#include <iostream> using namespace std; int f(int p, int q) { if (p > q) return p; else return q; } main() { int a = 5, b = 10; int k; bool x = true; bool y = f(a, b); k =((a * b) + (x + y)); cout << k; }

a) 55

b) 62

c) 52

d) 75

#include <iostream> using namespace std; int f(int p, int q) { if (p > q) return p; else return q; } main() { int a = 5, b = 10; int k; bool x = true; bool y = f(a, b); k =((a * b) + (x + y)); cout << k; }

a) 55

b) 62

c) 52

d) 75

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c

Explanation: In this question, value of x = true and value of y will be also true as f(a,b) will return a non-zero value. Now when adding these values with integers, the implicit type conversion takes place hence converting both x and y to 1(integer equivalent of bool true value). So expression (a*b) + (x+y) is evaluated to 52.