Financial Algebra

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A cashier earns \$15.50 an hour and works 32 hours. Find the gross pay.
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   \( 15.50 \times 32 = 496.00 \). There is no overtime because 32 is under 40. \$496.00

2. A worker earns \$22.25 an hour for exactly 40 hours. Find the gross pay.
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   Forty hours is the limit, so none is overtime: \( 22.25 \times 40 = 890.00 \). \$890.00

3. The regular rate is \$16.00 an hour. What is the overtime rate?
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   \( 1.5 \times 16.00 = 24.00 \). \$24.00 an hour

4. How many pay periods are there in a year when an employer pays biweekly?
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   Fifty-two weeks divided into two-week periods gives \( 52 \div 2 = 26 \). 26

5. A job pays \$21.50 an hour for 40 hours a week, 52 weeks a year. Find the annual gross pay.
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   A full-time year is \( 40 \times 52 = 2080 \) hours, and \( 21.50 \times 2080 = 44{,}720 \). \$44,720

6. A stock clerk earns \$14.80 an hour and works 44 hours in one week. Find the gross pay.
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   Regular: \( 40 \times 14.80 = 592.00 \). The overtime rate is \( 1.5 \times 14.80 = 22.20 \), and the 4 extra hours give \( 4 \times 22.20 = 88.80 \). Total \( 592.00 + 88.80 = 680.80 \). \$680.80

7. A biweekly paycheck for a 40-hour week with no overtime is \$1,120.00 gross. Find the hourly rate.
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   Two weeks of 40 hours is 80 hours, so the rate is \( \dfrac{1120}{80} = 14.00 \). \$14.00 an hour

8. A worker earning \$19.00 an hour works 52 hours in a week. Find the gross pay and the average hourly rate.
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   Regular: \( 40 \times 19.00 = 760.00 \). Overtime rate \( 28.50 \), for 12 hours: \( 12 \times 28.50 = 342.00 \). Gross \( 760.00 + 342.00 = 1{,}102.00 \). The average is \( \dfrac{1102}{52} = 21.19 \), between 19.00 and 28.50 as it must be. \$1,102.00; \$21.19 an hour

9. An employee has a \$52,000 annual salary. Find the gross pay per check if paid biweekly and if paid semimonthly, and say why they differ.
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   Biweekly: \( \dfrac{52000}{26} = 2000.00 \). Semimonthly: \( \dfrac{52000}{24} = 2166.67 \). The annual total is the same, but it is divided into 26 checks in one case and 24 in the other, so each semimonthly check is larger. \$2,000.00 biweekly; \$2,166.67 semimonthly

10. A student works 50 hours at \$12.00 an hour and computes the gross pay as \( 50 \times 1.5 \times 12 = 900 \). Find the error and the correct pay.
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    The student applied time and a half to all 50 hours. Only the 10 hours over 40 earn the premium. Regular: \( 40 \times 12 = 480 \). Overtime: \( 10 \times 18 = 180 \). Total \( 660 \). The wrong method overstated the pay by \$240. \$660.00

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. An annual salary of \$43,200 is paid biweekly. Find the gross pay per check.
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   \( \dfrac{43200}{26} = 1661.54 \), rounded to the cent at the end. \$1,661.54

2. A salesperson earns 8 percent commission on \$12,500 of sales. Find the commission.
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   \( 0.08 \times 12500 = 1000 \). \$1,000.00

3. A sewing worker is paid \$0.85 for each unit and finishes 640 units. Find the pay.
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   \( 0.85 \times 640 = 544.00 \). \$544.00

4. A customer leaves an 18 percent tip on a pre-tip bill of \$64.50. Find the tip.
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   \( 0.18 \times 64.50 = 11.61 \). \$11.61

5. Find the hourly equivalent of a \$62,400 salary for a 40-hour week.
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   \( \dfrac{62400}{2080} = 30.00 \). \$30.00 an hour

6. A salesperson earns a base of \$1,800 plus 4.5 percent of sales and sells \$27,000 in a month. Find the gross pay.
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   Commission: \( 0.045 \times 27000 = 1215 \). Total \( 1800 + 1215 = 3015 \). \$3,015.00

7. A graduated plan pays 2 percent on the first \$15,000 of sales and 5 percent on sales above that. Find the commission on \$38,000.
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   First tier: \( 0.02 \times 15000 = 300 \). The rest is \( 38000 - 15000 = 23000 \), and \( 0.05 \times 23000 = 1150 \). Total \( 300 + 1150 = 1450 \). \$1,450.00

8. A server earns \$16.00 an hour plus tips. In one week she works 30 hours and receives \$412 in tips. Find her average earnings per hour.
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   Wages \( 16.00 \times 30 = 480 \). Total \( 480 + 412 = 892 \). Per hour \( \dfrac{892}{30} = 29.73 \). \$29.73 an hour

9. Plan X pays \$2,000 plus 3 percent of sales. Plan Y pays \$1,200 plus 7 percent of sales. Find the sales at which they pay the same, and say which plan is better above that level.
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   Set \( 2000 + 0.03S = 1200 + 0.07S \). Then \( 800 = 0.04S \), so \( S = 20000 \). Above \$20,000 the plan with the larger commission rate, Plan Y, earns more with each extra dollar and ends ahead. Below it, Plan X's higher base wins. \$20,000 in sales; Plan Y is better above it

10. A student computes a 6 percent commission on \$3,000 of sales as \( 0.6 \times 3000 = 1800 \). Find the error.
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    Six percent is \( 0.06 \), not \( 0.6 \). The correct commission is \( 0.06 \times 3000 = 180 \). The wrong answer is ten times too large, and a commission of \$1,800 on \$3,000 of sales would be 60 percent, which no employer pays. \$180.00

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Gross pay is \$2,100.00 and total deductions are \$540.00. Find the net pay.
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   \( 2100 - 540 = 1560 \). \$1,560.00

2. Name two deductions that are required by law and two that are voluntary.
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   Required: any two of federal income tax, state income tax, Social Security, Medicare and (in California) SDI. Voluntary: any two of a retirement contribution, health insurance premium, union dues and a charitable gift. See the lists

3. A worker contributes 6 percent of a \$1,500.00 gross check to a 401(k). Find the contribution.
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   \( 0.06 \times 1500 = 90 \). \$90.00

4. Net pay is \$1,200.00 on a gross of \$1,600.00. What percent of gross is received?
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   \( \dfrac{1200}{1600} \times 100 = 75 \). 75 percent

5. A check is the ninth of the year, and the current Medicare deduction is \$24.00. What should the year-to-date Medicare column show if nothing has changed?
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   \( 9 \times 24.00 = 216.00 \). \$216.00

6. A check has gross pay of \$1,600.00 and a \$40.00 pre-tax health premium, with no other pre-tax deduction. Federal withholding is 10 percent of taxable wages. Find the federal withholding.
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   Taxable wages are \( 1600 - 40 = 1560 \), and \( 0.10 \times 1560 = 156.00 \). \$156.00

7. On the same check, find the Social Security tax at 6.2 percent, given that the health premium is exempt from payroll tax.
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   The payroll tax base is \( 1600 - 40 = 1560 \), and \( 0.062 \times 1560 = 96.72 \). \$96.72

8. Two workers each earn \$2,000.00 gross. One contributes 5 percent to a 401(k) and the other contributes nothing. Income tax withholding is 12 percent of taxable wages. How much less income tax is withheld from the contributor, and does her take-home fall by the full \$100 contribution?
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   The contribution is \( 0.05 \times 2000 = 100 \). Her taxable wages are \$1,900, so her withholding is \( 0.12 \times 1900 = 228 \), against \( 0.12 \times 2000 = 240 \). That is \$12 less income tax. Her take-home therefore falls by \( 100 - 12 = 88 \) dollars rather than \$100, since the tax saving offsets part of the contribution. \$12 less tax; take-home falls by \$88

9. A stub shows gross \$1,200.00, deductions of \$150.00, \$75.00, \$60.00 and \$40.00, and net pay of \$880.00. Decide whether the stub is consistent.
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   Total deductions are \( 150 + 75 + 60 + 40 = 325 \). Then \( 1200 - 325 = 875 \), not 880. The stub is off by \$5, so a line was misprinted or misread. Deductions plus net must equal gross. Not consistent: off by \$5

10. A student says, "My raise from \$1,400 to \$1,500 gross a check means I take home \$100 more." Explain the error.
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    The extra \$100 of gross is also subject to income tax, Social Security, Medicare and SDI. If those total 25 percent of each additional dollar, the take-home rises by \( 100 \times 0.75 = 75 \) dollars. The raise is always worth less than its gross amount, but it is never negative. Take-home rises by less than \$100, about \$75 at 25 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the \$170,000 Social Security wage base.

1. Find the Social Security and Medicare tax on a \$2,000.00 paycheck.
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   Social Security: \( 0.062 \times 2000 = 124.00 \). Medicare: \( 0.0145 \times 2000 = 29.00 \). \$124.00 and \$29.00

2. Find the Social Security and Medicare tax on a \$1,640.00 paycheck.
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   Social Security: \( 0.062 \times 1640 = 101.68 \). Medicare: \( 0.0145 \times 1640 = 23.78 \). \$101.68 and \$23.78

3. Find the combined FICA tax from the employee on a \$3,200.00 check.
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   The combined rate is \( 6.2 + 1.45 = 7.65 \) percent, and \( 0.0765 \times 3200 = 244.80 \). \$244.80

4. An employee is paid \$60,000 a year. Find the employer's total cost for wages and the employer's half of FICA.
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   The employer's half is \( 0.0765 \times 60000 = 4590 \). The total is \( 60000 + 4590 = 64590 \). \$64,590.00

5. An employee earns \$120,000 in a year. Find the annual Social Security tax.
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   The wages are below the \$170,000 base, so all of them are taxed: \( 0.062 \times 120000 = 7440 \). \$7,440.00

6. A worker is paid \$2,400.00 a week. Find the Social Security and Medicare tax on one check.
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   Social Security: \( 0.062 \times 2400 = 148.80 \). Medicare: \( 0.0145 \times 2400 = 34.80 \). \$148.80 and \$34.80

7. A worker is paid \$3,400 a week. In which week does cumulative pay first reach the \$170,000 Social Security base?
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   The number of checks needed is \( \dfrac{170000}{3400} = 50 \). Fifty checks total exactly \$170,000, so the ceiling is reached in week 50. Checks 51 and 52 pay no Social Security tax, and their Medicare tax continues. Week 50

8. A single employee earns \$250,000. Find the annual Medicare tax including the additional 0.9 percent on wages over \$200,000.
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   The basic tax is \( 0.0145 \times 250000 = 3625 \). The additional tax applies to \( 250000 - 200000 = 50000 \): \( 0.009 \times 50000 = 450 \). Total \( 3625 + 450 = 4075 \). \$4,075.00

9. An employee earning \$190,000 a year says her Social Security tax is \( 0.062 \times 190000 = 11{,}780 \). Find the error and the correct tax, and find her Medicare tax.
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   Only the first \$170,000 of wages is subject to Social Security. The correct tax is \( 0.062 \times 170000 = 10{,}540 \). Medicare has no ceiling: \( 0.0145 \times 190000 = 2{,}755 \). Her wages are below \$200,000, so there is no additional Medicare tax. \$10,540.00 Social Security; \$2,755.00 Medicare

10. A student says, "My employer pays half of FICA, so it costs me nothing." Explain in terms of total compensation.
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    The employer's half is part of what the employer spends to employ the worker, so the money could have been paid as wages. On a \$60,000 salary the employer's half is \$4,590. The employee does not see it on the pay stub, but the cost to hire is \$64,590, and the employee's compensation is effectively larger than the stub shows. The tax also buys the employee's future Social Security and Medicare benefits. It is a real cost of the job, even though it is not on the stub

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A salary of \$52,000 rises 4 percent. Find the new salary.
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   \( 52000 \times 1.04 = 54080 \). \$54,080

2. Find the percent change when an hourly wage goes from \$40 to \$46.
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   \( \dfrac{46 - 40}{40} \times 100 = 15 \). 15 percent increase

3. A price of \$100 is cut by 30 percent and then raised by 30 percent. Find the final price.
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   \( 100 \times 0.70 = 70 \), then \( 70 \times 1.30 = 91 \). The price is lower than the start because the second 30 percent is taken of \$70. \$91.00

