Question

Two accounts are opened at the same time. You deposit 1250 dollars into the first account, which earns interest at an effective rate of 6.1 percent. At the same time, you deposit 210 dollars into the second account, which earns an effective rate of 11.2 percent. How long will it take for the balance in the first account to be exactly twice the balance in the second account? (Assume compound interest at all times.)

Answer 1

Assuming a final balance of $3,000 for the second account, it would take 26,4 years of the first account to be exactly twice the balance in the second account.

Step-by-step explanation:

First, we need to determine a quantity for the second account. We use the compound interest formula:

A = P(1 + i/n)^n*t

where:

A = Final value

P = initial value

i = interest rate

n = number of times the interest rate is compounded in the period

t = number of periods elapsed

We will assume that we need to find the number of years it takes for the second account to give a balance of $3,000. Under this sceneario, our values will be:

A = $3,000

P = $210

i = 11.2% annually

n = 1 (the interest rate is an efective annual rate, therefore, it is compounded once in a year)

t = x (the number of periods is the incognita)

Next, we plug the amounts into the equation and solve:

210 (1 + 0.112)^X = 3,000

(1.1112)^X = 3,000 / 210

(1.112)^X = 14.3

Remember that we use logarithms to solve for an unknown exponent

X * Log 1.112 = Log 14.3

X = Log 14.3 / Log 1.112

X = 25.0 years

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Now, we need to find how long it takes the second account to give a balance that doubles 3,000. (6,000)

1,250 (1 + 0.061)^X = 6,000

(1.061)^X = 4.8

X*log 1.061 = log 4.8

X = log 4.8 / log 1.061

X = 26.49 years