Question

If you open a savings account that earns 6.5% simple interest per year, what is the minimum num- ber of years you must wait to triple your balance? Suppose you open another account that earns 6% interest compounded yearly. How many years will it take now to triple your balance?

Answer 1

Answer:

• 31 years
• 19 years

Explanation:

To triple your balance( principal ) using simple interest rate

A = 3 * P = 3P (equation 1)

A = new amount

p = principal

T = x

R = interest rate = 6.5%

to calculate the interest rate = $\frac{PTR}{100}$  ( equation 2 )

note A = principal + interest rate

        A = p + $\frac{PTR}{100}$   therefore A = p ( 1 + $\frac{TR}{100}$ ) ( equation 3 )

from (equation 1)  A = 3P substitute this into (equation 3)

equation 3 becomes: 3P = P ( 1 + $\frac{TR}{100}$ ) divide both sides of the equation by P

equation becomes: 3 = ( 1 + $\frac{T6.5}{100}$ ) therefore 3 =  1 + 0.065T

hence T = $\frac{3-1}{0.065}$ = $\frac{2}{0.065}$ = 30.76 ≈ 31 years

To triple your balance using the compound interest

R = 6%

A = 3P

using the compound interest formula

A = P ( 1 + $\frac{R}{100}$)^t

3P = P ( 1 + $\frac{6}{100}$ )^t

3 = ( 1.06 )^t

t = $\frac{log 3}{log 1.06}$ = 18.85 years ≈ 19 years

Answer 2

Answer:

6.5% at simple interest requires 30.77 years

6.0% at compounding of 6.5% requires 18.85 years

Explanation:

We want a PV of $1 to befome $3 in the future

$1 (1 + r \times n) = 3\\(1 + 0.065 \times n) = 3\\n = (3 - 1) / 0.065$

n = 30,769

Compounding interest:

$(1+r)^n = 3\\log _{1+r}3 = n\\n = \frac{log 3}{log (1+r)} \\n = \frac{log 3}{log (1+0.06)}$

n = 18.85417668

n = 18.85