Interest and compounding

Quick refresher

A rate has meaning only together with a period: “10% a year” and “10% over two years” are different terms.
Simple interest is calculated only on the original sum. Compound interest is calculated also on the interest already accumulated. The difference grows quickly with the term: at 10% a year over 30 years a sum grows 4 times under simple interest and 17.4 times under compound interest.
Under compound interest a sum doubles in about $72/p$ years, where $p$ is the rate in per cent (the rule of 72).
A nominal annual rate $j$ compounded $m$ times a year is equivalent to the effective rate $i = (1 + j/m)^{m} - 1$ . We compare rates with different compounding frequencies only as effective rates.
With continuous compounding (interest added at every instant) a sum grows as $A_0 e^{jn}$ , the effective rate is $e^{j} - 1$ , and the reverse conversion is $j = \ln(1+i)$ .
The real rate is the growth of purchasing power (what the money can buy): $1 + i = (1 + r)(1 + f)$ . The approximation $r \approx i - f$ is good only when inflation is low.

Key relationships

A_n = A_0\,(1 + n\,i)

simple interest: the sum on which interest is calculated does not change

A_n = A_0\,(1 + i)^{n}

compound interest; $(1+i)^{n}$ is the compounding factor

i = \left(1 + \dfrac{j}{m}\right)^{m} - 1

effective annual rate from the nominal rate $j$ with $m$ compoundings a year

A_n = A_0\,e^{jn}, \qquad i = e^{j} - 1

continuous compounding

1 + i = (1 + r)(1 + f)

nominal rate $i$ , real rate $r$ and rate of inflation $f$

Common pitfalls

Dividing the effective annual rate by 12 to get a monthly rate: the correct one is $(1+i)^{1/12} - 1$ .
Comparing rates with different compounding frequencies without converting them to effective rates.
Confusing the two meanings of the word “nominal”: a rate that ignores the compounding frequency, and a rate in money terms that includes inflation.
Assuming that interest withdrawn from the account every year grows by compound interest: without reinvestment the growth is simple.
Forecasting prices and costs decades ahead at a constant growth rate and not noticing how quickly the factor grows.

After this lesson you can

explain how compound interest differs from simple interest, and estimate the doubling time of a sum
convert a nominal annual rate (the rate as quoted) into an effective one (the rate that gives the same growth when compounded once a year) and back, for any compounding frequency, including continuous
compare offers with different compounding schemes by the effective rate
move sums and series of payments through time by compounding, step by step and directly
apply the same formula to the growth of prices and costs, and separate the real rate from the nominal one

Interest is the price of time

Money can be invested to earn an income, so whoever gives it up for use demands payment. This payment is called interest, and its ratio to the sum for a period is the interest rate. For the lender, interest is income. For the borrower, it is a cost. In project appraisal the investor plays the role of the lender: money put into a well could have earned an income elsewhere (time value of money).

A rate always has a period. To say “10%” without a period is to say nothing: 10% a quarter and 10% over two years are entirely different terms. So we state a rate together with its period, usually a year, and with the compounding scheme: when the interest is calculated and what happens to it next.

$A_0$: The original sum (principal), invested or borrowed at the initial moment.
$A_n$: The accumulated sum after $n$ periods. It is also called the future value (FV) or the terminal value.
$i$: The rate per period as a fraction: 10% is 0.10.
$n$: The number of compounding periods. The unit of time must match the period of the rate.

Simple interest

With simple interest, each period adds the same share of the original sum. The interest is not added to the base (the sum on which interest is calculated). It is paid out, or it piles up separately and earns nothing. So the sum grows linearly: by the same amount $A_0 i$ in each period.

Growth under simple interest

A_n = A_0\,(1 + n\,i)

where:

$A_n$ accumulated sum after $n$ periods, currency units
$A_0$ original sum, the same currency units
$i$ rate per period, fraction
$n$ number of periods, —

Units: any currency; the rate and the number of periods in consistent units of time

Holds when

interest is calculated only on the original sum and earns nothing in later periods
the rate is constant for the whole term
there are no additional deposits or withdrawals within the term

The factor $1 + ni$ grows in a straight line with the term: if the term doubles, the amount of interest doubles.

We find simple interest in short-term loans and in deposits without capitalisation (adding the interest to the sum). We find it in bonds whose coupon (the regular interest payment) is paid out rather than reinvested. And we use it for a part of a period: for example, if a deposit is closed in the middle of a period, the interest for the days that have passed is usually calculated in proportion to their number, by the rule written in the contract. In project appraisal, where the terms are years and decades, simple interest is hardly used.

Compound interest

With compound interest, the interest calculated for a period is added to the sum (this is called capitalisation), and in the next period the interest is calculated on the increased sum. After the first period $A_1 = A_0(1+i)$ , after the second $A_2 = A_1(1+i) = A_0(1+i)^2$ , and so on: each period multiplies the sum by the same factor $1 + i$ .

Growth under compound interest

A_n = A_0\,(1 + i)^{n}

where:

$A_n$ accumulated (future) sum after $n$ periods, currency units
$A_0$ original sum, the same currency units
$i$ rate per compounding period, fraction
$n$ number of periods; project calculations also allow a fractional value, —
$(1+i)^{n}$ compounding factor: how many times a unit of money grows, —

Units: any currency; an annual rate with a number of years, a monthly rate with a number of months

Holds when

the interest stays invested and earns income at the same rate
the rate is constant; if it changes, the factors of the individual periods are multiplied: $(1+i_1)(1+i_2)\cdots(1+i_n)$
interest is calculated once per period to which the rate refers

Growth is exponential: in each period the sum increases not by the same amount but by the same factor. So over a short term compound interest hardly differs from simple interest, and over a long one the two differ by several times.

A handy measure of the speed of growth is the doubling time: from $(1+i)^{n} = 2$ we get $n_2 = \ln 2/\ln(1+i)$ . For a quick estimate in your head there is the rule of 72: a sum doubles in about $72/p$ years, where $p$ is the rate in per cent. At 8% it is about 9 years (exactly 9.01), and at 12% about 6 (exactly 6.12). The same rule, read the other way round, shows in how many years the present value of a future payment is halved.

CalculatedGrowth of one unit of money at a rate of 10% a year: simple interest, compound interest compounded once a year (the dots are the values at the end of each year) and continuous compounding at a nominal rate of 10%. After 30 years: 4.0, 17.4 and 20.1.

Worked exampleSimple and compound interest over 10 years

Given (assumed data)

Original sum $A_0$: US$20 million
Rate $i$: 9% a year
Term $n$: 10 years

Solution

Simple interest: each year $0.09\cdot 20 = 1.8$ million dollars is added, 18 million over 10 years. The accumulated sum is $A_{10} = 20\,(1 + 10\cdot 0.09)$ $= 38.0$ million dollars.
Compound interest: the compounding factor is $1.09^{10} = 2.3674$ , and the accumulated sum is $A_{10} = 20\cdot 2.3674$ $= 47.35$ million dollars.
The difference is 9.35 million dollars. This is interest calculated on the interest of previous years.
Doubling time under compound interest: $n_2 = \ln 2/\ln 1.09$ $= 8.04$ years; the rule of 72 gives $72/9 = 8$ years. Under simple interest the sum doubles only after $1/0.09 = 11.1$ years.

Answer: After 10 years: $38.0 million with simple interest and $47.35 million with compound interest.

What it means. In the first year both schemes give the same result, but the gap grows every year, because under compound interest the base on which interest is calculated grows. The rule of 72 gives a good estimate at rates from about 5 to 15%. The data in this example are assumed for illustration.

Compounding frequency: nominal and effective rate

Banks often compound interest several times a year (every half-year, quarter, month or even daily) while announcing an annual rate. Such a rate is called the nominal annual rate, and we call it $j$ . It is the rate per compounding period multiplied by the number of periods in a year, $m$ . For example, 1.5% a month is a nominal 18% a year. In each period the sum is multiplied by $1 + j/m$ , and in a year by $(1 + j/m)^{m}$ .

Growth with compounding m times a year

A_n = A_0\left(1 + \frac{j}{m}\right)^{mn}

where:

$j$ nominal annual rate, fraction per year
$m$ number of compounding periods in a year: 2, 4, 12, 365…, —
$j/m$ rate per compounding period, fraction
$n$ term, years

Units: any currency

Holds when

the compounding periods are equal in length
the interest of each period is capitalised

At the same nominal rate, the more often the interest is compounded, the faster the sum grows, because the interest starts earning income sooner.

To compare schemes with different frequencies, we bring them to a single standard, the effective annual rate. This is the rate, compounded once a year, that gives the same growth over a year.

Effective annual rate

i = \left(1 + \frac{j}{m}\right)^{m} - 1, \qquad j = m\left[(1+i)^{1/m} - 1\right]

where:

$i$ effective annual rate, fraction per year
$j$ nominal annual rate with compounding $m$ times a year, fraction per year
$m$ number of compoundings in a year, —

Units: dimensionless fractions

Holds when

all interest is capitalised
one-off fees and bonuses are not part of the comparison; if there are any, we account for them separately, in the full cash flow

At $m = 1$ the effective rate equals the nominal rate; at $m > 1$ it is higher. The second formula is the reverse conversion: which nominal rate, compounded $m$ times a year, gives a given effective rate.

Effective annual rate at different compounding frequencies, %

Nominal rate: 6%
2 times a year: 6.09
4 times a year: 6.14
12 times a year: 6.17
Continuous: 6.18

Nominal rate: 10%
2 times a year: 10.25
4 times a year: 10.38
12 times a year: 10.47
Continuous: 10.52

Nominal rate: 20%
2 times a year: 21.00
4 times a year: 21.55
12 times a year: 21.94
Continuous: 22.14

Nominal rate: 30%
2 times a year: 32.25
4 times a year: 33.55
12 times a year: 34.49
Continuous: 34.99

Nominal rate	2 times a year	4 times a year	12 times a year	Continuous
6%	6.09	6.14	6.17	6.18
10%	10.25	10.38	10.47	10.52
20%	21.00	21.55	21.94	22.14
30%	32.25	33.55	34.49	34.99

The difference between the effective and the nominal rate grows roughly as the square of the rate: at 6% it is less than 0.2 percentage points at any frequency, at 30% with continuous compounding it is almost 5.

The table shows two patterns. First, the move from annual to monthly compounding is noticeable, but the move from monthly to continuous hardly is. A frequency higher than monthly adds practically nothing. Second, at low rates we can neglect the frequency, but at rates of tens of per cent a year we cannot. Such rates occur, for example, on loans in the national currency in periods of high inflation.

Worked exampleWhich bank offer is better for the depositor

Given (assumed data)

Offer A: 15.5% a year, compounded once a year
Offer B: 15.0% a year, monthly capitalisation
Offer C: 15.2% a year, capitalisation every half-year

Solution

A: interest is compounded once a year, so the effective rate equals the nominal one — 15.50%.
B: the monthly rate is $0.15/12 = 0.0125$ ; the effective rate is $1.0125^{12} - 1$ $= 16.08$ %.
C: the half-year rate is $0.152/2 = 0.076$ ; the effective rate is $1.076^{2} - 1$ $= 15.78$ %.
For comparison: with daily compounding the same 15.0% would give 16.180%, and with continuous compounding (the limit) 16.183%. Compounding more often than monthly is almost pointless.

Answer: Offer B is the best (16.08%), then C (15.78%) and A (15.50%), although by the nominal rate B looks the worst.

What it means. We can compare only effective rates. Frequent compounding favours the one who receives the interest and puts the one who pays it at a disadvantage. If a deposit or a loan has fees, we account for them together with the interest, as a full cash flow. The data in this example are assumed for illustration.

Continuous compounding

The more often interest is compounded, the larger the annual factor $(1 + j/m)^{m}$ , but the growth slows down and approaches a limit. When $m$ grows without bound ( $m \to \infty$ ), the factor approaches $e^{j}$ , where $e \approx 2.718$ is the base of natural logarithms. This limiting case is called continuous compounding (interest added at every instant).

Continuous compounding

A_n = A_0\,e^{jn}, \qquad i = e^{j} - 1, \qquad j = \ln(1 + i)

where:

$j$ nominal annual rate with continuous compounding (also called the force of interest or the continuous rate), fraction per year
$i$ equivalent effective annual rate, fraction per year
$n$ term, years
$\ln$ natural logarithm, —

Units: any currency; the rates as fractions per year

Holds when

interest is calculated and capitalised continuously
the rate is constant for the whole term

Continuous compounding is the upper limit for any frequency: a nominal 10% gives 10.00% effective with annual compounding, 10.47% with monthly and 10.52% with continuous. The inverse formula shows which continuous rate is equivalent to a given effective one: an effective 10% is $\ln 1.1 = 9.53$ % with continuous compounding.

Banks do not use continuous compounding for deposits and loans, and annual cash flow models use the effective annual rate. The continuous form is convenient in theory and in analytical calculations. The continuous rates of successive periods, multiplied by their lengths, simply add up, because they are the logarithms of the compounding factors. And formulas for payments spread over time are written with integrals. If a rate is given as continuous, convert it into an effective one before using it in an annual model.

Moving sums through time

We cannot add money of different dates directly, just as we cannot add metres and feet. First we bring all the sums to one date: compounding moves them forward, and discounting moves them back. For a series of payments this can be done in two ways, with the same result.

Two ways to compound a series of payments to one date

1
Step by step
Move the accumulated sum one period forward by multiplying it by $1+i$ , add the payment of that period, and repeat until you reach the date you need. This is how a bank account works. The intermediate balances are visible at every step.
2
Directly
Multiply each payment by its own compounding factor $(1+i)^{n-t}$ , where $n - t$ is the number of periods from the payment to the date of reference, and add the products. This is convenient for comparing the contributions of individual payments.

Accumulated sum of a series of payments

FV_n = \sum_{t=0}^{n} CF_t\,(1 + i)^{\,n-t}

where:

$FV_n$ the sum of all payments compounded to moment $n$ , currency units
$CF_t$ payment (cash flow) at moment $t$ : an investment or a receipt, with its own sign, currency units
$i$ rate per period, fraction
$n - t$ number of periods from the payment to the date of reference, —

Units: any currency, the same for all payments

Holds when

all payments are counted from one origin and assigned to moments by one convention
one rate for all periods and for balances of either sign

If the payments are the net cash flow of a project, $FV_n$ is the balance of the “project account” at moment $n$ . Brought back to the time origin, it equals the NPV (Net present value (NPV)).

Worked exampleWhat the development of a field costs by the first year of production

Given (assumed data)

Capital expenditure of the project from the lesson Project cash flow: 50, 90 and 20 million dollars in years 1, 2 and 3, in the middle of each year
Rate: 10% a year
Date of reference: the middle of year 3, the first year of production

Solution

Step by step: to the middle of year 2, $50\cdot 1.1 + 90 = 145.0$ million dollars; to the middle of year 3, $145.0\cdot 1.1 + 20 = 179.5$ million dollars.
Directly: $50\cdot 1.1^{2} = 60.5$ and $90\cdot 1.1 = 99.0$ million dollars; together with the costs of year 3, $60.5 + 99.0 + 20 = 179.5$ million dollars.
That is 19.5 million dollars, or 12.2%, on top of the 160 million dollars actually paid. This is what it costs to have the money tied up in construction for up to two years.

Answer: By the middle of the first year of production, the development costs 179.5 million dollars, allowing for the value of money at 10% a year.

What it means. Suppose the construction money were borrowed at 10% and the interest were capitalised. Then by this date the debt would be exactly this sum. Such an addition is called interest during construction. Both methods give the same answer. The step-by-step method also shows the intermediate balance, like a bank statement. The data in this example are assumed for illustration.

The step-by-step method is exactly the calculation of the balance of a bank account into which money is paid and from which it is withdrawn: the balance of the previous period with interest, plus deposits, minus withdrawals. The lesson on NPV follows the “project account” in the same way at the discount rate. The lesson on IRR follows it at the rate at which this account is brought to zero at the end of the project.

Growth of prices and costs, the real rate

The same formula describes the growth of prices and costs at a constant rate, which is called escalation. If a service costs $P_0$ today and rises in price by a fraction $f$ a year, in $n$ years it will cost $P_0(1+f)^{n}$ . This is how cost forecasts adjust the cost of drilling, services and equipment for price growth.

Price growth at a constant rate

P_n = P_0\,(1 + f)^{n}

where:

$P_n$ price after $n$ years, currency units
$P_0$ price in the base year, currency units
$f$ annual rate of price growth — general inflation or sector-specific escalation, fraction per year
$n$ number of years from the base year, years

Units: any currency

Holds when

the growth rate is constant; if it changes, the annual factors are multiplied
this is a forecast assumption, not a law: market prices, such as the oil price, do not change so smoothly

A constant rate over a long horizon gives surprisingly large factors: 5% a year over 20 years makes a price grow 2.65 times. We always check a forecast with constant escalation for decades ahead to see whether it is plausible.

When both the money in an account and the prices grow, what matters is how many more goods and services we can buy. If money grows at a rate $i$ and prices grow at a rate $f$ , purchasing power grows at the real interest rate $r$ . The rate $i$ , which already includes inflation, is in this sense also called nominal.

Relation between the nominal rate, the real rate and inflation

1 + i = (1 + r)(1 + f), \qquad r = \frac{i - f}{1 + f}

where:

$i$ rate in nominal money, that is, including inflation, fraction per year
$r$ real rate — the growth of purchasing power, fraction per year
$f$ rate of inflation — the growth of the general price level, fraction per year

Units: fractions for the same period

Holds when

all three quantities refer to the same period and are effective rates
inflation is measured by a price index suited to the task

The factors are multiplied, not added. The approximation $r \approx i - f$ works well at inflation of a few per cent, and it is noticeably off at high inflation. We discount nominal cash flows at a nominal rate, and flows in constant prices at a real rate (Inflation: nominal and real money).

Worked exampleThe real return on a deposit

Given (assumed data)

Deposit rate $i$ (effective): 14% a year
Inflation $f$: 9% a year

Solution

After a year, 100 tenge (KZT) on deposit become 114 tenge, while a basket of goods that cost 100 tenge will cost 109 tenge.
The purchasing power of the deposit grows by a factor of $114/109 = 1.0459$ , that is, the real rate is $r = 4.59$ %.
The same by the formula: $r = (0.14 - 0.09)/1.09 = 4.59$ %. The approximation $i - f$ gives 5%.
At a rate of 30% and inflation of 25% the approximation still gives 5%, while the exact real rate is $1.30/1.25 - 1 = 4.0$ %.

Answer: The real return on the deposit is about 4.59% a year.

What it means. A nominal 14% in fact means growth of purchasing power of less than 5% a year. The higher the inflation, the further the “rate minus inflation” approximation is from the truth. The data in this example are assumed for illustration.

Terms and symbols

Accumulated sum (FV): The sum to which an investment grows after a given term at a given rate. It is also called the future value.
Compounding factor: The factor $(1+i)^{n}$ : the number of times a unit of money grows in $n$ periods under compound interest.
Capitalisation of interest: Adding the interest to the sum, after which the interest earns interest itself.
Nominal annual rate $j$: The rate per compounding period multiplied by the number of periods in a year.
Effective annual rate $i$: The rate, compounded once a year, that gives the same growth as the given scheme. It is the basis of comparison.
Continuous compounding: The limiting case of infinitely frequent compounding. The compounding factor is $e^{jn}$ .
Escalation: Growth of a price or cost over time. At a constant rate $f$ it is described by the factor $(1+f)^{n}$ .
Real rate $r$: The growth of the purchasing power of money: $(1+i)/(1+f) - 1$ .

Common mistakes and misconceptions

Wrong

“1.5% a month is 18% a year.”

In fact

That is the nominal rate. If interest is capitalised monthly, over a year the sum grows $1.015^{12} = 1.1956$ times, which is an effective rate of 19.56%.

Wrong

“The monthly rate is the annual rate divided by 12.”

In fact

That is how the monthly rate is found from the nominal annual rate. From the effective annual rate, the monthly rate is found as $(1+i)^{1/12} - 1$ ; dividing by 12 overstates the annual growth.

Wrong

“At 10% a year a sum will grow fourfold in 30 years.”

In fact

Fourfold is for simple interest. With compound interest the factor is $1.1^{30} = 17.4$ . Most of the growth comes from interest calculated on interest.

Wrong

“The real rate is the nominal rate minus inflation.”

In fact

That is an approximation. Exactly: $r = (i - f)/(1 + f)$ . At 30% and inflation of 25% the approximation gives 5%, while the real rate is in fact 4%.

In depth: advanced detail

Where the number e and the rule of 72 come from

We take the logarithm of the annual factor: $\ln(1 + j/m)^{m} = m\ln(1 + j/m)$ . For large $m$ the quantity $j/m$ is small and $\ln(1 + x) \approx x$ , so $m\ln(1 + j/m) \to j$ , and the factor itself approaches $e^{j}$ . This is where continuous compounding comes from: over $n$ years the factor is $e^{jn}$ .

The doubling time under compound interest is $n_2 = \ln 2/\ln(1+i)$ . The expansion $\ln(1+i) \approx i - i^2/2$ gives $n_2 \approx \dfrac{0.693}{i}\,(1 + i/2)$ . At a rate of about 8% the factor $1 + i/2 = 1.04$ , and $69.3\cdot 1.04 \approx 72$ . Hence the rule of 72: $n_2 \approx 72/p$ . The number 72 is also convenient because it is divisible by 2, 3, 4, 6, 8, 9 and 12. With continuous compounding the exact rule is $69.3/p$ .

Comparing options by terminal value

We can compare investment options by compounding all their flows to one future date. If the options bring money at different times, we have to decide where the early receipts are invested until that date, and the answer depends on this reinvestment rate.

Worked exampleA lump sum later or a series of payments now

Given (assumed data)

Option X: US$20 million as a lump sum in 8 years
Option Y: US$2.3 million at the end of each year for 8 years
Money received: is invested at a rate $r$ until the end of year 8

Solution

The terminal value of Y is the sum of the compounded payments: $FV_Y = 2.3\,s_8(r)$ , where $s_8(r) = [(1+r)^{8} - 1]/r$ .
At $r = 0$ : $FV_Y = 2.3\cdot 8 = 18.4$ million dollars — X is better.
At $r = 3$ %: $FV_Y = 20.45$ ; at $r = 6$ %: $FV_Y = 22.76$ million dollars — Y is better.
The options are equivalent at $r = 2.37$ %: at this same rate their present values are equal too.

Answer: If the early receipts can be invested at a rate above 2.37%, Y is better; if below, X.

What it means. Comparison by terminal value at a reinvestment rate $r$ ranks the options in the same order as comparison by present value at the same rate, because all the sums differ by one factor $(1+r)^{8}$ . So in practice we compare present values and state the rate explicitly. The hidden assumption about the reinvestment rate is the main trap of the IRR indicator (Internal rate of return (IRR)). The data in this example are assumed for illustration.

Connections to other topics

Discounting and present value — compounding in reverse: present value and discount factors.
Net present value (NPV) — the sum of the project's discounted flows, and the “project account” at the discount rate.
Internal rate of return (IRR) — the rate at which NPV is zero, and the assumption about the reinvestment rate.
Annuities and discounted production — series of equal payments and their accumulated and present values.
Inflation: nominal and real money — nominal money, constant prices and conversion by price indices.
Project cash flow — the project's cash flow, which is compounded and discounted.

Check yourself

1
100 million tenge was put on deposit for 3 years at 10% a year, compounded once a year. How much will be in the account at the end of the term?
1. A.130.0 million tenge
2. B.110.0 million tenge
3. C.133.1 million tenge
4. D.134.8 million tenge
Show the answer
Correct answer: C. 133.1 million tenge
$100\cdot 1.1^{3} = 133.1$ million tenge. 130 is the result of simple interest, 134.8 is the result of monthly compounding at a nominal 10% ( $1.00833^{36}$ ), and 110 is the growth for one year only.
2
What is better for a depositor: 10% a year compounded monthly or 10.4% a year compounded once a year?
1. A.The same
2. B.10.4% compounded once a year
3. C.10% compounded monthly
Show the answer
Correct answer: C. 10% compounded monthly
The effective rate of the first offer is $(1 + 0.10/12)^{12} - 1 = 10.47$ %, of the second 10.40%. The difference is small, but only effective rates can be compared.
3
A cash flow model is run by months, and the discount rate is given as an effective annual rate of 12%. Which monthly rate should be used?
Show the answer
$(1.12)^{1/12} - 1 = 0.949$ % a month. If we used $12/12 = 1$ %, then $1.01^{12} - 1 = 12.68$ % would build up over a year, so the model would discount at a higher rate than the one given.
4
In how many years will a sum double at 6% a year compounded annually? Estimate in your head and check by calculation.
Show the answer
By the rule of 72: $72/6 = 12$ years. Exactly: $\ln 2/\ln 1.06 = 0.6931/0.05827 = 11.9$ years.
5
The deposit rate is 15% a year, inflation is 10% a year. What is the real rate?
1. A.4.5%
2. B.5.5%
3. C.25%
4. D.5.0%
Show the answer
Correct answer: A. 4.5%
$r = (1.15/1.10) - 1 = 0.0455$ , that is, 4.5%. 5% is the approximation “rate minus inflation”. The higher the inflation, the more it overstates the real rate.
6
A well repair costs 40 million tenge today. How much will it cost in 7 years if costs grow by 6% a year?
Show the answer
$40\cdot 1.06^{7} = 40\cdot 1.504 = 60.1$ million tenge, which is one and a half times more. This is a forecast under the assumption of a constant growth rate, not a promised price.

Updated: 19 October 2026