Compound interest: If 600 is invested at 8% compounded annually, find balance after 3 years.

[chijioke]

1. If [math]\sout{N}600[/math] is invested in an account that earns [math]8\%[/math] compounded annually, what will be the account balance after 3 years?​
2. Calculate the amount if [math]\sout{N} 100[/math] was invested at [math]4\%[/math] compounded quarterly after a period of 2 years

​

Solution​

1. [math]\text{A}=\text{600}(1+r)^n[/math]​

[math]\text{A} = \text{P}(1+\frac{\text{8}}{100})^3[/math]​

Account balance after years = [math]\sout{N} 755.83[/math]​

2. [math]\text{A} = \text{100}(1+\frac{\text{4}}{100})^2[/math]​

Amount after a period of 2 years = [math]\sout{N}108.16[/math]​

Now how do I deal with this. My challenge here now is that the value for A and P was not given.​

 

[JeffM]

Your first answer is correct. What is your question? You are supposed to compute A, and P is given. The formula is quite simple

A = amount earned plus principal returned
P = principal loaned
i = annual interest rate expressed as a number rather than a percentage
n = number of compounding periods in a year
y = number of years until payment

[math]A = P \left ( 1 + \dfrac{i}{n} \right )^{ny}[/math]
Your second answer is incorrect. You failed to consider that compounding is quarterly. How many compounding periods are involved? What is the interest rate per period?

 

[chijioke]

A = amount earned plus principal returned
P = principal loaned
i = annual interest rate expressed as a number rather than a percentage
n = number of compounding periods in a year
y = number of years until payment

[math]A = P \left ( 1 + \dfrac{i}{n} \right )^{ny}[/math]
Your second answer is incorrect. You failed to consider that compounding is quarterly. How many compounding periods are involved? What is the interest rate per period?

I think the second answer is correct, because it stated that it was compounded quarterly after the end of two years. At least two years was mentioned.
If really you think the second answer is not correct, please feel free to post the corrections.
Anyway what you answered was not what I was having problem with. I didn't know that it was not included in my post.l am just seeing it now.What I am having problem with, is :

1. After how many years will a sum of money be doubled if invested at [math]2 \frac{ 1 }{ 2 } \%[/math] per annum compound interest? - as you can see P and A were not given.​
2. The second one like it is : After how many years will a sum of money be increased by [math]60 \%[/math] if invested at $$ 4 \frac{ 1 }{ 2 } \% per annum compound interest?

 

[BigBeachBanana]

I think the second answer is correct, because it stated that it was compounded quarterly after the end of two years. At least two years was mentioned.
If really you think the second answer is not correct, please feel free to post the corrections.
Anyway what you answered was not what I was having problem with. I didn't know that it was not included in my post.l am just seeing it now.What I am having problem with, is :

1. After how many years will a sum of money be doubled if invested at [math]2 \frac{ 1 }{ 2 } \%[/math] per annum compound interest? - as you can see P and A were not given.​
2. The second one like it is : After how many years will a sum of money be increased by [math]60 \%[/math] if invested at $$ 4 \frac{ 1 }{ 2 } \% per annum compound interest?

You might not have a problem with #2 but we do. Let's talk about that. If you're compounding quarterly (4 times a year) for 2 years. That means there are 8 compounding periods.

\(\displaystyle A =100\left(1 + \frac{0.04}{4}\right)^{8} \approx 108.29\)

 

[Dr.Peterson]

What I am having problem with, is :

1. After how many years will a sum of money be doubled if invested at [imath]2 \frac{ 1 }{ 2 } \%[/imath] per annum compound interest? - as you can see P and A were not given.​
2. The second one like it is : After how many years will a sum of money be increased by [imath]60 \%[/imath] if invested at \( 4 \frac{ 1 }{ 2 } \%\) per annum compound interest?

The question supposes that you invest any amount of money, and it is doubled. So if the principal is P, the amount is 2P.

So you can do one of two things:

• Either choose a value for P, and use twice that for A (e.g. P=1, A=2);
• or leave P as a variable and replace A with 2P. You will soon be able to cancel P, so you'll see why the same answer works for any "sum of money invested".

The you can do something similar for the second problem.

 

[chijioke]

You might not have a problem with #2 but we do. Let's talk about that. If you're compounding quarterly (4 times a year) for 2 years. That means there are 8 compounding periods.

\(\displaystyle A =100\left(1 + \frac{0.04}{4}\right)^{8} \approx 108.29\)

Yeah I thought as much as that. But when I thought that compound interests are done in years not months, I settled for the year. Thank you for the correction.
So could we please proceed to the main problem?

 

[Steven G]

It doesn't matter how much you invest at a given interest rate and a given number of compounds per year for your money to double or go up 18% or to triple...

 

[Steven G]

You might not have a problem with #2 but we do. Let's talk about that. If you're compounding quarterly (4 times a year) for 2 years. That means there are 8 compounding periods.

