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Time Payments Question. (1 Viewer)

SoCal

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The question is:

"Mr. Read borrowed $7000 to buy a new car. He has to pay off the loan over 4 years in equal monthly instalments allowing for an interest rate of 15% p.a. which is compounded monthly. Find the amount of each instalment."

Now I keep getting an answer of $204.21 but the book says the answer is $194.82. So am I wrong or the book's answer:confused:? Thank you:).
 

slyball

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interest = 15/12 (the amount of months) = 1.25% = 0.0125
time is 12 (months) * 4 (years) = 48

m = monthly instalment

7000 * 1.0125^48 * M(1 + 1.0125 + 1.0125^2 + 1.0125^3 + ... + 1.0125^47)

bold is GP so...

a(r^n - 1) / r - 1 = 1(1.0125^48 - 1) / (1.0125 - 1) = 65.22838823

rearrange to find m,

m = (7000 * 1.0125^48) / 65.22838823

= $194.82 :D



i think.
 

SoCal

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Thank you slyball, you are right. After all that effort I was typing the wrong number into the calculator:mad:. I was typing 62.22838823 into the calculator for some reason, even though I had 65.22838823 written down:mad:. Thanks again:).
 

MrMiK

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i duno bout youz.. but i dislike that whole...
7000 * 1.0125^48 * M(1 + 1.0125 + 1.0125^2 + 1.0125^3 + ... + 1.0125^47)
line
i just shove a formula together which cuts out that step..
duno if i can type it proper tho

P(1+ r/100)^n = M [(1+ 1/100)^n - 1) / (r/100)]

(look MUCH neater when written because its over 2 lines)
but the answer is easy to get, simple substitution...
maybe you could at least use it to test answers..?

OH and its real easy to do with calculator.. u work out the right side like M(blah blah) shove the blah blah into Min.. then go do the other side and hit divide by MR.. and ur SET.

Would it b ok to use this in the test?

I WITHDRAW .. what i said looks untrue.. i just shoved question 10 b) ii) into the equation.. i got out 126250.. oof.. looks like im wrongety wrong wrong
 
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Supra

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in the hsc they mite mark u down for using the formula, its not showing skills in gp (which they're trying to assess u on)
 

iambored

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i agree.. they may want you to use the sequences and series formula
 

Lexicographer

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Look at the Notes from the Marking Centre for last year's exam. Towards the bottom you'll find the marking guidelines for each question. Basically, using your own formula doesn't demonstrate understanding of the concepts of geometric progression, as you are not taking the problem and properly expressing it in mathematical terms. Also, if there's an irregular question (eg three months interest free or something) your formula is worthless.

Do what all good teachers say: derive
 

SoCal

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Yes, I am sure you have to derive it because that is what they give the marks for. It is also why these kind of questions are worth a few marks. Also, sometimes they will give you a question with parts in it and say "Show after 3 years that the amount owing is:

A_3 = $7000 x 1.0125^3 - P(1 + 1.0125 + 1.0125^2)".:).
 

slyball

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part i of that question is unrelated to the 15 years as stated in the question. they just wanted you to derive an expression for A60 (and you'll make use of the answer in part iii and iv)...
 

young_gurl

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Time payments is so confusing i hope they ask for monthly installments....theres this other formula that makes it easy for me but i got the answer rite but teacher marked it worng/?! hmmm
 

SoCal

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Originally posted by young_gurl
Time payments is so confusing i hope they ask for monthly installments....theres this other formula that makes it easy for me but i got the answer rite but teacher marked it worng/?! hmmm
That is because you have to derive it. The marks are allocated for realising that it is a geometric series and working it out that way, not for just getting the correct answer:).
 

bMaN oNe

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Originally posted by slyball
7000 * 1.0125^48 * M(1 + 1.0125 + 1.0125^2 + 1.0125^3 + ... + 1.0125^47)
How do u get that line? is that the first line or is there more to it?
 

tempco

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Originally posted by bMaN oNe
How do u get that line? is that the first line or is there more to it?
Slyball got it from:

A1 = 7000(1.0125) - M
A2 = A1(1.0125) - M
= [7000(1.0125) - M](1.0125) - M
= 7000(1.0125)^2 - M(1.0125) - M
A3 = A2(1.0125) - M
= [7000(1.0125)^2 - M(1.0125) - M](1.0125) - M
= 7000(1.0125)^3 - M(1.0125)^2 - M(1.0125) - M
(Factorise M out of the last 3 terms)
= 7000(1.0125)^3 - M(1.0125^2 + 1.0125 + 1)
Therefore,
An = 7000(1.0125)^n - M(1 + 1.0125 + ... + 1.0125^n-1)

It's over 4 years, so that's 12 * 4 = 48. So,

A48 = 7000(1.0125)^48 - M(1 + 1.0125 + ... + 1.0125^47)

Hope it's clear enough for you.
 

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