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# 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 p

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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

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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 ------------------------------------------------------- 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
+92751
+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

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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