Financial Mathematics · applications

Applications: Financial Mathematics

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

01

A Prize Offered as Cash Today or a Larger Sum in Two Years

methodBring Both Offers to One Date Before Comparing Them, Carrying the Sum Available Today Forward by the Two-Year Multiplier and Bringing the Later Sum Back by Dividing by the Same Multiplier

A competition offers its winner $10000 today, or $11466 paid in two years' time. Money can be put into an account paying 5% a year, with the interest added once a year, and 1.052 = 1.1025. (a) Compare the two offers at the two-year date, and say which is worth more and by how much. (b) Compare the two offers at today's date instead, and check that the two answers agree.

the cash offer, carried forward$10000today1 year2 yearsthe later payment, brought backtoday1 year$114662 years$10000 today, or $11466 in two yearsboth have to be read at one date
The two offers fall on different dates, so neither can be read against the other until both are valued on one date.
Choose a date to compare them at. The two-year date is the natural one, because that is when the later payment arrives.
step 1 of 5

Two sums of money can only be compared when they fall on the same date, because money held today can be put to work and money promised later cannot. Choosing the date is the whole of the modeling step. Carrying a sum forward means multiplying by the growth multiplier, and bringing a sum back to today means dividing by that same multiplier. Either date gives the same decision, and the two gaps differ by exactly the multiplier, which is the check.

  1. Choose a date to compare them at. The two-year date is the natural one, because that is when the later payment arrives.
  2. Carry the cash offer forward two years. Interest is added once a year, so the multiplier for two years is 1.05 × 1.05 = 1.1025, and 10000 × 1.1025 = $11025.
  3. (a) At the two-year date the offers are worth $11025 and $11466, so the later payment is worth 11466 − 11025 = $441 more.
  4. Now compare them at today's date. A sum x today grows to 1.1025x in two years, so the sum that grows into $11466 is x = 11466 ÷ 1.1025 = $10400.
  5. (b) At today's date the offers are worth $10000 and $10400, so the later payment is worth $400 more. Check: carrying that gap forward gives 400 × 1.1025 = $441, which is the gap found in part (a), so the two dates agree.

answer(a) The later payment, because $10000 today is worth $11025 in two years and that is $441 short of $11466; (b) the later payment is worth $10400 today, which is $400 more than the cash offer

techniqueThe Time Value of Money

Common pitfalls

  • Comparing $10000 with $11466 as they stand and calling the later offer $1466 better. The two sums fall on different dates, and a dollar today is worth more than a dollar in two years because it can earn interest in between. Nothing may be compared until both sums are valued at the same date.
  • Discounting with 2 × 5% = 10%, so dividing by 1.1. The second year's interest is worked out on the first year's interest as well, so the two-year multiplier is 1.05 × 1.05 = 1.1025. Dividing 11466 by 1.1 gives $10423.64, which is $23.64 too high.
02

A Year of Free Credit at a Shop Against a Discount for Paying Today

methodPrice Each Offer at the Same Date by Dividing the Sum Due in a Year by the Savings Multiplier to Value It Today, and Multiplying the Cash Price by That Multiplier to Value It in a Year

A shop sells a sofa for $2600, with nothing to pay for one year. The same sofa costs 6% less if it is paid for today. Raj's savings account pays 4% a year. (a) Find the cash price, and compare the two offers in today's money. (b) Compare the two offers at the one-year date, and check that the two answers agree.

List price$2444$26006% of 2600 = $156cash price 2600 − 156 = $2444
The discount is 6100 × 2600 = $156, so the cash price is $2444.
The discount is 6% of $2600, which is 6100 × 2600 = $156, so the cash price is 2600 − 156 = $2444.
step 1 of 5

A year of credit is never free. The shop is asking the buyer to give up the discount in exchange for keeping the money in the bank for a year, so the question is whether a year of interest is worth more than the discount. Value the credit offer today by asking what sum set aside now would grow into the $2600 due in a year, and then compare that with the cash price. Valuing both offers again at the one-year date must give the same decision.

