The sum of all two-digit numbers divisible by 5 is:
Answer: C
The two-digit numbers divisible by 5 are 10, 15, 20, ..., 95. This is an arithmetic progression.
First term (a) = 10. Last term (l) = 95. Common difference (d) = 5.
Number of terms (n) = \((\frac{l-a}{d}) + 1 = (\frac{95-10}{5}) + 1 = 17 + 1 = 18\).
Sum = \(\frac{n}{2}(a+l) = \frac{18}{2}(10+95) = 9 \times 105 = 945\).