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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)
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`$8904.65`
`\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)}` |
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)
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`$12\ 803.59`
`\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` |
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)
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`$6411.70`
`\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)}` |
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
`C`
`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`
What amount must be invested now at 3% per annum, compounded annually, so that in two years it will have grown to $20 000? (2 marks)
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`$18\ 851.92`
`text(Using)\ \ FV = PV(1 + r)^n`
`r = 3text{%}, \ n=2`
| `20\ 000` | `= PV(1 + 0.03)^(2)` |
| `:.PV` | `= (20\ 000)/1.03^(2)` |
| `= $18\ 851.92` |
The inflation rate over the year from January 2019 to January 2020 was 2%.
The cost of a school jumper in January 2020 was $122.
Calculate the cost of the jumper in January 2019 assuming that the only change in the cost of the jumper was due to inflation. (2 marks)
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`$119.61`
| `FV` | `=PV(1+r)^n` |
| `122` | `=C_(2019)(1+0.02)^1` |
| `C_2019 xx 1.02` | `= 122` |
| `C_2019` | `= frac(122)(1.02)` |
| `= $119.61` |
What amount must be invested now at 4% per annum, compounded quarterly, so that in five years it will have grown to $60 000?
`C`
`text(Using)\ \ FV = PV(1 + r)^n`
| `r` | `= text(4%)/4` | `= text(1%) = 0.01\ text(per quarter)` |
| `n` | `= 5 xx 4` | `= 20\ text(quarters)` |
| `60\ 000` | `= PV(1 + 0.01)^(20)` |
| `:.PV` | `= (60\ 000)/1.01^(20)` |
| `= $49\ 172.66…` |
`⇒ C`