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Mitch and Bill are the same age. When Mitch is 25 years old, he begins depositing $100 every month into a long-term savings plan. He consistently makes these deposits for 10 years, at which point he is forced to stop making deposits of any kind. However, he leaves his money in this same account for the next 40 years (where it continues to earn interest that is compounded monthly). Bill, on the other hand, doesn’t start saving until he is 35 years old. At that point in his life (age 35) and for the next 40 years after that, he makes consistent monthly deposits of $100. Assume that both accounts earn interest at an annual percentage rate of 7% and interest in both accounts is compounded monthly.

Guest May 14, 2015

Best Answer 

 #1
avatar+92751 
+10

I do not actually see any question here   

I suppose you want to know how much they each have at age 75 years ?

i = (0.07/12)

 

 

Mitch

How much after 10 years

$${\mathtt{100}}{\mathtt{\,\times\,}}\left({\frac{\left({\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)\right)}^{\left({\mathtt{120}}\right)}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}}\right) = {\mathtt{17\,308.480\: \!743\: \!353\: \!610\: \!059\: \!6}}$$

After a further 40 years

$${\mathtt{17\,308.48}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}^{{\mathtt{480}}} = {\mathtt{282\,325.739\: \!551\: \!243\: \!036\: \!749\: \!5}}$$

 

So it seems that Mitch will have    $282326

 

What about Bill

 

$${\frac{{\mathtt{100}}{\mathtt{\,\times\,}}\left({\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}^{{\mathtt{480}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}} = {\mathtt{262\,481.339\: \!833\: \!333\: \!259\: \!683\: \!2}}$$

 

Bill will only have $262 481

 

So Mitch will have more at age 75 years

Melody  May 14, 2015
 #1
avatar+92751 
+10
Best Answer

I do not actually see any question here   

I suppose you want to know how much they each have at age 75 years ?

i = (0.07/12)

 

 

Mitch

How much after 10 years

$${\mathtt{100}}{\mathtt{\,\times\,}}\left({\frac{\left({\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)\right)}^{\left({\mathtt{120}}\right)}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}}\right) = {\mathtt{17\,308.480\: \!743\: \!353\: \!610\: \!059\: \!6}}$$

After a further 40 years

$${\mathtt{17\,308.48}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}^{{\mathtt{480}}} = {\mathtt{282\,325.739\: \!551\: \!243\: \!036\: \!749\: \!5}}$$

 

So it seems that Mitch will have    $282326

 

What about Bill

 

$${\frac{{\mathtt{100}}{\mathtt{\,\times\,}}\left({\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}^{{\mathtt{480}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\frac{{\mathtt{0.07}}}{{\mathtt{12}}}}\right)}} = {\mathtt{262\,481.339\: \!833\: \!333\: \!259\: \!683\: \!2}}$$

 

Bill will only have $262 481

 

So Mitch will have more at age 75 years

Melody  May 14, 2015

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