SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

v1 Financial Maths, STD2 F4 2008 HSC 27c

A laptop depreciated in value by 20% per annum. Three years after it was purchased, it had depreciated to a value of $2048, using the declining balance method.

What was the purchase price of the laptop?   (2 marks)

Show Answers Only

`$4000`

Show Worked Solution

`S = V_0 (1 – r)^n`

`2048` `= V_0 (1-0.20)^3`
`2048` `= V_0 (0.80)^3`
`V_0` `= 2048 / 0.512`
  `= 4000`

 

`:.\ text(The purchase price) = $4000`

v1 Financial Maths, STD2 F4 2005 HSC 26a

A new high-end coffee machine is purchased for $25 000 in January 2020.

At the end of each year, starting in 2021, the machine depreciates in value by 15% per annum, using the declining balance method of depreciation.

In which year will the value of the machine first fall below $15 000? (2 marks)

Show Answers Only

`text(The value falls below $15 000 in the fourth year)`

`text{which will be during 2024.}`

Show Worked Solution

`text(Using the formula:)\ \ S = V_0(1-r)^n`

`text(where) \ \ V_0 = 25\ 000,\ \ r = 0.15`

`text(If)\ n = 1\ \text{(2021)}`

`S` `= 25\ 000(0.85)^1`
  `=21\ 250`

 

`text(If)\ n = 2\ \text{(2022)}`

`S` `=25\ 000(0.85)^2`
  `=25\ 000(0.7225)`
  `=18\ 062.50`

 

`text(If)\ n = 3\ \text{(2023)}`

`S` `=25\ 000(0.85)^3`
  `=25\ 000(0.614125)`
  `=15\ 353.13`

 

`text(If)\ n = 4\ \text{(2024)}`

`S` `=25\ 000(0.85)^4`
  `=25\ 000(0.52200625)`
  `=13\ 050.16`

 

`:.\ \text{The value first falls below $15 000 in the fourth year}`

`text{which will be during 2024.}`

v1 Financial Maths, STD2 F4 2019 HSC 37

A machine is purchased for $32 800. Each year the value of the machine is depreciated by the same percentage.

The table shows the value of the machine, based on the declining-balance method of depreciation, for the first three years.

\[ \begin{array} {|c|c|} \hline \textit{End of year} & \textit{Value} \\ \hline 1 & \$27\,056.00 \\ \hline 2 & \$22\,888.16 \\ \hline 3 & \$19\,377.82 \\ \hline \end{array} \]

What is the value of the machine at the end of 10 years?  (3 marks)

Show Answers Only

`$4783.78`

Show Worked Solution

`text(Find the depreciation rate:)`

`S` `= V_0(1-r)^n`
`27\ 056` `= 32\ 800(1-r)^1`
`1-r` `= \frac{27\ 056}{32\ 800} = 0.82488`
`r` `= 0.17512`

 

`:.\ \text(Value after 10 years)`

`= 32\ 800(1-0.17512)^{10}`

`= 32\ 800(0.82488)^{10}`

`= 32\ 800 × 0.1458`

`= $4783.78`

v1 Financial Maths, STD2 F4 2018 HSC 26h

A piece of machinery is purchased for $18,500.

The value of the machine depreciates by 14% each year using the declining-balance method.

What is the value of the machine after three years?  (2 marks)

Show Answers Only

`$11\ 767\ \ (text(nearest dollar))`

Show Worked Solution
`S` `= V_0(1 – r)^n`
  `= 18\ 500(1 – 0.14)^3`
  `= 18\ 500(0.86)^3`
  `= 18\ 500 × 0.636056`
  `= 11\ 767.04`
  `= $11\ 767\ \ (text(nearest dollar))`

v1 Financial Maths, STD2 F4 2014 HSC 9 MC

A laptop is purchased for $2500. It depreciates at a rate of 25% per annum using the declining balance method.

What will be the salvage value of the laptop after 2 years, to the nearest dollar?

  1. $1406
  2. $1250
  3. $1681
  4. $1875
Show Answers Only

`A`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 2500 (1-25/100)^2`
  `= 2500 (0.75)^2`
  `= 2500 × 0.5625`
  `= $1406.25`

 

`Rightarrow \ text(To the nearest dollar, the salvage value is **$1406**)`

`Rightarrow A`

v1 Financial Maths, STD2 F4 2021 HSC 4 MC

A machine was purchased for $3200 four years ago. It depreciates at a rate of 12% per year using the declining-balance method.

What is the machine’s current value, to the nearest dollar?

  1. $1850
  2. $1919
  3. $1945
  4. $2010
Show Answers Only

`B`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 3200 (1-0.12)^4`
  `= 3200 (0.88)^4`
  `= 3200 × 0.5997`
  `= 1919`

⇒ `B`

v1 Financial Maths, STD2 F4 2021 HSC 26

Mila plans to invest $42 000 for 1.5 years. She is offered two different investment options.

Option A:  Interest is paid at 5% per annum compounded monthly.

Option B:  Interest is paid at `r` % per annum simple interest.

  1. Calculate the future value of Mila's investment after 1.5 years if she chooses Option A. (2 marks)
  2. Find the value of `r` in Option B that would give Mila the same future value after 1.5 years as for Option A. Give your answer correct to two decimal places. (2 marks)
Show Answers Only
  1. `$45\ 264.08`
  2. `5.18text(%)`
Show Worked Solution
a.   `r` `= text(5%)/12 = text(0.4167%) = 0.004167\ \text(per month)`
  `n` `= 12 × 1.5 = 18`
`FV` `= PV(1 + r)^n`
  `= 42\ 000(1 + 0.004167)^{18}`
  `= $45\ 264.08`

 

b.   `I` `= Prn`
  `3\ 264.08` `= 42\ 000 × r × 1.5`
  `r` `= 3\ 264.08 / (42\ 000 × 1.5)`
    `= 0.0518…`
    `= 5.18\ \text{% (to 2 d.p.)}`

v1 Financial Maths, STD2 F4 2015 HSC 26d

A laptop currently costs $850.

Assuming a constant annual inflation rate of 3.2%, calculate the cost of the same laptop in 4 years’ time.  (2 marks)

Show Answers Only

`$962.38\ \text{(nearest cent)}`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 850(1.032)^4`
  `= 850(1.132216)`
  `= 962.3836…`
  `= $962.38\ \text{(nearest cent)}`

v1 Financial Maths, STD2 F4 2022 HSC 11 MC

In eight years, the future value of an investment will be $120 000. The interest rate is 6% per annum, compounded half-yearly.

