Simple and Compound Interest · applications

Applications: Simple and Compound Interest

10 question types · Secondary 3 · each worked step by step with a figure that follows the steps

SAT · GCSE Higher

01

A Savings Account at Simple Interest: the Interest Year by Year and the Year a Target Is Reached

methodWork One Year's Interest Out from the Money Put In and Nothing Else, Add That Same Amount for Each Year, and Solve One Linear Inequality for the Number of Years a Target Needs

Priya puts $2500 into an account that pays simple interest at 4% a year. Simple interest is worked out on the money she put in, and never on the interest already paid. (a) Find the interest after 6 years, and her balance then. (b) Priya wants a balance of at least $3150. Find the first whole year in which she has it, and her balance in that year.

the money put ininterest012345674% of $2500 is $100simple interest adds $100 every year
Simple interest is worked out on the $2500 alone, so one year pays 4100 × 2500 = $100.
One year's interest is 4% of $2500, which is 4100 × 2500 = $100.
step 1 of 6

Simple interest pays the same amount every year, because it is always the same percentage of the amount put in at the start. So the first move is to work out one year's interest, and the interest after any number of years is that amount multiplied by the number of years. The balance then grows in equal steps, which makes the year a target is reached a single linear inequality to solve, and the answer has to be rounded up to a whole year because the interest is only paid at the end of each year.

  1. One year's interest is 4% of $2500, which is 4100 × 2500 = $100.
  2. Simple interest pays that same $100 every year, so after n years the interest is 100n dollars and the balance is 2500 + 100n dollars.
  3. (a) After 6 years the interest is 6 × 100 = $600, so the balance is 2500 + 600 = $3100.
  4. For a balance of at least $3150, solve 2500 + 100n ≥ 3150. Subtract 2500 from both sides to get 100n ≥ 650, and then divide both sides by 100 to get n ≥ 6.5.
  5. The interest is paid at the end of each year, so n has to be a whole number. The first whole number that is at least 6.5 is 7.
  6. (b) In year 7 the balance is 2500 + 7 × 100 = $3200. Check: after 6 years she had $3100, which is below the target, and after 7 years she has $3200, which is above it.

answer(a) $600 of interest, so the balance is $3100; (b) year 7, when the balance is $3200

techniqueSimple Interest

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Working the second year's interest out as 4% of $2600. That is what compound interest does. Simple interest is always a percentage of the $2500 that was put in at the start, so every year pays exactly $100.
  • Rounding n ≥ 6.5 down to 6 years. At 6 years the balance is only $3100, which is short of the target, so the number of whole years has to be rounded up to 7.
02

The Same Deposit at Simple and at Compound Interest, and How the Gap Between Them Widens

methodAdd the Same Interest Every Year for Simple and Multiply by the Same Number Every Year for Compound, Then Subtract the Two Balances at Each Date to See the Gap Grow

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.

simple interestcompound interest3 years5 years10 yearssimple: 6% of $2000 is $120 a yearbalance = 2000 + 120n
Aaron gains $120 every year, so his slices are all the same depth: 2000 + 120n dollars after n years.
Aaron's interest each year is 6% of $2000, which is 6100 × 2000 = $120, so his balance after n years is 2000 + 120n dollars.
step 1 of 6

The two accounts start together and are driven by two different rules. Simple interest adds a fixed amount each year, so Aaron's balance is a starting amount plus a multiple of that fixed amount. Compound interest multiplies by the same number each year, so Ben's balance is the starting amount times a power of that multiplier. Working both out at the same three dates and subtracting shows that the gap is small at first and then grows quickly, because Ben's interest is itself earning interest.

  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.
  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.
  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.
  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.
  5. After 10 years Aaron has 2000 + 1200 = $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.

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

techniqueSimple versus Compound Interest · Simple Interest · Compound Interest

examsSAT · GCSE Higher

Common pitfalls

  • 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.
03

A Bond That Pays Every Six Months, Read as a Yearly Rate

methodApply the Period Rate Once for Each Period, Then Read the Effective Annual Rate Off the One-Year Multiplier Instead of Multiplying the Period Rate by the Number of Periods

A savings bond pays 4% interest every six months, and the interest is added to the balance each time. Mr Tan puts in $6000 and reads the offer as 8% a year. (a) Find his balance after one year, and the effective annual rate of the bond. (b) Find his balance after two years, and how much more that is than an account paying 8% a year with the interest added once a year. Give every amount to the nearest cent.

