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Kimberly borrows 1000 dollars from Lucy, who charged interest of 5% per month (which compounds monthly). What is the least integer number of months after which Kimberly will owe more than twice as much as she borrowed?

mathtoo  Aug 28, 2018

Best Answer 

 #1
avatar
+1

5% / 1200 =0.004166667 - monthly interest rate.

2,000 /1,000 =1.004166667^n, where n = number of months to double the money

2 = 1.004166667^n   take the log of both sides

n =Log(2) / Log(1.004166667)

n =166.7 months to double the money.Or:166 months - the least integer number of months.

n = 167 months when Kimberly will owe: $2,002.48

Note: I have taken your interest rate to mean "5% annual rate compounded monthly." Your question says "5% compounded monthly", which is equivalent to 79.59% effective annual rate.

If you DO MEAN 5% compounded monthly, then the above calculation will look like this:

2,000/1,000 =1.05^n

2 = 1.05^n

n = Log(2) / Log(1.05)

n =14.21 months, or: 14 - the least integer number of months, after which:

n = 15 months when Kimberly will owe $2,078.93

Guest Aug 28, 2018
edited by Guest  Aug 28, 2018
edited by Guest  Aug 28, 2018
 #1
avatar
+1
Best Answer

5% / 1200 =0.004166667 - monthly interest rate.

2,000 /1,000 =1.004166667^n, where n = number of months to double the money

2 = 1.004166667^n   take the log of both sides

n =Log(2) / Log(1.004166667)

n =166.7 months to double the money.Or:166 months - the least integer number of months.

n = 167 months when Kimberly will owe: $2,002.48

Note: I have taken your interest rate to mean "5% annual rate compounded monthly." Your question says "5% compounded monthly", which is equivalent to 79.59% effective annual rate.

If you DO MEAN 5% compounded monthly, then the above calculation will look like this:

2,000/1,000 =1.05^n

2 = 1.05^n

n = Log(2) / Log(1.05)

n =14.21 months, or: 14 - the least integer number of months, after which:

n = 15 months when Kimberly will owe $2,078.93

Guest Aug 28, 2018
edited by Guest  Aug 28, 2018
edited by Guest  Aug 28, 2018

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