Which equation will give the present value `(PV)` of the investment?

  1. `PV=(120\ 000)/((1+0.06)^(8))`
  2. `PV=(120\ 000)/((1+0.03)^(8))`
  3. `PV=(120\ 000)/((1+0.03)^(16))`
  4. `PV=(120\ 000)/((1+0.06)^(16))`
Show Answers Only

`C`

Show Worked Solution

`text{Compounding periods} = 8 xx 2 = 16`

`text{Compounding rate} = (6text{%}) / 2 = 3text{%} = 0.03`

`PV = (120\ 000) / ((1 + 0.03) ^{16})`

`=> C`

v1 Financial Maths, STD2 F1 2021 HSC 19

Sophie bought a set of gym equipment four years ago. It depreciated by $1800 each year using the straight-line method of depreciation. The equipment is now valued at $6200.

Find the initial value of the gym equipment.  (2 marks)

Show Answers Only

`$13\ 400`

Show Worked Solution

`text{Find initial value}\ (V_0):`

`S` `=V_0-Dn`  
`6200` `=V_0-1800 xx 4`  
`V_0` `=6200 + 7200`  
  `=$13\ 400`  

v1 Financial Maths, STD2 F1 2017 HSC 11 MC

A car was bought for $22 500 and one year later its value had depreciated to $18 450.

What is the approximate depreciation, expressed as a percentage of the purchase price?

  1. 18%
  2. 22%
  3. 78%
  4. 82%
Show Answers Only

`A`

Show Worked Solution
`text(Net Depreciation)` `= 22\ 500-18\ 450`
  `= $4050`

 

`:. %\ text(Depreciation)` `= 4050 / (22\ 500) xx 100`
  `= 18%`

`=> A`

EXAMCOPY Algebra, STD2 A4 2022 HSC 9 MC

An object is projected vertically into the air. Its height, `h` metres, above the ground after `t` seconds is given by  `h=-5 t^2+80 t`.
 

For how long is the object at a height of 300 metres or more above the ground?

  1. 4 seconds
  2. 6 seconds
  3. 8 seconds
  4. 10 seconds
Show Answers Only

`A`

Show Worked Solution

`text{Object reaches 300 m when}\ \ t=6\ text{seconds.}`

`text{Object drops back below 300 m when}\ \ t=10\ text{seconds.}`

`text{Time at 300 m or above}\ = 10-6=4\ text{seconds}`

`=>A`

Bivariate Data, SM-Bank 016

The scatterplot below shows the rainfall (in mm) and the percentage of clear days for each month of 2023. 
 

An equation of the line of best fit for this data set is

\(\textit{rainfall}\ = 131-2.68 \times\ \textit{percentage of clear days} \)

  1. Using coordinates at the graph extremities or otherwise, draw this line on the scatterplot.  (2 marks)

  2. Describe this association in terms of strength and direction.  (1 mark)

Show Answers Only

i.   
       

ii.    \(\text{Strength: moderate (data points are moderately close to the LOBF)}\)

\(\text{Direction: negative (as percentage of clear days ↑, rainfall ↓)}\)

Show Worked Solution

i.    \(\text{Calculate values at the limits of the graph:}\)

\((0, 131)\ \ \Rightarrow \ \ y \text{-intercept}\ = 131 \)

\( (30, 50.6)\ \ \Rightarrow \ \text{At}\ x=30, \ y=131-2.68 \times 30 = 50.6 \)
 

ii.    \(\text{Strength: moderate (data points are moderately close to the LOBF)}\)

\(\text{Direction: negative (as percentage of clear days ↑, rainfall ↓)}\)

Bivariate Data, SM-Bank 015

The heights (in cm) and ages (in months) of a random sample of 15 boys have been plotted in the scatterplot below.

A line of best fit has been fitted to the data.

 

  1. State the independent variable in the graph.   (1 mark)

  2. Describe this association in terms of strength and direction.   (2 marks)
  3. Determine the gradient of the line of best fit, giving your answer correct to one decimal place.   (2 marks)
Show Answers Only

i.    \(\text{Independent variable: age (months)}\)

ii.    \(\text{Association: strong and positive}\)

iii.   \(0.5\)

Show Worked Solution

i.    \(\text{Independent variable}\ \ \Rightarrow\ \ x\text{-axis variable} \)

\(\text{age (months)}\)
 

ii.    \(\text{Association:} \)

\(\text{Strength: strong (data points are tightly gathered to the LOBF)}\)

\(\text{Direction: positive (as age ↑, height ↑)}\)
 

iii.   \(\text{LOBF passes through (15, 83.25) and (35, 94)} \)

\(\text{Gradient}\ = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{94-83.25}{35-15} = 0.5375 = 0.5\ \text{(1 d.p.)} \)

Bivariate Data, SM-Bank 014

The scatterplot below displays the mean age, in years, and the mean height, in centimetres, of 648 women from seven different age groups.

In an initial analysis of the data, a line of best fit is drawn, as shown.

 

  1. Describe this association in terms of strength and direction.   (2 marks)
  2. Determine the mean height predicted for a group of 65 year old women.   (1 mark)
Show Answers Only

i.    \(\text{Association: strong and negative}\)

ii.   \(\text{Predicted height = 160.5 cm}\)

Show Worked Solution

i.    \(\text{Association:} \)

\(\text{Strength: strong (data points are tightly gathered to the LOBF)}\)

\(\text{Direction: negative (as mean age ↑, mean height ↓)}\)
 

ii.   \(\text{Mean age = 65}\ \ \Rightarrow \ \ \text{Predicted height = 160.5 cm}\)

Bivariate Data, SM-Bank 013

The scatterplot below plots male life expectancy (male) against female life expectancy (female) in 1950 for a number of countries.

A line of best fit has been fitted to the scatterplot as shown.
 

  1. State the dependent variable in the graph.   (1 mark)
  2. Determine the age at which males and females have the same life expectancy.   (1 mark)
Show Answers Only

i.    \(\text{Male life expectancy}\)

ii.   \(\text{Life expectancy the same at 30 years of age.}\)

Show Worked Solution

i.    \(\text{Dependent variable}\ \ \Rightarrow \ \ y \text{-axis variable} \)

\(\text{Male life expectancy}\) 
 

ii.   \(\text{Life expectancy the same at 30 years of age.}\)

Bivariate Data, SM-Bank 012

A teacher analysed the class marks of 15 students who sat two tests.

