Question

IN PSEUDOCODE and C++

Program 3: Give a baby $5,000! Did you know that, over the last century, the stock market has returned an average of 10%? You may not care, but you’d better pay attention to this one. If you were to give a newborn baby $5000, put that money in the stock market and NOT add any additional money per year, that money would grow to over $2.9 million by the time that baby is ready for retirement (67 years)! Don’t believe us? Check out the compound interest calculator from MoneyChimp and plug in the numbers!

To keep things simple, we’ll calculate interest in a simple way. You take the original amount (called the principle) and add back in a percentage rate of growth (called the interest rate) at the end of the year. For example, if we had $1,000 as our principle and had a 10% rate of growth, the next year we would have $1,100. The year after that, we would have $1,210 (or $1,100 plus 10% of $1,100). However, we usually add in additional money each year which, for simplicity, is included before calculating the interest.

Your task is to design (pseudocode) and implement (source) for a program that 1) reads in the principle, additional annual money, years to grow, and interest rate from the user, and 2) print out how much money they have each year. Task 3: think about when you earn the most money!

Lesson learned: whether it’s your code or your money, save early and save often…

Sample run 1:

Enter the principle: 2000

Enter the annual addition: 300

Enter the number of years to grow: 10

Enter the interest rate as a percentage: 10

Year 0: $2000

Year 1: $2530

Year 2: $3113

Year 3: $3754.3

Year 4: $4459.73

Year 5: $5235.7

Year 6: $6089.27

Year 7: $7028.2

Year 8: $8061.02

Year 9: $9197.12

Year 10: $10446.8

Sample run 2 (yeah, that’s $9.4MM):

Enter the principle: 5000

Enter the annual addition: 1000

Enter the number of years to grow: 67

Enter the interest rate as a percentage: 10

Year 0: $5000

Year 1: $6600

Year 2: $8360

Year 3: $10296

Year 4: $12425.6

Year 5: $14768.2

.

.

Year 59: $4.41782e+06

Year 60: $4.86071e+06

Year 61: $5.34788e+06

Year 62: $5.88376e+06

Year 63: $6.47324e+06

Year 64: $7.12167e+06

Year 65: $7.83493e+06

Year 66: $8.61952e+06

Year 67: $9.48258e+06

Answer 1

PseudoCode:

start

input principle,annual_addition,years,rate_of_interest

initialize amt:=principle

for i in 0....n+1

do

print "Year "+i+": $"+amt

amt=amt+addition

amt=amt+rate/100*amt

done

end

Solution(Source code):

#include<iostream>

using namespace std;

int main()

{

    long principle,addition,years;

    int i;

    double amt,rate;

    cout<<"Enter the principle: ";

    cin>>principle;

    cout<<"Enter the annual addition: ";

    cin>>addition;

    cout<<"Enter the number of years to grow: ";

    cin>>years;

    cout<<"Enter the interest rate as a percentage: ";

    cin>>rate;

    amt=principle;

    for(i=0;i<=years;i++)

    {

        cout<<"Year "<<i<<": $"<<amt<<endl;

        amt=amt+addition;

        amt=amt+(double)((rate/100)*amt);

    }

    return 0;

}

Output: