Compound Interest

Interest paid on the interest already earned.

Interest that earns interest

With simple interest, every year of interest is worked out on the principal, the money first put in. With compound interest, each year of interest is added to the balance and stays there, and the next year of interest is worked out on the whole balance, earlier interest included. So the interest earns interest of its own.

Year by year

Save $1000 at 10% a year, compound. In year 1 the interest is 10% of $1000, which is $100, and the balance grows to $1100.

In year 2 the interest is 10% of $1100, not of $1000. That is $110: the $1000 earns $100 again, and the $100 of interest from year 1 earns 10% of itself, $10. The balance grows to $1210.

In year 3 the interest is 10% of $1210, which is $121, and the balance becomes $1331. Each year the interest is bigger than the year before, because it is 10% of a bigger balance.

startyear 1year 2year 3interest$100$110$121balance$1000$1100$1210$1331

The yearly interest grows from $100 to $110 to $121, so the balance rises by more each year: $1000, $1100, $1210, $1331.

year 1: $100year 2: $110

Each part is $10. The interest in year 1 is 10 parts, $100. In year 2 it is 11 parts: the same $100 earned by the $1000, and one gold part, the $10 earned by year 1's interest.

One multiplication a year

Adding 10% to a balance keeps all of it, 100%, and adds 10% more, so the new balance is 110% of the old one. That is one multiplication by 1.1, the multiplier for a 10% rise. In the same way a rise of r% is a multiplication by 1 + r/100.

So each year multiplies the balance by 1.1 once more: 1000, 1000 × 1.1, 1000 × 1.1², 1000 × 1.1³, and after n years 1000 × 1.1ⁿ. This is exponential growth with a factor of 1.1, and the exponent counts the years.

For any principal P at r% a year, with the interest added once a year, the amount after n years is A = P(1 + r/100)ⁿ. The interest earned is the amount minus the principal, A − P.

nA

The balance A at the end of each year n, on the curve A = 1000 × 1.1ⁿ, from $1000 at the start to $2593.74 after 10 years. Each year's rise is bigger than the one before, so the points bend upward.

Using the formula

$5000 is saved at 4% a year, compound, for 10 years. The multiplier is 1.04, and 1.04¹⁰ = 1.480244, to 6 decimal places. So A = 5000 × 1.480244 = 7401.22 dollars, to the nearest cent, and the interest earned is 7401.22 − 5000 = 2401.22 dollars.

The multiplier also gives the total growth as a percentage. 1.1³ = 1.331, so three years at 10% raise a balance by 33.1%, not by 3 × 10% = 30%: each 10% is a share of a bigger balance than the one before.

The usual mistakes

Paying 10% of the principal every year. That is simple interest, and it takes $1000 to $1300 in 3 years. Compound interest takes 10% of the growing balance, so the balance is $1331.

Stopping a year early. $1210 is the balance after 2 years; the third year multiplies by 1.1 once more, to $1331.

Worked example: How Long a Deposit Takes to Double, With the Rule of 72 Checked Against the True Answer

Question Mr Lim puts $4000 into an account paying compound interest at 8% a year, with the interest added at the end of each year. A rule of thumb says that money doubles in about 72 divided by the rate as a percentage. Take 1.089 = 1.999005 and 1.0810 = 2.158925, each correct to six decimal places. (a) Use the rule to estimate the number of years the money takes to double, and find the balance after that many years. (b) Find the first whole year in which the balance is at least $8000, and the balance then. Give amounts to the nearest cent.

  1. 1.Doubling means the balance reaches 2 × 4000 = $8000, so the multiplier 1.08n has to reach 2.

    the money put ininterest0246810doubling means 2 × 4000 = $8000the dashed line is $8000
    the money put ininterest0246810doubling means 2 × 4000 = $8000the dashed line is $8000
    The deposit has doubled when a column reaches the dashed line at $8000.

  2. 2.(a) The rule of thumb gives 72 ÷ 8 = 9, so the estimate is 9 years.

    the money put ininterest0246810the rule of thumb: 72 divided by 8 = 9so try 9 years
    the money put ininterest0246810the rule of thumb: 72 divided by 8 = 9so try 9 years
    (a) The rule of thumb gives 72 ÷ 8 = 9 years, so year 9 is where to look.

  3. 3.After 9 years the balance is 4000 × 1.999005 = $7996.02. The multiplier 1.999005 is just under 2, so the money has not quite doubled: the balance is 8000 − 7996.02 = $3.98 short.

    the money put ininterest02468$7996.02101.089= 1.999005, just under 24000 × 1.999005 = $7996.02, $3.98 short
    the money put ininterest02468$7996.02101.089= 1.999005, just under 24000 × 1.999005 = $7996.02, $3.98 short
    Year 9 is still under the line: 4000 × 1.999005 = $7996.02, short of $8000 by $3.98.

