Annual Compounding Interest

**magentarita** · December 25th 2008, 04:54 AM

Jane has $6 and Sarah has $8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.

a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95

I just want the answer.

Thanks

**masters** · December 25th 2008, 05:40 AM

Originally Posted by magentarita

Jane has $6 and Sarah has $8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.

a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95

I just want the answer.

Thanks

It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.

$6(1+.11)^t=8(1+.08)^t$

$\log 6(1.11)^t=\log 8(1.08)^t$

Etc. and so forth to find that $t \approx 10.4997388$

**Soroban** · December 25th 2008, 06:53 AM

Hello, magentarita!

I got a different result . . .

Jane has $6 and Sarah has $8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.

$a)\;\$10.50 \qquad b)\;\text{never equal}\qquad c)\;\$17.23 \qquad d)\;\$17.95$

At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.

So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d )}$

**masters** · December 25th 2008, 08:26 AM

Originally Posted by Soroban

Hello, magentarita!

I got a different result . . .

At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.

So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d )}$

You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.

**magentarita** · December 28th 2008, 07:21 AM

Originally Posted by Soroban

Hello, magentarita!

I got a different result . . .

At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.

So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$

Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$

. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$

They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.

At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d )}$

Yes, the answer is (d) and I want to thank you.

**magentarita** · December 28th 2008, 07:23 AM

Originally Posted by masters

It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.

$6(1+.11)^t=8(1+.08)^t$

$\log 6(1.11)^t=\log 8(1.08)^t$

Etc. and so forth to find that $t \approx 10.4997388$

I got the answer from Soroban but I thank you for your effort.

**magentarita** · December 28th 2008, 07:24 AM

Originally Posted by masters

You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.

Don't feel bad. No one is perfect except God and He is not taking any math courses lately.

Thread: Annual Compounding Interest

LinkBack

Thread Tools

Display

Annual Compounding Interest

yes...

ok..

yes...

Similar Math Help Forum Discussions

compounding interest

semi-annual compounding question

Compounding interest problem help

Compounding Interest Continuously

Interest compounding with a twist

Search Tags