Arithmetic

Lessons

Senior 4 · Basic Mathematics

Arithmetic

5 lessons

Lesson 5 of 5

Interest, investment and arc elasticity of demand

25 min

Money that grows

Uwamahoro leaves 500 000 Frw in a savings account that pays 8% interest a year. After one year she has earned $0.08 \times 500\,000 = 40\,000$ Frw. What happens in the second year depends on the kind of interest: if she earns interest only on the 500 000 Frw she put in, she gets another 40 000 Frw; if she also earns interest on last year's interest, she gets $0.08 \times 540\,000 = 43\,200$ Frw. The difference looks small, but over 20 years it is more than a million francs. This lesson applies the arithmetic of the unit (percentages, powers and logarithms) to saving, borrowing and prices.

Definition

Interest: money paid for the use of money. The amount lent or deposited is the principal $P$ ; the rate $r$ is the interest per year as a decimal (8% gives $r = 0.08$ ); the time is $t$ years.

Simple interest is paid on the principal only:

I = Prt, \qquad A = P + I = P(1 + rt),

where $A$ is the amount (final value) at the end.

Example: 500 000 Frw at 8% simple interest for 3 years: $I = 500\,000 \times 0.08 \times 3 = 120\,000$ Frw, so $A = 620\,000$ Frw.

Definition

Compound interest: the interest of each period is added to the principal, and the next period's interest is calculated on the new total. After $n$ years at rate $r$ per year, compounded yearly, the final value of an investment $P$ is

A = P (1 + r)^{n} .

If interest is compounded $k$ times a year ( $k = 4$ quarterly, $k = 12$ monthly), the rate per period is $\frac{r}{k}$ and there are $k t$ periods:

A = P\left(1 + \frac{r}{k}\right)^{kt} .

Simple interest grows by the same 40 000 Frw every year, a straight line. Compound interest grows by 8% of an ever larger amount, so its curve bends upwards: after 20 years it has earned more than twice as much interest as the simple account.

Worked example 1: compounding yearly and quarterly

Find the final value of 500 000 Frw invested for 3 years at 8% a year, (a) compounded yearly, (b) compounded quarterly.

Solution

$A = 500\,000 \times 1.08^3 = 500\,000 \times 1.259712 = 629\,856$ Frw.
Rate per quarter $\frac{0.08}{4} = 0.02$ , number of quarters $4 \times 3 = 12$ : $A = 500\,000 \times 1.02^{12} \approx 634\,121$ Frw.

Compare with 620 000 Frw from simple interest: the more often interest is compounded, the more it earns.

Worked example 2: how long, and how much now?

(a) How many whole years does it take 500 000 Frw at 8% compounded yearly to reach 1 000 000 Frw? (b) How much must be deposited now at 7% compounded yearly to have 2 000 000 Frw after 5 years?

Solution

We need $500\,000 \times 1.08^n \ge 1\,000\,000$ , so $1.08^n \ge 2$ . With logarithms: $n \ge \frac{\log 2}{\log 1.08} \approx 9.006$ . So 10 years. (After 9 years the account holds 999 502 Frw, just short.)
$P \times 1.07^5 = 2\,000\,000$ , so $P = \frac{2\,000\,000}{1.07^5} \approx 1\,425\,972$ Frw. This is the present value of 2 000 000 Frw in five years.

Worked example 3: saving a fixed amount every month

A member of a savings group deposits 20 000 Frw at the end of every month for a year, and the group pays 1% interest a month, compounded monthly. What is the final value of the savings after the 12th deposit?

Solution

The last deposit earns nothing, the one before earns one month's interest, …, the first earns 11 months'. So the final value is

20\,000\left(1 + 1.01 + 1.01^2 + \dots + 1.01^{11}\right) = 20\,000 \times \frac{1.01^{12} - 1}{0.01} \approx 253\,650 \text{ Frw},

using the sum of a geometric sequence. She paid in 240 000 Frw and earned about 13 650 Frw of interest.

Warning

Match the rate to the period. A rate "per year" compounded monthly is divided by 12, and $n$ counts months, not years. Always write the rate as a decimal (8% is 0.08, not 8). And do not add up yearly percentages under compound interest: 8% a year for 3 years is a 25.97% increase, not 24%.

Arc elasticity of demand

When the price of a good rises, people usually buy less of it. Economists ask how much less: the price elasticity of demand compares the percentage change in quantity with the percentage change in price. Between two points on a demand curve, $(P_{1}, Q_{1})$ and $(P_{2}, Q_{2})$ , the percentage changes are measured from the average of the two values, so the answer is the same whichever point you start from. This is the arc (midpoint) elasticity.

