Two accounts
Two accounts each open with a principal of $1000 at 10% a year. One pays simple interest: every year it adds 10% of the principal. The other pays compound interest: every year it adds 10% of the balance, earlier interest included.
In the first year the two cannot differ. Both balances are still the $1000 principal, so both pay 10% of $1000, which is $100, and both end the year at $1100.
The simple account: equal steps
The simple account gains the same $100 every year: $1100, then $1200, then $1300. After n years its balance is 1000 + 100n dollars. It grows by the same amount at every step, which is linear growth, so on a graph its balances lie on a straight line.
The compound account: growing steps
The compound account multiplies by 1.1 every year: $1100, then $1210, then $1331. Its gains are $100, then $110, then $121, bigger every year. After n years its balance is dollars. It grows by the same factor at every step, which is exponential growth, so on a graph its balances lie on a curve that bends upward.
The two balances are equal at the start and after year 1. After that the compound account is ahead, by $10 after year 2 and by $31 after year 3.
The gold curve is the compound balance, , and the white line is the simple balance, 1000 + 100n, after n years. They start together and are still together after year 1. After 10 years the compound account holds $2593.74 and the simple account $2000.
Why the gap keeps growing
The whole gap is interest earned on interest. In year 2 the compound account earns 10% of year 1's $100 of interest, which is $10, and the simple account earns nothing on its interest. So after year 2 the compound account is $10 ahead.
In year 3 the compound account has been paid $210 of interest so far, and it earns 10% of that, $21, on top of the $100 both accounts earn. The gap grows by $21, to $31. In year 4 the interest so far is $331, the extra is $33.10, and the gap grows to $64.10.
The interest already paid keeps growing, so the extra it earns keeps growing too, and the gap widens faster every year. For every whole number of years after the first, the compound balance is ahead.
Over the long run
After 10 years at 10% the gap is $593.74. After 20 years the simple account holds 1000 + 100 × 20 = 3000 dollars and the compound account holds dollars, to the nearest cent, more than twice as much. After 30 years it is $4000 against $17,449.40.
An account that adds the same amount every year falls further and further behind one that multiplies by the same factor every year, because the amount added stays fixed while the amount the factor adds keeps growing.
simple interest adds 8% of 100 = 8 every year, a straight line; compound interest adds 8% of a growing balance, so it curves away — the gap this year is 2
Find the first year the balance has doubled
Here $100 is saved at 8% a year. After 3 years the compound balance is dollars and the simple balance is 100 + 3 × 8 = 124 dollars, a gap of $1.97, which the figure rounds to whole dollars. Drag the year to the first year the compound balance has doubled, to at least $200. It is year 10, when the compound balance is $215.89 and the simple balance only $180.
The usual mistakes
Expecting equal rates to pay the same. Both accounts pay 10%, but from year 2 on the compound account takes 10% of a bigger balance.
Giving a whole year of interest as the gap. After 2 years the compound account is $10 ahead, not $100: the gap is only the interest on year 1's interest.
Using the simple formula for a compound account. 10 years at 10% is not 1000 + 10 × 100 = 2000 dollars on a compound account; it is dollars.
Worked example: The Same Deposit at Simple and at Compound Interest, and How the Gap Between Them Widens
Question Aaron and Ben each put $2000 into an account for ten years. Aaron's account pays simple interest at 6% a year. Ben's account pays compound interest at 6% a year, so his interest is added to his balance and earns interest itself. Take 1.065 = 1.338226 and 1.0610 = 1.790848, each correct to six decimal places. (a) Find how much more Ben has than Aaron after 3 years and after 5 years. (b) Find how much more Ben has after 10 years. Give every amount to the nearest cent.
1.Aaron's interest each year is 6% of $2000, which is 6100 × 2000 = $120, so his balance after n years is 2000 + 120n dollars.
Aaron gains $120 every year, so his slices are all the same depth: 2000 + 120n dollars after n years. 2.Ben's balance is multiplied by 1 + 6100 = 1.06 each year, so after n years it is 2000 × 1.06n dollars. Building the multiplier up: 1.062 = 1.1236 and 1.063 = 1.1236 × 1.06 = 1.191016.