  1. The discount is 6% of $2600, which is 6100 × 2600 = $156, so the cash price is 2600 − 156 = $2444.
  2. To meet the credit offer, Raj sets a sum aside today and lets it grow. A sum x today grows to 1.04x in a year, so x = 2600 ÷ 1.04 = $2500.
  3. (a) In today's money the credit offer costs $2500 and the cash offer costs $2444, so paying today is 2500 − 2444 = $56 cheaper.
  4. At the one-year date: had Raj kept the $2444 in the account instead of spending it, it would have grown to 2444 × 1.04 = $2541.76, and the credit offer needs $2600 on that date.
  5. (b) Paying today is 2600 − 2541.76 = $58.24 better at the one-year date. Check: the gap of $56 today carried forward is 56 × 1.04 = $58.24, so the two dates agree.

answer(a) The cash price is $2444, and the credit offer costs $2500 in today's money, so paying today is $56 cheaper; (b) at the one-year date the cash buyer still holds $2541.76 against the $2600 due, so paying today is $58.24 better

techniqueThe Time Value of Money

Common pitfalls

  • Taking the credit because it is advertised as costing nothing. The year of credit is paid for by giving up the $156 discount, and a year of interest on the $2500 set aside is only $100 at 4%, so the credit costs $56 in today's money.
  • Comparing the cash price of $2444 with the $2600 due in a year and calling the saving $156. Those two sums fall a year apart. The $2600 is met by setting aside only $2500 today, so the saving in today's money is $56.
03

A Pay Rise Smaller Than the Rise in Prices

methodDivide the New Salary by the Price Multiplier Rather Than Subtracting One Percentage From the Other, and Read the Real Change by Comparing the Result With the Salary Before the Rise

Mei earns $41600 a year. Her employer offers a rise of 3%. Over the same year prices rise by 4%. (a) Find her new salary, and what that salary is worth in the money of today. (b) Find the rise that would have left her exactly as well off, and say how far the offer falls short of it.

dollarstoday's moneynow4160041600a 3% risea 4% risea 3% rise multiplies by 1.0341600 × 1.03 = $42848
A rise of 3% multiplies the salary by 1.03: 41600 × 1.03 = $42848.
A rise of 3% multiplies the salary by 1.03, so the new salary is 41600 × 1.03 = $42848.
step 1 of 5

A salary is only worth what it buys. When prices rise, every dollar of next year's money buys less than a dollar of today's money, so the new salary has to be divided by the price multiplier before it can be set beside the old one. Dividing, not subtracting, is the whole of the method: subtracting the two percentages is a shortcut that is close but wrong, and the difference is worth stating.

  1. A rise of 3% multiplies the salary by 1.03, so the new salary is 41600 × 1.03 = $42848.
  2. Prices are 1.04 times what they were, so a dollar of next year's money buys what 1 ÷ 1.04 of a dollar buys today. To value the new salary in today's money, divide it by 1.04.
  3. (a) 42848 ÷ 1.04 = $41200. The new salary buys what $41200 buys today, so its real value has fallen by 41600 − 41200 = $400.
  4. A rise that exactly keeps pace with prices multiplies the salary by 1.04, giving 41600 × 1.04 = $43264.
  5. (b) The offer is 43264 − 42848 = $416 below that. Check: $416 of next year's money is 416 ÷ 1.04 = $400 of today's money, which is the fall found in part (a).

answer(a) Her new salary is $42848, which is worth $41200 in today's money, a real fall of $400; (b) a rise of 4%, or $43264, would have kept her as well off, so the offer is $416 short

techniqueReal Value After Inflation

Common pitfalls

  • Subtracting the percentages, 3% − 4% = −1%, and calling the real loss 1% of $41600, or $416, in today's money. Growth is divided by prices, never reduced by them: the real multiplier is 1.03 ÷ 1.04 = 0.99038 to five decimal places, so the fall in today's money is $400. The $416 is a sum in next year's money.
  • Undoing a rise of 4% in prices by taking 4% off, so multiplying by 0.96. That gives 42848 × 0.96 = $41134.08, which is $65.92 too low. A rise of 4% is undone by dividing by 1.04, not by multiplying by 0.96.
04