  4. 4.One more year multiplies by 1.08 again, and 1.0810 = 2.158925, which is past 2. The balance is 4000 × 2.158925 = $8635.70.

    the money put ininterest02468$7996.0210one more year: 1.0810= 2.158925that is past 2
    the money put ininterest02468$7996.0210one more year: 1.0810= 2.158925that is past 2
    One more year multiplies by 1.08 again, and 1.0810 = 2.158925 is past 2.

  5. 5.(b) So the balance first reaches $8000 in year 10, when it is $8635.70. The rule of 72 was one year short here, which is what it is for: a quick estimate that a proper calculation then settles.

    the money put ininterest02468$8635.70104000 × 2.158925 = $8635.70the balance first passes $8000 in year 10
    the money put ininterest02468$8635.70104000 × 2.158925 = $8635.70the balance first passes $8000 in year 10
    (b) Year 10 is the first column above the line, at $8635.70.

Answer: (a) an estimate of 9 years, after which the balance is $7996.02; (b) year 10, when the balance is $8635.70

Common mistakes

  • Taking $4000 at 8% to double in 100 ÷ 8 = 12.5 years, as though the interest were simple. Simple interest would indeed need 12.5 years; compound interest earns on its own interest, which is why 10 years is enough.
  • Reading the rule of 72 as an exact answer. Here it gives 9 years, and after 9 years the balance is $7996.02, which is below $8000. An estimate has to be tested before it is used.

More simple and compound interest problems, worked step by step →

Worked example: A Balance Read Backwards to the Rate, and Then Forwards to a Later Year

Question A deposit of $8000 was left in an account for 2 years. Interest was added at the end of each year at the same rate, and the balance was then $8820. (a) Find the annual rate of interest. (b) The deposit is left for a third year at the same rate. Find the balance then, and show that it is the first year in which the balance is over $9200.

  1. 1.Interest is added once a year at the same rate, so the deposit is multiplied by the same number twice. Call one year's multiplier m: then 8000 × m2 = 8820.

    $8000start?after 1 year$8820after 2 years× ?× ?the same multiplier twice takes 8000 to 8820what number is it?
    $8000start?after 1 year$8820after 2 years× ?× ?the same multiplier twice takes 8000 to 8820what number is it?
    Interest is added once a year at the same rate, so the same multiplier acts twice: 8000 × m2 = 8820.

  2. 2.Divide both sides by 8000: m2 = 88208000 = 1.1025.

    $8000start?after 1 year$8820after 2 years× 1.10258820 out of 8000 is 1.1025so the multiplier twice over is 1.1025
    $8000start?after 1 year$8820after 2 years× 1.10258820 out of 8000 is 1.1025so the multiplier twice over is 1.1025
    Divide both sides by 8000: m2 = 88208000 = 1.1025.

  3. 3.(a) Take the square root. Since 1.05 × 1.05 = 1.1025, the multiplier is m = 1.05, so the rate is 5% a year. Check: 8000 × 1.05 = $8400 after one year, and 8400 × 1.05 = $8820 after two.

    $8000start$8400after 1 year$8820after 2 years× 1.05× 1.051.05 × 1.05 = 1.1025the rate is 5% a year
    $8000start$8400after 1 year$8820after 2 years× 1.05× 1.051.05 × 1.05 = 1.1025the rate is 5% a year
    (a) 1.05 × 1.05 = 1.1025, so the rate is 5% a year, and $8000 passes through $8400.

  4. 4.A third year multiplies by 1.05 once more, so the multiplier is 1.053 = 1.1025 × 1.05 = 1.157625, and the balance is 8000 × 1.157625 = $9261.

    $8000start$8400after 1 year$8820after 2 years?after 3 years× 1.05× 1.05× 1.05a third year multiplies by 1.05 again1.053= 1.157625
    $8000start$8400after 1 year$8820after 2 years?after 3 years× 1.05× 1.05× 1.05a third year multiplies by 1.05 again1.053= 1.157625
    A third year multiplies by 1.05 once more: 1.053 = 1.157625.

  5. 5.(b) After 2 years the balance was $8820, which is below $9200, and after 3 years it is $9261, which is above it. So year 3 is the first year in which the balance is over $9200.

    $8000start$8400after 1 year$8820after 2 years$9261after 3 years× 1.05× 1.05× 1.058000 × 1.157625 = $92618820 was under 9200, and 9261 is over it
    $8000start$8400after 1 year$8820after 2 years$9261after 3 years× 1.05× 1.05× 1.058000 × 1.157625 = $92618820 was under 9200, and 9261 is over it
    (b) The balance is 8000 × 1.157625 = $9261, the first year over $9200.

Answer: (a) 5% a year; (b) $9261 after 3 years, which is the first year over $9200

Common mistakes

  • Taking the total growth of $820 over the two years as a rate of 8208000 = 10.25% and halving it to 5.125%. Compound rates are multiplied, not added, so the two years have to be undone by a square root and not by halving.
  • Guessing the square root of 1.1025 and leaving it. Any candidate multiplier can be tested in one line: only 1.05 takes $8000 to $8400 and then to $8820, so the check settles it.

More simple and compound interest problems, worked step by step →

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