Bunuel wrote:

Set X consists of eight consecutive integers. Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X. How many more integers are there in set Y than in set X ?

A. 0

B. 4

C. 8

D. 12

E. 16

Since set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X, it must have 8 integers each is 4 more than the ones in X and other 8 integers each is 4 less than the ones in X, unless there are overlaps in the elements obtained in each step. To verify that there are no overlaps, let the smallest element in X be x. After adding 4 to each element in the set, the smallest element obtained in this way is x + 4. If x is the smallest element in X, then x + 7 is the largest element in X (since there are 8 consecutive integers in X). Thus, the largest element obtained after subtracting 4 from each element is x + 3. In other words, there are no overlaps and Y has 8 + 8 = 16 integers, so it has 16 - 8 = 8 integers more than X.

Alternate Solution:

Let’s assume that set X contains 1, 2, 3, 4, 5, 6, 7, 8. Adding 4 to each term yields 5, 6, 7, 8, 9, 10, 11, 12.

Now, we do the same thing, but instead we subtract 4 from each term of the original set; thus, we have -3, -2, -1, 0, 1, 2, 3, 4.

We see that we have a total of 8 + 4 + 4 = 16 elements in set Y. This is 16 - 8 = 8 more elements than the number of elements in set X.

Answer: C

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