A Pension Fixed in Dollars, Measured Against a Price Index

methodMultiply by the Ratio of the Two Index Numbers, the Old Index Over the New to Value Today's Money at the Old Date and the New Over the Old to Find the Payment That Would Keep Pace

Mr Lim retired in 2010 on a pension of $2400 a month, and the amount has never changed since. The consumer price index, which stood at 100 in 2010, stands at 150 today. (a) Find what the pension is worth today in 2010 money, and how much of its buying power has gone. (b) Find the pension that would buy today what $2400 bought in 2010, and say how far the real pension falls short of it.

The pension$2400a basket at $100 in 2010 costs $150 todayprices are 1.5 times what they were
The index prices one basket at two dates: $100 in 2010 and $150 today, so prices are 1.5 times what they were.
The index says that what cost $100 in 2010 costs $150 today, so today's prices are 150 ÷ 100 = 1.5 times the 2010 prices.
step 1 of 5

A price index is a basket of goods priced at two dates. Here a basket that cost $100 in 2010 costs $150 today, so prices are 1.5 times what they were. A payment fixed in dollars therefore buys less every year, and the loss is found by multiplying by the ratio of the index numbers. Which way up the ratio goes is decided by the date the answer is wanted in: the old index on top to value today's money at the old date, the new index on top to find the payment that keeps pace.

  1. The index says that what cost $100 in 2010 costs $150 today, so today's prices are 150 ÷ 100 = 1.5 times the 2010 prices.
  2. To value today's dollars in 2010 money, multiply by 100150, which is 23.
  3. (a) 2400 × 23 = $1600. The pension buys today what $1600 bought in 2010, so 2400 − 1600 = $800 a month of buying power has gone, which is a third of it.
  4. To keep pace with prices, the pension would have to rise in the same ratio as the index, so multiply by 150100 = 1.5.
  5. (b) 2400 × 1.5 = $3600 a month. The pension is 3600 − 2400 = $1200 a month short of that. Check: $1200 of today's money is 1200 × 23 = $800 of 2010 money, which is the monthly loss found in part (a).

answer(a) The pension is worth $1600 a month in 2010 money, so $800 a month of buying power has gone, a third of it; (b) it would have to be $3600 a month, so the pension falls $1200 a month short

techniqueReal Value After Inflation

Common pitfalls

  • Reading a rise of 50% in prices as a loss of 50% in buying power, so $1200 a month. The pension still buys $1600 of 2010 goods, so a third of its buying power has gone, not a half. A rise of 50% in prices is undone by a fall of a third, because 1.5 × 23 = 1.
  • Taking $2400 down by the index itself, as 2400 − 150 = $2250. An index number is not a sum of money. It is only the ratio 150100 that carries any meaning, and the pension is divided by that ratio.
05

A Loan Repaid in Three Equal Payments, Read Line by Line

methodCharge Each Year's Interest on the Amount Still Owed at the Start of That Year, Take the Rest of the Payment Off the Balance, and Carry the New Balance Into the Next Line

Nadia borrows $3310 at 10% a year and repays it in three equal yearly payments of $1331. Interest is charged at the end of each year on the amount still owed at the start of that year, and whatever is left of the payment comes off the loan. (a) Find the interest, the amount repaid and the balance left for the first payment and for the second. (b) Do the same for the third payment, and find the total interest paid over the whole loan.

