+0  
 
+3
2
766
1
avatar+44 

Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for him, but it pays 8% annual interest. If there is $500 more in the account that pays 6%, and the total interest from both accounts is $114, how much money is in each of the accounts? PLEASE HELP

cbenes  Sep 21, 2014

Best Answer 

 #2
avatar+94203 
+8

Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for him, but it pays 8% annual interest. If there is $500 more in the account that pays 6%, and the total interest from both accounts is $114, how much money is in each of the accounts? 

 

It depends on  how often the interest is paid but lets assume it is paid yearly, bank interest is compound interest.  

 

 

$$\\$interest from $8\% $ account $=I_1 = P(1.08)^n -P\\
$interest from $\% $ account $= I_2= (P+500)(1.06)^n -(P+500)\\\\
I_1+I_2=P(1.08)^n -P+(P+500)(1.06)^n -(P+500)=114\\\\
P(1.08)^n -P+(P+500)(1.06)^n -(P+500)=114\\\\
P(1.08)^n -P+(P+500)(1.06)^n -P-500=114\\\\
P(1.08)^n +(P+500)(1.06)^n -2P=614\\\\
P(1.08)^n +P(1.06)^n+500(1.06)^n -2P=614\\\\
P[(1.08)^n +(1.06)^n-2]+500(1.06)^n=614\\\\
P=\frac{614-500(1.06)^n}{(1.08)^n +(1.06)^n-2}\\\\$$

 

THE 8% ACCOUNT HAS $P  AND THE 6% ACCOUNT HAS $(P+500)

This is saying that the original principals will depend on how long the money is invested for.   

-------------------------------------------------------------------------------

I'm just going to try it with a couple of values

1 YEAR

$${\frac{\left({\mathtt{614}}{\mathtt{\,-\,}}{\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{1.06}}}^{{\mathtt{1}}}\right)}{\left({{\mathtt{1.08}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.06}}}^{{\mathtt{1}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}} = {\mathtt{600}}$$

Time 1 year,  amounts $600 and $1100       Interest = $48+$66=$114

-------------------------------------------------------

2 YEARS

$${\frac{\left({\mathtt{614}}{\mathtt{\,-\,}}{\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{1.06}}}^{{\mathtt{2}}}\right)}{\left({{\mathtt{1.08}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.06}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}} = {\mathtt{180}}$$

Time 2 years, amounts $180 and $680            Interest = $29.952+$84.048=$114

-------------------------------------------

3 YEARS

$${\frac{\left({\mathtt{614}}{\mathtt{\,-\,}}{\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{1.06}}}^{{\mathtt{3}}}\right)}{\left({{\mathtt{1.08}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.06}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}} = {\mathtt{41.026\: \!960\: \!827\: \!816\: \!332\: \!7}}$$

Time 3 years, amounts $41.0269 and $541.0269         etc

 

YES, THAT WORKS!

THAT QUESTION WAS  A BIT DIFFERENT FROM THE NORM.  

Melody  Sep 21, 2014
 #2
avatar+94203 
+8
Best Answer

Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for him, but it pays 8% annual interest. If there is $500 more in the account that pays 6%, and the total interest from both accounts is $114, how much money is in each of the accounts? 

 

It depends on  how often the interest is paid but lets assume it is paid yearly, bank interest is compound interest.  

 

 

$$\\$interest from $8\% $ account $=I_1 = P(1.08)^n -P\\
$interest from $\% $ account $= I_2= (P+500)(1.06)^n -(P+500)\\\\
I_1+I_2=P(1.08)^n -P+(P+500)(1.06)^n -(P+500)=114\\\\
P(1.08)^n -P+(P+500)(1.06)^n -(P+500)=114\\\\
P(1.08)^n -P+(P+500)(1.06)^n -P-500=114\\\\
P(1.08)^n +(P+500)(1.06)^n -2P=614\\\\
P(1.08)^n +P(1.06)^n+500(1.06)^n -2P=614\\\\
P[(1.08)^n +(1.06)^n-2]+500(1.06)^n=614\\\\
P=\frac{614-500(1.06)^n}{(1.08)^n +(1.06)^n-2}\\\\$$

 

THE 8% ACCOUNT HAS $P  AND THE 6% ACCOUNT HAS $(P+500)

This is saying that the original principals will depend on how long the money is invested for.   

-------------------------------------------------------------------------------

I'm just going to try it with a couple of values

1 YEAR

$${\frac{\left({\mathtt{614}}{\mathtt{\,-\,}}{\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{1.06}}}^{{\mathtt{1}}}\right)}{\left({{\mathtt{1.08}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.06}}}^{{\mathtt{1}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}} = {\mathtt{600}}$$

Time 1 year,  amounts $600 and $1100       Interest = $48+$66=$114

-------------------------------------------------------

2 YEARS

$${\frac{\left({\mathtt{614}}{\mathtt{\,-\,}}{\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{1.06}}}^{{\mathtt{2}}}\right)}{\left({{\mathtt{1.08}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.06}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}} = {\mathtt{180}}$$

Time 2 years, amounts $180 and $680            Interest = $29.952+$84.048=$114

-------------------------------------------

3 YEARS

$${\frac{\left({\mathtt{614}}{\mathtt{\,-\,}}{\mathtt{500}}{\mathtt{\,\times\,}}{{\mathtt{1.06}}}^{{\mathtt{3}}}\right)}{\left({{\mathtt{1.08}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.06}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}} = {\mathtt{41.026\: \!960\: \!827\: \!816\: \!332\: \!7}}$$

Time 3 years, amounts $41.0269 and $541.0269         etc

 

YES, THAT WORKS!

THAT QUESTION WAS  A BIT DIFFERENT FROM THE NORM.  

Melody  Sep 21, 2014

20 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.