$60000$62406 mo× 1.044% every six months: multiply by 1.046000 × 1.04 = $6240
The interest joins the balance at every tick. After six months 6000 × 1.04 = $6240.
Each half-year multiplies the balance by 1 + 4100 = 1.04, and there are two half-years in a year.
step 1 of 5

Interest added twice a year is two separate payments, and the second one is worked out on a balance that already holds the first. So the year's growth is not 4% + 4% but a multiplier of 1.04 applied twice. The number that multiplier comes to over one year is what the effective annual rate reports, and it is always a little above the rate a saver gets by adding the periods up. Two years is simply four of those half-year steps.

  1. Each half-year multiplies the balance by 1 + 4100 = 1.04, and there are two half-years in a year.
  2. After six months the balance is 6000 × 1.04 = $6240, and after another six months it is 6240 × 1.04 = $6489.60.
  3. (a) The one-year multiplier is 1.042 = 1.0816, so the year adds 8.16% and the effective annual rate is 8.16%, not 8%. The balance after one year is $6489.60.
  4. Two years is four half-years, so the multiplier is 1.044 = 1.08162 = 1.16985856, and the balance is 6000 × 1.16985856 = $7019.15.
  5. (b) An account paying 8% a year once a year would hold 6000 × 1.082 = 6000 × 1.1664 = $6998.40, so the bond is 7019.15 − 6998.40 = $20.75 better over the two years.

answer(a) $6489.60, and an effective annual rate of 8.16%; (b) $7019.15, which is $20.75 more than the $6998.40 that 8% a year added once a year would give

techniqueCompounding More Than Once a Year · Compound Interest

examsSAT · GCSE Higher

Common pitfalls

  • Reading 4% every six months as 8% a year. The second half-year's 4% is worked out on $6240 and not on $6000, so the year pays $489.60 rather than $480.
  • Finding a rate by taking the total interest and sharing it over the four half-years. A rate is a multiplier applied again and again, so the effective annual rate comes from 1.042 and never from total interest shared over time.
04

A Credit Card Charging 2 Per Cent a Month on an Unpaid Balance

methodCompound the Monthly Rate Twelve Times for the Year's Multiplier, and Take the Effective Annual Rate From That Multiplier Rather Than From Twelve Times the Monthly Rate

A credit card charges interest of 2% a month on whatever is owed at the end of the month. The advertisement calls this 24% a year. Sara owes $1000 and pays nothing back for a whole year. Take 1.0212 = 1.268242, correct to six decimal places. (a) Find what she owes after one year, to the nearest dollar, and how much of that is interest. (b) Find the effective annual rate, correct to one decimal place, and say how far the advertisement is out.

the $1000 owedinterest charged0$1020369122% a month: multiply by 1.02 each monthmonth 1: 1000 × 1.02 = $1020
The debt is multiplied by 1.02 at the end of every month: 1000 × 1.02 = $1020.
Each month multiplies the debt by 1 + 2100 = 1.02, so after twelve months the debt is 1000 × 1.0212 dollars.
step 1 of 5

A monthly charge on whatever is owed at the end of the month is compound interest with twelve periods in the year. Each month multiplies the debt by the same number, so twelve months multiply it by that number to the power twelve. The advertisement adds the twelve monthly rates instead, which ignores the interest charged on earlier interest, and the difference between the two figures is what the cardholder actually pays for.

  1. Each month multiplies the debt by 1 + 2100 = 1.02, so after twelve months the debt is 1000 × 1.0212 dollars.
  2. Building the multiplier up shows what is happening: 1.022 = 1.0404 and 1.023 = 1.0404 × 1.02 = 1.061208, already more than 1.06. The whole year is 1.0212 = 1.268242.
  3. (a) The debt is 1000 × 1.268242 = 1268.242, which is $1268 to the nearest dollar, and 1268 − 1000 = $268 of it is interest.
  4. The effective annual rate is what one year's multiplier says: 1.268242 − 1 = 0.268242, which is 26.8242%.
  5. (b) To one decimal place the effective annual rate is 26.8%. The advertised 24% is 12 × 2%, which leaves out the interest charged on earlier interest, so the true figure is 26.8 − 24 = 2.8 percentage points higher.

