James purchased a bond for $3200, and ten months later he sold for $3700. What annual rate would he have to earn in a savings compounded monthly, to earn the same money on his investment?
I have to read the question more carefully.
It clearly said 10 months and I used 12 months
so let's just make a simple change
(1+i)^10 = 1.15625
take the 10th root
1+i = 1.15625^(1/10)
1+i = 1.014624..
i = .014624
annual rate = 12(.014624) = .1755 or appr 17.55%
let the monthly rate be i
3200(1+i)^12 = 3700
(1+i)^12 = 1.15625
take 12th root
1+i = 1.01217..
i = .01217..
annual rate compounded monthly = .14606
or appr 14.6%
The tight answer is 17.55% but I want to know how to get that the solution
The right answer is 17.55% but I want to know how to get that the solution please help
To calculate the annual rate required to earn the same money through a savings account compounded monthly, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = final amount
P = principal amount
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = number of years
In this case, we know:
P = $3200
A = $3700
n = 12 (compounded monthly)
t = 10/12 (since it's ten months)
Let's plug in the values we know and solve for r:
$3700 = $3200(1 + r/12)^(12 * (10/12))
To isolate r, we need to rearrange the formula:
(1 + r/12)^(10/12) = 3700/3200
[(1 + r/12)^(1/12)]^10 = 37/32
Now, to get rid of the exponent, we need to take the natural logarithm (ln) of both sides:
ln[(1 + r/12)^(1/12)]^10 = ln(37/32)
Using the property of logarithms, we can bring down the exponent:
(1/12) * 10 * ln(1 + r/12) = ln(37/32)
Simplifying further:
ln(1 + r/12) = 12 * [ln(37/32)] / 10
Now we can solve for r by taking the exponential (e) of both sides:
1 + r/12 = e^(ln(37/32) * 12/10)
Subtracting 1:
r/12 = e^(ln(37/32) * 12/10) - 1
Multiplying by 12:
r = 12 * [e^(ln(37/32) * 12/10) - 1]
Using a calculator, we can evaluate the right-hand side of the equation to find the annual rate James would have to earn.