The test 1 mark and test 2 mark, all whole number values, are shown in the scatterplot below.

A line of best fit has been fitted to the scatterplot.
 

  1. If a student scored 34 in the first test, what is their expected mark in the second test.   (1 mark)
  2. The line of best fit shows the predicted test 2 mark for each student based on their test 1 mark.
  3. Determine the number of students whose actual test 2 mark was within two marks of that predicted.   (1 mark)

Show Answers Only

i.    \(46\)

ii.   \(\text{5 values are within 1 grid height (measured vertically), or 2 marks,}\)

\(\text{from the LOBF.}\)

Show Worked Solution

i.    \(\text{1st test mark = 34}\ \ \Rightarrow\ \ \text{2nd test mark = 46} \)
 

ii.   \(\text{5 values are within 1 grid height (measured vertically), or 2 marks,}\)

\(\text{from the LOBF.}\)

Bivariate Data, SM-Bank 011

The scatterplot below displays the resting pulse rate, in beats per minute, and the time spent exercising, in hours per week, of 16 students.

A line of best fit has been fitted to the data.
 

  1. If a student spends 8 hours exercising per week, determine the resting pulse rate predicted by the line of best fit.   (1 mark)

  2. Provide TWO descriptions of the association between the variables time spent exercising and resting pulse rate.   (2 marks)

Show Answers Only

i.    \(\text{8 hours exercising}\ \ \Rightarrow\ \ \text{pulse rate = 59.5} \)

ii.   \(\text{Association should include two of the following:} \)

\(\text{Linear (straight line)}\)

\(\text{Negative (as time spent exercising ↑, resting pulse rate ↓)}\)

\(\text{Strong (data points are found tightly around the LOBF)}\)

Show Worked Solution

i.    \(\text{8 hours exercising}\ \ \Rightarrow\ \ \text{pulse rate = 59.5} \)
 

ii.   \(\text{Association should include two of the following:} \)

\(\text{Linear (straight line)}\)

\(\text{Negative (as time spent exercising ↑, resting pulse rate ↓)}\)

\(\text{Strong (data points are found tightly around the LOBF)}\)

Bivariate Data, SM-Bank 010

The scatterplot below shows the wrist circumference and ankle circumference, both in centimetres, of 13 people.

A line of best fit been drawn with ankle circumference as the independent variable.
 

  1. If a person has a wrist circumference of 18.5 centimetres, estimate the ankle circumference that is predicted by the line of best fit.   (1 mark)

  2. Explain why the \(y\)-intercept of this graph has no meaning in this context.   (1 mark)

Show Answers Only

i.    \(\text{18.5 cm wrist}\ \ \Rightarrow \ \ \text{Ankle circumference ≈ 24.5 cm}\)

ii.    \(y \text{-intercept occurs when ankle circumference = 0 cm, which is}\)

\(\text{meaningless in this context.}\)

Show Worked Solution

i.    \(\text{18.5 cm wrist}\ \ \Rightarrow \ \ \text{Ankle circumference ≈ 24.5 cm}\)
 

ii.    \(y \text{-intercept occurs when ankle circumference = 0 cm, which is}\)

\(\text{meaningless in this context.}\)

Bivariate Data, SM-Bank 009

The height (in cm) and foot length (in cm) for each of eight Year 12 students were recorded and displayed in the scatterplot below.

A line of best fit has been fitted to the data as shown.
 

  1. Determine the predicted foot size of a student who is 176 centimetres tall.   (1 mark)
  2. Calculate the gradient of the line of best fit, giving your answer correct to two decimal places.   (2 marks)
  3. What is the equation of the line of best fit?   (2 marks)
Show Answers Only

i.    \(\text{Height 176 cm}\ \ \Rightarrow\ \ \text{Foot length = 27 cm} \)

ii.    \(1.29 \)

iii.    \(\textit{height}\ = 141.2 + 1.29 \times \textit{foot length} \)

Show Worked Solution

i.    \(\text{Height 176 cm}\ \ \Rightarrow\ \ \text{Foot length = 27 cm} \)
 

ii.    \(\text{LOBF passes through (20, 167) and (34, 185):}\)

\(\text{Gradient}\ = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{185-167}{34-20} = \dfrac{18}{14} = 1.285… = 1.29 \)
 

iii.   \(\text{Equation}\ \ \Rightarrow \ \ \text{Gradient = 1.29, passes through (20, 167)} \)

\(y-y_1 \) \(=m(x-x_1) \)  
\(y-167\) \(=1.29(x-20) \)  
\(y\) \(=1.29x + 141.2\)  

 
\(\textit{height}\ = 141.2 + 1.29 \times \textit{foot length} \)

Bivariate Data, SM-Bank 008 MC

A line of best fit has been fitted to the scatterplot above to enable distance, in kilometres, to be predicted from time, in minutes.

The equation of this line is closest to

  1. distance `= 3.5 + 1.6 ×`time
  2. time `= 3.5 + 1.6 ×`distance
  3. distance `= 1.6 + 3.5 ×`time
  4. time `= 1.8 + 3.5 ×`distance
Show Answers Only

`A`

Show Worked Solution

`text{Line passes through  (0, 3.5) and (50, 82)`

\(\text{Gradient}\ = \dfrac{y_2-y_1}{x_2-x_1} \approx \dfrac{82-3.5}{50-0} \approx 1.57 \)
 

`text{Distance is the dependent variable}\ (y)\ \text{and the}`

`y text(-intercept is approximately 3.5.)`

`=> A`

Bivariate Data, SM-Bank 007 MC

Dr Chris measures the weights (in grams) and lengths (in cm) of 12 baby pythons.

The results were recorded and plotted in the scatterplot below. The line of best fit that enables the weight of the pythons to be predicted from their length has also been plotted.
 

The line of best fit predicts that the weight, in grams, of a python of length 30 cm would be closest to

  1. `240`
  2. `252`
  3. `262`
  4. `274`
Show Answers Only

`C`

Show Worked Solution

`text{The line of best fit crosses the 30cm length (on the}`

`xtext{-axis) at approx 262.}`

`=>C`

Bivariate Data, SM-Bank 006

The lengths and diameters, in millimetres, of a sample of jellyfish selected were recorded and displayed in the scatterplot below.