4. After a 4 percent raise a salary is \$54,080. Find the salary before the raise.
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   Divide by the factor: \( \dfrac{54080}{1.04} = 52000 \). \$52,000

5. A monthly rent falls from \$25 to \$20 in a worked example of a discount. Find the percent change.
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   \( \dfrac{20 - 25}{25} \times 100 = -20 \). 20 percent decrease

6. An employee gets raises of 3 percent, 4 percent and 2 percent in three years. Find the total percent increase.
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   \( 1.03 \times 1.04 \times 1.02 = 1.092624 \), a total increase of 9.26 percent. Adding the three rates would give 9 percent, which understates the growth. 9.26 percent

7. A job pays \$27.50 an hour for 40 hours and 52 weeks. Another pays \$56,000 a year. Find the annual pay of the first and the difference.
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   \( 27.50 \times 2080 = 57{,}200 \). The difference is \( 57200 - 56000 = 1200 \) in favor of the hourly job. \$57,200; \$1,200 more

8. A price is reduced by 20 percent and then increased by 20 percent. Find the final price from \$100, and find what percent increase would restore the original price after the cut.
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   \( 100 \times 0.80 \times 1.20 = 96 \). To restore \$100 from \$80, the increase is \( \dfrac{100 - 80}{80} \times 100 = 25 \) percent. The restoring percent is larger than the cut because it is taken of a smaller amount. \$96.00; a 25 percent increase

9. A rate rises from 5 percent to 7 percent. State the change in percentage points and in percent.
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   The rise is \( 7 - 5 = 2 \) percentage points. As a percent change, the increase is \( \dfrac{7 - 5}{5} \times 100 = 40 \) percent of the original rate. 2 percentage points; a 40 percent increase

10. A worker gets a 4 percent raise in a year when prices rise 3 percent. Find her real raise, meaning the change in what her pay buys.
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    Pay is multiplied by 1.04 and prices by 1.03, so buying power is multiplied by \( \dfrac{1.04}{1.03} = 1.00971 \), a gain of about 0.97 percent. Subtracting the rates to get 1 percent is close but not exact. About 0.97 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use \$0.65 a mile and the 15.3 percent on 92.35 percent rule.

1. A freelancer invoices \$3,200 and has \$840 in business expenses. Find the net earnings.
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   \( 3200 - 840 = 2360 \). \$2,360.00

2. Find the self-employment tax on net earnings of \$12,000.
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   \( 0.9235 \times 12000 = 11082 \), and \( 0.153 \times 11082 = 1695.55 \). \$1,695.55

3. A courier drives 1,200 business miles. Find the mileage deduction.
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   \( 1200 \times 0.65 = 780.00 \). \$780.00

4. A self-employed worker sets aside income for four equal quarterly payments totaling \$4,800 for the year. Find each payment.
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   \( \dfrac{4800}{4} = 1200 \). \$1,200.00

5. A designer has \$28,000 of revenue and \$6,500 of expenses. Find the net earnings and the self-employment tax.
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   Net \( 28000 - 6500 = 21500 \). The tax base is \( 0.9235 \times 21500 = 19855.25 \), and the tax is \( 0.153 \times 19855.25 = 3037.85 \). \$21,500.00 net; \$3,037.85 tax

6. Using the driver's figures in the example, find her hourly rate after self-employment tax if she is able to cut the waiting time from 11 hours to 5, for 39 hours in all.
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   Net after the tax is \( 897 - 126.74 = 770.26 \). Over 39 hours, \( \dfrac{770.26}{39} = 19.75 \). Cutting 6 hours of waiting raises the true rate by about \$2.63 an hour. \$19.75 an hour

7. An employee earns \$18.00 an hour for 2,080 hours. Find the annual wages and the employer's half of FICA.
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   \( 18 \times 2080 = 37440 \). The employer's half is \( 0.0765 \times 37440 = 2864.16 \), so the job costs the employer \$40,304.16. \$37,440.00 wages; \$2,864.16 employer FICA

8. A freelancer wants to match an employee's \$18.00 an hour for 45 hours, all of which she counts. How much gross must she collect in a week, before expenses and taxes?
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   The minimum is \( 18 \times 45 = 810 \). This is only the starting point: she also needs to cover the expenses, the self-employment tax and her own benefits, so the real target is higher. At least \$810.00

9. A contractor has \$52,000 of revenue and \$9,000 of business expenses. Find the self-employment tax and the half of it that is deductible.
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   Net earnings are \( 52000 - 9000 = 43000 \). The taxable amount is \( 0.9235 \times 43000 = 39710.50 \), and the tax is \( 0.153 \times 39710.50 = 6075.71 \). The deductible half is \( 6075.7065 \div 2 = 3037.85 \). \$6,075.71 tax; \$3,037.85 deductible

10. A student says, "I earned \$1,000 driving this week, so I earn \$25 an hour because I worked 40 hours." List three things the \$25 leaves out.
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    The \$1,000 is revenue, not earnings, so vehicle costs (or mileage), phone and any fees must come off first. The 15.3 percent self-employment tax on what remains also comes off. And any unpaid hours spent waiting or driving to pickups must be counted among the 40 or added to them. Each of these lowers the true hourly rate. Expenses, self-employment tax and unpaid hours

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A health plan costs \$7,200 a year and the employer pays 75 percent. Find the value of the benefit to the employee.
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   \( 0.75 \times 7200 = 5400 \). \$5,400.00

2. An employer matches 3 percent of a \$62,000 salary. Find the match, assuming the employee contributes at least 3 percent.
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   \( 0.03 \times 62000 = 1860 \). \$1,860.00

3. Twelve paid vacation days are worth how much at \$25 an hour, 8 hours a day?
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   \( 12 \times 8 \times 25 = 2400 \). Paid time off already sits inside a salary, so this is the pay for hours not worked. \$2,400.00

4. A job pays \$48,000 and the employer contributes \$4,200 for health insurance and \$1,440 to a retirement plan. Find the total compensation.
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   \( 48000 + 4200 + 1440 = 53640 \). \$53,640.00

5. A worker has 10 vacation days and 9 holidays, all paid. Find the hours actually worked in a 2,080-hour year.
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   The paid days off total \( 10 + 9 = 19 \), or \( 19 \times 8 = 152 \) hours, so \( 2080 - 152 = 1928 \). 1,928 hours

6. For that worker on a \$47,000 salary, find the effective hourly rate per hour worked.
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   \( \dfrac{47000}{1928} = 24.38 \). The simple rate on 2,080 hours is \( \dfrac{47000}{2080} = 22.60 \), so the paid time off raises the effective rate by about \$1.78. \$24.38 an hour

7. An employee earning \$50,000 contributes 6 percent and her employer matches 100 percent of contributions up to 4 percent of salary. Find the employer's match.
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   The match stops at 4 percent of salary: \( 0.04 \times 50000 = 2000 \). Her own contribution is \( 0.06 \times 50000 = 3000 \). The last 2 percent of her contribution, \$1,000, earns no match. \$2,000.00

8. An employer matches 50 percent of contributions up to 6 percent of salary. An employee earning \$60,000 contributes 6 percent. Find the match and the return on what she put in.
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   Her contribution is \( 0.06 \times 60000 = 3600 \). The match is 50 percent of that, \( 0.5 \times 3600 = 1800 \). The immediate return is \( \dfrac{1800}{3600} \times 100 = 50 \) percent before any investment growth. \$1,800.00; a 50 percent immediate return

9. Job A offers \$61,000 with no benefits. Job B offers \$56,000 plus \$4,800 of employer health insurance. Which has the higher total, and by how much?
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   Job B's total is \( 56000 + 4800 = 60800 \). Job A's is \$61,000. Job A is higher by \( 61000 - 60800 = 200 \) dollars, which is too small a gap to decide on, so other factors (hours, commute, growth, how much the worker values the coverage) should settle it. Job A by \$200.00

10. A student says, "The 401(k) match is free money, so I should contribute as much as I can." Find the flaw in the conclusion and state what is true.
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    The match is free only up to the employer's limit. On a match of 100 percent up to 4 percent of salary, contributing 4 percent captures the whole match, and contributions above 4 percent earn no match. They may still be worth making for saving and tax reasons, but they should be judged on those reasons, not as free money. Also, the match is paid only when the worker contributes, and some plans make the worker wait before the match is fully hers. The match is free only up to the limit

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. All use the teaching schedule in the method.

1. What is the marginal rate for a single filer with taxable income of \$30,000?
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   \$30,000 lies between \$12,000 and \$48,000, in the 12 percent bracket. 12 percent

2. Find the tax on taxable income of \$10,000.
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   All of it is in the first bracket: \( 0.10 \times 10000 = 1000 \). \$1,000

3. Find the tax on taxable income of \$12,000.
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   Exactly the top of the first bracket: \( 0.10 \times 12000 = 1200 \). \$1,200

4. Find the tax on taxable income of \$40,000.
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   The first \$12,000 gives \$1,200. The remaining \( 40000 - 12000 = 28000 \) is taxed at 12 percent: \( 0.12 \times 28000 = 3360 \). Total \( 1200 + 3360 = 4560 \). \$4,560

5. How many of the dollars in \$62,000 of taxable income are taxed at 22 percent?
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   Only those above \$48,000: \( 62000 - 48000 = 14000 \). \$14,000

6. Find the tax on taxable income of \$85,000.
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   Shortcut: \( 5520 + 0.22 \times (85000 - 48000) = 5520 + 0.22 \times 37000 = 5520 + 8140 = 13660 \). \$13,660

7. Find the tax on taxable income of \$130,000.
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   The income reaches the 24 percent bracket, which starts at \$100,000 with tax \$16,960. Then \( 16960 + 0.24 \times 30000 = 16960 + 7200 = 24160 \). \$24,160

8. A filer with taxable income of \$47,000 receives a \$2,000 raise, all of it taxable. Find the extra tax and the amount kept.
   Show the full solution

   The raise straddles \$48,000. The first \$1,000 (up to \$48,000) is taxed at 12 percent: \$120. The last \$1,000 is taxed at 22 percent: \$220. Extra tax is \( 120 + 220 = 340 \), and the amount kept is \( 2000 - 340 = 1660 \). \$340 tax; \$1,660 kept

9. A student says, "With taxable income of \$50,000 I'm in the 22 percent bracket, so my tax is \( 0.22 \times 50000 = 11{,}000 \)." Find the error and the correct tax.
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   The 22 percent applies only to the \( 50000 - 48000 = 2000 \) above \$48,000. The correct tax is \( 5520 + 0.22 \times 2000 = 5520 + 440 = 5960 \). The student overstated the tax by \$5,040 by applying the marginal rate to every dollar. \$5,960

10. A filer owes \$4,560 in tax. Find the taxable income.
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    \$4,560 exceeds \$1,200, the tax at \$12,000, so the income is in the 12 percent bracket. The tax above \$1,200 is \( 4560 - 1200 = 3360 \), which came from \( \dfrac{3360}{0.12} = 28000 \) dollars in that bracket. Income is \( 12000 + 28000 = 40000 \). \$40,000

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the teaching schedule and round rates to two decimal places.

