(c) 73
The required numbers are 11, 15, 19, ..., 299.
This is an AP in which a = 11, d = (15 - 11) = 4 and Tn = 299
Now, Tn = 299
⇒ a + (n - 1)d = 299
⇒ 11 + (n - 1) ⨯ 4 = 299
⇒ 4n = 292
⇒ n = 73
Hence, 73 numbers, which when divided by 4 leave a remainder 3, lie in between 10 and 300.