The correct option is
C 73
The first number greater 10 which when divided by 4 leaves a remainder 3 is 11.
So, the next number will be 11+4=15
The other numbers in this will be =15+4=19;19+4=23;23+4=27
So, the AP =11,15,19,23,27,32......
The last term of this AP will be 299.
We have to find the total numbers that lie between 10 and 300, which when divided by 4 leaves a remainder 3.
⇒ an=a+(n−1)d
Here, an=299,a=11 and d=4
⇒ 299=11+(n−1)4
⇒ 299=11+4n−4
⇒ 299−11+4=4n
⇒ 292=4n
⇒ n=2924
∴ n=73
So, total 73 numbers lie between 10 to 300 which when divided by 4 leaves a remainder 3.