Question

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?

Answer 1

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