A line of best fit for this data is shown.
 

  1. Determine the expected length of a jellyfish with a diameter of 12 millimetres.   (1 mark)

  2. Determine the expected diameter of a jellyfish with a length of 16 millimetres.   (1 mark)

Show Answers Only

i.    \(\text{A diameter of 12 mm (x-axis)}\ \ \Rightarrow \ \ \text{length 14 mm}\)

ii.    \(\text{Length of 16 mm (y-axis)}\ \ \Rightarrow \ \ \text{diameter 14.4 mm (approx)}\)

Show Worked Solution

i.    \(\text{A diameter of 12 mm (x-axis)}\ \ \Rightarrow \ \ \text{length 14 mm}\)

ii.    \(\text{Length of 16 mm (y-axis)}\ \ \Rightarrow \ \ \text{diameter 14.4 mm (approx)}\)

Bivariate Data, SM-Bank 005

Each member of a group of males had his height and foot length measured and recorded. The results were graphed and a line of fit drawn.
 

  1. Why does the value of the `y`-intercept have no meaning in this situation?  (1 mark)

  2. George is 10 cm taller than his brother Harry. Use the line of fit to estimate the difference in their foot lengths.  (1 mark)

Show Answers Only
  1. `text(The y-intercept occurs when)\ x = 0.\ text(It has`
    `text(no meaning to have a height of 0 cm.)`
  2. `text(A 10 cm height difference means George should)`
    `text(have a 3 cm longer foot.)`
Show Worked Solution

i.  `text(The y-intercept occurs when)\ x = 0.\ text(It has)`

`text(no meaning to have a height of 0 cm.)`

 

ii.  `text(A 20 cm height difference results in a foot length)`

`text(difference of 6 cm.)`

`:.\ text(A 10 cm height difference means George should)`

`text(have a 3 cm longer foot.)`

Bivariate Data, SM-Bank 004

The shoe size and height of ten students were recorded.

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Shoe size} \rule[-1ex]{0pt}{0pt} & \text{6} & \text{7} & \text{7} & \text{8} & \text{8.5} & \text{9.5} & \text{10} & \text{11} & \text{12} & \text{12} \\
\hline \rule{0pt}{2.5ex} \text{Height} \rule[-1ex]{0pt}{0pt} & \text{155} & \text{150} & \text{165} & \text{175} & \text{170} & \text{170} & \text{190} & \text{185} & \text{200} & \text{195} \\
\hline
\end{array}

  1. Complete the scatter plot AND draw a line of best fit by eye.  (2 marks)
     
     
  2. Use the line of best fit to estimate the height difference between a student who wears a size 7.5 shoe and one who wears a size 9 shoe.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `text{13 cm (or close given LOBF drawn)}`
Show Worked Solution

i.    
     

ii.    `text{Shoe size 7.5 gives a height estimate of 162 cm (see graph)}`

`text{Shoe size 9 gives a height estimate of 175 cm (see graph)}`

`:.\ text(Height difference)` `= 175-162`
  `= 13\ text{cm  (or close given LOBF drawn)}`

Bivariate Data, SM-Bank 003

Ahmed collected data on the age (`a`) and height (`h`) of males aged 11 to 16 years.

He created a scatterplot of the data and constructed a line of best fit to model the relationship between the age and height of males.
 

  1. Determine the gradient of the line of best fit shown on the graph.   (1 mark)

  2. Determine the equation of the line of best fit shown on the graph.  (2 marks)

  3. Use the line of best fit to predict the height of a typical 17-year-old male.   (1 mark)

  4. Why would this model not be useful for predicting the height of a typical 45-year-old male?   (1 mark)

Show Answers Only
  1. `text(Gradient = 6)`
  2. `h = 6a + 80`
  3. `text(182 cm)`
  4. `text(People slow and eventually stop growing)`

  5.  

    `text(after they become adults.)`

Show Worked Solution

i.    `text{Gradient}\ =(176-146)/(16-11)=30/5=6`
 

ii.   `text{Gradient = 6,  Passes through (11, 146)}`

`y-y_1` `=m(x-x_1)`
`h-146` `=6(a-11)`
`:. h` `=6a-66+146`
  `=6a + 80`
♦♦ Mean marks of 38% and 25% respectively for parts (i)-(ii).
COMMENT: Choose extreme points for calculating gradient.

 
iii.   `text{Substitue}\ \ a=17\ \ \text{into equation from part (ii):}`

`h=(6 xx 17) +80=182`

`:.\ text{A typical 17 year old is expected to be 182cm.}`
  

iv.    `text(People slow and eventually stop growing)`
  `text(after they become adults.)`

Bivariate Data, SM-Bank 002

People are placed into groups to complete a puzzle. There are 9 different groups.

The table shows the number of people in each group and the amount of time, in minutes, for each group to complete the puzzle.

\begin{array} {|l|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & 2 & 2 & 3 & 5 & 5 & 7 & 7 & 7 & 8 \\
\hline
\rule{0pt}{2.5ex} \text{Time taken (min)} \rule[-1ex]{0pt}{0pt} & 28 & 30 & 26 & 19 & 21 & 12 & 13 & 11 & 8 \\
\hline
\end{array}

  1. Complete the scatterplot by adding the last four points from the table.  (1 mark)
     
       
  2. Add a line of best fit by eye to the graph in part (a).  (1 mark)

  3. The graph in part (a) shows the association between the time to complete the puzzle and the number of people in the group.
  4. Identify the form (linear or non-linear), the direction and the strength of the association.  (2 marks)

Show Answers Only

a.

b.


 

c.    `text(Form: linear)

`text{Direction: negative}`

`text{Strength: strong}`

Show Worked Solution

a.

b.


 

c.    `text{Form: linear (i.e. straight line)}`

`text{Direction: negative}`

`text{Strength: strong}`

Bivariate Data, SM-Bank 001

The graph shows a line of best fit describing the life expectancy of people born between 1900 and 2000.
 


