Suppose you deposit $5,000 in savings account that earns 3% annual interest. If you make no other withdrawals or deposits, how many years will it take the account balance to reach at least $6,000?
a) 10 years
b) 6 years
c) 7 years
d) 4 years
We can use the formula for compound interest to solve this problem:
A = P(1 + r/n)^(nt)
where:
A = the final amount (in this case, $6,000)
P = the principal amount (in this case, $5,000)
r = the annual interest rate (3%)
n = the number of times the interest is compounded per year (we'll assume it's compounded annually, so n = 1)
t = the number of years
Plugging in the values we know, we get:
6,000 = 5,000(1 + 0.03/1)^(1t)
Simplifying:
1.2 = (1.03)^t
Taking the natural logarithm of both sides:
ln(1.2) = t ln(1.03)
Dividing both sides by ln(1.03):
t = ln(1.2) / ln(1.03) ≈ 7.22
So it will take about 7 years for the account balance to reach at least $6,000. The answer is c) 7 years.
To find the number of years it will take for the account balance to reach at least $6,000, we can use the formula for compound interest:
A = P(1 + r/n)^(n*t)
Where:
A = the final account balance
P = the initial deposit amount
r = the annual interest rate (decimal form)
n = the number of times interest is compounded per year
t = the number of years
In this case, the initial deposit amount (P) is $5,000, the annual interest rate (r) is 3% or 0.03, and we want to find the number of years (t) it will take for the account balance to reach at least $6,000.
Let's plug in these values into the formula and solve for t:
$6,000 = $5,000(1 + 0.03/n)^(n*t)
To simplify the equation, we can assume that the interest is compounded annually, meaning n = 1:
$6,000 = $5,000(1 + 0.03/1)^(1*t)
$6,000 = $5,000(1 + 0.03)^t
Divide both sides of the equation by $5,000:
6,000 / 5,000 = (1 + 0.03)^t
1.2 = (1.03)^t
To solve for t, we need to take the logarithm of both sides:
log(1.2) = log(1.03)^t
Using the logarithmic property log(a^b) = b*log(a), we can rewrite the equation:
log(1.2) = t * log(1.03)
Now, divide both sides by log(1.03) to solve for t:
t = log(1.2) / log(1.03)
Using a scientific calculator or an online logarithm calculator, we can find the value of t.
Calculating this, we find:
t ≈ 9.42
Since t represents the number of years, we can round up to the nearest whole number as we cannot have partial years. Therefore, it will take approximately 10 years for the account balance to reach at least $6,000.
Therefore, the correct answer is:
a) 10 years.
To solve this problem, we can use the formula for compound interest:
A = P(1 + r/n)^(n*t)
Where:
A is the final amount
P is the principal amount (initial deposit)
r is the annual interest rate (in decimal form)
n is the number of times interest is compounded per year
t is the number of years
In this case, the principal amount (P) is $5,000 and the final amount (A) is $6,000. The annual interest rate (r) is 3% which is 0.03 in decimal form.
Let's test each option by substituting the values into the formula:
a) 10 years
A = 5000(1 + 0.03/1)^(1*10)
A ≈ 5000(1.03)^10 ≈ $6,383.35
b) 6 years
A = 5000(1 + 0.03/1)^(1*6)
A ≈ 5000(1.03)^6 ≈ $5,943.62
c) 7 years
A = 5000(1 + 0.03/1)^(1*7)
A ≈ 5000(1.03)^7 ≈ $6,118.53
d) 4 years
A = 5000(1 + 0.03/1)^(1*4)
A ≈ 5000(1.03)^4 ≈ $5,665.50
The option that results in at least $6,000 in the account balance is option a) 10 years. The account balance will reach $6,000 in approximately 10 years.