I know that any multiple of 5 must end in either 0 or 5, so I can ignore the ones that end in 0. That means I only need to count the ones that end in 5, which are 5, 15, 25, ..., 95. I can use a formula to find the number of terms in this arithmetic sequence. The formula is n = (a_n - a_1) / d + 1, where n is the number of terms, a_n is the last term, a_1 is the first term, and d is the common difference. In this case, a_n = 95, a_1 = 5, and d = 10. Plugging these values into the formula, I get n = (95 - 5) / 10 + 1 = 90 / 10 + 1 = 9 + 1 = 10. So there are 10 positive multiples of 5 that are less than 100 and have a units digit of 5. 
How many positive multiples of 5 that are less than 100 have a units digit of 5?