Question

Assume that an int variable age has been given a value. Assume further that the user has just been presented with the following menu: S: hangar steak, red potatoes, asparagus T: whole trout, long rice, brussel sprouts B: cheddar cheeseburger, steak fries, cole slaw (Yes, this menu really IS a menu!) Write some code that reads the string (S or T or B) that the user types in into a variable choice and prints out a recommended accompanying drink as follows: if the value of age is 21 or lower, the recommendation is "vegetable juice" for steak, "cranberry juice" for trout, and "soda" for the burger. Otherwise, the recommendations are "cabernet", "chardonnay", and "IPA" for steak, trout, and burger respectively. Regardless of the value of age, your code should print "invalid menu selection" if the character read into choice was not S or T or B.

Answer 1

Answer:

No programming language stated.

I'll answer the question using C++ programming language

Explanation:

The code goes this;

.#include <iostream>

using namespace std;

int main()

{

string order; int age;

cout<<"Press S for hangar steak, red potatoes, asparagus"<<endl;

cout<<"Press T for whole trout, long rice, brussel sprouts"<<endl;

cout<<"Press B for cheddar cheeseburger, steak fries, cole slaw"<<endl;

cout<<"Please make your order";

cin>>order;

cout<<"Please what's your age";

cin>>age;

if(order == "S" || order == "T" || order == "B")

{

if(age <= 21)

{

if(order == "S")

{

cout<<"The Recommendation is Vegetable Juice";

}

else if(order == "T")

{

cout<<"The Recommendation is Cranberry Juice";

}

else if(order == "B")

{

cout<<"The Recommendation is Soda";

}

}

else

{

if(order == "S")

{

cout<<"The Recommendation is Cabernet";

}

else if(order == "T")

{

cout<<"The Recommendation is Chardonnay";

}

else if(order == "B")

{

cout<<"The Recommendation is IPA";

}

}

}

else

{

cout<<"Invalid Menu Selection";

}

return 0;

}