Henry Devine bought a new dishwasher for 320. He paid 20 down and made 10 monthly payments of 34. What actual yearly interest rate did Henry pay?

Question

using this formula r=2mI/p(n+1)
i got this r=2(12)(20)/300(10+1)=

but i don't know what to do next . can you please show me how to do it and correct me it am wrong.

Answer 1

Wow, haven't seen or used that formula in almost 50 years.

I don't think you are substituting the correct values.
First of all since Henry paid $20 down the actual principal is 300

in r=2mI/p(n+1) m is the possible number of payments per year, in this case m=12
I is the actual interest paid, in this case it is 10(34) - 300 = 40
and n is the actual number of payments
p is the principal

so rate = 2(12)(40)/(300(11))
= .2909 or 29.09%

Answer 2

I would say that the rate is 12.5%.

btw, this task is just nonsense

Answer 3

He borrowed 300 dollars to buy the 320 dollar washer.(since he paid 20 down)
Using the mortgage payment formula:
payment = principal * r/[ 1-(1+r)^-n]
where r is the monthly rate and n is ten months
34 = 300 *r / [1-(1+r)^-10]
[1-(1+r)^-10] = 8.82 r
1 = 8.82 r + (1+r)^-10
I have to solve that by iteration
table of r versus right side, must be 1
.10 .882 + .386 = 1.26
.09 .794 + .422 = 1.22
.08 .706 + .463 = 1.16
.07 .617 + .508 = 1.12
.05 .441 + .614 = 1.05
.04 .353 + .676 = 1.02
.03 .265 + .744 = 1.01
.02 .176 + .820 = .996
so around r = .025 per month or .3 per year
30%
wicked high
check
pmt = 300 (.025 /(1 - (1.025)^-10) ) ??
= 34.27 close enough

Answer 4

I guess it doesn't hurt to have one more opinion.

I will try to work from basic principles, where the principal will accumulate over the 10 months, and the payments will accumulate at the same rate of interest, separately. At the end of 10 months, the future value of the principal and the payments should be equal.

Let
P=principal, $300
r=rate of interest per month, about 0.025
R=1+r, about 1.025
n=number of periods, 10
A=monthly payment, $34

Future value of the principal
= PR^n

Future value (FV) of the first payment (payable after at the end of the first month)
= AR^(n-1)
FV of second payment
= AR^(n-2)
...
Future value of last payment
=A
Thus future value of monthly payments
=A(1+R+R^2+....+R^(n-1))
=A(R^n-1)/(R-1)

Equating future values,

PR^n = A(R^n-1)/(R-1)
Substituting numerical values:
300R^10=34(R^10 -1)/(R-1)

This can be solved by iteration if we arrange the equation as:

R=(34(R^10-1)/(300(R-1)))^(1/10)
With an initial value of R=1.025, we get successively
R=1.02417
1.02377
1.02359
1.02350
...
and finally
R=1.0234293 at the 8th iteration.
or 2.343% per month, which translates to 1.0234293^12
=1.3204, or
32.4% p.a. simple interest

Check using Damon's formula
monthly payment
= principal * r/[ 1-(1+r)^-n]
=300*.02343/(1-1/1.02343^10)
=34.000
OK