Math annunity question (1 Viewer)

[girlworld_club]

I tried doing this question but kept failing on numerous attempts.

Farmer brown puts aside part of the farm's earnings at the beginning of each month to buy a new truck in 10 years time when the old one wears out. He invests 400 each month at 9% p.a and estimates the cost of a new truck to be $80,000. Will the investment earn enough money to buy the new truck? what is the difference?

 

[girlworld_club]

if anyone wants i can upload my work and you can check it, but if your super nice please post your own answer.

 

[Bobbo1]

No, he'll have $79,489, assuming that interest is compounded yearly.

 

[asianese]

My final answer was $77,986.25366 ie he is not able to buy it. (compounded monthly)

Please upload your work.

 

[omgiloverice]

 

[braintic]

You are both correct. Your answer is correct based on monthly interest, the other is correct based on yearly interest. The question did not state how often interest is compounding, only how often money is invested. Both answers should be marked correct in an exam.

 

[omgiloverice]

You are both correct. Your answer is correct based on monthly interest, the other is correct based on yearly interest. The question did not state how often interest is compounding.

I understand where you are coming from, but this question constantly mentions the word 'monthly' so it is safe to assume that the interest is compounded monthly. Plus this question is from Maths in Focus so I wouldn't so concerned about any small details.

 

[braintic]

I'm saying only that both answers would have to be marked correct in an exam. That is hardly being concerned about small details.

 

[omgiloverice]

I'm saying only that both answers would have to be marked correct in an exam. That is hardly being concerned about small details.

haha fair enough :L

 

[asianese]

Just a note omgiloverice, you need to derive the full formula before just writing down the long series.

So you need to say: Money at the end of 1st month = 400 * (1.0075)
Money at the end of 2nd month = (400*1.0075 + 400)*1.0075 = 400*(1.0075)^2+400*1.0075
Money at the end of 3rd month = (400*(1.0075)^2+400*1.0075+400)*1.0075 = 400*(1.0075)^3+400*(1.0075)^2+400*1.0075 = 400(1.0075+1.0075^2+1.0075^3)
... ...
Money at the end of 120th month = 400(1.0075^120+1.0075^119+...+1.0075^2+1.0075) which forms a geometric series with T_1 = 1.0075, r=1.0075, and n=120

Money at the end of 120th month = 400* (1.0075(1-1.0075^120)/(1-1.0075))= 77985 whatever it was.

 

[omgiloverice]

:S I was never taught that, but thanks ^^

 

[girlworld_club]

Can someone validate what he said.

and show working if it was compounded monthly vs if it was compounded yearly. This would be totally appreciated.

 

[girlworld_club]

Just a note omgiloverice, you need to derive the full formula before just writing down the long series.

So you need to say: Money at the end of 1st month = 400 * (1.0075)
Money at the end of 2nd month = (400*1.0075 + 400)*1.0075 = 400*(1.0075)^2+400*1.0075
Money at the end of 3rd month = (400*(1.0075)^2+400*1.0075+400)*1.0075 = 400*(1.0075)^3+400*(1.0075)^2+400*1.0075 = 400(1.0075+1.0075^2+1.0075^3)
... ...
Money at the end of 120th month = 400(1.0075^120+1.0075^119+...+1.0075^2+1.0075) which forms a geometric series with T_1 = 1.0075, r=1.0075, and n=120

Money at the end of 120th month = 400* (1.0075(1-1.0075^120)/(1-1.0075))= 77985 whatever it was.

I think you are doing the wrong working, this question is based on annuity, meaning that a certain amount is invested and left to mature each on it's own independently.

However, You are right in saying omgiloverice should provide more working

A1 = 400 (1.0075)^200
A2 = 400 (1.0075) ^119
.
.
.
A120 = 400 (1.0075) ^1

Then you simply add them.

 

[braintic]

I think you are doing the wrong working, this question is based on annuity, meaning that a certain amount is invested and left to mature each on it's own independently.

This is a valid alternative method for this question.

 

[Timothy6340]

When is the intrest compounded an that method is wrong

 

[girlworld_club]

^^ if interest was compounded yearly can you show me working?

 

[braintic]

Sorry - watching the tennis.

 

[Timothy6340]

u need more info for yearly but with what u told us i think its monthly so
amount=400 r= 0.0075
A1= 400(1.0075)^120
A2=400(1.0075)^119
A3=400(1.0075)^118
.
.
.
A120+400(1.00075)

Atotal=A1+A2+A3+...+A120
=400(1.0075)^120+400(1.0075)^119+....400(1.00075)
=400(1.0075)((1.0075)^119+(1.0075)^118+...1)
=400(1.0075)((1-(1.0075)^120)/1-1.0075)
i hope thats right i havnt done it for a while
the answer is 77986 took so many attemps :/