YearPaidInterestRepaidOwed033101133133110002310% of 3310 = $331 of interest1331 − 331 = $1000 comes off the loan
Year 1 opens with $3310 owed, so the interest is $331 and only $1000 of the payment comes off the loan.
Year 1 opens with $3310 owed. The interest is 10% of $3310, which is $331, so of the $1331 payment only 1331 − 331 = $1000 comes off the loan.
step 1 of 5

Every payment is the same size, but it is not split the same way. The interest is worked out on what is still owed, which falls with every payment, so the interest part of each payment shrinks and the part that reduces the loan grows. The table is built one line at a time: interest first, then the amount repaid, then the new balance, which becomes the opening balance of the next line. A loan that is properly worked out ends at a balance of exactly nothing.

  1. Year 1 opens with $3310 owed. The interest is 10% of $3310, which is $331, so of the $1331 payment only 1331 − 331 = $1000 comes off the loan.
  2. (a) The balance is 3310 − 1000 = $2310. Year 2 opens with that, so its interest is 10% of $2310, which is $231, the amount repaid is 1331 − 231 = $1100, and the balance is 2310 − 1100 = $1210.
  3. Year 3 opens with $1210 owed. The interest is $121, so the amount repaid is 1331 − 121 = $1210, and the balance is 1210 − 1210 = 0: the loan is cleared exactly.
  4. (b) The interest paid is 331 + 231 + 121 = $683. Check: the three payments come to 3 × 1331 = $3993, and 3993 − 3310 = $683, which is the same total.
  5. Read down the table: the interest falls $331, $231, $121 while the amount repaid rises $1000, $1100, $1210. The first payment is mostly interest and the last is almost all loan.

answer(a) Year 1 pays $331 interest and repays $1000, leaving $2310; year 2 pays $231 interest and repays $1100, leaving $1210; (b) year 3 pays $121 interest and repays $1210, leaving nothing owed, and the interest over the loan is $683

techniqueAmortizing a Loan

Common pitfalls

  • Splitting the loan into three equal parts of $1103.33 and charging 10% on each. The interest is charged on the balance still owed, and that balance falls as the loan is repaid, so the interest falls from $331 to $231 to $121 while the payment stays at $1331.
  • Charging 10% of the original $3310 for each of the three years, giving 3 × 331 = $993 of interest. That would be the charge if nothing were repaid until the end. Because the balance falls each year, the true total is $683.
06

The Same Mortgage Over Twenty Years and Over Twenty-Five

methodMultiply Each Monthly Payment by the Number of Payments to Find the Cash Handed Over, Subtract the Amount Borrowed to Leave the Interest, and Set the Smaller Payment Against the Larger Total

A bank lends $240000 at 5% a year, with the interest charged each month on the amount still owed. The bank quotes a monthly payment of $1583.89 over 20 years, which is 240 payments, and $1403.02 over 25 years, which is 300 payments. (a) Find the total paid and the total interest for the 20-year loan. (b) Do the same for the 25-year loan, and say how much more interest the longer loan costs and how much smaller its monthly payment is.

what was borrowedinterest, 20 yearsinterest, 25 years20 years25 years20 years is 240 payments of $1583.89240 × 1583.89 = $380133.60
Every payment is the same, so the cash handed over is 240 × 1583.89 = $380133.60.
Over 20 years there are 12 × 20 = 240 payments of $1583.89, so the cash handed over is 240 × 1583.89 = $380133.60.
step 1 of 5

A longer loan is easier to pay each month and dearer in the end, and the second half of that sentence is the one borrowers miss. Every payment is the same size, so the cash handed over is simply the payment multiplied by the number of payments. Whatever is left after the amount borrowed is taken out is interest. Setting the two loans side by side then turns a choice about a monthly payment into a choice about a total.

