Question

Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for it to double? Round to the nearest tenth of a year. Show your work.
How long will it take to triple? Round to nearest tenth of a year. Show your work.
Kristen has a choice to invest her money at 6.5% interest compounded monthly for 5 years or invest her money compounding quarterly at a rate of 6.75% for 5 years. What option would be best for Kristen? Explain and show your work.

Answer 1

In order to figure out when the investment will double, we only need to look at the interest rate. The equation to use to find when the investment has doubled is years = ln(2)/ln(1+interest rate), where “ln” is the natural log, and the interest rate is 0.065. Years = ln(2) / ln(1.065) Years = 11.007 = 11.0 years Alternatively, If we want to approximate with the “rule of 72”, (which is a faster, slightly less approximate method), we just divide 72 by the interest rate percentage: Years = 72/6.5 = 11.077 = 11.1 years Alternativley, If we want to approximate with the “rule of 72”, (which is a faster, slightly less approximate method), we just divide 72 by the interest rate percentage: Years = 72/6.5 = 11.077 = 11.1 years

Answer 2

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

$A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},$

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.$

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},$

$3=(1.0054)^{12t},$

$12t=\log_{1.0054}3,$

$t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.$

3. 1 choice: P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \ 7936.39.$

2 choice: P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \ 8032.58.$

The best will be 2nd option, because $8032.58>$7936.39