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Virat invested $5000 into an account that earned interest compounding fortnightly.
The effective annual interest rate for Virat's investment was 4.51%.
Assume that there are exactly 26 fortnights in one year.
After five years, the amount of interest earned by Virat was closest to
\(B\)
\(\text{Find nominal interest rate (by CAS):}\)
\(\operatorname{nom}(4.51,26)=4.415 \%\)
\(\text{Solve for} \ FV \ \text {(by CAS):}\)
| \(N\) | \(=5 \times 26=130\) |
| \(I(\%)\) | \(=4.415\) |
| \(PV\) | \(=-5000\) |
| \(PMT\) | \(=0\) |
| \(FV\) | \(=\boldsymbol{6233.89}\) |
| \(PY\) | \(=CY=26\) |
\(\therefore \text{Interest}=6233.89-5000=\$1233.89\)
\(\Rightarrow B\)
Edo invests $10 000 in an account earning 3% interest per annum compounding monthly.
Question 19
The value, \(V_n\), of Edo's investment after \(n\) months is given by
Question 20
The effective interest rate for Edo's investment is closest to
\(\text{Question 19:}\ A\)
\(\text{Question 20:}\ E\)
\(\text{Question 19}\)
\(\text{Interest rate}\ = \dfrac{3.0\%}{12}=0.25\%\ \text{per month}\)
\(\text{After \(n\) months:}\)
\(V_n=10\,000 \times 1.0025^{n}\)
\(\Rightarrow A\)
\(\text{Question 20}\)
\(\text{Let \(r\) = annual bank rate = 3.0%} \)
\(r_{effective} = \left[\left(1+\dfrac{3.0}{100 \times 12}\right)^{12}-1\right] \times 100\%=3.04\% \)
\(\Rightarrow E\)
Emi invested profits of $10 000 into a savings account that earns interest compounding fortnightly, for one year. The effective interest rate, rounded to two decimal places, is 5.07%. Assume that there are exactly 26 fortnights in a year. --- 1 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- a. \(4.95\%\ \text{(2 d.p.)}\) b. \(\text{Nominal and effective interest rates are equal if there is only one}\) \(\text{compounding period per year. If there are more compounding}\) \(\text{periods (as in this example) the effective rate will be higher than}\) \(\text{the nominal rate.}\) a. \(\text{Using CAS: FINANCE}\ \rightarrow \ \text{Interest Conversion}\ \rightarrow \ \text{Nominal Interest Rate}\) \(\text{nom}(5.07, 26) = 4.95037\%\approx 4.95\%\ \text{(2 d.p.)}\) \(\text{Using formula:}\) \(\text{compounding period per year. If there are more compounding}\) \(\text{periods (as in this example) the effective rate will be higher than}\) \(\text{the nominal rate.}\)
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Mean mark (a) 51%.
\(r_e\)
\(=\left[\left(1+\dfrac{r}{100n}\right)^n-1\right]\times 100\%\)
\(\dfrac{5.07}{100}\)
\(=\left[\left(1+\dfrac{r}{2600}\right)^{26}-1\right]\)
\(\left(1+\dfrac{r}{2600}\right)^{26}\)
\(=1.0507\)
\(\dfrac{r}{2600}\)
\(=(1.0507)^\frac{1}{26}-1\)
\(r\)
\(=\left((1.0507)^\frac{1}{26}-1\right)\times 2600\)
\(=4.95036\dots\approx 4.95\%\ \text{(2 d.p.)}\)
b. \(\text{Nominal and effective interest rates are equal if there is only one}\)
♦♦♦ Mean mark (b) 26%.
Consider the following four statements regarding nominal and effective interest rates as they apply to compound interest investments and loans:
How many of these four statements are true?
\(D\)
\(\text{Statement 1}\)
\(\text{Let}\ \ n = 1\ \ \text{and}\ \ r = 12\% :\)
\(R_{eff}=\Big{(}1 + \dfrac{12}{100 \times 1}\Big{)}^1-1=12\%\ \ \checkmark\)
\(\text{Statement 2}\)
\(\text{Let}\ \ n = 2\ \ \text{and}\ \ r = 12\% :\)
\(R_{eff}=\Big{(}1 + \dfrac{12}{100 \times 2}\Big{)}^2-1=12.36\%\ \ \checkmark \)
\(\text{Statement 3}\)
\(\dfrac{12\text{%}}{12\ \text{months}} = 1\text{%} \ \text{per month}\ \ \checkmark\)
\(\text{Statement 4}\)
\(\text{Statements 1 and 2 make the 4th statement incorrect.}\)
\(\Rightarrow D\)
Enrico invests $3000 in an account that pays interest compounding monthly.
After four years, the balance of the account is $3728.92
The effective annual interest rate for this investment, rounded to two decimal places, is
`D`
`text{By TVM Solver:}`
| `N` | `= 48` |
| `I(%)` | `= ?` |
| `PV` | `= -3000` |
| `PMT` | `= 0` |
| `FV` | `= 3728.92` |
| `text(P/Y)` | `= text(C/Y) = 12` |
`=> I(%) = 5.45text(%)`
| `r_text{effective}` | `= [(1 + {5.45}/{100 xx 12})^12 -1] xx 100text(%)` |
| `= 5.588 …%` |
`=> D`
The nominal interest rate for a loan is 8% per annum.
When rounded to two decimal places, the effective interest rate for this loan is not
`C`
`text(Consider option C:)`
| `r_text(effective)` | `= [(1 + 8/(100 xx 26))^26 – 1] xx 100` |
| `~~ 8.3154` | |
| `~~ 8.32%` |
`=> C`
Millie invested $20 000 in an account at her bank with interest compounding monthly.
After one year, the balance of Millie’s account was $20 732.
The difference between the rate of interest per annum used by her bank and the effective annual rate of interest for Millie’s investment is closest to
`B`
`text(Let)\ \ r = text(bank annual rate)`
| `20\ 000 (1 + r/12)^12` | `= 20\ 732` |
| `(1 + r/12)^12` | `= (20\ 732)/(20\ 000)` |
| `1 + r/12` | `= 1.003` |
| `r` | `= 0.03600` |
| `= 3.6text(%)` |
| `r_text(effective)` | `= [(1 + r/(100 xx 12))^12 – 1] xx 100text(%)` |
| `= 3.66text(%)` |
`:.\ text(Difference) = 0.06text(%)`
`=> B`