  1. According to the graph, what is the life expectancy of a person born in 1932?  (1 mark)

  2. Determine the value of the gradient of the line of best fit.  (2 marks)

Show Answers Only
  1. \(\text{68 years}\)
  2. \(0.25\)
Show Worked Solution

i.    \(\text{68 years}\)

ii.    \(\text{Using (1900,60), (1980,80):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{80-60}{1980-1900}\)
  \(= 0.25\)

Depreciation, SMB-004

A new electric vehicle costs $46 000 and is depreciated using the reducing balance method by 15% of its value each year.

Find its value after four years, giving your answer to the nearest dollar.   (2 marks)

Show Answers Only

`$24\ 012`

Show Worked Solution

`V_o = 46\ 000,\ \ \ r = 15% = 0.15,\ \ \ n = 4`

`S` `= V_0 (1-r)^n`
  `= 46\ 000 (1-0.15)^4`
  `=46\ 000 xx 0.85^4`
  `= $24\ 012\ \ \text{(nearest dollar)}`

 

Depreciation, SMB-003 MC

A new car costs $30 000 and is depreciated using the reducing balance method by 20% of its value each year.

After three years its value is

  1. `$6000`
  2. `$12\ 000`
  3. `$15\ 360`
  4. `$24\ 000`
Show Answers Only

`C`

Show Worked Solution

`V_0= 30\ 000,\ \ \ r = 20% = 0.2,\ \ \ n = 3`

`S` `= V_0 (1-r)^n`
  `= 30\ 000 (1-0.2)^3`
  `=30\ 000 xx 0.8^3`
  `= $15\ 360`

 
`=>  C`

Depreciation, SMB-002

Peter installed his new pool fence on 1 January 2013 at a cost of $12 000.

On 1 January of each year after 2013 its value is depreciated by 15% using the reducing balance method.

The value of the pool fence will be below $4000 for the first time on 1 January of what year? Show all working.   (3 marks)

Show Answers Only

`text{Test the values of different years using:}\ \ S=V_0(1-r)^n`

`S \text{(Jan17)} = 12\ 000(1-0.15)^4= 5000 xx 0.85^4= $2560`

`S \text{(Jan19)} = 12\ 000 xx 0.85^6= $4525.79`

`S \text{(Jan20)} = 12\ 000 xx 0.85^7= $3846.93`

`\text{2020 is the 1st year its value is below $4000 on 1 January.}`

Show Worked Solution

`r=15% = 0.15`

`text{Test the values of different years using:}\ \ S=V_0(1-r)^n`

`S \text{(Jan17)} = 12\ 000(1-0.15)^4= 5000 xx 0.85^4= $2560`

`S \text{(Jan19)} = 12\ 000 xx 0.85^6= $4525.79`

`S \text{(Jan20)} = 12\ 000 xx 0.85^7= $3846.93`

`\text{2020 is the 1st year its value is below $4000 on 1 January.}`

Depreciation, SMB-001 MC

A new air-conditioning unit was purchased for $5000 on 1 January 2017.

On 1 January of each year after 2017 its value is depreciated by 20% using the reducing balance method.

The value of the air conditioner will be below $1500 for the first time on 1 January

  1. 2020
  2. 2021
  3. 2022
  4. 2023
Show Answers Only

`D`

Show Worked Solution

`r=20% = 0.2`

`text{Test the values of given options using:}\ \ S=V_0(1-r)^n`

`S \text{(Jan20)} = 5000(1-0.2)^3= 5000 xx 0.8^3= $2560`

`S \text{(Jan22)} = 5000 xx 0.8^5= $1638.40`

`S \text{(Jan23)} = 5000 xx 0.8^6= $1310.72`

 `=>  D`

Compound Interest, SMB-031

Camilla is considering a term deposit that pays 4.8% interest per annum, compounded monthly.

She calculates that if she invests in the term deposit, her money will be worth $9800 in 2 years' time.

Determine the amount that Camilla is planning to invest, giving your answer to the nearest cent.   (2 marks)

Show Answers Only

`$8904.65`

Show Worked Solution

`\text{Interest rate}\ (r) = \frac{0.048}{12} = 0.004\ \ \text{(per month)}`

`\text{Compounding periods}\ (n) = 2 xx 12 = 24`

`FV` `=PV(1+r)^n`
`9800` `= PV(1 + 0.004)^(24)`
`:.PV` `= \frac{9800}{1.004^{24}}`
  `= $8904.65\ \ \text{(nearest cent)}`

Compound Interest, SMB-030

Michelle decided to invest some of her money at a higher interest rate. She deposited $3000 in an account paying 8.2% per annum, compounding half yearly.

  1. Write down an expression involving the compound interest formula that can be used to find the value of Michelle’s $3000 investment at the end of two years. Find this value correct to the nearest cent.  (2 marks)

  2. How much interest will the $3000 investment earn over a four-year period?

     

    Write your answer correct to the nearest cent.  (2 marks)

Show Answers Only
  1. `$3523.09`
  2. `$1137.40`
Show Worked Solution

a.   `text{Interest rate}\ (r) = \frac{8.2%}{2} = 4.1% = 0.041\ \ \text{(per 6 months)}`

`text(Compounding periods)\ (n) = 2 xx 2 = 4`

`FV` `= PV(1+r)^n`
  `= 3000(1.041)^4`
  `= 3523.093…`
  `= $3523.09\ \ text{(nearest cent)}`

 

b.   `text{Compounding periods}\ (n) = 4 xx 2 = 8`

`FV` `= 3000(1.041)^8`
  `= 4137.396…`

 

`:.\ text(Interest earned)` `= 4137.40-3000`
  `= $1137.40\ \ text{(nearest cent)}`

Compound Interest, SMB-029

The golf club’s social committee has $3400 invested in an account which pays interest at the rate of 4.4% per annum compounding quarterly.

  1. Show that the interest rate per quarter is 1.1%.  (1 mark)

  2. Determine the value of the $3400 investment after three years.

     

    Write your answer in dollars correct to the nearest cent.  (1 mark)

  3. Calculate the interest the $3400 investment will earn over six years.

     

    Write your answer in dollars correct to the nearest cent.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `$3876.97`
  3. `$1020.86`
Show Worked Solution

a.   `text{Interest rate}\ = \frac{4.4}{4} = 1.1text{%  (per quarter)}`
 

b.   `text{Compounding periods}\ (n) = 3 xx 4 =12`

`FV` `= PV(1+r)^n`
  `= 3400(1.011)^12`
  `= 3876.973…`
  `= $3876.97`

 

c.   `text{Compounding periods}\ (n) = 6 xx 4 = 24`

`FV` `= 3400(1.011)^24`
  `= 4420.858…`

 
`text(Interest earned over 6 years)`

`= 4420.86-3400`

`= $1020.86\ \ text{(nearest cent)}`

Compound Interest, SMB-028

It is estimated that inflation will average 3.5% per annum over the next eight years.

