Given the following: int funcOne(int n) { n *= 2; return n; } int funcTwo(int &n) { n *= 10; return n; } What will the following code output? int n = 30; n = funcOne(funcTwo(n)); cout << "num1 = " << n << endl; Group of answer choices Error num1 = 60 num 1 = 30 num1 = 600 num1 = 300

num1 = 600

Explanation:

Given

The attached code snippet

Required

The output of n = funcOne(funcTwo(n));  when n = 30

The inner function is first executed, i.e.

funcTwo(n)

When n = 30, we have:

funcTwo(30)

This returns the product of n and 10 i.e. 30 * 10 = 300

So:

funcTwo(30) = 300

n = funcOne(funcTwo(n)); becomes: n = funcOne(300);

This passes 300 to funcOne; So, we have:

n = funcOne(300);

This returns the product of n and 2 i.e. 300 * 2 = 600

Hence, the output is 600

i don't think so. although it would be nice. i am pretty sure that was an option an update or two ago. i guess they want you to stay 14 forever. haha.