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[h] IB Mathematics AA HL Flashcards- Compound interest

[q] Compound Interest

[a]

In terms of real-world usage, this may be the most helpful subtopic in IB mathematics courses. It will be useful for: mortgages, loans, investments, banking, and more.

UNDERSTANDING // If we take the simple case of an investment accruing compound interest once per year, it may not be immediately clear how geometric sequences are involved. If we have 5% interest p.a. (per annum/yr.), some may want to find 5% of our investment, then add it, then repeat.

[q]

This would work, but wouldn’t be the most efficient method. If, instead, we essentially find 105% of our amount with quick multiplication, i.e. \( \times 1.05 \) or \( \times \left( 1 + \frac{5}{100} \right) \). If we simply do one multiplication year after year, this is essentially a geometric sequence.

[a]

E.G.1 // Find the value of a $5,000 investment if compounded annually with an interest rate of 4.2% p.a., after 5 years: We can increase something by 4.2% by multiplying by 1.042.

∴ Final Value = \( 5000 \times (1.042)^5 = \$6141.98 \)

The interest may be compounded quarterly or monthly as well. For example, ‘6% p.a. compounded quarterly’ means it increases by 1.5% every 3 months.

[q]

E.G.2 // €260,000 with 7.2% p.a. interest compounded monthly for 4 years:

Each month we have \( \frac{7.2}{12} \% \) growth = 0.6% growth. There are 48 months in 4 years ∴

\( 260000 \times (1.006)^{48} = €346,474 \approx. €346,000 \) to 3 s.f.

[a]

The dividing of the rate, and multiplying of the time periods gives us:

\[

\text{F.V.} = \text{P.V.} \times \left( 1 + \frac{r}{100k} \right)^{kn}

\]

F.V. = Future Value / P.V. = Present Value

k = no. of periods per yr. / r% = rate / n = years

We may also need to find another one of the unknowns, not just the F.V.:

[q]

E.G.3 // A £4,500 investment gives £5,533.04 if compounded quarterly for 4 years at what % rate per annum?

Use \( \text{F.V.} = \text{P.V.} \times \left( 1 + \frac{r}{100k} \right)^{kn} \):

\[

5533.04 = 4500 \left( 1 + \frac{r}{400} \right)^{16}

\]

\[

\frac{5533.04}{4500} = \left( 1 + \frac{r}{400} \right)^{16}

\]

\[

1.2296 = \left( 1 + \frac{r}{400} \right)^{16}

\]

\[

1.013 = 1 + \frac{r}{400}

\]

\[

0.013 = \frac{r}{400}

\]

\( r = 0.013 \times 400 = 5.2\% \text{ interest rate p.a.} \)

[a] Standard form:

A number in the format , \( \pm a \times10^k \) , where \(1\leqslant a\leqslant 10 \) and \(k\in Z\) (k is an integer)

[q] Standard Form – Solved Examples 1

The diameter of a spherical planet is 6 × 10^{4} km .

(a) Write down the radius of the planet. [1]

The volume of the planet can be expressed in the form π(a × 10^{k}) km^{3} where 1 ≤ a <10 and k ∈ \(\mathbb{Z}\)

(b) Find the value of a and the value of k .

[a] Answer – Solved Examples 1

(a) \(3 \times 10^4\)

(b) \(\frac{4}{3}\pi(3 \times 10^4)^3\)

\(=\frac{4}{3}\pi\times 27 \times 10^{12}\)

\(=\pi(3.6 \times 10^{13})(km)^3\)

Hence

\(a= 3.6 \) , \(k =13\)

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