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

#2**+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.

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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

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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

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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**+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