1. The tax on taxable income of \$40,000 is \$4,560. Find the effective rate.
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   \( \dfrac{4560}{40000} \times 100 = 11.4 \). 11.4 percent

2. Find the effective rate on taxable income of \$85,000, given the tax of \$13,660.
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   \( \dfrac{13660}{85000} \times 100 = 16.07 \). 16.07 percent

3. Find the tax and the effective rate on taxable income of \$55,000.
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   Tax: \( 5520 + 0.22 \times 7000 = 5520 + 1540 = 7060 \). Effective rate: \( \dfrac{7060}{55000} \times 100 = 12.84 \). \$7,060; 12.84 percent

4. A single filer has gross income of \$70,000 and a standard deduction of \$14,000. The tax on her taxable income is \$7,280. Find her effective rate on gross income.
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   \( \dfrac{7280}{70000} \times 100 = 10.4 \). On her taxable income of \$56,000 the rate would be 13 percent, so always say which income you used. 10.4 percent

5. State the marginal rate for taxable income of \$56,000.
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   \$56,000 is in the bracket from \$48,000 to \$100,000. 22 percent

6. Find the tax and the effective rate for taxable income of \$130,000, and compare the effective rate with the marginal rate.
   Show the full solution

   Tax: \( 16960 + 0.24 \times 30000 = 24160 \). Effective: \( \dfrac{24160}{130000} \times 100 = 18.58 \) percent. The marginal rate is 24 percent, more than five points above the effective rate. \$24,160; 18.58 percent against 24 percent

7. Taxable income rises from \$95,000 to \$100,000. Find the extra tax, and the amount kept of the extra \$5,000.
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   All of the increase is inside the 22 percent bracket, which runs up to \$100,000. The extra tax is \( 0.22 \times 5000 = 1100 \), and \( 5000 - 1100 = 3900 \) is kept. \$1,100 tax; \$3,900 kept

8. Taxable income rises from \$95,000 to \$105,000. Find the extra tax and the amount kept.
   Show the full solution

   The raise straddles \$100,000. The first \$5,000 is taxed at 22 percent (\$1,100) and the last \$5,000 at 24 percent (\$1,200). Extra tax \( 1100 + 1200 = 2300 \); kept \( 10000 - 2300 = 7700 \). \$2,300 tax; \$7,700 kept

9. Compare the effective rates on taxable incomes of \$30,000 and \$60,000, and say what the comparison shows.
   Show the full solution

   Tax on \$30,000: \( 1200 + 0.12 \times 18000 = 3360 \), an effective rate of 11.2 percent. Tax on \$60,000: \( 5520 + 0.22 \times 12000 = 8160 \), an effective rate of 13.6 percent. The higher income pays a higher share, which is what progressive means. 11.2 percent and 13.6 percent: a progressive tax

10. A worker says, "If I take a \$4,000 raise I'll move from the 12 percent to the 22 percent bracket and take home less." His taxable income is \$47,000. Test the claim.
    Show the full solution

    Taxable income becomes \$51,000. The tax rises from \$5,400 to \( 5520 + 0.22 \times 3000 = 6180 \), an increase of \$780. He keeps \( 4000 - 780 = 3220 \). The first \$1,000 is taxed at 12 percent (\$120) and the last \$3,000 at 22 percent (\$660). He takes home more, never less. False: he keeps \$3,220

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the teaching schedule and a \$14,000 standard deduction.

1. A filer in the 22 percent bracket takes an extra \$1,000 deduction. Find the tax saved.
   Show the full solution

   \( 1000 \times 0.22 = 220 \). \$220

2. A filer in the 12 percent bracket takes an extra \$2,500 deduction. Find the tax saved.
   Show the full solution

   \( 2500 \times 0.12 = 300 \). \$300

3. A single filer has wages of \$52,000 and takes the standard deduction. Find the taxable income.
   Show the full solution

   \( 52000 - 14000 = 38000 \). \$38,000

4. The tax on \$38,000 of taxable income is \$4,320. A \$500 credit applies. Find the final tax.
   Show the full solution

   The credit comes off the tax: \( 4320 - 500 = 3820 \). \$3,820

5. A filer has itemized expenses of \$6,200, \$4,100 and \$1,500. Should she itemize?
   Show the full solution

   The total is \( 6200 + 4100 + 1500 = 11800 \), which is less than \$14,000. No: take the standard deduction

6. A filer has wages of \$68,000 and itemized deductions of \$16,500. Find the tax saved by itemizing instead of taking the standard deduction.
   Show the full solution

   Standard: taxable \$54,000, tax \( 5520 + 0.22 \times 6000 = 6840 \). Itemized: taxable \$51,500, tax \( 5520 + 0.22 \times 3500 = 6290 \). Saving \( 6840 - 6290 = 550 \), which equals \( 2500 \times 0.22 \). \$550

7. In the 22 percent bracket, which saves more: a \$1,500 credit or a \$1,500 deduction? By how much?
   Show the full solution

   The credit saves \$1,500. The deduction saves \( 1500 \times 0.22 = 330 \). The credit saves \( 1500 - 330 = 1170 \) more. The credit, by \$1,170

8. A filer's tax before credits is \$1,800. She qualifies for a \$2,500 credit. What happens if the credit is nonrefundable, and what if it is refundable?
   Show the full solution

   A nonrefundable credit reduces the tax only to zero, so she owes nothing and the other \( 2500 - 1800 = 700 \) is lost. A refundable credit reduces the tax below zero, so she receives the \$700 as a payment. Nonrefundable: tax \$0, \$700 lost. Refundable: she receives \$700

9. A filer has taxable income of \$41,000. A \$2,000 deduction lowers it to \$39,000. Find the tax saved.
   Show the full solution

   Tax on \$41,000: \( 1200 + 0.12 \times 29000 = 4680 \). Tax on \$39,000: \( 1200 + 0.12 \times 27000 = 4440 \). The saving is \$240, which is \( 2000 \times 0.12 \). \$240

10. A student says, "A \$1,000 deduction and a \$1,000 credit are the same thing, because they are both \$1,000." Correct the statement using a filer in the 12 percent bracket.
    Show the full solution

    The deduction comes off income, so it saves the marginal rate on it: \( 1000 \times 0.12 = 120 \). The credit comes off the tax, so it saves \$1,000. The credit is worth \( 1000 \div 0.12 = 8{,}333.33 \) dollars of deduction in that bracket. The deduction saves \$120; the credit saves \$1,000

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the teaching schedule and a \$14,000 standard deduction.

1. A filer has wages of \$38,000 and \$150 of interest. Find her total income.
   Show the full solution

   \( 38000 + 150 = 38150 \). \$38,150

2. With an adjustment of \$400, find the AGI and the taxable income for that filer.
   Show the full solution

   AGI is \( 38150 - 400 = 37750 \). Taxable income is \( 37750 - 14000 = 23750 \). AGI \$37,750; taxable income \$23,750

3. The tax on \$23,750 is \$2,610. She has a \$300 credit and \$2,600 was withheld. Find the refund.
   Show the full solution

   Liability \( 2610 - 300 = 2310 \). Refund \( 2600 - 2310 = 290 \). \$290

4. A filer has taxable income of \$47,000 and \$5,200 was withheld. There are no credits. Does she receive a refund or owe? How much?
   Show the full solution

   Tax: \( 1200 + 0.12 \times 35000 = 5400 \). Withheld \$5,200 is less than \$5,400, so she owes \( 5400 - 5200 = 200 \). She owes \$200

5. A student has wages of \$9,000 and takes the \$14,000 standard deduction. Find the taxable income and the tax.
   Show the full solution

   The deduction exceeds income, so taxable income is zero, not negative. The tax is \$0, and any tax withheld is refunded. \$0 taxable income; \$0 tax

6. A filer has taxable income of \$30,000 and \$2,500 of tax withheld, with no credits. Find the refund or the amount owed.
   Show the full solution

   Tax: \( 1200 + 0.12 \times 18000 = 3360 \). Then \( 3360 - 2500 = 860 \) is owed. She owes \$860

7. Find the tax on taxable incomes of \$9,500 and \$12,500 and explain why the second is not a whole \$3,000 times 10 percent more.
   Show the full solution

   Tax on \$9,500: \( 0.10 \times 9500 = 950 \). Tax on \$12,500: \( 1200 + 0.12 \times 500 = 1260 \). The increase is \$310, not \$300, because the extra \$500 above \$12,000 is taxed at 12 percent rather than 10 percent. \$950 and \$1,260

8. A filer has wages of \$44,000, \$1,200 of interest, an \$800 adjustment, a credit of zero, and \$3,000 withheld. Find the AGI, the taxable income, the tax, and the refund or balance due.
   Show the full solution

   Total income \( 44000 + 1200 = 45200 \). AGI \( 45200 - 800 = 44400 \). Taxable income \( 44400 - 14000 = 30400 \). Tax \( 1200 + 0.12 \times 18400 = 3408 \). Withheld \$3,000 is less than \$3,408, so she owes \( 3408 - 3000 = 408 \). AGI \$44,400; taxable income \$30,400; tax \$3,408; she owes \$408

9. For that filer, find the effective rate on taxable income.
   Show the full solution

   \( \dfrac{3408}{30400} \times 100 = 11.21 \). 11.21 percent

10. A friend says, "I got a \$1,500 refund, so the government gave me \$1,500." Explain what a refund actually is, and what a better goal is.
    Show the full solution

    A refund is money the filer overpaid during the year. The government held it without paying interest. If \$1,500 was overwithheld evenly, it is the same as lending \$1,500 to the government, for about half a year on average. The goal is to have withholding match the liability, so the money is in the filer's hands all year and the refund or balance due is near zero. A refund is the return of an interest-free loan; aim for a refund near zero

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the sales tax on a \$40 purchase at a rate of 7.75 percent.
   Show the full solution

   \( 0.0775 \times 40 = 3.10 \). \$3.10

2. A meal costs \$26.50 before tax and the tax rate is 8.75 percent. Find the tax and the total.
   Show the full solution

   Tax: \( 0.0875 \times 26.50 = 2.32 \). Total: \( 26.50 + 2.32 = 28.82 \). \$2.32 tax; \$28.82 total

3. A receipt total is \$108.25 at 8.25 percent tax. Find the price before tax.
   Show the full solution

   \( \dfrac{108.25}{1.0825} = 100.00 \). \$100.00

4. A home has an assessed value of \$480,000 and a 1.1 percent property tax rate. Find the annual tax.
   Show the full solution

   \( 0.011 \times 480000 = 5280 \). \$5,280

5. Gasoline is taxed \$0.54 a gallon. Find the tax on 14 gallons.
   Show the full solution

   \( 0.54 \times 14 = 7.56 \). \$7.56

6. A state sales tax is 6 percent and the local tax is 2.25 percent. Find the combined rate and the tax on a \$1,200 purchase.
   Show the full solution

   Combined \( 6 + 2.25 = 8.25 \) percent. Tax \( 0.0825 \times 1200 = 99.00 \). 8.25 percent; \$99.00

7. A property has an assessed value of \$350,000 and a tax rate of 14 mills. Find the annual tax.
   Show the full solution

   A mill is \$1 per \$1,000, so the tax is \( 14 \times \dfrac{350000}{1000} = 14 \times 350 = 4900 \). \$4,900

8. A worker earning \$26,000 a year spends \$18,000 on items taxed at 8.75 percent. Another earning \$90,000 spends \$35,000 on taxed items. Find each person's sales tax paid and its share of income.
   Show the full solution

   First: \( 0.0875 \times 18000 = 1575 \), a share of \( \dfrac{1575}{26000} \times 100 = 6.06 \) percent. Second: \( 0.0875 \times 35000 = 3062.50 \), a share of \( \dfrac{3062.50}{90000} \times 100 = 3.40 \) percent. The lower earner pays the smaller amount but the larger share. \$1,575 (6.06 percent); \$3,062.50 (3.40 percent)

9. A \$240 jacket is 15 percent off and the sales tax is 8.25 percent. Find the total.
   Show the full solution

   Discount price \( 240 \times 0.85 = 204 \). Tax and total: \( 204 \times 1.0825 = 220.83 \). The tax is computed on the discounted price, so the discount also lowers the tax. \$220.83

10. A student finds the pre-tax price of a \$54.05 receipt at 8.25 percent by computing \( 54.05 \times 0.9175 = 49.59 \). Find the error and the correct price.
    Show the full solution

    The tax is 8.25 percent of the pre-tax price, not of the total, so subtracting 8.25 percent of the total takes off too much. The total is 1.0825 times the price, so the price is \( \dfrac{54.05}{1.0825} = 49.93 \). Check: \( 49.93 \times 1.0825 = 54.05 \). \$49.93

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A filer had \$6,400 withheld and owes \$5,900 in tax. Find the refund.
   Show the full solution

   \( 6400 - 5900 = 500 \). \$500 refund

2. A filer had \$4,900 withheld and owes \$5,400 in tax. Find the balance due.
   Show the full solution

   \( 5400 - 4900 = 500 \). \$500 owed

3. A tax liability of \$9,100 is withheld evenly over 26 paychecks. Find the withholding per check.
   Show the full solution

   \( \dfrac{9100}{26} = 350 \). \$350.00

4. A freelancer expects to owe \$12,800 in tax. Find the equal quarterly payment.
   Show the full solution

   \( \dfrac{12800}{4} = 3200 \). \$3,200

5. Last year's tax was \$7,400. This year's is expected to be \$8,800. Find the safe-harbor quarterly payment.
   Show the full solution

