Question

Cpp challenge

Description

The purpose of this challenge is to use various flow control structures. This challenge uses techniques for displaying formatted output, simulating a payment structure for paying off outstanding debt such as a credit card or any kind of interest-bearing loan.

Requirements

1. Declare double variables called balance, monthly, apr, apr_monthly, start_bal, interest, end_bal
2. Declare an integer called months
3. Prompt the user to enter values for balance, monthly, and apr
4. Use a while loop to display the payment breakdown and how long it will take to pay off the balance. We use a while loop since we don’t know how many months it will take based on the input values. This loop will continue to run as long as there is a balance remaining
5. Use these formulas to calculate the payment breakdown
   • apr_monthly = (apr / 12) / 100
   • interest = start_bal * apr_monthly
   • end_bal = start_bal + interest – monthly
   • end_bal becomes start_bal the following payment cycle
6. Include <iomanip> and call cout << fixed << setprecision(2) to format your output
7. Use the setw(x) call in your cout stream to format your output. x is an integer indicating the width of the displayed column. For example, setw(5) will set a column width of 5. See below on how to display the table data

   cout << setw(5) << months << setw(10) << start_bal ...
   (of course, you can't actually use ellipses in the code. That's just an indicator that you can continue with whatever data and formatting you need)
   

8. Note that some combination of input values may create an infinite loop, meaning the balance never gets paid off. It can, in fact, increase if the monthly payment cannot cover the interest payment (try balance = 2000, monthly = 25, apr = 24 during your test runs)

Do Not Use

You may not use the break statement to terminate any loops

Sample Interaction / Output

What is the outstanding balance? 2000
What is the APR? 24
How much do you pay every month? 125

The table shows how long it will take to pay:

Month     Start Interest        Monthly     End
    1   2000.00    40.00         125.00 1915.00
    2   1915.00    38.30         125.00 1828.30
    3   1828.30    36.57         125.00 1739.87
    4   1739.87    34.80         125.00 1649.66
    5   1649.66    32.99         125.00 1557.66
    6   1557.66    31.15         125.00 1463.81
    7   1463.81    29.28         125.00 1368.09 
    8   1368.09    27.36         125.00 1270.45
    9   1270.45    25.41         125.00 1170.86
   10   1170.86    23.42         125.00 1069.27
   11   1069.27    21.39         125.00  965.66
   12    965.66    19.31         125.00  859.97
   13    859.97    17.20         125.00  752.17
   14    752.17    15.04         125.00  642.22
   15    642.22    12.84         125.00  530.06
   16    530.06    10.60         125.00  415.66
   17    415.66     8.31         125.00  298.97
   18    298.97     5.98         125.00  179.95
   19    179.95     3.60         125.00   58.55
   20     58.55     1.17         125.00  -65.28

Answer 1

If you have any doubts, please give me comment...

#include<iostream>

#include<iomanip>

using namespace std;

int main(){

    double balance, monthly, apr, apr_monthly, start_bal, interest, end_bal;

    int months;

    cout<<"What is the outstanding balance? ";

    cin>>balance;

    cout<<"What is the APR? ";

    cin>>apr;

    cout<<"How much do you pay every month? ";

    cin>>monthly;

    start_bal = balance;

    months = 1;

    cout<<"\nThe table shows how long it will take to pay:\n"<<endl;

    cout<<setprecision(2)<<fixed;

    cout<<setw(5)<<"Month"<<setw(10)<<"Start"<<setw(10)<<"Interest"<<setw(10)<<"Monthly"<<setw(10)<<"End"<<endl;

    while(start_bal>0){

        apr_monthly = (apr/12.0)/100;

        interest = start_bal * apr_monthly;

        end_bal = start_bal + interest - monthly;

        cout<<setw(5)<<months<<setw(10)<<start_bal<<setw(10) <<interest<<setw(10)<<monthly<<setw(10)<<end_bal<<endl;

        start_bal = end_bal;

        months++;

    }

    return 0;

}