Ben’s balance is multiplied by 1.06 every year, so his slices widen: 2000 × 1.06n dollars after n years. 3.After 3 years Aaron has 2000 + 360 = $2360 and Ben has 2000 × 1.191016 = $2382.03, so Ben is ahead by 2382.03 − 2360 = $22.03.
After 3 years Aaron has $2360 and Ben has $2382.03, a gap of $22.03. 4.(a) After 5 years Aaron has 2000 + 600 = $2600 and Ben has 2000 × 1.338226 = $2676.45, so Ben is ahead by 2676.45 − 2600 = $76.45.
(a) After 5 years Aaron has $2600 and Ben has $2676.45, a gap of $76.45. 5.After 10 years Aaron has 2000 + 1200 = $3200 and Ben has 2000 × 1.790848 = $3581.70.
After 10 years Aaron has $3200 and Ben has 2000 × 1.790848 = $3581.70. 6.(b) Ben is ahead by 3581.70 − 3200 = $381.70. The gap was $22.03, then $76.45, then $381.70: it widens because each year Ben earns interest on interest that Aaron never earns.
(b) The gap after 10 years is $381.70, more than seventeen times the gap after 3 years.
Answer: (a) Ben is ahead by $22.03 after 3 years and by $76.45 after 5 years; (b) after 10 years Aaron has $3200 and Ben has $3581.70, so Ben is ahead by $381.70
Common mistakes
- Working Ben's ten years out as 2000 × 1.6, from 10 × 6% = 60%. Percentages for successive years are multiplied, not added, so ten years of 6% is 1.0610 = 1.790848 and not 1.6.
- Reading the small gap after 3 years as proof that the two accounts are much the same. The gap is a difference of a growing amount and a fixed amount: after 10 years it is $381.70, which is more than seventeen times the gap after 3 years.
More simple and compound interest problems, worked step by step →
Worked example: A Season Ticket Whose Price Rises by the Same Percentage Every Year
Question A rail season ticket costs $1500 today, and its price rises by 3% every year. Take 1.036 = 1.194052 and 1.037 = 1.229874, each correct to six decimal places. (a) Find the price 5 years from now, and how much more that is than one rise of 15% would give. (b) Find the first year from now in which the price is over $1800, and the price then. Give amounts to the nearest cent.
1.A rise of 3% multiplies the price by 1 + 3100 = 1.03, and each year multiplies again, so after n years the price is 1500 × 1.03n dollars.
A rise of 3% multiplies the price by 1.03: after one year it is $1545. 2.Building the multiplier up: 1.032 = 1.0609, 1.033 = 1.092727, 1.034 = 1.12550881 and 1.035 = 1.15927407 to eight decimal places.
Each year multiplies again, so after n years the price is 1500 × 1.03n dollars, and 1.035 = 1.15927407. 3.After 5 years the price is 1500 × 1.15927407 = $1738.91.
After 5 years the price is 1500 × 1.15927407 = $1738.91. 4.(a) Five rises of 3% added together would be 15%, and 1500 × 1.15 = $1725. The true price is 1738.91 − 1725 = $13.91 more, because each rise is worked out on a price that has already risen.
(a) The dashed line is $1725, what one rise of 15% would give. The true price stands $13.91 above it. 5.Step on: after 6 years the price is 1500 × 1.194052 = $1791.08, which is still under $1800.
The dashed line is now $1800. After 6 years the price is $1791.08, still below it. 6.(b) After 7 years the price is 1500 × 1.229874 = $1844.81, which is over $1800. So year 7 is the first year in which the ticket costs more than $1800.
(b) Year 7 is the first column above the line, at $1844.81.
Answer: (a) $1738.91, which is $13.91 more than the $1725 that a single rise of 15% would give; (b) year 7, when the price is $1844.81
Common mistakes
- Turning five rises of 3% into one rise of 15%. The percentages are of different amounts: the fifth rise is 3% of a price that has already risen four times, so the true multiplier is 1.035 = 1.15927407 and not 1.15.
- Stopping at year 6 because $1791.08 rounds to $1800 to the nearest hundred dollars. The question asks for a price over $1800, and $1791.08 is under it, so the first year that qualifies is year 7.
More simple and compound interest problems, worked step by step →