Logarithms and Exponentials: Applications 1

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Name	Status	Author	Last Modified
Logarithms and Exponentials: Applications 1	Ready to use	Ben McGovern	10/10/2022 12:59
Logarithms and Exponentials: Applications 2	Ready to use	Ben McGovern	10/10/2022 12:59

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Question statement

A savings account gives $\var{int}\%$ interest per year. The amount of money in the account is modelled as $S=£\var{start} \times \var{1+int/100}^t$ , where $t$ is the number of years. Interest is paid into the account monthly.

Give any introductory information the student needs.

			Name	Type	Generated Value

start	integer	350
int	integer	4
n	integer	3
target	number	595
sol1	number	356.93
sol2	decimal	dec("364.00")
sol3	decimal	dec("393.70")
yt	decimal	dec("13")
mt	decimal	dec("6.36")
tt	decimal	dec("1.3529286145011790392770939327446761171e+1")
tt2	decimal	dec("13.53")
answer	list	[ dec("13"), dec("7") ]
grammar	string	months

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(300..800 #50)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

350

→ Used by:

sol1
sol2
sol3
target
tt

Statement
Advice

Parts

Part a) Gap-fill
Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
  Add
2. Gap Gap 1. Mathematical expression
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

What is the total amount in the savings account after:

6 months: £Gap 0.

1 year: £Gap 1.

$\var{n}$ years: £Gap 2.

Give your answers to the nearest penny.

Advice

Part a:

To calculate how much money will be in the savings account after a certain amount of time, we substitute the amount of time for $t$ in the equation

$S =£\var{start} \times \var{1+int/100}^t .$

For 6 months, $t=0.5$ , therefore

$\begin{split} S &\,= £\var{start} \times \var{1+int/100}^{0.5} \\ &\,= £\var{sol1} \,.\end{split}$

For 1 year:

$\begin{split} S &\,= £\var{start} \times \var{1+int/100} \\ &\,= £\var{sol2} \,.\end{split}$

For $\var{n}$ years:

$\begin{split} S &\,= £\var{start} \times \var{1+int/100}^\var{n} \\ &\,= £\var{sol3} \,. \end{split}$

Part b:

To find out how long it will take for the amount of savings to exceed $£\var{target}$ , we need to solve the inequality

$\var{start} \times \var{1+int/100}^t > \var{target}$ for

$t$ :

$\begin{split} \var{start} \times \var{1+int/100}^t\, &\,> \var{target} \\ \var{1+int/100}^t &\,> \var{target/start} \\ t \log(\var{1+int/100}) &\,> \log(\var{target/start}) \\t &\,> \frac{\log(\var{target/start})}{\log(\var{1+int/100})} \qquad \text{(Be careful when rearranging inequalities. Since $\log(\var{1+int/100})>0$, this will not change the inequality in this case.)} \\ t &\,> \var{tt2} \text{ years (2 d.p.)}\,.\end{split}$

Since $\var{tt2}$ years is equivalent to $\var{yt}$ years and $\var{mt}$ months, it will take $\var{answer[0]}$ years and $\var{answer[1]}$ {grammar} for the savings to exceed $£\var{target}$ .

Give a worked solution to the whole question.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 10 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

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Logarithms and Exponentials: Applications 1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`