but not both is 4+12=16.

Answer: B.
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IMO B.

Set X=10
Set Y=18

both X&Y = 6

(Either X or Y or both) = (X) + (Y) - (both X&Y) = 10+18-6 = 22

Now we want a set of integers from either X or Y but not from both X and Y
X@Y = (Either X or Y or both) - (Both X&Y) = 22-6 = 16.
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Re: If X and Y are sets of integers, X@Y denotes the set of inte [#permalink]

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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If 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 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


As per Set theory :
A@B= A + B - 2(A n B), so 10 + 18-2*6 = 16
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Re: If X and Y are sets of integers, X@Y denotes the set of inte [#permalink]

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If 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 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
Attachment:
untitled1.PNG
untitled1.PNG [ 3.39 KiB | Viewed 5508 times ]


Sol: Look at above figure.
Now X@Y = Number of elements in X and Y which are not present in Both.

So X@Y= 10-6+18-6= 16 Ans B
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If 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 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

Exactly 1 = X + Y - 2(X&Y)

When you add X and Y the intersection gets added twice hence we have to deduct it twice :)

Exactly 1 = 10 + 18 - 12 = 16

Answer B
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SOLUTION

If 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 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

The number of integers that belong to set X ONLY is 10-6=4;
The number of integers that belong to set Y ONLY is 18-6=12;

The number of integers that belong to set X or set Y, but not both is 4+12=16.

Answer: B.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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New post 28 May 2015, 19:03
Is it possible to solve this problem using a matrix?
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cg0588 wrote:
Is it possible to solve this problem using a matrix?


Hi cg0588,

The question asks us the number of integers which belong to set X or Set Y but not both. This would be equal to the number of integers which belong to only set X + number of integers which belong to only set Y

Please find below the matrix diagram of the solution

Image

We are given that set X consists of 10 integers out of which there are 6 integers which are common to set Y. Hence integers which belong to only set X = 10 - 6 = 4

Similarly, we know that set Y consists of 18 integers. As there are 6 integers which are common to set X, we will have 18 - 6 = 12 integers which belong to only set Y.

Thus number of integers which belong to set X or set Y but not both = 4 + 12 = 16

Hope it's clear :)

Regards
Harsh
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Re: If X and Y are sets of integers, X@Y denotes the set of inte [#permalink]

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New post 02 Feb 2017, 15:52
Its a very simple yet tricky concept to understand. Think of it as a set for example
X = 1,2,3,4,5,6,7,8,9,10
5,6,7,8,9,10
Y=5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,28

Essentially the set of six numbers is being double counted....... so you can add X and Y and than subtract 2*(the set of shared integers which in this case is 6)
Leaving you with 16. Hope this makes it clear.
KUDOS +1 if you like my explanation ;)
Re: If X and Y are sets of integers, X@Y denotes the set of inte   [#permalink] 02 Feb 2017, 15:52
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