   Ninety percent of this year's is \( 0.9 \times 8800 = 7920 \). The safe harbor is the smaller of \$7,920 and \$7,400, which is \$7,400, and \( \dfrac{7400}{4} = 1850 \). \$1,850 per quarter

6. A refund of \$1,560 was taken evenly through the year. If the money could have earned 4.5 percent for an average of half a year, find the lost interest.
   Show the full solution

   Interest on \$1,560 at 4.5 percent for half a year: \( 1560 \times 0.045 \times 0.5 = 35.10 \). About \$35.10

7. A worker files and owes \$2,100. Did she necessarily pay a penalty?
   Show the full solution

   No. Owing at filing means withholding was less than the liability. A penalty applies only if total payments during the year were below the safe harbor, which is the smaller of 90 percent of this year's tax or 100 percent of last year's. Not necessarily

8. A worker adds \$40 of extra withholding to each of 26 paychecks. Is that an increase or a decrease in the tax paid in, and by how much over the year?
   Show the full solution

   Additional withholding of \$40 per check increases the tax paid in, by \( 40 \times 26 = 1040 \) dollars over the year. To lower withholding she would request less, not more. An increase of \$1,040 a year

9. A freelancer's liability is \$8,000 and his payments during the year total \$6,400. Find what he owes at filing, and whether his payments reached 90 percent of this year's tax.
   Show the full solution

   He owes \( 8000 - 6400 = 1600 \). Ninety percent of \$8,000 is \$7,200, and \$6,400 is \( 7200 - 6400 = 800 \) below it. Unless last year's tax was \$6,400 or less, he may owe a penalty. Owes \$1,600; \$800 below 90 percent of this year's tax

10. A student says, "I want the biggest refund I can get, so I'll have extra withheld." Explain why that is not the best plan.
    Show the full solution

    Extra withholding gives the government an interest-free loan: the money leaves each check months before it is owed and is returned without interest. On a \$1,560 refund that is about \$35 of lost interest, and the student also has less to spend or save each pay period. The better goal is a refund near zero, with the extra money saved or used during the year. A refund is money you were not free to use all year

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A flat tax of 9 percent applies. Find the tax on \$48,000 and on \$90,000.
   Show the full solution

   \( 0.09 \times 48000 = 4320 \) and \( 0.09 \times 90000 = 8100 \). Both take 9 percent of income. \$4,320 and \$8,100

2. Find the sales tax on \$500 of purchases at 8.75 percent.
   Show the full solution

   \( 0.0875 \times 500 = 43.75 \). \$43.75

3. A person earning \$30,000 buys 900 gallons of gasoline a year and another earning \$90,000 also buys 900 gallons. At \$0.50 a gallon in gas tax, find the share of income each pays.
   Show the full solution

   Each pays \( 0.50 \times 900 = 450 \). Shares: \( \dfrac{450}{30000} = 1.5 \) percent and \( \dfrac{450}{90000} = 0.5 \) percent. 1.5 percent and 0.5 percent: regressive

4. A 12 percent flat tax is imposed. Find the tax on \$80,000.
   Show the full solution

   \( 0.12 \times 80000 = 9600 \). \$9,600

5. Name the three types of tax by how the share of income changes with income.
   Show the full solution

   If the share stays the same the tax is flat (proportional). If the share rises with income it is progressive. If the share falls it is regressive. Flat, progressive, regressive

6. A state taxes the first \$40,000 of taxable income at 5 percent and the rest at 9 percent. Find the tax and the effective rate on \$70,000.
   Show the full solution

   \( 0.05 \times 40000 = 2000 \) and \( 0.09 \times 30000 = 2700 \), a total of \$4,700. The effective rate is \( \dfrac{4700}{70000} \times 100 = 6.71 \) percent, below the top rate of 9 percent. \$4,700; 6.71 percent

7. Social Security tax is 6.2 percent on wages up to \$170,000. Find the share of wages paid by a \$200,000 earner and a \$50,000 earner, and classify the tax at those levels.
   Show the full solution

   The higher earner pays \( 0.062 \times 170000 = 10540 \), a share of \( \dfrac{10540}{200000} = 5.27 \) percent. The lower earner pays \( 0.062 \times 50000 = 3100 \), a share of 6.2 percent. The share falls as income rises above the ceiling, so the tax is regressive at those levels. 5.27 percent against 6.2 percent: regressive

8. Under the teaching income tax with a \$14,000 deduction, compare the shares of total income taken from a person earning \$30,000 and one earning \$60,000.
   Show the full solution

   The \$30,000 earner has taxable income \$16,000 and tax \$1,680, a share of 5.6 percent. The \$60,000 earner has taxable income \$46,000 and tax \$5,280, a share of 8.8 percent. 5.6 percent and 8.8 percent: progressive

9. A student says, "A sales tax is fair because everyone pays the same rate." Use numbers to show what the statement leaves out.
   Show the full solution

   The rate is the same, but the base is spending, and spending is a smaller share of income for higher earners. At 8 percent, a person earning \$30,000 who spends \$28,000 on taxed goods pays \$2,240, which is 7.47 percent of income. A person earning \$120,000 who spends \$70,000 pays \$5,600, which is 4.67 percent. Equal rates do not mean equal burdens. The share of income is higher for the lower earner

10. A high earner pays more sales tax in dollars than a low earner. A student concludes the tax is progressive. Find the flaw.
    Show the full solution

    Dollar amounts are the wrong comparison. A tax is progressive only if the share of income rises with income. A tax can take more dollars from the high earner and still a smaller share, as in the last problem, where \$5,600 is 4.67 percent and \$2,240 is 7.47 percent. Always divide by income. Compare shares of income, not dollar amounts

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Net pay is \$2,400 and planned spending is \$2,250. Find the balance.
   Show the full solution

   \( 2400 - 2250 = 150 \), a surplus. \$150 surplus

2. Net pay is \$2,800 a month. Find the 50/30/20 amounts.
   Show the full solution

   Needs \( 0.50 \times 2800 = 1400 \). Wants \( 0.30 \times 2800 = 840 \). Savings \( 0.20 \times 2800 = 560 \). The three sum to \$2,800. \$1,400; \$840; \$560

3. Rent is \$900 and net pay is \$2,400. Find rent as a percent of net pay.
   Show the full solution

   \( \dfrac{900}{2400} \times 100 = 37.5 \). 37.5 percent

4. A worker saves 20 percent of a \$2,800 net monthly pay. Find the annual savings.
   Show the full solution

   Monthly: \$560. Annual: \( 560 \times 12 = 6720 \). \$6,720

5. Monthly expenses are \$820, \$240, \$310, \$150, \$90 and \$180, and net pay is \$1,900. Find the balance.
   Show the full solution

   Expenses total \( 820 + 240 + 310 + 150 + 90 + 180 = 1790 \). The balance is \( 1900 - 1790 = 110 \). \$110 surplus

6. A worker with net pay of \$2,400 has needs of \$1,140. Find the needs as a percent of net pay and compare with the 50 percent guideline.
   Show the full solution

   \( \dfrac{1140}{2400} \times 100 = 47.5 \) percent, which is \( 2.5 \) points under the guideline. 47.5 percent: within the guideline

7. Dining out costs \$120 a month. A worker cuts it by a quarter. Find the saving in a year.
   Show the full solution

   One quarter of \$120 is \$30 a month, and \( 30 \times 12 = 360 \). \$360 a year

8. A worker with a \$150 monthly surplus gets a raise that adds \$150 to her net pay. She spends nothing more. Find the new surplus.
   Show the full solution

   \( 150 + 150 = 300 \). \$300 a month

9. A worker with net pay of \$2,600 saves 20 percent each month. How much does she have after 6 months, ignoring interest?
   Show the full solution

   Monthly savings \( 0.20 \times 2600 = 520 \), and \( 520 \times 6 = 3120 \). \$3,120

10. A worker earns \$3,900 gross and \$3,120 net, and plans \$3,300 of spending because "I make \$3,900." Find the error and the shortfall.
    Show the full solution

    The budget used gross pay, but only the net pay of \$3,120 is received. Planned spending of \$3,300 exceeds it by \( 3300 - 3120 = 180 \) dollars every month, which would be borrowed or drawn from savings. Budget from net pay; the plan is \$180 short

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. An annual cost of \$480 is spread over 12 months. Find the monthly amount.
   Show the full solution

   \( \dfrac{480}{12} = 40 \). \$40 a month

2. Electric bills for three months are \$212, \$198 and \$245. Find the average.
   Show the full solution

   \( \dfrac{212 + 198 + 245}{3} = \dfrac{655}{3} = 218.33 \). \$218.33

3. A streaming and phone bundle costs \$65 a month. Find the cost in a year.
   Show the full solution

   \( 65 \times 12 = 780 \). \$780

4. Fixed costs are \$1,230 and net pay is \$3,120. Find fixed costs as a percent of net pay.
   Show the full solution

   \( \dfrac{1230}{3120} \times 100 = 39.42 \). 39.42 percent

5. A worker needs \$2,400 a year for car repairs. How much goes into the sinking fund each month?
   Show the full solution

   \( \dfrac{2400}{12} = 200 \). \$200 a month

6. Car insurance is paid in four installments of \$168.50 a year. Find the annual cost and the monthly equivalent.
   Show the full solution

   Annual \( 4 \times 168.50 = 674 \). Monthly \( \dfrac{674}{12} = 56.17 \). \$674.00 a year; \$56.17 a month

7. Six months of water bills are \$86, \$94, \$101, \$77, \$65 and \$71. Find the average and say why an average is used in the budget.
   Show the full solution

   The total is \( 86 + 94 + 101 + 77 + 65 + 71 = 494 \), and \( \dfrac{494}{6} = 82.33 \). A variable cost cannot be known in advance, so the average of recent months is the best single estimate to plan with. \$82.33 a month

8. Yearly costs are \$240, \$360 and \$150. Find the monthly sinking fund.
   Show the full solution

   The total is \( 240 + 360 + 150 = 750 \), and \( \dfrac{750}{12} = 62.50 \). \$62.50 a month

9. Classify as fixed, variable or irregular: rent, groceries, car registration, a phone plan, holiday gifts, gasoline.
   Show the full solution

   Rent and a phone plan are the same every month, so fixed. Groceries and gasoline change from month to month, so variable. Car registration and holiday gifts come once a year, so irregular. Fixed: rent, phone. Variable: groceries, gasoline. Irregular: registration, gifts

10. A student's budget shows a monthly surplus of \$400, but it leaves out \$2,532 of yearly costs. Find the true monthly surplus.
    Show the full solution

    The yearly costs are \( \dfrac{2532}{12} = 211 \) dollars a month. The true surplus is \( 400 - 211 = 189 \) dollars, less than half of what the budget showed. \$189 a month

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A register shows a balance of \$1,250.00. A \$300.00 deposit and a \$120.50 check are recorded. Find the new balance.
   Show the full solution

   \( 1250.00 + 300.00 - 120.50 = 1429.50 \). \$1,429.50

2. Two checks, \$45.10 and \$62.75, have not cleared. Find the total outstanding.
   Show the full solution

   \( 45.10 + 62.75 = 107.85 \). \$107.85

3. A statement shows \$900.00, with a \$150.00 deposit in transit and \$107.85 of checks outstanding. Find the adjusted bank balance.
   Show the full solution

   \( 900.00 + 150.00 - 107.85 = 942.15 \). \$942.15

4. A register shows \$1,420.30. The statement lists a \$12.00 fee and \$3.45 of interest that are not in the register. Find the adjusted register balance.
   Show the full solution

   \( 1420.30 - 12.00 + 3.45 = 1411.75 \). \$1,411.75

5. A statement shows \$2,310.75 and there is no deposit in transit. Checks totaling \$650.25 are outstanding. Find the adjusted bank balance.
   Show the full solution

   \( 2310.75 - 650.25 = 1660.50 \). \$1,660.50

6. For that account the register shows \$1,671.65 before recording a \$12.00 fee and \$0.85 interest. Does it reconcile?
   Show the full solution

   Adjusted register: \( 1671.65 - 12.00 + 0.85 = 1660.50 \). This matches the adjusted bank balance of \$1,660.50. Yes, both are \$1,660.50

7. Suppose instead the adjusted register balance is \$1,651.50 against \$1,660.50. The difference is divisible by 9. Suggest a likely cause.
   Show the full solution

   The difference is \( 1660.50 - 1651.50 = 9.00 \). A difference divisible by 9 often means two digits were swapped. The register is lower than the bank, so a payment was recorded for too much, for example \$54.00 entered for a check that was really \$45.00. A transposed figure, such as 54 for 45

8. A check for \$98.00 was recorded in the register as \$89.00, and the register balance is \$742.10. Find the corrected balance.
   Show the full solution

   The register subtracted \$9.00 too little, so its balance is \$9.00 too high: \( 742.10 - 9.00 = 733.10 \). \$733.10

9. A statement shows \$1,100.00, a deposit of \$250.00 is in transit and a \$90.00 check is outstanding. Find the adjusted bank balance.
   Show the full solution

   \( 1100.00 + 250.00 - 90.00 = 1260.00 \). \$1,260.00

10. A student uses the example's figures and gets \$1,622.14 by adding the outstanding checks to the statement balance and ignoring the deposit. Find the errors and the correct balance.
    Show the full solution

    The student computed \( 1284.50 + 337.64 = 1622.14 \). There are two mistakes: outstanding checks must be subtracted, because the money is already committed, and the \$420.00 deposit in transit must be added. The correct adjusted balance is \( 1284.50 + 420.00 - 337.64 = 1366.86 \). \$1,366.86

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Round APY to two decimal places unless told otherwise.