If a new machine costs $60 000 now, calculate the cost of a similar new machine in eight years time, adjusted for inflation. Assume no other cost change.

Write your answer correct to the nearest dollar.  (2 marks)

Show Answers Only

`$79\ 008.54`

Show Worked Solution
`FV` `= PV(1+r)^n`
  `= 60\ 000(1.035)^8`
  `= $79\ 008.54\ \ \text{(nearest cent)}`

Compound Interest, SMB-027 MC

An amount of $8000 is invested for a period of 4 years.

The interest rate for this investment is 7.2% per annum compounding quarterly.

The interest earned by the investment in the fourth year (in dollars) is given by

  1. `8000 xx 1.018^4 - 8000 xx 1.018^3`
  2. `8000 xx 1.072^4 - 8000 xx 1.072^3`
  3. `8000 xx 1.018^16 - 8000 xx 1.018^12`
  4. `8000 xx 1.072^16 - 8000 xx 1.072^12`
Show Answers Only

 `C`

Show Worked Solution

`text{Interest earned in 4th year}\ = FV\text{(after 4 years)}-FV\text{(after 3 years)}`

`text{Interest rate}\ (r) = \frac{7.2%}{4} = \frac{0.072}{4} = 0.018`

`text{Compounding periods (3 years)}\ = 3 xx 4 = 12`

`text{Compounding periods (4 years)}\ = 4 xx 4 = 16`
 

`FV\text{(after 4 years)}\ = PV(1+r)^n = 8000(1.018)^16`

`FV\text{(after 3 years)}\ = 8000(1.018)^12`

`:.\ text(Interest earned in 4th year) = 8000 xx 1.018^(16)-8000 xx 1.018^(12)`

`=>  C`

Compound Interest, SMB-026 MC

$10 000 is invested for five years. Interest is earned at a rate of 8% per annum, compounding quarterly.

Which one of the following calculations will give the total interest earned, in dollars, by this investment?

  1. `10\ 000 xx 1.02^5-10\ 000`  
  2. `10\ 000 xx 1.02^20-10\ 000`  
  3. `10\ 000 xx 1.08^5 -10\ 000`
  4. `10\ 000 xx 1.08^20-10\ 000`
Show Answers Only

`B`

Show Worked Solution

`text{Interest rate}\ (r) = \frac{8%}{4} =2% = 0.02\ text{(per quarter)}`

`text{Compounding periods}\ (n) =5 xx4=20`

`FV=PV(1+r)^n = 10\ 000(1.02)^20`

`:.\ text(Interest earned)= 10\ 000 xx 1.02^20-10\ 000`

`=>  B`

Compound Interest, SMB-025

$15 000 is invested for 12 months.

For the first six months the interest rate is 6.1% per annum compounding monthly.

After six months the interest rate increases to 6.25% per annum compounding monthly.

Calculate the total interest earned by this investment over 12 months, giving your answer to the nearest dollar.   (4 marks)

Show Answers Only

`$953`

Show Worked Solution

`text{In the first 6 months:}`

`text{Interest rate}\ (r) = \frac{0.061}{12}`

`FV(\text{after 6 months})` `=PV(1+r)^n`  
  `=15\ 000(1 + \frac{0.061}{12})^6`   
  `=$15\ 463.35…`  

 
`text{In the second 6 months:}`

`text{Interest rate}\ (r) = \frac{0.0625}{12}`

`FV(\text{after 2nd 6 months})` `=15\ 463.35(1 + \frac{0.0625}{12})^6`  
  `=$15\ 952.91…`   

 

`:.\ text(Interest)` `= 15\ 952.91-15\ 000`
  `= $953\ \ \text{(nearest dollar)}`

Compound Interest, SMB-024

$9000 is invested at a rate of 8% per annum compounding half yearly.

Determine the value, to the nearest dollar, of this investment after four years.   (2 marks)

Show Answers Only

`$12\ 317`

Show Worked Solution

`text(Interest rate)\ (r) = \frac{8%}{2} = \frac{0.08}{2} = 0.04\ \ \text{(per 6 months)}`

`text{Compounding periods}\ (n) =4 xx 2=8`

`FV` ` = PV(1+r)^n`
  `= 9000 xx (1+0.04)^(8)`
  `= $12\ 317.12`
  `=$12\ 317\ \ \text{(nearest dollar)}`

Compound Interest, SMB-023 MC

$10 000 is invested at a rate of 10% per annum compounding half yearly.

The value, in dollars, of this investment after five years, is given by

  1. `10\ 000 xx 0.10 xx 5`
  2. `10\ 000 xx 0.05 xx 10`
  3. `10\ 000 xx 0.05^10`
  4. `10\ 000 xx 1.05^10`
Show Answers Only

`D`

Show Worked Solution

`text(Interest rate)\ (r) = \frac{10%}{2} = \frac{0.10}{2} = 0.05\ \ \text{(per 6 months)}`

`text{Compounding periods}\ (n) =5 xx 2=10`

`FV` ` = PV(1+r)^n`
  `= 10\ 000 xx (1+0.05)^(10)`
  `=10\ 000(1.05)^10`

 
`=>  D`

Compound Interest, SMB-022 MC

The points on the graph below show the balance of an investment at the start of each quarter for a period of six years.

The same rate of interest applied for these six years.
 

   

In relation to this investment, which one of the following statements is true?

  1. interest is compounding annually and is credited annually
  2. interest is compounding annually and is credited quarterly
  3. simple interest is paid on the opening balance and is credited annually
  4. simple interest is paid on the opening balance and is credited quarterly
Show Answers Only

`A`

Show Worked Solution

`text{By elimination:}`

`text{From the graph, as balance increases after each year, interest is}`

`text{credited annually (eliminate B and D).}`

`text{The difference of the balances between successive years is increasing}`

`text{which indicates that interest is compounding (eliminate D).}`

`=>  A`

Compound Interest, SMB-021

Callum invests $20 000 in a term deposit account that adds 3.8% interest annually, calculated on the account balance at the end of each year.