  1. Over 20 years there are 12 × 20 = 240 payments of $1583.89, so the cash handed over is 240 × 1583.89 = $380133.60.
  2. (a) Of that, $240000 is the amount borrowed, so the interest is 380133.60 − 240000 = $140133.60.
  3. Over 25 years there are 12 × 25 = 300 payments of $1403.02, so the cash handed over is 300 × 1403.02 = $420906, and the interest is 420906 − 240000 = $180906.
  4. (b) The longer loan costs 180906 − 140133.60 = $40772.40 more in interest, and its monthly payment is 1583.89 − 1403.02 = $180.87 smaller.
  5. Check by following the borrower: for the first 240 months the longer loan saves $180.87 a month, which is 240 × 180.87 = $43408.80, and then it runs on for 60 months more at 60 × 1403.02 = $84181.20. The difference is 84181.20 − 43408.80 = $40772.40, the same extra interest.

answer(a) $380133.60 is paid in all, of which $140133.60 is interest; (b) $420906 is paid in all, of which $180906 is interest, so the longer loan costs $40772.40 more in interest for a payment $180.87 a month smaller

techniqueAmortizing a Loan · The Time Value of Money

Common pitfalls

  • Choosing the longer loan because it costs less. It costs $180.87 less each month and $40772.40 more in all, so the two sentences describe the same loan and the choice has to be made with both in view.
  • Working the interest out as 5% of the whole $240000 for every year, giving 25 × 12000 = $300000. The interest is charged on the amount still owed, which falls with every payment, so the true total over 25 years is $180906.
07

A Prize Paid as Twenty Yearly Payments Against a Single Cash Sum

methodValue the Run of Equal Payments by Multiplying One Payment by the Annuity Factor for That Rate and That Number of Years, and Compare the Result With the Single Sum Offered Today

A lottery pays its prize as $50000 at the end of each year for 20 years, or as a single payment of $650000 today. Money is worth 4% a year. At that rate $1 paid at the end of each year for 20 years is worth $13.590326 today, and 1.0420 = 2.191123. (a) Find the value today of the twenty payments, and say which offer is worth more and by how much. (b) The prize is advertised as $1000000. Find how much of that is lost to the waiting.

15101520each of the twenty payments, valued todaythe first is worth 50000 divided by 1.04= $48076.92, and the last only $22819.35
Each column is one payment: the gold part is what it is worth today and the faint part is what the waiting takes. The dashed line is the $50000 written on every payment.
The first payment arrives in a year and is worth 50000 ÷ 1.04 = $48076.92 today; the twentieth arrives in 20 years and is worth 50000 ÷ 2.191123 = $22819.35 today. Every payment in between lies between those two.
step 1 of 5

Each of the twenty payments falls on a different date, so each is worth less today than the $50000 written on it, and the further off it is the less it is worth. Adding twenty separate divisions is the honest way and a slow one; the annuity factor is that sum already done for one dollar a year, so one multiplication does the work. Only once both offers are valued at today's date can they be compared.

  1. The first payment arrives in a year and is worth 50000 ÷ 1.04 = $48076.92 today; the twentieth arrives in 20 years and is worth 50000 ÷ 2.191123 = $22819.35 today. Every payment in between lies between those two.
  2. The annuity factor adds all twenty of those values together for one dollar a year, so the twenty payments are worth 50000 × 13.590326 = $679516.30 today.
  3. (a) The single payment is $650000, so the yearly payments are worth 679516.30 − 650000 = $29516.30 more today.
  4. The cash the yearly payments actually hand over is 20 × 50000 = $1000000, which is the advertised prize.
  5. (b) Of that advertised million, 1000000 − 679516.30 = $320483.70 is lost to the waiting. Check: the last payment alone keeps 22819.3550000 of its face value, which is under half, and the twenty payments together keep about 68% of theirs.

answer(a) The twenty payments are worth $679516.30 today, so they beat the single payment of $650000 by $29516.30; (b) $320483.70 of the advertised million is lost to the waiting

techniqueThe Value of an Annuity · The Time Value of Money

Common pitfalls

  • Comparing the advertised $1000000 with the $650000 and taking the yearly payments by a wide margin. The million is handed over across twenty years, and the last of it waits twenty years to arrive, so it is worth $679516.30 today, not a million.
  • Multiplying $50000 by twenty and then dividing once by 2.191123, giving $456386.98. That would be right only if all twenty payments arrived together in year 20. They arrive one a year, and the early ones are discounted far less, so the true value is higher.
08

