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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

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

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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

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.

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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02 Jan 2014, 23:53

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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 [ 3.39 KiB | Viewed 5166 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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Re: If X and Y are sets of integers, X@Y denotes the set of inte [#permalink]

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03 Jan 2014, 03:03

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This post received KUDOS

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

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.

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

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

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

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