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?

Step-by-step explanation:

Given that X consists of 8 consecutive integers

say X = {n, n+1, ... n+8}

Set Y is formed by adding 4 to each and also subtracting 4 from each element of X

Numbers got by addition = n+4, n+5, ... n+12

Numbers got by subtraction = n-4, n-3, n-2, n, n+1, n+2, n+3, n+4

We find that n+4 is repeated in both.

Hence Y = {n-4, n-3, ... n+4, ... n+12}

n (Y) = 15

Y has 7 integers more than X