1. Find the APY of an account paying 3 percent compounded monthly.
   Show the full solution

   \( \left(1 + \dfrac{0.03}{12}\right)^{12} - 1 = (1.0025)^{12} - 1 = 0.030416 \). 3.04 percent

2. An account pays 1.5 percent APY. Find the interest on \$8,000 after one year.
   Show the full solution

   \( 8000 \times 0.015 = 120 \). \$120.00

3. Find the balance after one year on \$12,000 at 4.2 percent APY.
   Show the full solution

   \( 12000 \times 1.042 = 12504 \). \$12,504.00

4. Find the APY of an account paying 6 percent compounded monthly.
   Show the full solution

   \( (1.005)^{12} - 1 = 0.061678 \). 6.17 percent

5. Which is better: 4.5 percent compounded quarterly, or 4.55 percent compounded once a year?
   Show the full solution

   The quarterly account has an APY of \( (1.01125)^4 - 1 = 0.045765 \), or 4.58 percent. The annual account has an APY of 4.55 percent. The quarterly account is better. 4.5 percent quarterly (4.58 percent APY)

6. Find the balance after two years on \$2,500 at 4.4 percent APY.
   Show the full solution

   The APY applies once each year, so \( 2500 \times (1.044)^2 = 2500 \times 1.089936 = 2724.84 \). \$2,724.84

7. A checking account charges \$12 a month and pays 4.0 percent APY on an \$800 balance. Find the net result over a year.
   Show the full solution

   Interest: \( 800 \times 0.04 = 32 \). Fees: \( 12 \times 12 = 144 \). Net \( 32 - 144 = -112 \). The account loses \$112 a year. A loss of \$112.00

8. Find the monthly rate that compounds to an APY of 5.0 percent.
   Show the full solution

   Solve \( (1 + i)^{12} = 1.05 \): \( 1 + i = 1.05^{1/12} = 1.004074 \), so \( i = 0.4074 \) percent a month. This is a little under one-twelfth of 5 percent (0.4167), because the interest compounds. About 0.4074 percent a month

9. Compare the APY of 3 percent compounded monthly, quarterly and daily.
   Show the full solution

   Monthly: 3.0416 percent. Quarterly: \( (1.0075)^4 - 1 = 3.0339 \) percent. Daily: \( \left(1 + \dfrac{0.03}{365}\right)^{365} - 1 = 3.0453 \) percent. More frequent compounding pays more, but the gap from quarterly to daily is only 0.0114 of a point. 3.0339, 3.0416 and 3.0453 percent

10. A student says, "APY and the stated rate are the same thing." Use 6 percent compounded monthly on \$10,000 to show the difference in dollars.
    Show the full solution

    The stated rate of 6 percent would give \( 10000 \times 0.06 = 600 \) if the interest were paid once at year end. With monthly compounding the APY is 6.1678 percent, so the interest is \( 10000 \times 0.061678 = 616.78 \). The APY gives \$16.78 more, because interest added each month earns interest. APY is higher: \$616.78 against \$600.00

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A monthly fee of \$8.95 is charged every month. Find the annual cost.
   Show the full solution

   \( 8.95 \times 12 = 107.40 \). \$107.40

2. A customer is charged three overdraft fees of \$34. Find the total.
   Show the full solution

   \( 3 \times 34 = 102 \). \$102

3. An overdraft fee of \$34 was caused by a \$12 purchase. Express the fee as a percent of the purchase.
   Show the full solution

   \( \dfrac{34}{12} \times 100 = 283.33 \). 283.33 percent

4. Each outside ATM withdrawal costs \$2.50 from the bank and \$3.00 from the owner. Find the cost of four withdrawals.
   Show the full solution

   \( 4 \times (2.50 + 3.00) = 22.00 \). \$22.00

5. Find the yearly cost if a customer makes four outside withdrawals every month at those fees.
   Show the full solution

   Each month costs \$22.00, and \( 22 \times 12 = 264 \). \$264.00

6. Account A charges \$15 a month with no minimum balance. Account B is free if the balance stays at \$2,000, which the customer could otherwise invest at 4.0 percent. Find the annual cost of each and the difference.
   Show the full solution

   Account A costs \( 15 \times 12 = 180 \). Account B costs the interest given up, \( 2000 \times 0.04 = 80 \). B is cheaper by \( 180 - 80 = 100 \) dollars a year. A: \$180; B: \$80; B saves \$100

7. A free account requires a \$2,500 minimum balance. Money could otherwise earn 4.4 percent. Find the yearly cost of the requirement.
   Show the full solution

   \( 2500 \times 0.044 = 110 \). The account is not free; it costs about \$110 a year in lost interest. \$110.00

8. A customer overdraws three times a month at \$34 each. Find the yearly cost.
   Show the full solution

   \( 3 \times 34 \times 12 = 1224 \). This is more than many people save in a year. \$1,224

9. A card charges \$0.35 for each debit purchase. A customer makes 8 purchases a month. Find the monthly and yearly cost.
   Show the full solution

   Monthly \( 8 \times 0.35 = 2.80 \). Yearly \( 2.80 \times 12 = 33.60 \). \$2.80 a month; \$33.60 a year

10. A student says, "A \$34 overdraft fee is small compared with my \$1,000 balance." Explain why that is the wrong comparison, using a \$4.50 coffee.
    Show the full solution

    The fee should be measured against what triggered it. On a \$4.50 purchase, \$34 is 755.56 percent, a rate no lender could charge openly. It can also be charged again on the next transaction. Comparing it with the balance hides both facts. The fee is 755.56 percent of the purchase

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Ignore interest.

1. Essential expenses are \$2,200 a month. Find a three-month emergency fund.
   Show the full solution

   \( 3 \times 2200 = 6600 \). \$6,600

2. How many months does it take to save \$6,600 at \$275 a month?
   Show the full solution

   \( \dfrac{6600}{275} = 24 \). 24 months

3. A worker wants \$4,800 in 16 months. Find the monthly deposit.
   Show the full solution

   \( \dfrac{4800}{16} = 300 \). \$300 a month

4. A saver has \$1,850 toward a \$7,400 goal. Find the percent of the goal reached.
   Show the full solution

   \( \dfrac{1850}{7400} \times 100 = 25 \). 25 percent

5. A worker wants \$9,000 in 18 months. Find the monthly deposit.
   Show the full solution

   \( \dfrac{9000}{18} = 500 \). \$500 a month

6. A goal is \$6,000 and \$1,750 is already saved. At \$250 a month, how many months remain?
   Show the full solution

   Still needed \( 6000 - 1750 = 4250 \), and \( \dfrac{4250}{250} = 17 \). 17 months

7. Essential expenses are \$2,450 a month. Find a four-month fund, and the months needed at \$350 a month.
   Show the full solution

   Fund \( 4 \times 2450 = 9800 \). Months \( \dfrac{9800}{350} = 28 \). \$9,800; 28 months

8. A saver wants \$2,400 for a down payment and saves \$150 a month plus \$75 from occasional income. How many months are needed?
   Show the full solution

   Monthly saving is \( 150 + 75 = 225 \). Then \( \dfrac{2400}{225} = 10.67 \), which rounds up to 11 months, since 10 months gives only \$2,250. 11 months

9. A worker with \$2,400 of net pay a month saves 20 percent. Find the annual savings.
   Show the full solution

   Monthly \( 0.20 \times 2400 = 480 \). Annual \( 480 \times 12 = 5760 \). \$5,760

10. A student sizes an emergency fund as three months of income, \$3,120 a month, getting \$9,360, although essential expenses are \$1,985. Find the error and the right target.
    Show the full solution

    An emergency fund replaces the expenses that cannot be skipped, not the income. The right target is \( 3 \times 1985 = 5955 \). The income-based figure is \( 9360 - 5955 = 3405 \) dollars too large, which adds months of saving for no added protection. \$5,955, not \$9,360

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the simple interest on \$1,200 at 5 percent for 2 years.
   Show the full solution

   \( 1200 \times 0.05 \times 2 = 120 \). \$120.00

2. Find the total amount after 2 years on that loan.
   Show the full solution

   \( 1200 + 120 = 1320 \). \$1,320.00

3. Find the simple interest on \$3,000 at 6 percent for 9 months.
   Show the full solution

   Nine months is \( \dfrac{9}{12} = 0.75 \) of a year: \( 3000 \times 0.06 \times 0.75 = 135 \). \$135.00

4. Find the simple interest on \$500 at 4 percent for 90 days.
   Show the full solution

   \( t = \dfrac{90}{365} = 0.24658 \), and \( 500 \times 0.04 \times 0.24658 = 4.93 \). \$4.93

5. Find the rate if \$2,000 earns \$180 in 3 years.
   Show the full solution

   \( r = \dfrac{180}{2000 \times 3} = 0.03 \). 3 percent

6. Find the time for \$4,000 at 4.5 percent to earn \$360.
   Show the full solution

   \( t = \dfrac{360}{4000 \times 0.045} = \dfrac{360}{180} = 2 \). 2 years

7. Find the principal that earns \$270 at 3 percent in 2.5 years.
   Show the full solution

   \( P = \dfrac{270}{0.03 \times 2.5} = \dfrac{270}{0.075} = 3600 \). \$3,600.00

8. \$5,000 grows to \$5,600 in 2 years at simple interest. Find the annual rate.
   Show the full solution

   Interest is \( 5600 - 5000 = 600 \), and \( r = \dfrac{600}{5000 \times 2} = 0.06 \). 6 percent

9. A borrower takes \$2,500 at 12 percent simple interest for 6 months. Find the amount repaid.
   Show the full solution

   \( I = 2500 \times 0.12 \times 0.5 = 150 \), and \( A = 2500 + 150 = 2650 \). \$2,650.00

10. A student finds the interest on \$8,000 at 4.5 percent for 18 months as \( 8000 \times 0.045 \times 18 = 6480 \). Find the error.
    Show the full solution

    The rate is annual, so the time must be in years: 18 months is 1.5 years. The correct interest is \( 8000 \times 0.045 \times 1.5 = 540 \). The student's answer is 12 times too large, and more than the principal itself. \$540.00

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the value of \$1,000 at 5 percent compounded annually for 2 years.
   Show the full solution

   \( 1000 \times (1.05)^2 = 1000 \times 1.1025 = 1102.50 \). \$1,102.50

2. Find the value of \$2,500 at 4 percent compounded annually for 10 years.
   Show the full solution

   \( 2500 \times (1.04)^{10} = 2500 \times 1.480244 = 3700.61 \). \$3,700.61

3. Find the value of \$3,000 at 6 percent compounded monthly for 5 years.
   Show the full solution

   The rate per month is \( \dfrac{0.06}{12} = 0.005 \) and there are 60 periods: \( 3000 \times (1.005)^{60} = 3000 \times 1.348850 = 4046.55 \). \$4,046.55