answer(a) $1268 owed, of which $268 is interest; (b) an effective annual rate of 26.8%, which is 2.8 percentage points more than the 24% advertised

techniqueCompounding More Than Once a Year · Compound Interest

examsSAT · GCSE Higher

Common pitfalls

  • Taking 12 × 2% = 24% and charging $240 of interest. That would be right only if the 2% were worked out on the original $1000 every month, and it is worked out on whatever is owed, which grows.
  • Writing the effective annual rate as 1.268242%. The multiplier 1.268242 is the whole debt, original amount and all; the rate is what is left after taking the original amount away, so it is 0.268242, or 26.8%.
05

A Car Loan Quoted as Flat Interest on the Whole Amount but Repaid Month by Month

methodWork the Quoted Interest Out on the Whole Loan for the Whole Term, Then Compare It With the Average Amount Actually Owed to Find the Rate That Is Really Being Charged

A dealer offers a car loan of $9000 over 2 years at a quoted flat rate of 6% a year. The interest is worked out on the whole $9000 for both years, added to the loan, and the total is repaid in 24 equal monthly payments. (a) Find the interest charged, the total repaid and the monthly payment. (b) An equal share of the loan is repaid each month, so the amount still owed falls in equal steps from $9000 at the start of the first month to $375 at the start of the last. Find the average amount owed over the 24 months, and the yearly rate that the interest really works out at.

The loan$9000flat 6% on the whole $9000one year $540, two years $1080
A flat rate charges 6% of the whole $9000 for each year: 2 × 540 = $1080.
One year's flat interest is 6% of the whole $9000, which is 6100 × 9000 = $540, and two years of it is 2 × 540 = $1080.
step 1 of 5

A flat rate charges interest on the whole loan for the whole term, as though none of it were ever repaid. In fact the borrower repays an equal share of the loan every month, so the amount owed falls steadily and is on average about half the loan. Comparing the interest charged with the amount the borrower really had the use of, over the time they had it, gives the rate that is actually being paid, and it is close to twice the quoted figure.

  1. One year's flat interest is 6% of the whole $9000, which is 6100 × 9000 = $540, and two years of it is 2 × 540 = $1080.
  2. (a) The total repaid is 9000 + 1080 = $10080, spread over 24 months, so each payment is 10080 ÷ 24 = $420.
  3. Of each payment, the loan itself is repaid at 9000 ÷ 24 = $375 a month, so the amount still owed falls by $375 each month: $9000, then $8625, then $8250, and so on down to $375 in the last month.
  4. Those 24 amounts go down in equal steps, so their average is halfway between the first and the last: 9000 + 3752 = $4687.50.
  5. (b) The borrower is charged $1080 for the use of an average of $4687.50 over 2 years, so the yearly rate is 10804687.50 × 2 = 10809375 = 0.1152, which is 11.52% a year. Check: that is almost twice the 6% quoted, because the quote charges for $9000 that the borrower does not have for most of the term.

answer(a) $1080 of interest, $10080 repaid in all, and $420 a month; (b) an average of $4687.50 owed, so the real rate is 11.52% a year

techniqueSimple Interest

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Believing that a flat 6% and a 6% rate on the amount still owed cost the same. They agree only if nothing is repaid until the end. Here half the loan has been repaid by the halfway point, so the same interest is being charged for the use of much less money.
  • Taking the average amount owed as 90002 = $4500. The amounts owed run from $9000 down to $375, not down to $0, because the last payment still has $375 of the loan to clear, so the average is $4687.50.
06

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

methodEstimate the Doubling Time by Dividing 72 by the Rate as a Percentage, Then Test the Estimate by Working the Balance Out in That Year and in the Next One

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.

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.
Doubling means the balance reaches 2 × 4000 = $8000, so the multiplier 1.08n has to reach 2.
step 1 of 5

Doubling means the balance reaches twice the amount put in, so the multiplier has to reach 2. The rule of thumb turns that into a quick division, and it is worth having because it needs no powers at all. Testing it is a matter of working the multiplier out in the estimated year and in the next one: if the multiplier is below 2 the money has not doubled yet, and the first year whose multiplier is at least 2 is the true answer.

