+0

# interest

+1
105
1

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?

Aug 28, 2018

#1
+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

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

#1
+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