To determine the output of the given C code, let's go through it step by step:

    #include<stdio.h>

    int main() 
    {
        int x = 10;
        int *p = &x;
        *p = *p + 5;
        printf("%d %d", x, *p);
        return 0;
    }
    

1. The code starts by including the standard input-output header file <stdio.h>, which is essential for using the printf function.
2. In the main function, an integer variable x is initialized with the value 10.
3. An integer pointer p is created and initialized with the address of x using the address-of operator &.
4. The statement *p = *p + 5; is executed. Here, *p dereferences the pointer, accessing the value of x. Thus, *p + 5 results in 10 + 5 = 15. This value is then assigned back to x through the pointer.
5. The printf("%d %d", x, *p); function call prints the values of x and *p. Since *p is still pointing to x, both x and *p print the value 15.

Therefore, the output of this code will be 15 15.

Conclusion: The correct answer is 15 15. This confirms that the pointer p successfully modified the value of x to 15, and both x and *p reflect this change.