A Scholarship Fund Paying Out Only What It Earns

methodRead the Yearly Payment as the Interest the Fund Earns, so That the Fund Needed Is the Payment Divided by the Rate and the Fund Itself Is Never Touched

A school holds a scholarship fund of $250000, invested at 4% a year. Each year the interest is paid out as scholarships and the fund itself is left untouched, so the payments can go on year after year without end. (a) Find the amount paid out each year. (b) The school would like to pay out $15000 a year on the same terms. Find the fund that would be needed, and show what happens in the first year if $15000 is paid out of the $250000 instead.

the fund stays at $250000, year after year$10000year 1$10000year 2year 3year 4and on4% of 250000 = $10000 in a yearpay out just that and the fund is whole
The fund earns 4100 × 250000 = $10000 in a year, and paying out exactly that leaves the fund itself untouched.
The fund earns 4% of $250000 in a year, which is 4100 × 250000 = $10000.
step 1 of 5

A fund that pays out only its interest is the same size at the start of every year, so it can pay the same amount without end. That one observation turns the question into a single equation: the yearly payment is the rate multiplied by the fund. Read forwards it gives the payment from the fund; read backwards it gives the fund from the payment. Paying out more than the interest breaks the condition, and the fund then falls every year.

  1. The fund earns 4% of $250000 in a year, which is 4100 × 250000 = $10000.
  2. (a) If exactly that $10000 is paid out, the fund is back at $250000 at the start of the next year, so $10000 a year can be paid out year after year without end.
  3. Let the fund needed for $15000 a year be F. The same condition gives 4100 × F = 15000.
  4. (b) Multiply both sides by 100 and then divide both sides by 4: F = 15000 × 1004 = $375000, which is 375000 − 250000 = $125000 more than the school holds.
  5. If instead $15000 is paid out of the $250000, the first year leaves 250000 × 1.04 − 15000 = $245000. The fund has fallen by $5000, it earns less the next year, and it falls faster every year after that.

answer(a) $10000 a year; (b) a fund of $375000 is needed, which is $125000 more than the school holds, and paying $15000 from $250000 leaves $245000 after one year

techniqueThe Value of an Annuity

Common pitfalls

  • Paying $15000 a year out of a fund of $250000 and calling the payments endless. Only the interest may be paid out if the fund is to stay whole, and 4% of $250000 is $10000, so $15000 a year eats into the fund from the first year.
  • Reading payments without end as needing a fund without end. The fund is finite, because the payment is met by the interest and not by the fund itself: $375000 invested at 4% earns exactly $15000 every year.
09

Saving the Same Amount Every Year Toward a Deposit

methodMultiply the Yearly Saving by the Factor That Says What One Dollar Saved at the End of Each Year Grows To, and Test the Following Year When the Target Is Missed

Ana wants $20000 for a deposit on a flat. She can save $3000 at the end of each year into an account paying 6% a year. At that rate $1 saved at the end of each year grows to $5.637093 after 5 years and to $6.975319 after 6 years. (a) Find what she has after 5 years, and how far short of $20000 that leaves her. (b) Find what she has after 6 years, and say whether that reaches the target and by how much.

what she has savedthe target, $20000123456each deposit earns only for the years after itthe fifth earns nothing at all
Each column is what she holds after that year’s deposit. The first $3000 earns for four years and the fifth earns nothing.
The deposits are made at the end of each year, so the first $3000 earns interest for 4 years, the next for 3 years, and the fifth earns nothing. The factor 5.637093 is those five deposits added up for one dollar a year.
step 1 of 5

Each deposit earns interest only for the years that follow it, so the first is the hardest worker and the last earns nothing at all. The factor given is that whole run of deposits added up for one dollar a year, which turns the problem into a single multiplication. When the target is missed, the next year is tested rather than the saving recalculated, because the question asks when the deposit is ready.