Calculate the interest paid in the third year of Callum's investment.   (3 marks)

Show Answers Only

`$818.86`

Show Worked Solution

`r=3.8% = 0.038`

`text(Using)\ \ FV=PV(1+r)^n :`

`text(Value after 2 years)` `= 20\ 000(1.038)^2`
  `= $21\ 548.88`
 `text(Value after 3 years)` `= 20\ 000(1.038)^3`
  `= $22\ 367.74`

 

 `:.\ text(Interest paid in 3rd year)` `= 22\ 367.74-21\ 548.88`
  `= $818.86`

Compound Interest, SMB-020

Tim invests $3000 in a term deposit account that adds 6.5% interest annually, calculated on the account balance at the end of each year.

The interest paid in the fourth year is

  1. `$221.16`
  2. `$235.55`
  3. `$3623.85`
  4. `$3859.40`
Show Answers Only

`B`

Show Worked Solution

`r=6.5% = 0.065`

`text(Using)\ \ FV=PV(1+r)^n :`

`text(Value after 3rd year)` `= 3000(1.065)^3`
  `= $3623.85`
 `text(Value after 4th year)` `= 3000(1.065)^4`
  `= $3859.40`

 

 `:.\ text(Interest paid in 4th year)` `= 3859.40-3623.85`
  `= $235.55`

`=>  B`

Compound Interest, SMB-019

Kelly invests $8 000 at an interest rate of 7.5% per annum, compounding annually.

After how many years will her investment first be more than double its original value, giving your answer to the nearest year?   (3 marks)

Show Answers Only

`10\ text{years}`

Show Worked Solution

`FV=PV(1+r)^n`

`r=7.5% = 0.075, PV=8000`

`text(Find)\ n\ text(when)\ FV>16\ 000:`

`8000(1+0.075)^n` `> 16\ 000`
`1.075^n` `> 2`

 
`text(Testing possible values:)`

`1.075^10 = 2.06`

`1.075^9 = 1.92`

`:.\ \text{Kelly’s investment will first double after 10 years.}`

Compound Interest, SMB-018 MC

Gerry invests $10 000 at an interest rate of 5.5% per annum, compounding annually.

After how many years will his investment first be more than double its original value?

  1. 12
  2. 13
  3. 14
  4. 15
Show Answers Only

`B`

Show Worked Solution

`FV=PV(1+r)^n`

`r=5.5% = 0.055, PV=10\ 000`

`text(Find)\ n\ text(when)\ FV>20\ 000:`

`10\ 000(1+0.055)^n` `> 20\ 000`
`1.055^n` `> 2`

 
`text(Test answer options:)`

`1.055^12 = 1.90`

`1.055^13 = 2.005`

`=>  B`

Compound Interest, SMB-017

Ekamjot invests $13 000 for two years.

Interest is calculated at the rate of 7.2% per annum, compounding quarterly.

How much interest does Ekamjot earn from this investment, giving your answer to the nearest cent?   (2 marks)

Show Answers Only

`$1994.28`

Show Worked Solution

`text{Annual interest rate}\ = 7.2% = 0.072`

`text(Quarterly interest rate)\ (r) = \frac{0.072}{4} = 0.018`

`text(Compounding periods)\ (n) = 2 xx 4 =8`

`FV(text{after 2 years})` `= PV(1+r)^n`
  `=13\ 000(1 + 0.018)^8`
  `=$14\ 994.278….`
  `=$14\ 994.28`

 
`text{Interest earned}\ =14\ 994.28-13\ 000=$1994.28`

Compound Interest, SMB-016 MC

Mary invests $1200 for two years.

Interest is calculated at the rate of 3.35% per annum, compounding monthly.

The amount of interest she earns in two years is closest to

  1. `$6.71`
  2. `$80.40`
  3. `$81.75`
  4. `$83.03`
Show Answers Only

`D`

Show Worked Solution

`text{Annual interest rate}\ = 3.35% = 0.0335`

`text(Monthly interest rate)\ (r) = \frac{0.0335}{12}`

`text(Compounding periods)\ (n) = 2 xx 12 =24`

`FV(text{after 2 years})` `= PV(1+r)^n`
  `=1200(1 + \frac{0.0335}{12})^24`
  `=$1283.03…`

 
`text{Interest earned}\ =1283.03-1200=$83.03`

`=> D`

Compound Interest, SMB-015

Amy invests  $15 000 for 150 days.

Interest is calculated at the rate of 4.60% per annum, compounding daily.

Assuming that there are 365 days in a year, find the value of her investment after 150 days, giving your answer to the nearest dollar.   (3 marks)

Show Answers Only

`$15\ 286`

Show Worked Solution

`text{Annual interest rate}\ =4.60%=0.046`

`text{Daily interest rate}\ = \frac{0.046}{365}`

`FV` `= 15\ 000 xx (1 + \frac{0.046}{365})^150`
  `= 15\ 000 xx (1.000126…)^150`
  `= 15\ 000 xx (1.01908…)`
  `= $15\ 286\ \ \text{(nearest dollar)}`

Compound Interest, SMB-014

Sam and Charlie each invest $5000 for three years.

Sam’s investment earns simple interest at the rate of 7.5% per annum.

Charlie’s investment earns interest at the rate of 7.5% per annum compounding annually.

At the conclusion of three years, determine which investment has the highest value and by how much.   (4 marks)

Show Answers Only

 `text{Charlie’s investment is worth $86.48 more than Sam’s.}`

Show Worked Solution

`text(Sam’s Investment:)`

`I=Prn = 5000 xx \frac{7.5}{100} xx 3 = $1125`

`text{Investment value}\ = 5000 + 1125 = $6125`

  
`text(Charlie’s Investment:)`

`FV` `= PV(1+r)^n`
  `= 5000 xx 1.075^3`
  `= $6211.48`

 
`text{Difference}\ = 6211.48-6125= $86.48`

`:.\ \text{Charlie’s investment is worth $86.48 more than Sam’s.}`

Compound Interest, SMB-013

What amount must be invested now at 6% per annum, compounded quarterly, so that in eighteen months it will have grown to `$14\ 000`? Give your answer to the nearest cent.  (2 marks)

Show Answers Only

`$12\ 803.59`

Show Worked Solution

`\text{Interest rate}\ (r) = \frac{0.06}{4} = 0.015\ \ \text{(per quarter)}`

`\text{Compounding periods}\ (n) = \frac{18}{3} = 6`
 

`FV` `=PV(1+r)^n`
`14\ 000` `= PV(1 + 0.015)^(6)`
`:.PV` `= (14\ 000)/1.015^(6)`
  `= $12\ 803.59`

Compound Interest, SMB-012

Louise's investment earns 3.6% per annum, compounded quarterly.

