Consider the following C program. #include int main (void) i<=99) printf ("%d", (i-i/10*10) * 10+i/10); else printf ("out of range"); return 0; a - What is the output if the input i = 9? b - What is the output if the input i = 151?

When i=9 it is 90, when i=151 it is 25.

Explanation:

The program asks the user to enter the value of the i. Then, i is checked. If i is greater than 10 or smaller than or equal to 99, the value of the equation, (i-i/10*10) * 10+i/10), is printed. Otherwise, "out of range" is printed.

For i=9, since the value is smaller than or equal to 99, the if part will be executed. If you substitute 9 for the i in the equation, the output will be 90.

→ (9-9/10*10) * 10+9/10) (Since both numbers are integers in 9/10, it is 0, not 0.9)

→ (9-0*10) * 10+0)

→ (9) * 10

→ 90

For i=151, since the value is greater than 10, again the if part will be executed. If you substitute 151 for the i in the equation, the output will be 25.

→ (151-151/10*10) * 10+151/10) (Since both numbers are integers in 151/10, it is 15, not 15.1)

→ (151-15*10) * 10+15)

→ (151-150) * 10+15)

→ 1*10+15

→ 25