Answer:
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