In a set of integers between 18 and 95, inclusive, how many are divisible by 6 but not 12?

18≤12n≤95  divide all term by 12

1.5≤n≤7.9

n=[2,7]

So multiples of 12 are 7-2+1=6

18≤6n≤95 divide all terms by 6

3≤n≤15.8

3≤n≤15

So multiples of 6 are 15-3+1=13

So there are 13-6=7 multiples of 6 that are not multiples of 12 in the set of integers [18,95]

1.5 pounds because you put 9over 6 and then divide by 6