Question 81460: How much must you deposit in an account that pays 6% annual interest compounded yearly to have a balance of $ 1000 after 10 years Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! The equation that is used to solve this problem is:
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in which the variables are:
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P is the future value of the invested money
C is the initial deposit
r is the interest rate expressed as a decimal
n is the number of times a year that the interest is compounded
t is the number of years of the investment
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For this problem the variables are given as follows:
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P is $1000
C is the amount of the initial deposit and you are to find C
r is 0.06 (the interest rate)
n is 1 compounding per year
t is 10 years
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Substituting these values into the equation results in:
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This simplifies to:
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and this further simplifies to:
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Calculator time. Use your calculator to raise 1.06 to the 10th power. Your calculator
may have a key for . If it does, enter 1.06 then press the key, and
next enter 10 and then press the = key. You should get 1.790847697. You can also get
this by entering 1.06 on your calculator and then multiplying it 9 times by 1.06.
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Substitute 1.790847697 for and the equation becomes:
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Finally solve for C by dividing both sides by 1.790847697 to get:
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After doing the division on the left side the equation simplifies to:
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Rounding this off to the next greatest cent tells you that to end up with at least $1000
you need to invest $558.40 at 6% interest compounded yearly for 10 years.
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(Note that if you invest $558.39 under the same circumstances you end up with $999.99 to the
nearest cent but by investing $558.40 you get $1000.01 to the nearest cent.)
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Hope this helps you to understand the problem a little better.
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