HELLLPP! - Sequences and series question!!! (2 Viewers)

[Suuup2]

Hi can someone help me with this question, I cant seem to get the correct answer.
" the smith family buys a car for $38 000, paying a 10% deposit and taking a loan out for the balance. if the loan is over 5yrs with interest of 1.5% monthly, find amount of each monthly repayment.

Answer= $592

Thank you in advance !!

 

[braintic]

What answer are you getting? I get $868.

 

[obliviousninja]

What answer are you getting? I get $868.

this

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[bongoli]

this

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this

 

[dim-sims]

Is this from MiF?

 

[gwc]

" the smith family buys a car for $38 000, paying a 10% deposit and taking a loan out for the balance. if the loan is over 5yrs with interest of 1.5% monthly, find amount of each monthly repayment.

Answer= $592

2 years late but for those that are looking for the answer to this in the future, here's how to do it

P = monthly repayment
r = 0.015/12 = 0.00125
n = 5x12 = 60
Loan = $38 000
but, since they payed a 10% deposit, the amount that you work with is 38 000x0.9 = $34 200

A1 = 34 200(1.00125) - P
A2 = A1(1.00125) - P
= [34 200(1.00125) - P](1.00125) - P
= 34 200(1.00125^2) - P(1.00125 + 1)
A3 = A2(1.00125) - P
= [34 200(1.00125^2) - P(1.00125 + 1)](1.00125) - P
= 34 200(1.00125^3) - P(1.00125^2 + 1.00125 + 1)
...
An = 34 200(1.00125^n) - P[1+ 1.00125 + 1.00125^2 +...+ 1.00125^(n-1)]
therefore, A60 = 34 200(1.00125^60) - P(1+ 1.00125 + 1.00125^2 +...+ 1.00125^59)
however after 60 months, A60 = 0
0 = 34 200(1.00125^60) - P(1+ 1.00125 + 1.00125^2 +...+ 1.00125^59)
P = [34 200(1.00125^60)]/[1+ 1.00125 + 1.00125^2 +...+ 1.00125^59]
= [34 200(1.00125^60)]/[a(r^n - 1)/r-1]

G.S.
a = 1
r = 1.00125
n = 60​

= [34 200(1.00125^60)]/[1(1.00125^60 - 1)/1.00125-1]
= $591.99

If you still need help, let me know!

 

[pikachu975]

2 years late but for those that are looking for the answer to this in the future, here's how to do it

P = monthly repayment
r = 0.015/12 = 0.00125
n = 5x12 = 60
Loan = $38 000
but, since they payed a 10% deposit, the amount that you work with is 38 000x0.9 = $34 200

A1 = 34 200(1.00125) - P
A2 = A1(1.00125) - P
= [34 200(1.00125) - P](1.00125) - P
= 34 200(1.00125^2) - P(1.00125 + 1)
A3 = A2(1.00125) - P
= [34 200(1.00125^2) - P(1.00125 + 1)](1.00125) - P
= 34 200(1.00125^3) - P(1.00125^2 + 1.00125 + 1)
...
An = 34 200(1.00125^n) - P[1+ 1.00125 + 1.00125^2 +...+ 1.00125^(n-1)]
therefore, A60 = 34 200(1.00125^60) - P(1+ 1.00125 + 1.00125^2 +...+ 1.00125^59)
however after 60 months, A60 = 0
0 = 34 200(1.00125^60) - P(1+ 1.00125 + 1.00125^2 +...+ 1.00125^59)
P = [34 200(1.00125^60)]/[1+ 1.00125 + 1.00125^2 +...+ 1.00125^59]
= [34 200(1.00125^60)]/[a(r^n - 1)/r-1]

G.S.
a = 1
r = 1.00125
n = 60​

= [34 200(1.00125^60)]/[1(1.00125^60 - 1)/1.00125-1]
= $591.99

If you still need help, let me know!

You do not divide the interest by 12 since it says 1.25% monthly already

 

[gwc]

You do not divide the interest by 12 since it says 1.25% monthly already

Did you mean 1.5% monthly not 1.25% monthly? If so, I tried with r=0.015 and I got $513 which isn't the answer at the back of the textbook. It works with my original working out though. I think the authors worded the question poorly and meant '1.5%p.a. monthly'. Either that or I miraculously got the right answer with the wrong working out

 

[pikachu975]

Did you mean 1.5% monthly not 1.25% monthly? If so, I tried with r=0.015 and I got $513 which isn't the answer at the back of the textbook. It works with my original working out though. I think the authors worded the question poorly and meant '1.5%p.a. monthly'. Either that or I miraculously got the right answer with the wrong working out

Yeah 1.5% the question was probably worded poorly because both yours and the textbook's answer divide 1.5% by 12.