  1. The deposits are made at the end of each year, so the first $3000 earns interest for 4 years, the next for 3 years, and the fifth earns nothing. The factor 5.637093 is those five deposits added up for one dollar a year.
  2. After 5 years Ana has 3000 × 5.637093 = $16911.28, to the nearest cent.
  3. (a) That is 20000 − 16911.28 = $3088.72 short of the target.
  4. After 6 years she has 3000 × 6.975319 = $20925.96.
  5. (b) That passes the target, by 20925.96 − 20000 = $925.96. Check line by line: the $16911.28 earns 6% and becomes $17925.96, and the sixth deposit of $3000 brings it to $20925.96.

answer(a) She has $16911.28, which is $3088.72 short of the target; (b) she has $20925.96, which passes $20000 by $925.96

techniqueThe Value of an Annuity

Common pitfalls

  • Adding the deposits up as 5 × 3000 = $15000 and then adding 6% once, giving $15900. Each deposit earns for a different number of years, so they cannot be added first and grown afterwards; the factor 5.637093 is what does that work correctly.
  • Giving the fifth deposit a year of interest as well, so multiplying $16911.28 by 1.06. The fifth deposit is paid in at the end of the fifth year, so it has earned nothing by then, and 5 × 1.06 years of interest is not what the factor counts.
10

A School Fee Rising Every Year Against a Fund Saved For It

methodCarry the Cost Forward at the Rate Prices Rise and the Fund Forward at the Rate It Earns, Then Compare the Two at the Date the Money Is Needed Rather Than Today

A school fee is $12000 a year today, and fees rise by 4% a year. Hui's daughter starts there in 5 years. Hui puts $11000 into an account paying 6% a year now and adds nothing more. Take 1.045 = 1.216653 and 1.065 = 1.338226. (a) Find the fee in 5 years' time. (b) Find what the fund is worth then, and say whether it covers the fee and by how much.

the fund, less the fee it is saving for$1000 behind012345today the fee is $12000, the fund $11000the fund starts $1000 behind
Each column is the fund less the fee of that year. Today the fund is $1000 below the fee it has to meet.
A rise of 4% multiplies the fee by 1.04 each year, so after 5 years the fee is 12000 × 1.045.
step 1 of 5

Two amounts are moving at once, and they move at different rates, so neither today's figures nor a single rate will answer the question. Carry the fee forward at the rate fees rise and the fund forward at the rate the account pays, and compare them on the date the fee falls due. The fund starts behind and the question is whether the gap in the rates closes that gap in the amounts in time.

  1. A rise of 4% multiplies the fee by 1.04 each year, so after 5 years the fee is 12000 × 1.045.
  2. (a) 12000 × 1.216653 = $14599.84, to the nearest cent.
  3. The account multiplies the fund by 1.06 each year, so after 5 years the fund is 11000 × 1.338226 = $14720.49.
  4. (b) The fund covers the fee, with 14720.49 − 14599.84 = $120.65 to spare.
  5. The fund started 12000 − 11000 = $1000 behind and finished ahead, because 6% a year beats 4% a year. Check: had the account paid only 4%, the fund would have grown by the same multiplier as the fee and stayed behind by 1000 × 1.216653 = $1216.65.

answer(a) The fee is $14599.84; (b) the fund is worth $14720.49, so it covers the fee with $120.65 to spare

techniqueReal Value After Inflation · The Time Value of Money

Common pitfalls

  • Comparing the $11000 with today's fee of $12000 and calling the fund $1000 short. The fee is not paid today. Both amounts have to be carried forward to the date the fee falls due, and there the fund is ahead.
  • Taking the rise in the fee as 5 × 4% = 20%, giving 12000 × 1.2 = $14400. Each year's rise is worked out on the fee that already includes the earlier rises, so the multiplier is 1.045 = 1.216653 and the fee is $14599.84.
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