To find how many numbers between 1 and 100 are of the form 7x+5, where x is a positive integer, we can set up an inequality:

1≤7x+5≤100

Now, let's solve for x:

For the lower bound: 1≤7x+5

Subtract 5 from both sides: −4≤7x

Divide both sides by 7 (remember, x is a positive integer): −4/7≤x

Since x must be a positive integer, the smallest possible value of x is 1. So, we have x≥1 for the lower bound.

For the upper bound: 7x+5≤100

Subtract 5 from both sides: 7x≤95

Divide both sides by 7 (rounding up because x is a positive integer): x≤95/7≈13.57

Since x must be a positive integer, the largest possible value of x is 13. So, we have x≤13 for the upper bound.

Now, we need to count how many positive integers satisfy 1≤x≤13. These are the values of x for which 7x+5 falls between 1 and 100.

The positive integers between 1 and 13 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.

There are 13 positive integers that satisfy the condition.

So, there are 13 numbers between 1 and 100 of the form 7x+5, where x is a positive integer. The correct answer is 13.