4. Find the growth factor for one quarter at 5 percent compounded quarterly.
   Show the full solution

   \( 1 + \dfrac{0.05}{4} = 1.0125 \). 1.0125

5. Find the interest earned on \$1,500 at 3 percent compounded annually for 4 years.
   Show the full solution

   \( 1500 \times (1.03)^4 = 1500 \times 1.125509 = 1688.26 \), so the interest is \( 1688.26 - 1500 = 188.26 \). \$188.26

6. How much must be deposited now at 5 percent compounded annually to have \$10,000 in 8 years?
   Show the full solution

   \( P = \dfrac{10000}{(1.05)^8} = \dfrac{10000}{1.477455} = 6768.39 \). \$6,768.39

7. Find the value of \$2,000 at 5 percent compounded quarterly for 6 years.
   Show the full solution

   The rate per quarter is 0.0125 and there are 24 periods: \( 2000 \times (1.0125)^{24} = 2000 \times 1.347349 = 2694.70 \). \$2,694.70

8. Find the value of \$5,000 at 7 percent for 30 years compounded annually, and with simple interest. Find the ratio.
   Show the full solution

   Compound: \( 5000 \times (1.07)^{30} = 5000 \times 7.612255 = 38061.28 \). Simple: \( 5000(1 + 0.07 \times 30) = 15500 \). The ratio is \( \dfrac{38061.28}{15500} = 2.46 \), so compounding gives about two and a half times as much. \$38,061.28 against \$15,500.00

9. At 8 percent compounded annually, by what factor does money grow in 10 years?
   Show the full solution

   \( (1.08)^{10} = 2.158925 \). Money a little more than doubles in ten years. About 2.16

10. A student finds the value of \$1,000 at 5 percent for 10 years as \( 1000 \times 1.05 \times 10 = 10{,}500 \). Find the error and the correct value.
    Show the full solution

    The student multiplied by the number of years instead of raising the factor to that power. Each year multiplies the balance by 1.05, so ten years gives \( 1000 \times (1.05)^{10} = 1628.89 \). The wrong answer is more than six times too large. \$1,628.89

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the value of \$3,000 at 4 percent compounded monthly for 6 years.
   Show the full solution

   The rate per month is \( \dfrac{0.04}{12} = 0.003333 \), for 72 periods: \( 3000 \times \left(1 + \dfrac{0.04}{12}\right)^{72} = 3812.23 \). \$3,812.23

2. Find the value of \$3,000 at 4 percent compounded continuously for 6 years.
   Show the full solution

   \( 3000 \times e^{0.04 \times 6} = 3000 \times e^{0.24} = 3000 \times 1.271249 = 3813.75 \). \$3,813.75

3. Find the value of \$1,000 at 5 percent for 1 year compounded daily, and annually.
   Show the full solution

   Daily: \( 1000 \times \left(1 + \dfrac{0.05}{365}\right)^{365} = 1051.27 \). Annually: \( 1000 \times 1.05 = 1050.00 \). Daily compounding earns \$1.27 more. \$1,051.27 and \$1,050.00

4. Find the value of \$12,000 at 3.6 percent compounded semiannually for 4 years.
   Show the full solution

   The rate per period is \( \dfrac{0.036}{2} = 0.018 \) and there are 8 periods: \( 12000 \times (1.018)^8 = 13840.87 \). \$13,840.87

5. Find the value of \$2,000 at 4.5 percent compounded continuously for 8 years.
   Show the full solution

   \( 2000 \times e^{0.36} = 2000 \times 1.433329 = 2866.66 \). \$2,866.66

6. Compare \$7,500 at 5 percent for 3 years compounded quarterly and monthly. Find the difference.
   Show the full solution

   Quarterly: \( 7500 \times (1.0125)^{12} = 8705.66 \). Monthly: \( 7500 \times \left(1 + \dfrac{0.05}{12}\right)^{36} = 8711.04 \). The difference is \( 8711.04 - 8705.66 = 5.38 \) dollars over three years. \$5.38 more with monthly

7. Compute \( \left(1 + \dfrac{1}{1000}\right)^{1000} \) and compare it with \( e \).
   Show the full solution

   \( (1.001)^{1000} = 2.7169 \), slightly below \( e = 2.7183 \). The values climb toward \( e \) as \( n \) grows but never pass it. 2.7169, just under \( e \)

8. For \$10,000 at 6 percent over 10 years, how much more does daily compounding earn than monthly?
   Show the full solution

   Daily \$18,220.29 minus monthly \$18,193.97 is \$26.32, which is a tiny 0.14 percent of the balance. The jump from annual to monthly is worth \$285.49. \$26.32

9. Compare \$1,000 for one year at 5 percent compounded continuously with 5.1 percent compounded annually.
   Show the full solution

   Continuous: \( 1000 \times e^{0.05} = 1051.27 \). Annual at 5.1: \( 1000 \times 1.051 = 1051.00 \). Continuous at 5 percent beats a higher stated rate of 5.1 percent compounded once, by \$0.27. Continuous is better by \$0.27

10. A student puts \( r = 6 \) in \( A = Pe^{rt} \) for \$10,000 over 10 years. Find the error and what answer results.
    Show the full solution

    The rate must be a decimal: 0.06, not 6. The student computed \( 10000 \times e^{60} \), about \( 1.1 \times 10^{30} \) dollars, more money than exists. The correct value is \( 10000 \times e^{0.6} = 18221.19 \). An answer that large is a signal to recheck the decimal. \$18,221.19

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Round rates to two decimal places.

1. Find the APY of a 5 percent rate compounded quarterly.
   Show the full solution

   \( (1.0125)^4 - 1 = 0.050945 \). 5.09 percent

2. Find the effective annual rate of an 18 percent APR compounded monthly.
   Show the full solution

   \( (1.015)^{12} - 1 = 0.195618 \). 19.56 percent

3. Find the interest on \$1,000 for a year at 12 percent compounded monthly.
   Show the full solution

   \( 1000 \times (1.01)^{12} = 1126.83 \), so the interest is \$126.83 and the APY is 12.68 percent. \$126.83

4. Find the APY of 4.75 percent compounded daily.
   Show the full solution

   \( \left(1 + \dfrac{0.0475}{365}\right)^{365} - 1 = 0.048643 \). 4.86 percent

5. Find the APY of 6 percent compounded annually.
   Show the full solution

   With \( n = 1 \) there is no extra compounding: \( 1.06 - 1 = 0.06 \). 6.00 percent

6. A loan of \$2,000 at 15 percent compounded monthly is left unpaid for a year. Find the interest owed.
   Show the full solution

   \( 2000 \times (1.0125)^{12} = 2321.51 \), so the interest is \( 2321.51 - 2000 = 321.51 \), an effective rate of 16.08 percent. \$321.51

7. Which earns more: 7.9 percent compounded semiannually or 7.85 percent compounded monthly?
   Show the full solution

   The first has APY \( (1.0395)^2 - 1 = 8.056 \) percent. The second has APY \( (1.006542)^{12} - 1 = 8.139 \) percent. The monthly account earns more despite its lower stated rate. 7.85 percent monthly (8.14 percent APY)

8. Compare the effective rates of a 29.99 percent card compounded daily and monthly.
   Show the full solution

   Daily: \( \left(1 + \dfrac{0.2999}{365}\right)^{365} - 1 = 34.96 \) percent. Monthly: \( (1.024992)^{12} - 1 = 34.48 \) percent. Daily compounding costs 0.48 of a point more. 34.96 percent and 34.48 percent

9. What nominal rate compounded monthly gives an APY of 6 percent?
   Show the full solution

   Solve \( (1 + i)^{12} = 1.06 \): \( i = 1.06^{1/12} - 1 = 0.004868 \). The nominal annual rate is \( 12 \times 0.004868 = 0.05841 \). About 5.84 percent

10. A student says, "An APR of 12 percent means I pay 12 percent." Use \$1,000 compounded monthly to correct this.
    Show the full solution

    At 1 percent a month, \$1,000 becomes \( 1000 \times (1.01)^{12} = 1126.83 \), so the interest is \$126.83, which is 12.68 percent. The stated 12 percent would be \$120.00. The effective cost is higher than the APR, and the gap widens as the rate rises. The effective rate is 12.68 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Use the rule of 72 to estimate the years to double at 4 percent.
   Show the full solution

   \( \dfrac{72}{4} = 18 \). 18 years

2. Use the rule of 72 to estimate the years to double at 9 percent.
   Show the full solution

   \( \dfrac{72}{9} = 8 \). 8 years

3. What rate doubles money in 6 years, by the rule of 72?
   Show the full solution

   \( \dfrac{72}{6} = 12 \). 12 percent

4. Find the exact time for money to double at 7 percent compounded annually.
   Show the full solution

   \( t = \dfrac{\ln 2}{\ln 1.07} = \dfrac{0.693147}{0.067659} = 10.24 \). The rule gives \( 72 \div 7 = 10.29 \). 10.24 years

5. Find the time for \$12,000 to grow to \$20,000 at 4.5 percent compounded annually.
   Show the full solution

   \( (1.045)^t = \dfrac{20000}{12000} = 1.6667 \), so \( t = \dfrac{\ln 1.6667}{\ln 1.045} = \dfrac{0.510826}{0.044017} = 11.61 \). 11.61 years

6. Find the time for \$2,500 to grow to \$4,000 at 6 percent compounded monthly.
   Show the full solution

   \( (1.005)^{12t} = 1.6 \), so \( 12t = \dfrac{\ln 1.6}{\ln 1.005} = \dfrac{0.470004}{0.004988} = 94.23 \) months. Then \( t = \dfrac{94.23}{12} = 7.85 \) years. 7.85 years

7. Estimate and then find exactly the time to quadruple at 6 percent compounded annually.
   Show the full solution

   Quadrupling is two doublings, so the estimate is \( 2 \times 12 = 24 \) years. Exactly, \( t = \dfrac{\ln 4}{\ln 1.06} = \dfrac{1.386294}{0.058269} = 23.79 \). About 24 years; exactly 23.79 years

8. Find the time for money to double at 5 percent compounded continuously.
   Show the full solution

   Solve \( e^{0.05t} = 2 \): \( t = \dfrac{\ln 2}{0.05} = 13.86 \). 13.86 years

9. Compare the rule of 72 with the exact doubling time at 3 percent.
   Show the full solution

   The rule gives \( 72 \div 3 = 24 \) years. The exact time is \( \dfrac{\ln 2}{\ln 1.03} = 23.45 \) years. The rule is off by about half a year, a 2.3 percent overestimate. 24 years against 23.45 years

10. A student estimates the time to double at 6 percent as \( 72 \div 0.06 = 1200 \) years. Find the error and the correct estimate.
    Show the full solution

    The rule of 72 takes the rate as a percent number, so it is \( 72 \div 6 \), not \( 72 \div 0.06 \). The correct estimate is 12 years. An answer of 1,200 years is absurd for a 6 percent account and should trigger a check of the units. 12 years

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. An item costs \$40 today. Find the price in 10 years at 3 percent inflation.
   Show the full solution

   \( 40 \times (1.03)^{10} = 40 \times 1.343916 = 53.76 \). \$53.76

2. A car costs \$25,000. Find the price in 5 years at 2.5 percent inflation.
   Show the full solution

   \( 25000 \times (1.025)^5 = 25000 \times 1.131408 = 28285.21 \). \$28,285.21

3. Find the buying power, in today's dollars, of \$500 received in 15 years at 2 percent inflation.
   Show the full solution

   \( \dfrac{500}{(1.02)^{15}} = \dfrac{500}{1.345868} = 371.51 \). \$371.51

4. Find the real rate when the nominal rate is 3 percent and inflation is 2 percent.
   Show the full solution

   \( \dfrac{1.03}{1.02} - 1 = 0.009804 \). 0.98 percent

5. College costs \$12,000 a year and rise 4 percent a year. Find the cost in 8 years.
   Show the full solution

   \( 12000 \times (1.04)^8 = 12000 \times 1.368569 = 16422.83 \). \$16,422.83

6. A salary of \$50,000 does not change for 10 years while inflation is 2.5 percent. Find the salary's buying power in today's dollars at the end.
   Show the full solution

   \( \dfrac{50000}{(1.025)^{10}} = \dfrac{50000}{1.280085} = 39059.92 \). The worker has taken a cut of about 22 percent in buying power. \$39,059.92

