Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. After c years, Calvin has $658.80. Makayla deposits $300 in a different savings account that accrues 6% interest compounded quarterly. After m years, Makayla has $613.04. What is the approximate difference in the number of years that Calvin and Makayla have their money invested?
400(1+.05/12)^(12c) = 658.8
300(1+.06/4)^(4m) = 613.04
m-c = 2
Let's start by solving for c, the number of years Calvin has his money invested.
The formula for compound interest is: A = P(1 + r/n)^(nt), where:
A: the final amount
P: the principal amount (initial deposit)
r: annual interest rate (as a decimal)
n: number of times the interest is compounded per year
t: number of years
For Calvin's savings account:
P = $400
r = 5% = 0.05 (as a decimal)
n = 12 (compounded monthly)
A = $658.80
Using the formula, we can rearrange it to solve for t:
A = P(1 + r/n)^(nt)
658.80 = 400(1 + 0.05/12)^(12c)
658.80/400 = (1.0041667)^(12c)
1.647 = 1.0041667^(12c)
Now, let's solve for c using logarithms:
log(1.647) = log(1.0041667^(12c))
log(1.647) = 12c * log(1.0041667)
c = log(1.647)/(12 * log(1.0041667))
Using a calculator, we find:
c ≈ 2.5
Now, let's solve for m, the number of years Makayla has her money invested.
Using the same formula for compound interest:
P = $300
r = 6% = 0.06 (as a decimal)
n = 4 (compounded quarterly)
A = $613.04
613.04 = 300(1 + 0.06/4)^(4m)
613.04/300 = (1.015)^(4m)
Again, using logarithms:
log(2.0434667) = 4m * log(1.015)
m = log(2.0434667)/(4 * log(1.015))
Using a calculator, we find:
m ≈ 4
Finally, we can calculate the approximate difference in the number of years:
Difference = c - m
Difference ≈ 2.5 - 4
Difference ≈ -1.5
The approximate difference in the number of years that Calvin and Makayla have their money invested is -1.5 years.
To find the approximate difference in the number of years that Calvin and Makayla have their money invested, we need to solve the given equations:
For Calvin:
A = P(1 + r/n)^(nt)
Where:
A = the final amount ($658.80)
P = the principal amount ($400)
r = the annual interest rate (5% or 0.05)
n = the number of times interest is compounded per year (12, monthly compounding)
t = the number of years (unknown, denoted as c)
For Makayla:
A = P(1 + r/n)^(nt)
Where:
A = the final amount ($613.04)
P = the principal amount ($300)
r = the annual interest rate (6% or 0.06)
n = the number of times interest is compounded per year (4, quarterly compounding)
t = the number of years (unknown, denoted as m)
Let's solve the equations to find the values of c and m.
For Calvin:
$658.80 = $400(1 + 0.05/12)^(12c)
For Makayla:
$613.04 = $300(1 + 0.06/4)^(4m)
Unfortunately, there isn't a straightforward algebraic solution to determine the values of c and m. Instead, we can use approximation methods or numerical methods to find their approximate values.
One approach is to use trial and error, plugging in different values for c and m until the equations match the given amounts, $658.80 and $613.04.
Starting with Calvin's equation:
$658.80 ≈ $400(1 + 0.05/12)^(12c)
By trial and error, we can try different values for c:
- c = 2
- c = 3
- c = 4
- c = 5
...
Similarly, we can try different values for m in Makayla's equation:
$613.04 ≈ $300(1 + 0.06/4)^(4m)
- m = 2
- m = 3
- m = 4
- m = 5
...
By trying different values and plugging them into the equations, we eventually find the values of c and m that satisfy the equations.
Once we have the values of c and m, we can calculate the approximate difference in the number of years by subtracting them:
Difference = c - m
Therefore, the approximate difference in the number of years that Calvin and Makayla have their money invested is Difference.