a loan of 4000 was repaid at the end of 10 months with a check for 4270. what annual rate of intreste was charged.
4270=4000(1+i)^10/12
take the log of each side
log 4270=log4000+ 10/12 log(1+i)
find log(1+i)=??
then take the antilog, solve for i.
TRAP is a trapezoid and XY is the median find the length of TP
R A is 22.6
x Y is 45.36
T P
To find the annual rate of interest charged, we can use the formula for simple interest:
Simple Interest = Principal * Rate * Time
Given that a loan of $4000 was repaid at the end of 10 months with a check for $4270, the interest charged would be $4270 - $4000 = $270.
Plugging in the given values into the formula, we have:
270 = 4000 * Rate * (10/12)
Simplifying the equation:
270 = 3333.33 * Rate
Now, we can solve for the annual rate of interest:
Rate = 270 / 3333.33
Rate = 0.081, which is equivalent to 8.1%.
Therefore, an annual rate of interest of 8.1% was charged on the loan.
To find the annual rate of interest charged on the loan, we need to use the formula for simple interest:
Simple Interest = (Principal * Rate * Time) / 100
In this case, the principal amount (loan) is $4000 and it was repaid after 10 months. The total amount repaid (including interest) is $4270.
From the given information, we can calculate the simple interest obtained on the loan:
Interest = Total Amount Repaid - Principal
Interest = $4270 - $4000
Interest = $270
Now, using the formula for simple interest, we can find the annual rate of interest charged:
Rate = (Interest * 100) / (Principal * Time)
Rate = ($270 * 100) / ($4000 * (10/12))
Rate ≈ 67.5%
Therefore, the annual rate of interest charged on the loan is approximately 67.5%.