7. Inflation this year was 3.2 percent. What salary does a \$48,000 worker need next year to keep the same buying power?
   Show the full solution

   \( 48000 \times 1.032 = 49536 \). \$49,536

8. An account pays 1.5 percent while inflation is 3 percent. Find the real rate.
   Show the full solution

   \( \dfrac{1.015}{1.03} - 1 = -0.014563 \). The balance rises but its buying power falls 1.46 percent a year. Negative 1.46 percent

9. By the rule of 72, in how many years do prices double at 3 percent inflation?
   Show the full solution

   \( \dfrac{72}{3} = 24 \) years. A basket that costs \$100 today costs about \$200 then. 24 years

10. A student keeps \$10,000 in a drawer for 20 years and says, "It is the same \$10,000, so nothing is lost." Find the error using 3 percent inflation.
    Show the full solution

    The number of dollars is unchanged, but what they buy is not. At 3 percent inflation the \$10,000 buys what \( \dfrac{10000}{(1.03)^{20}} = 5536.76 \) buys today. The loss in buying power is \( 10000 - 5536.76 = 4463.24 \) dollars. A loss of \$4,463.24 in buying power

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the APY of 3.9 percent compounded monthly.
   Show the full solution

   \( \left(1 + \dfrac{0.039}{12}\right)^{12} - 1 = (1.00325)^{12} - 1 = 0.039705 \). 3.97 percent

2. Compare \$12,000 for 3 years at 4.1 percent compounded quarterly with 4.12 percent compounded annually.
   Show the full solution

   Quarterly: \( 12000 \times (1.01025)^{12} = 13562.12 \). Annual: \( 12000 \times (1.0412)^3 = 13545.15 \). The quarterly account earns \$16.97 more. \$13,562.12 against \$13,545.15

3. An account with a \$60 annual fee pays 4.5 percent APY on \$5,000. Find the net gain and the net rate.
   Show the full solution

   Interest \( 5000 \times 0.045 = 225 \). Net \( 225 - 60 = 165 \). The net rate is \( \dfrac{165}{5000} \times 100 = 3.3 \) percent. \$165.00; 3.30 percent

4. An account requires a \$10,000 minimum and pays 4.6 percent APY. Find the interest for one year.
   Show the full solution

   \( 10000 \times 0.046 = 460 \). \$460.00

5. Find the value of \$7,000 at 4.4 percent compounded monthly for 3 years.
   Show the full solution

   \( 7000 \times \left(1 + \dfrac{0.044}{12}\right)^{36} = 7000 \times 1.140833 = 7985.83 \). \$7,985.83

6. A two-year CD pays 4.8 percent APY. Find the value of an \$8,000 deposit at the end.
   Show the full solution

   \( 8000 \times (1.048)^2 = 8000 \times 1.098304 = 8786.43 \). \$8,786.43

7. That CD charges an early-withdrawal penalty of three months' interest at the stated rate. Find the penalty on \$8,000 at 4.8 percent.
   Show the full solution

   Three months is a quarter of a year: \( 8000 \times 0.048 \times 0.25 = 96 \). \$96.00

8. If the saver cashes out the CD at the end of the term and pays the penalty, find the net gain.
   Show the full solution

   Gain before the penalty: \( 8786.43 - 8000 = 786.43 \). After the penalty: \( 786.43 - 96 = 690.43 \). If the account is held to the term, the penalty does not apply; it is charged only on an early withdrawal. \$690.43 with the penalty

9. Compare \$6,000 for 3 years at 4.1 percent compounded monthly with 4.15 percent compounded annually.
   Show the full solution

   Monthly: \( 6000 \times \left(1 + \dfrac{0.041}{12}\right)^{36} = 6783.88 \). Annual: \( 6000 \times (1.0415)^3 = 6778.43 \). The monthly account wins by \$5.45 despite a lower stated rate. 4.1 percent monthly, by \$5.45

10. A student chooses 5.00 percent compounded annually over 4.95 percent compounded daily because "5.00 is higher." Find the error.
    Show the full solution

    The APY of 4.95 percent compounded daily is \( \left(1 + \dfrac{0.0495}{365}\right)^{365} - 1 = 5.0742 \) percent, which exceeds 5.00 percent. On \$10,000 for a year the gap is \$7.42. The stated rates cannot be compared when compounding differs; only APYs can. 4.95 percent daily is better (5.07 percent APY)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Deposits are at the end of each period and compounding matches the deposits.

1. Find the future value of \$100 deposited monthly for 5 years at 6 percent compounded monthly.
   Show the full solution

   \( i = 0.005 \), \( n = 60 \). The factor is \( \dfrac{(1.005)^{60} - 1}{0.005} = 69.7700 \), so \( 100 \times 69.7700 = 6977.00 \). \$6,977.00

2. Find the future value of \$500 deposited quarterly for 10 years at 4 percent compounded quarterly.
   Show the full solution

   \( i = 0.01 \), \( n = 40 \). The factor is \( \dfrac{(1.01)^{40} - 1}{0.01} = 48.8864 \), and \( 500 \times 48.8864 = 24443.19 \). \$24,443.19

3. Find the future value of \$250 deposited monthly for 20 years at 3 percent compounded monthly, and the interest earned.
   Show the full solution

   \( i = 0.0025 \), \( n = 240 \). Future value \( 250 \times \dfrac{(1.0025)^{240} - 1}{0.0025} = 82075.50 \). Contributions are \( 250 \times 240 = 60000 \), so the interest is \$22,075.50. \$82,075.50; interest \$22,075.50

4. Find the future value of \$1,200 deposited once a year for 25 years at 5 percent compounded annually.
   Show the full solution

   \( 1200 \times \dfrac{(1.05)^{25} - 1}{0.05} = 1200 \times 47.7271 = 57272.52 \). \$57,272.52

5. A saver deposits \$100 a month for 5 years in an account that pays no interest. Find the balance.
   Show the full solution

   With no interest, the balance is the contributions: \( 100 \times 60 = 6000 \). The interest-bearing account in problem 1 held \$977 more. \$6,000.00

6. Find the future value of \$200 a month for 30 years at 8 percent compounded monthly, and the share that is interest.
   Show the full solution

   \( i = \dfrac{0.08}{12} \), \( n = 360 \). \( 200 \times \dfrac{(1.006667)^{360} - 1}{0.006667} = 298071.89 \). Contributions are \( 200 \times 360 = 72000 \), so interest is \( 226071.89 \), which is \( \dfrac{226071.89}{298071.89} = 75.8 \) percent of the balance. \$298,071.89; about three quarters is interest

7. Which gives more at 6 percent compounded monthly: doubling the deposit to \$400 a month for 10 years, or keeping \$200 a month for 20 years?
   Show the full solution

   Doubling the deposit doubles the result of the \$200 case: \( 2 \times 32775.87 = 65551.74 \). Doubling the time gives \( 200 \times \dfrac{(1.005)^{240} - 1}{0.005} = 92408.18 \). Time wins, because the earlier deposits have many years to compound. \$200 for 20 years: \$92,408.18, against \$65,551.74

8. Four deposits of \$1,000 are made at the end of each year at 10 percent compounded annually. Add up the growth of each deposit and check against the formula.
   Show the full solution

   The first grows for 3 years: \( 1000 \times 1.331 = 1331 \). The second for 2 years: 1210. The third for one year: 1100. The fourth not at all: 1000. The sum is \( 1331 + 1210 + 1100 + 1000 = 4641 \). The formula gives \( 1000 \times \dfrac{(1.1)^4 - 1}{0.1} = 4641 \). \$4,641.00

9. What is the annuity factor for \( i = 0.005 \) and \( n = 120 \), and what does it mean?
   Show the full solution

   \( \dfrac{(1.005)^{120} - 1}{0.005} = 163.88 \). It means that each dollar deposited monthly for 10 years at this rate becomes \$163.88, although only \$120 was put in. 163.88

10. A student estimates the future value of \$200 a month for 10 years at 6 percent as \( 200 \times 120 \times 1.005 = 24{,}120 \). Find the error.
    Show the full solution

    The student added one month of interest to the total contributions. Each deposit earns interest for every month it stays in the account, which is up to 119 months. The correct value is \( 200 \times 163.8793 = 32775.87 \). The estimate is about \$8,650 too low. \$32,775.87

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the present value of \$10,000 due in 5 years at 4 percent compounded annually.
   Show the full solution

   \( \dfrac{10000}{(1.04)^5} = \dfrac{10000}{1.216653} = 8219.27 \). \$8,219.27

2. Find the present value of \$5,000 due in 3 years at 6 percent compounded monthly.
   Show the full solution

   \( \dfrac{5000}{(1.005)^{36}} = \dfrac{5000}{1.196681} = 4178.22 \). \$4,178.22

3. Find the present value of \$8,000 due in 10 years at 7 percent compounded annually.
   Show the full solution

   \( \dfrac{8000}{(1.07)^{10}} = \dfrac{8000}{1.967151} = 4066.79 \). \$4,066.79

4. Find the present value of \$1,000 due in one year at 10 percent.
   Show the full solution

   \( \dfrac{1000}{1.10} = 909.09 \). \$909.09

5. How much must be invested now at 4.5 percent compounded annually to have \$25,000 in 12 years?
   Show the full solution

   \( \dfrac{25000}{(1.045)^{12}} = \dfrac{25000}{1.695881} = 14741.60 \). \$14,741.60

6. You can have \$50,000 now or \$60,000 in 4 years. At 5 percent compounded annually, which is better, and by how much in today's dollars?
   Show the full solution

   The present value of \$60,000 is \( \dfrac{60000}{(1.05)^4} = 49362.15 \). That is \$637.85 less than \$50,000, so take the \$50,000 now. \$50,000 now, by \$637.85

7. An offer pays \$12,000 in 6 years, or \$9,000 now. At 6 percent compounded annually, which is better?
   Show the full solution

   \( \dfrac{12000}{(1.06)^6} = \dfrac{12000}{1.418519} = 8459.53 \). This is less than \$9,000, so the \$9,000 now is better. \$9,000 now

8. Find the annual rate at which \$900 now equals \$1,000 in two years.
   Show the full solution

   Solve \( 900(1 + r)^2 = 1000 \): \( (1 + r)^2 = 1.1111 \), so \( 1 + r = 1.054093 \) and \( r = 5.41 \) percent. 5.41 percent

9. Find the present value of \$10,000 due in 10 years at 3 percent and at 9 percent, and say what the difference shows.
   Show the full solution

   At 3 percent: \( \dfrac{10000}{(1.03)^{10}} = 7440.94 \). At 9 percent: \( \dfrac{10000}{(1.09)^{10}} = 4224.11 \). The higher rate makes the present value much smaller, since a smaller deposit reaches the same target. \$7,440.94 and \$4,224.11

10. A student says the present value of \$1,000 in 2 years at 5 percent is \( 1000 - 2 \times 50 = 900 \). Find the error.
    Show the full solution

    The student subtracted simple interest on the future amount. The present value is the amount that would grow to \$1,000, and it earns interest on itself, so it is found by dividing: \( \dfrac{1000}{1.1025} = 907.03 \). Check: \( 907.03 \times 1.1025 = 1000 \). \$907.03

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Deposits are at the end of each period and compounding matches the deposits.

