Finding the number of years

Finance and economics, grade 11 (3 units) Lesson 7 of 14Learning

Finding the number of years

Trying values

When the question is “after how many years?”, the unknown $\displaystyle n$ is in the exponent. It is found by trying: work out the value for a few numbers of years, until the sum is passed.

Shaked deposited 10,000 ₪ at a compound interest of 5% a year.
After how many years will the deposit hold more than 15,000 ₪?

Years	The sum in the deposit
7	$\displaystyle 10000\cdot1.05^{7}\approx14071.00$
8	$\displaystyle 10000\cdot1.05^{8}\approx14774.55$
9	$\displaystyle 10000\cdot1.05^{9}\approx15513.28$

After 8 years there is not yet 15,000 ₪, and after 9 years there is: the answer is 9 years.

The number of years on the graph board

On the graph, the sum 15,000 ₪ is the straight line $\displaystyle {y}={15000}$ . The point where the deposit’s graph crosses it gives the number of years.

\displaystyle y=10000\cdot1.05^{x}

\displaystyle y=15000

\displaystyle A=(8.31,15000)

The graphs meet when $\displaystyle x\approx8.3$ .
The interest joins the deposit only at the end of each year, so the sum passes 15,000 ₪ at the end of the ninth year.

When the value falls to half

In decay, look for when the value falls below the sum in the same way.

A car that cost 100,000 ₪ loses 15% of its value every year.
When will it be worth less than half the price?

After 4 years: $\displaystyle 100000\cdot0.85^{4}\approx52200.63$ ₪ – still more than half.
After 5 years: $\displaystyle 100000\cdot0.85^{5}\approx44370.53$ ₪ – already less than half.

The rule of 72: the doubling time

The rule of 72 gives a quick estimate of the time in which a sum doubles, growing by $\displaystyle p$ percent a year:

\displaystyle n\approx\frac{72}{p}

At 6% a year: $\displaystyle {\frac{72}{6}}={12}$ years. Check: $\displaystyle 1.06^{12}\approx2.01$ .
At 4% a year: $\displaystyle {\frac{72}{4}}={18}$ years. Check: $\displaystyle 1.04^{18}\approx2.03$ .

The estimate is good for small percents – up to about 10%.

Exercise 1

Alon deposited 4,000 ₪ at a compound interest of 8% a year.
He wants to know after how many years he will have at least 6,000 ₪.

a

Work out the sum after 5 years. Round to agorot.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

Work out the sum after 6 years. Round to agorot.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

After how many years will Alon have at least 6,000 ₪?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

An electric scooter was bought for 2,000 ₪, and its value falls by 10% every year.
On the board: the graph of its value, and the line $\displaystyle {y}={1000}$ .
Mark the point where the graphs meet.

\displaystyle y=2000\cdot0.9^{x}

\displaystyle y=1000

a

After how many whole years will the scooter be worth less than half its price? Check by working it out.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

A savings fund gives a return of 9% a year, compounded.

a

By the rule of 72, after about how many years will the money in the fund double?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

Check: how many times does the money grow in the number of years you found? Round to two digits after the point.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

10,000 ₪ in a deposit:
Offer A – a simple interest of 6% a year: $\displaystyle {y}={10000+600x}$ .
Offer B – a compound interest of 5% a year: $\displaystyle {y}={10000\cdot1.05^{x}}$ .
Mark the points where the graphs meet.

\displaystyle y=10000+600x

\displaystyle y=10000\cdot1.05^{x}

a

From which year does offer B give more? Check the year before it and the year itself by working them out.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: finance and economics, grade 11 (3 units)

Growth and decay

Simple interest – $\displaystyle A_0$ the principal, $\displaystyle p$ the interest rate a year in percent, $\displaystyle n$ the number of years, $\displaystyle A$ the sum at the end
$\displaystyle A=A_0+n\cdot A_0\cdot\frac{p}{100}$
The growth factor – a rise of $\displaystyle p$ percent
$\displaystyle q=1+\frac{p}{100}$
The decay factor – a fall of $\displaystyle p$ percent
$\displaystyle q=1-\frac{p}{100}$
Growth and decay – $\displaystyle A_0$ the value at the start, $\displaystyle q$ the factor for a period, $\displaystyle n$ the number of periods, $\displaystyle A$ the value at the end
$\displaystyle A$ $\displaystyle =$ $\displaystyle A_0\cdot q^{n}$
The factor of two changes, one after the other
$\displaystyle q_1\cdot q_2$
Today’s value of a sum $\displaystyle A$ in $\displaystyle n$ years, when prices grow $\displaystyle q$ times every year
$\displaystyle \frac{A}{q^{n}}$
The rule of 72 – about how many years until a sum doubles, growing by $\displaystyle p$ percent a year
$\displaystyle n\approx\frac{72}{p}$

The mean and the spread

The mean of $\displaystyle n$ data
$\displaystyle \bar{x}=\frac{x_1+x_2+\dots+x_n}{n}$
The range
$\displaystyle x_{\max}-x_{\min}$
The variance – the mean of the squares of the deviations from the mean
$\displaystyle \sigma^{2}=\frac{(x_1-\bar{x})^{2}+(x_2-\bar{x})^{2}+\dots+(x_n-\bar{x})^{2}}{n}$
The standard deviation
$\displaystyle \sigma=\sqrt{\sigma^{2}}$

Quartiles and deciles

The place of the median in an ordered list of $\displaystyle n$ data
$\displaystyle \frac{n+1}{2}$
The interquartile range – $\displaystyle Q_1$ the lower quartile, $\displaystyle Q_3$ the upper quartile
$\displaystyle Q_3-Q_1$
The gap between the top and the bottom decile – how many times the mean income in decile 10 is that in decile 1
$\displaystyle \frac{\bar{x}_{10}}{\bar{x}_{1}}$