\(\displaystyle A =100\left(1 + \frac{0.04}{4}\right)^{8} \approx 108.29\)

Yeah I thought as much as that. But when I thought that compound interests are done in years not months, I settled for the year. Thank you for the correction.
So could we please proceed to the main problem?

Yes, you are correct! The interest in 2 years has been compounded 8 times--nothing about months was mentioned!

 

[chijioke]

The question supposes that you invest any amount of money, and it is doubled. So if the principal is P, the amount is 2P.

So you can do one of two things:

• Either choose a value for P, and use twice that for A (e.g. P=1, A=2);
• or leave P as a variable and replace A with 2P. You will soon be able to cancel P, so you'll see why the same answer works for any "sum of money invested".

The you can do something similar for the second problem.

1. A= 2P, P=P, [math]r=2.5[/math][math]1.025 ^ { n } = \frac{ 2P }{ P }[/math][math]n= \frac{ \log ( 2 )}{ \log ( 1.025 )}= 28~ years[/math]Is this what you are saying concerning #1?

 

[JeffM]

1. A= 2P, P=P, [math]r=2.5[/math][math]1.025 ^ { n } = \frac{ 2P }{ P }[/math][math]n= \frac{ \log ( 2 )}{ \log ( 1.025 )}= 28~ years[/math]Is this what you are saying concerning #1?

Yes, exactly. Well done.

For future students, the detailed steps for chiojioke’s solution are:

[math]A = 2P \text { and } A = P(1.025)^n \implies\\ 2P = P(1.025)^n \implies \\ 1.025^n = \dfrac{2P}{P} = 2 \implies \\ \log(2) = \log(1.025^n) = n \ \log(1.025) \implies \\ n = \dfrac{\log(2)}{\log(1.025} \approx 28.1.[/math]

 

[lookagain]

[math]r=2.5[/math]

r is the rate, so r = 2.5% = 0.025 to be used in the formula below.

The commonly used formula in texts, where r (and t) are used, is
\(\displaystyle \ A \ = \ P \bigg(1 \ + \ \dfrac{r}{n} \bigg)^{nt}.\)

The variables represent the same as before, except "r" takes the place of "i," and "t"
takes the place of "y" for the time in years until the payment.

 

[chijioke]

Yes, exactly. Well done.

For future students, the detailed steps for chiojioke’s solution are:

[math]A = 2P \text { and } A = P(1.025)^n \implies\\ 2P = P(1.025)^n \implies \\ 1.025^n = \dfrac{2P}{P} = 2 \implies \\ \log(2) = \log(1.025^n) = n \ \log(1.025) \implies \\ n = \dfrac{\log(2)}{\log(1.025} \approx 28.1.[/math]

Why did you choose to round up to 1 decimal place instead of be just having as whole number just like I did?

 

[chijioke]

Yes, exactly. Well done.

For future students, the detailed steps for chiojioke’s solution are:

[math]A = 2P \text { and } A = P(1.025)^n \implies\\ 2P = P(1.025)^n \implies \\ 1.025^n = \dfrac{2P}{P} = 2 \implies \\ \log(2) = \log(1.025^n) = n \ \log(1.025) \implies \\ n = \dfrac{\log(2)}{\log(1.025} \approx 28.1.[/math]

For #2, [math]A=160\% \text{,}~ P=100 \%\text{,}~ r = 4 \frac{ 1 }{ 2 } \%[/math][math]\log ( 1 + r ) ^ { n } = \log \left (\frac{ A }{ P}\right)[/math] [math]n = \frac{ \log 1.6 )}{ \log 1.045 } = 11[/math]Thus number of years = 11
Right?

 

[BigBeachBanana]

For #2, [math]A=160\% \text{,}~ P=100 \%\text{,}~ r = 4 \frac{ 1 }{ 2 } \%[/math][math]\log ( 1 + r ) ^ { n } = \log \left (\frac{ A }{ P}\right)[/math] [math]n = \frac{ \log 1.6 )}{ \log 1.045 } = 11[/math]Thus number of years = 11
Right?

You can verify your own solution by picking an arbitrary principal amount and see after 11 years, you get 1.6x of your principal amount.

 

[JeffM]

Why did you choose to round up to 1 decimal place instead of be just having as whole number just like I did?

Because it will not happen in exactly 28 years. Saying 28 when you know full well that it will not happen in 28 years is what gives lawyers huge paydays. I was a banker for 43 years, and there are only two things you sell as a banker: you keep your word (but not necessarily one micron beyond) and you can count. No one expects bankers to be particularly intelligent or to be kind hearted, but if they say “double your money in 28 years,” they better deliver to the second. Juries do not like banks. When it comes to the mathematics of finance, you are not dealing with the pristine idealizations of mathematics, but the realities of regulators and hostile lawyers and an ignorant public.