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

As per Set theory :
A@B= A + B - 2(A n B), so 10 + 18-2*6 = 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


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

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

Hi

I know its silly question but can you please clear my understanding?
Why cannot I do.... 10+18-6??? am I not deducting both from X and Y by doing this??

That way you'll get the total number of elements in X and y, while we need the number of elements that belong to set X or set Y, but not both.

6 elements belong to both X and y, thus there are 10-6=4 unique elements in X and 18-6=12 unique elements in Y. Thus there are total of 4 + 12 = 16 unique elements in X and Y.

Hope it's clear.

X #Y represents the shaded region as shown in diagram

Answer = 16 = B

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



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

Note that the 6 numbers belonging to both sets must be subtracted from set X and again from set Y.

We can use the equation:

X@Y= set X - both + set Y - both

X@Y = 10 - 6 + 18 - 6 = 16

Answer: B

Overlapping sets question staring us right in the face.

X = 10
Y = 18
X and Y = 6

10 - 6 = 4
18 - 6 = 12

4 + 12 = 16

B.

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