  1. Doubling means the balance reaches 2 × 4000 = $8000, so the multiplier 1.08n has to reach 2.
  2. (a) The rule of thumb gives 72 ÷ 8 = 9, so the estimate is 9 years.
  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.
  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.
  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.

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

techniqueCompound Interest

examsSAT · GCSE Higher

Common pitfalls

  • 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.
07

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

methodDivide the Later Balance by the Earlier One for the Multiplier Over the Whole Period, Take the Root That Matches the Number of Years for One Year's Multiplier, and Check It by Multiplying Forward Again

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.

$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.
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.
step 1 of 5

Two years of compound interest multiply the deposit by the same number twice, so the growth over the two years is that number squared. Dividing the later balance by the earlier one recovers the squared multiplier, and its square root is one year's multiplier, which names the rate. Once the rate is known, going forward is ordinary compound interest, and the answer is checked by multiplying the balance back up year by year.

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

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

techniqueCompound Interest

examsSAT · GCSE Higher

Common pitfalls

  • 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.
08

Two One-Year Deposits, One Added Yearly and One Added Every Quarter

methodTurn Each Offer Into the Multiplier It Gives Over One Whole Year Before Comparing Them

A saver has $16000 to put away for one year. Bank A pays 10.3% a year, added once at the end of the year. Bank B pays 10% a year, added every three months, which is 2.5% each quarter. Take 1.0254 = 1.103813, correct to six decimal places. (a) Find the balance after one year at each bank. (b) Find the effective annual rate at each bank, correct to two decimal places, and say which offer is better and by how much.

Bank A: added once a year$16000start$17648after 1 year× 1.103Bank A adds 10.3% once: multiply by 1.10316000 × 1.103 = $17648
Bank A adds its interest at one moment, the end of the year: 16000 × 1.103 = $17648.
Bank A adds its interest once, so the multiplier for the year is 1 + 10.3100 = 1.103, and the balance is 16000 × 1.103 = $17648.
step 1 of 5

Two offers can only be compared once they are written in the same form, and the form that works is the multiplier each one gives over a single year. A rate added once a year is that rate and no more. A rate added four times a year is a quarter of the rate applied four times, and because each quarter earns on the quarters before it the year comes to a little more than the advertised figure. The effective annual rate is exactly that one-year multiplier, reported as a percentage.

  1. Bank A adds its interest once, so the multiplier for the year is 1 + 10.3100 = 1.103, and the balance is 16000 × 1.103 = $17648.
  2. Bank B divides its rate over four quarters: 10% ÷ 4 = 2.5% each quarter, a multiplier of 1.025 applied four times.
  3. (a) Building it up, 1.0252 = 1.050625, and squaring that gives 1.0254 = 1.103813. The balance is 16000 × 1.103813 = $17661.01.
  4. The effective annual rate is what each one-year multiplier says. Bank A: 1.103 − 1 = 0.103, so 10.30%. Bank B: 1.103813 − 1 = 0.103813, so 10.38% to two decimal places.
  5. (b) Bank B is the better offer even though its advertised rate is lower, by 17661.01 − 17648 = $13.01 on $16000 over the year.

answer(a) $17648 at Bank A and $17661.01 at Bank B; (b) effective annual rates of 10.30% and 10.38%, so Bank B is better by $13.01

techniqueCompounding More Than Once a Year · Compound Interest

examsSAT · GCSE Higher

Common pitfalls

  • Choosing Bank A because 10.3% is more than 10%. An advertised rate says nothing until the number of times a year the interest is added is known, and here the four quarterly payments lift Bank B's year to 10.38%.
  • Working Bank B out as 16000 × 1.1 ÷ 4 × 4 or as 16000 × 1.025. The quarterly multiplier has to be applied once for every quarter in the term, so one year is 1.0254 and not 1.025.
09

A Charge Taken Once at the Start Against the Same Money Charged Every Year

methodTake the One-Off Charge Off the Amount Invested and Compound the Full Rate on What Is Left, Compound the Reduced Rate on the Whole Amount for the Yearly Charge, and Compare the Two Balances Year by Year

