Answer:
7
Explanation:
Given the codes as follows:
In line 1, the value of 1 is assigned to variable i and then the variable i is assigned to variable j. This means j will hold the value of 1 as well.
In line 2, j will be multiplied with 5 prior to the addition with j itself. So, j + j * 5 -> 1 + (1 * 5) -> 1 + 5 -> 6
Next, value of 6 will be added with i. i = i + 6 -> i = 1 + 6 -> 7#
Eventually, value of 7 will be printed out (Line 3).