Definition

Arc elasticity of demand

E = \frac{\dfrac{Q_2 - Q_1}{\frac{1}{2}(Q_1 + Q_2)}}{\dfrac{P_2 - P_1}{\frac{1}{2}(P_1 + P_2)}} = \frac{Q_2 - Q_1}{Q_1 + Q_2} \times \frac{P_1 + P_2}{P_2 - P_1} .

For ordinary goods $E$ is negative. Demand is elastic if $|E| \gt 1$ (quantity reacts strongly), inelastic if $|E| \lt 1$ , and unit elastic if $∣ E ∣ = 1$ .

Worked example 4: bus tickets and tomatoes

(a) A bus company raises its fare from 1 000 Frw to 1 200 Frw, and its daily passengers fall from 5 000 to 4 600 (illustrative). (b) A market trader cuts the price of tomatoes from 400 Frw to 300 Frw per kg, and her daily sales rise from 200 kg to 300 kg (illustrative). Find the arc elasticity in each case, and the effect on daily revenue.

Solution

$E = \frac{4600 - 5000}{9600} \times \frac{2200}{200} = \frac{-400}{9600} \times 11 \approx -0.46$ . $|E| \lt 1$ : inelastic. Revenue rises from $1000 \times 5000 = 5\,000\,000$ to $1200 \times 4600 = 5\,520\,000$ Frw.
$E = \frac{300 - 200}{500} \times \frac{700}{300 - 400} = 0.2 \times (-7) = -1.4$ . $|E| \gt 1$ : elastic. Revenue rises from $400 \times 200 = 80\,000$ to $300 \times 300 = 90\,000$ Frw.

The pattern is general: when demand is inelastic, raising the price raises revenue; when it is elastic, lowering the price raises revenue.

Tip

Read $∣ E ∣$ as "the percentage change in quantity for each 1% change in price". $E = - 1.4$ means a 1% price cut sells about 1.4% more. Goods with few substitutes (salt, bus travel to work) tend to be inelastic; goods with many substitutes (one trader's tomatoes, when the next stall sells them too) tend to be elastic.

Mathispower4u: Compounded Interest Formula, quarterly compounding. Watch how the yearly rate is divided by 4 and the number of years multiplied by 4.

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Practice

Try each one before opening the answers. They get harder as you go.

Find the simple interest on 250 000 Frw at 6% a year for 4 years.
A loan of 350 000 Frw costs 84 000 Frw of simple interest over 6 years. Find the yearly rate.
Find the final value of 300 000 Frw after 6 years at 5% compounded yearly.
Find the final value of 1 000 000 Frw after 2 years at 12% a year compounded quarterly.
How many whole years does an investment at 10% compounded yearly take to triple?
When the price of a good rises from 50 Frw to 60 Frw, the quantity demanded falls from 1 000 to 900 units. Find the arc elasticity and say whether demand is elastic.
A cinema raises its ticket from 2 000 Frw to 2 500 Frw and sales fall from 300 to 150 tickets. Find the arc elasticity. Was the price rise good for revenue?

Show answersHide answers

$250\,000 \times 0.06 \times 4 = 60\,000$ Frw.
$r = \frac{84\,000}{350\,000 \times 6} = 0.04$ , so 4% a year.
$300\,000 \times 1.05^6 \approx 402\,029$ Frw.
$1\,000\,000 \times 1.03^8 \approx 1\,266\,770$ Frw.
$1.1^n \ge 3$ : $n \ge \frac{\log 3}{\log 1.1} \approx 11.5$ , so 12 years.
$E = \frac{-100}{1900} \times \frac{110}{10} \approx -0.58$ : inelastic.
$E = \frac{-150}{450} \times \frac{4500}{500} = -3$ : elastic. Revenue fell from 600 000 Frw to 375 000 Frw, so the rise was bad for revenue.

Quick check

400 000 Frw is invested at 5% a year for 2 years. Which statement is correct?

Key points

Simple interest: $I = P r t$ , $A = P (1 + r t)$ . Compound interest: $A = P\left(1 + \frac{r}{k}\right)^{kt}$ .
Use logarithms to find the time: $n = \frac{\log\left(\frac{A}{P}\right)}{\log(1 + r)}$ , rounded up to a whole period.
Present value $P = \frac{A}{(1 + r)^n}$ .
Arc elasticity $E = \frac{Q_2 - Q_1}{Q_1 + Q_2} \times \frac{P_1 + P_2}{P_2 - P_1}$ ; $|E| \gt 1$ elastic, $|E| \lt 1$ inelastic.

1Money that grows

2Arc elasticity of demand

Money that grows

Arc elasticity of demand