Mrs Rahman has $10000 to invest, and both funds she is offered earn 6% a year before charges. Fund A takes an entry fee of 3% of the money at the start and takes nothing after that. Fund B takes no entry fee but keeps 1% a year, so the saver earns 5% a year. (a) Find the value of each fund after 3 years, and say which is worth more. (b) Find the value of each after 4 years, and say from which year on Fund A is worth more. Give amounts to the nearest cent.

the entry feeFund AFund BAByear 0year 1year 2year 3year 43% of 10000 is a $300 entry feeFund A invests $9700 and grows at 6%
Fund A’s entry fee is the top block: 3% of $10000 is $300, so only $9700 is invested.
Fund A's entry fee is 3% of $10000, which is $300, so only $9700 is invested and it grows at 6% a year. Its value after n years is 9700 × 1.06n dollars.
step 1 of 6

A charge taken once makes the amount invested smaller but leaves the growth rate alone, so it is a fixed multiplier in front of a full-rate compounding. A charge taken every year leaves the amount invested alone but lowers the growth rate, so it is a smaller multiplier applied once for each year. The one-off charge therefore costs the same proportion forever while the yearly charge costs more and more, and comparing the two at successive years shows where they cross.

  1. Fund A's entry fee is 3% of $10000, which is $300, so only $9700 is invested and it grows at 6% a year. Its value after n years is 9700 × 1.06n dollars.
  2. Fund B charges nothing at the start, but the saver's money grows at 5% a year on the whole $10000. Its value after n years is 10000 × 1.05n dollars.
  3. For three years, 1.062 = 1.1236 and 1.063 = 1.1236 × 1.06 = 1.191016, so Fund A is worth 9700 × 1.191016 = $11552.86.
  4. (a) And 1.052 = 1.1025 and 1.053 = 1.1025 × 1.05 = 1.157625, so Fund B is worth 10000 × 1.157625 = $11576.25. After 3 years Fund B is ahead by 11576.25 − 11552.86 = $23.39.
  5. For a fourth year, 1.064 = 1.12362 = 1.26247696 and 1.054 = 1.10252 = 1.21550625, so Fund A is worth 9700 × 1.26247696 = $12246.03 and Fund B is worth 10000 × 1.21550625 = $12155.06.
  6. (b) After 4 years Fund A is ahead by 12246.03 − 12155.06 = $90.97, and it stays ahead, because the entry fee is paid once while the 1% is taken from a balance that keeps growing. So Fund A is worth more from year 4 on.

answer(a) $11552.86 for Fund A and $11576.25 for Fund B, so Fund B is ahead by $23.39; (b) $12246.03 and $12155.06, so Fund A is ahead by $90.97 and is worth more from year 4 on

techniqueCompound Interest

examsSAT · GCSE Higher

Common pitfalls

  • Deciding on the charges alone, and calling a 3% fee worse than a 1% yearly charge because 3 is more than 1. The two charges are of different kinds: one is paid once and one is paid every year, so only the balances after a stated number of years can be compared.
  • Working Fund A out as 10000 × 1.06n and taking the 3% off at the end. The fee is taken at the start, so it never earns anything: 9700 is what grows, and 9700 × 1.064 is not 0.97 × 10000 × 1.064 plus any later saving.
10

A Season Ticket Whose Price Rises by the Same Percentage Every Year

methodMultiply the Price by the Same Number Once for Each Year Rather Than Adding the Yearly Percentages Together, and Step Forward Year by Year to Find When a Level Is Passed

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.

the price todaythe rises since0$15451234567a rise of 3% multiplies by 1.031500 × 1.03 = $1545
A rise of 3% multiplies the price by 1.03: after one year it is $1545.
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.
step 1 of 6

A price that rises by the same percentage every year is compound growth, with the price in place of a balance. Each rise is worked out on a price that has already risen, so the yearly percentages are multiplied rather than added, and the answer comes out above what adding them would suggest. Finding the year a level is passed is then a matter of stepping the multiplier forward until the price crosses that level, and naming the first year that does.

  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.
  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.
  3. 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.
  5. Step on: after 6 years the price is 1500 × 1.194052 = $1791.08, which is still under $1800.
  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.

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

techniqueCompound Interest · Simple versus Compound Interest

examsSAT · GCSE Higher

Common pitfalls

  • 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.
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