She calculates that her investment will be worth $7400 in 4 years.

Determine the amount that Louise initially invests, giving your answer to the nearest cent.   (2 marks)

Show Answers Only

`$6411.70`

Show Worked Solution

`\text{Interest rate}\ (r) = \frac{0.036}{4} = 0.009\ \ \text{(per quarter)}`

`\text{Compounding periods}\ (n) = 4 xx 4 = 16`

`FV` `=PV(1+r)^n`
`7400` `= PV(1 + 0.009)^(16)`
`:.PV` `= \frac{7400}{1.009^{16}}`
  `= $6411.70\ \ \text{(nearest cent)}`

Compound Interest, SMB-011 MC

Marshall's investment earns 5% per annum, compounded annually.

He calculates that his investment will be worth $1100 in 3 years, to the nearest dollar.

The amount Marshall invests now is closest to

  1. $892
  2. $928
  3. $950
  4. $1008
Show Answers Only

`C`

Show Worked Solution

`FV = PV(1 + r)^n`

`r` `=\ text(5%)` `= 0.05\ text(per annum)`
`n` `=3`  

 

`1100` `= PV(1 + 0.05)^(3)`
`:.PV` `= \frac{1100}{1.05^{3}}`
  `= $950`

 
`=> C`

Compound Interest, SMB-010

A bank's Compound Saver Account pays interest at 4% per annum, compounded quarterly.

Sacha deposits $6500 when he opens his Compound Saver Account and makes no further deposits or withdrawals.

What will be the balance in the account at the end of 1.5 years, to the nearest dollar?   (2 marks)

Show Answers Only

`$6900`

Show Worked Solution

`text{Compounding periods}\ (n) =1.5 xx 4=6`

`text{Compounding rate}\ (r) = \frac{0.04}{4} = 0.01`

`FV` `= PV(1 + r)^n`
  `= 6500(1 + 0.01)^6`
  `= 6899.88…`
  `=$6900\ \ \text{(nearest dollar)}`

Compound Interest, SMB-009

Jill opens a Smartsave Account that pays interest at 5% per annum, compounded quarterly.

Jill deposits $700 when she opens the account and makes no further deposits or withdrawals.

What will be the balance in the account at the end of 3 years, to the nearest dollar?   (2 marks)

Show Answers Only

`$813`

Show Worked Solution

`text{Compounding periods}\ (n) =3 xx 4=12`

`text{Compounding rate}\ (r) = \frac{0.05}{4} = 0.0125`

`FV` `= PV(1 + r)^n`
  `= 700(1 + 0.0125)^12`
  `= 812.528…`
  `=$813\ \ \text{(nearest dollar)}`

Compound Interest, SMB-008 MC

A bank's Maxi Saver Account pays interest at 6% per annum, compounded quarterly.

Jen opens a Maxi Saver Account and deposits $3000 into it. 

Assuming no further deposits or withdrawals are made, what will be the balance in the account at the end of two years?

  1. $3120.00
  2. $3360.00
  3. $3379.48
  4. $4781.54
Show Answers Only

`=> C`

Show Worked Solution

`text{Compounding periods}\ (n) =2 xx 4=8`

`text{Compounding rate}\ (r) = \frac{0.06}{4} = 0.015`

`FV` `= PV(1 + r)^n`
  `= 3000(1 + 0.015)^8`
  `= $3379.48`

 
`=> C`

Compound Interest, SMB-007

Shannon invests $25 000 in an account that earns interest at 6% per annum, compounded monthly.

What is the future value of Shannon's investment, to the nearest dollar, after 2 years?   (2 marks)

Show Answers Only

`$28\ 179`

Show Worked Solution

`text{Annual interest rate = 6.0% = 0.06}` 

`text(Monthly interest rate) = \frac(0.06)(12) = 0.005`

`n = 2 xx 12 = 24`
  

`FV` `= PV(1 + r)^n`
  `= 25\ 000 (1 + 0.005)^24`
  `= 28\ 178.99`
  `=$28\ 179\ \ \text{(nearest dollar)}`

Compound Interest, SMB-006

Natalie is saving for a netball hoop and invests $2200 in an account that earns interest at 5.5% per annum, compounded monthly.

What is the future value of Natalie's investment, to the nearest dollar, after 1.5 years?   (2 marks)

Show Answers Only

`$2389`

Show Worked Solution

`text{Annual interest rate = 5.5% = 0.055}` 

`text(Monthly interest rate) = \frac(0.055)(12)`

`n = 1.5 xx 12 = 18`
  

`FV` `= PV(1 + r)^n`
  `= 2200 (1 + frac(0.055)(12))^18`
  `= 2388.74…`
  `=$2389`

Compound Interest, SMB-005 MC

Roger invests $1400. He earns interest at 4% per annum, compounded monthly.

What is the future value of Roger's investment after 2.5 years?

  1. $1540.00
  2. $1546.98
  3. $3080.00
  4. $4540.76
Show Answers Only

`B`

Show Worked Solution

`text{Annual interest rate = 4% = 0.04}` 

`text(Monthly interest rate) = \frac(0.04)(12)`

`n = 2.5 xx 12 = 130`
  

`FV` `= PV(1 + r)^n`
  `= 1400 (1 + frac(0.04)(12))^30`
  `= $1546.98`

 
`=> \ B`

Compound Interest, SMB-004

In Japan, a bowl of ramen cost 1000 Japanese yen in 2005. The cost of ramen has increased by 0.5% per annum since then. 

Determine the cost of the same bowl of ramen in 2020 in Japanese yen, to the nearest yen.   (2 marks)

Show Answers Only

`1078\ text{yen}`

Show Worked Solution

`r =\ text(0.5%)\ = 0.005`

`n = 15\ text(years)`

`FV` `=PV(1+r)^n`  
  `=1000(1 + 0.005)^15`  
  `=1077.68…`  
  `=1078\ text{yen}`