 July 17, 2019, 04:24:26 pm

### AuthorTopic: Modelling linear growth and decay  (Read 189 times) Tweet Share

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#### yo091

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« on: April 18, 2019, 06:18:07 pm »
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The amount which should be invested at 4.5% per annum simple interest, if you require $15 700 in three years time, is closest to #### AlphaZero ##### Re: Modelling linear growth and decay « Reply #1 on: April 18, 2019, 09:31:34 pm » 0 The question is. Help please. The amount which should be invested at 4.5% per annum simple interest, if you require$15 700 in three years time, is closest to

I didn't do Further Maths in VCE, so if someone could please check if my method is what is intended by the course that would be great.

Using  $F=P(1+rt)\implies P=\dfrac{F}{1+rt}$,  we have $P=\frac{\15\,700}{1+0.045\!\times\! 3}=\13\,832.60,\ \ \text{correct to the nearest cent.}$
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### fish58

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• School: haileybury ##### Re: Modelling linear growth and decay
« Reply #2 on: April 19, 2019, 12:34:42 pm »
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A simple way to do this is to realise that 3 x 4.5 % = 13.5 %.
Therefore 15700 is 1.135 times the original amount.
Divide 15700 by 1.135 and you get the correct answer.

#### yo091

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« Reply #3 on: April 19, 2019, 01:06:01 pm »
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What is this formula called?

I didn't do Further Maths in VCE, so if someone could please check if my method is what is intended by the course that would be great.

Using  $F=P(1+rt)\implies P=\dfrac{F}{1+rt}$,  we have $P=\frac{\15\,700}{1+0.045\!\times\! 3}=\13\,832.60,\ \ \text{correct to the nearest cent.}$

#### yo091

•  