1. Find the monthly deposit needed to have \$10,000 in 5 years at 6 percent compounded monthly.
   Show the full solution

   \( i = 0.005 \), \( n = 60 \). \( D = \dfrac{10000 \times 0.005}{(1.005)^{60} - 1} = \dfrac{50}{0.348850} = 143.33 \). \$143.33 a month

2. Find the monthly deposit needed to have \$50,000 in 10 years at 5 percent compounded monthly.
   Show the full solution

   \( i = \dfrac{0.05}{12} \), \( n = 120 \). \( (1.004167)^{120} - 1 = 0.647009 \). \( D = \dfrac{50000 \times 0.004167}{0.647009} = 321.99 \). \$321.99 a month

3. Find the monthly deposit needed to have \$20,000 in 4 years at 3 percent compounded monthly.
   Show the full solution

   \( i = 0.0025 \), \( n = 48 \). \( (1.0025)^{48} - 1 = 0.127328 \). \( D = \dfrac{20000 \times 0.0025}{0.127328} = \dfrac{50}{0.127328} = 392.69 \). \$392.69 a month

4. Find the monthly deposit needed to have \$1,000,000 in 40 years at 7 percent compounded monthly.
   Show the full solution

   \( i = \dfrac{0.07}{12} \), \( n = 480 \). \( (1.005833)^{480} - 1 = 15.3114 \). \( D = \dfrac{1000000 \times 0.005833}{15.3114} = 380.98 \). Total contributions are only \( 380.98 \times 480 = 182{,}870 \). \$380.98 a month

5. Find the quarterly deposit needed to have \$15,000 in 6 years at 4 percent compounded quarterly.
   Show the full solution

   \( i = 0.01 \), \( n = 24 \). \( (1.01)^{24} - 1 = 0.269735 \). \( D = \dfrac{15000 \times 0.01}{0.269735} = 556.10 \). \$556.10 a quarter

6. A saver has \$5,000 now and wants \$30,000 in 4 years at 4.8 percent compounded monthly. Find the monthly deposit.
   Show the full solution

   The \$5,000 grows to \( 5000 \times (1.004)^{48} = 6056.03 \). The deposits must make up \( 30000 - 6056.03 = 23943.97 \). With \( n = 48 \), \( D = \dfrac{23943.97 \times 0.004}{(1.004)^{48} - 1} = 453.47 \). \$453.47 a month

7. With no interest, how much a month saves \$12,000 in 4 years, and how does 4.8 percent change it?
   Show the full solution

   With no interest, \( \dfrac{12000}{48} = 250 \). At 4.8 percent compounded monthly, \( D = \dfrac{12000 \times 0.004}{(1.004)^{48} - 1} = 227.27 \). Interest lowers the deposit by \$22.73 a month. \$250.00 against \$227.27

8. A saver deposits \$250 a month for 4 years at 6 percent compounded monthly. Does she reach \$14,000?
   Show the full solution

   \( 250 \times \dfrac{(1.005)^{48} - 1}{0.005} = 250 \times 54.0978 = 13524.46 \). That is \( 14000 - 13524.46 = 475.54 \) short. No: she is \$475.54 short

9. Find how much a month a saver would need for the \$30,000 goal at 4.8 percent compounded monthly if she had 10 years instead of 8.
   Show the full solution

   \( n = 120 \): \( D = \dfrac{30000 \times 0.004}{(1.004)^{120} - 1} = 195.27 \), against \$256.95 for 8 years, a saving of \$61.68 a month. \$195.27 a month

10. A student plans for \$30,000 in 96 months by dividing: \( 30000 \div 96 = 312.50 \). Find the error and how much too much that saves.
    Show the full solution

    The division ignores the interest that the deposits earn. At 4.8 percent compounded monthly the deposit that reaches \$30,000 is \$256.95. The division overstates it by \( 312.50 - 256.95 = 55.55 \) dollars a month, which is safe but more than needed. The correct deposit is \$256.95

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use 7 percent compounded monthly.

1. A single deposit of \$1,000 earns 7 percent compounded monthly. Find its value after 30 years and after 40 years.
   Show the full solution

   After 30 years: \( 1000 \times (1.005833)^{360} = 8116.50 \). After 40 years: \( 1000 \times (1.005833)^{480} = 16311.41 \). Ten more years doubled the result. \$8,116.50 and \$16,311.41

2. Find the value of \$100 a month for 20, 30 and 40 years.
   Show the full solution

   20 years: \$52,092.67. 30 years: \$121,997.10. 40 years: \$262,481.34. The 20-year amounts and 30-year amounts differ by a factor of 2.3, and the 30-year and 40-year by a factor of 2.2. \$52,092.67; \$121,997.10; \$262,481.34

3. Find the cost of delaying a \$100-a-month plan from 40 years to 30 years.
   Show the full solution

   \( 262481.34 - 121997.10 = 140484.24 \). The delay cost more than the entire 30-year balance of the later plan. \$140,484.24

4. A saver deposits \$300 a month for 5 years and then stops, leaving it to grow for 35 more years. Find the final balance.
   Show the full solution

   After 5 years: \( 300 \times \dfrac{(1.005833)^{60} - 1}{0.005833} = 300 \times 71.5929 = 21477.87 \). Growing 35 years (420 months) multiplies by \( (1.005833)^{420} = 11.5062 \), so \( 21477.87 \times 11.5062 = 247127.64 \). \$247,127.64

5. That saver contributed \$18,000. Find the balance as a multiple of the contributions.
   Show the full solution

   \( \dfrac{247127.64}{18000} = 13.73 \). 13.73 times

6. How much a month must a saver who starts at 35 deposit for 30 years to match Ana's \$280,968.48?
   Show the full solution

   \( D = \dfrac{280968.48 \times 0.005833}{(1.005833)^{360} - 1} = \dfrac{1638.98}{7.1165} = 230.31 \). Ana deposited \$200 for 10 years; the late starter must deposit \$230.31 for 30. \$230.31 a month

7. Ben contributed 3 times as much as Ana. Explain in terms of the dates why his balance is smaller.
   Show the full solution

   Ana's deposits were made in years 1 through 10 and had 30 to 40 years to grow. Ben's were made in years 11 through 40 and on average had only 15 to 20 years. The growth factor for 30 years is 8.1 and for 15 years is about 2.8, so an early dollar is worth far more. Early dollars have longer to compound

8. Find the value of \$200 a month for 25 years and for 35 years.
   Show the full solution

   25 years: \( 200 \times 810.0717 = 162014.34 \). 35 years: \( 200 \times 1801.0546 = 360210.92 \). \$162,014.34 and \$360,210.92

9. How much did the extra 10 years add in the last problem, and how much more was put in?
   Show the full solution

   The gain is \( 360210.92 - 162014.34 = 198196.58 \) dollars. The extra contributions were \( 200 \times 120 = 24000 \). The added decade turned \$24,000 into \$198,196.58. \$198,196.58 from \$24,000 more

10. A student says, "Ben saved for 30 years and Ana for 10, and Ben put in three times as much, so Ben must have more at 65." Correct the claim with the balances.
    Show the full solution

    The claim compares contributions, but the question is about balances at 65. Ana has \$280,968.48 and Ben has \$243,994.20. Ben has \$36,974.28 less, despite contributing \$48,000 more, because Ana's money was in the account longer. Ana has more

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. An employer matches 100 percent of contributions up to 3 percent of a \$60,000 salary. Find the maximum match in a year.
   Show the full solution

   \( 0.03 \times 60000 = 1800 \). \$1,800

2. An employee earning \$48,000 contributes 5 percent. Find the annual contribution.
   Show the full solution

   \( 0.05 \times 48000 = 2400 \). \$2,400

3. An employer matches 50 percent of contributions up to 6 percent of pay. An employee earning \$60,000 contributes 6 percent. Find the match.
   Show the full solution

   Her contribution is \( 0.06 \times 60000 = 3600 \), and the match is \( 0.50 \times 3600 = 1800 \). \$1,800

4. Find the balance after 30 years of \$300 a month at 7 percent compounded monthly.
   Show the full solution

   \( 300 \times \dfrac{(1.005833)^{360} - 1}{0.005833} = 300 \times 1219.971 = 365991.30 \). \$365,991.30

5. Find the balance after 30 years if the same worker's deposit, with match, is \$450 a month.
   Show the full solution

   \( 450 \times 1219.971 = 548986.95 \). The match of \$150 a month is worth \( 548986.95 - 365991.30 = 182995.65 \). \$548,986.95

6. A \$1,000 pre-tax contribution is made by a worker whose marginal rate is 22 percent. Find the cost to take-home pay.
   Show the full solution

   \( 1000 \times (1 - 0.22) = 780 \). \$780

7. An employee earning \$50,000 contributes 2 percent and the employer matches 100 percent up to 4 percent. Find the match received and the match missed.
   Show the full solution

   The match received is \( 0.02 \times 50000 = 1000 \). The maximum is \( 0.04 \times 50000 = 2000 \), so \$1,000 is missed. The extra 2 percent contribution (\$1,000) would earn \$1,000 immediately, a 100 percent return. \$1,000 received; \$1,000 missed

8. An employer's match of \$1,800 a year vests 25 percent for each year of service. An employee leaves after 2 years. How much of the match is hers?
   Show the full solution

   Two years of matching total \( 2 \times 1800 = 3600 \). After two years, 50 percent has vested: \( 0.50 \times 3600 = 1800 \). Her own contributions are hers in full. \$1,800

9. Explain why the employer match is a better return than any investment.
   Show the full solution

   A 100 percent match doubles every dollar contributed, up to the limit, the moment it is deposited. No investment reliably returns 100 percent in a day. The match is also unaffected by market swings, which makes it the highest priority for savings. It is a guaranteed 100 percent return

10. A student says, "The account's interest is small; I'd have saved the money anyway." Use the example to correct this.
    Show the full solution

    In the example the worker contributes \$113,400 and the account holds \$486,284.74, so the growth is \( 486284.74 - 113400 = 372884.74 \), which is \( \dfrac{372884.74}{486284.74} = 76.68 \) percent of the balance. More than three quarters of the account came from growth, not from deposits. About 77 percent of the balance is growth

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A saver in the 24 percent bracket considers a \$5,000 traditional contribution. Find the Roth contribution that costs the same take-home pay.
   Show the full solution

   \( 5000 \times (1 - 0.24) = 3800 \). \$3,800

2. Find the value of \$3,000 in 25 years at 6 percent compounded annually, and its value after a 12 percent tax at withdrawal.
   Show the full solution

   \( 3000 \times (1.06)^{25} = 12875.61 \). After the tax, \( 12875.61 \times 0.88 = 11330.54 \). \$12,875.61 before tax; \$11,330.54 after

3. Find the tax saved this year by a \$6,000 traditional contribution in the 24 percent bracket.
   Show the full solution

   \( 6000 \times 0.24 = 1440 \). This is the same as the gap between the \$6,000 contribution and its take-home cost of \$4,560. \$1,440

4. Find the value of the \$4,560 Roth contribution at 7 percent for 30 years.
   Show the full solution

   \( 4560 \times 7.6123 = 34711.88 \), tax-free. \$34,711.88

5. An account holding \$8,000 of pre-tax money grows at 7 percent for 20 years and is taxed at 15 percent at withdrawal, while the Roth version of the same take-home cost is \( 8000 \times 0.78 \). Find the after-tax value of each.
   Show the full solution

   Growth factor \( (1.07)^{20} = 3.8697 \). Traditional: \( 8000 \times 3.8697 \times 0.85 = 26313.85 \). Roth: \( 6240 \times 3.8697 = 24146.83 \). The traditional account wins because the tax rate at withdrawal, 15 percent, is lower than the 22 percent at contribution. Traditional \$26,313.85; Roth \$24,146.83

6. Explain why the two accounts tie when the tax rate is the same at contribution and at withdrawal.
   Show the full solution

   Let the tax rate be \( t \) and the growth factor \( g \). The traditional account gives \( P \times g \times (1 - t) \) after tax. The Roth contribution that costs the same is \( P(1 - t) \), which grows to \( P(1 - t) \times g \). The two expressions are the same product in a different order. \( P \cdot g \cdot (1 - t) \) in both cases

7. Find the growth of a \$2,000 Roth contribution in 5 years at 7 percent, and the tax due on withdrawal.
   Show the full solution

   \( 2000 \times (1.07)^5 = 2805.10 \). A qualified Roth withdrawal is not taxed, so the tax due is \$0 and the saver keeps all \$2,805.10. \$2,805.10; no tax

8. A saver expects to pay 28 percent tax on withdrawals from a traditional account of \$100,000. Find the tax and the amount kept.
   Show the full solution

   \( 0.28 \times 100000 = 28000 \), and \( 100000 - 28000 = 72000 \). A balance of \$100,000 in a traditional account is not \$100,000 of spending money. \$28,000 tax; \$72,000 kept

9. A traditional balance is \$120,000 and the withdrawal tax rate is 15 percent. Find the after-tax value.
   Show the full solution

   \( 120000 \times 0.85 = 102000 \). \$102,000

10. A student says, "Roth is always better because the growth is tax-free." Test the claim against the 12 percent case in the example.
    Show the full solution

    Growth is tax-free in the Roth account, but the contribution was taxed first. At equal take-home cost and a 12 percent rate at withdrawal, the traditional account nets \$40,192.71 against the Roth's \$35,625.35. The Roth wins only when the rate at withdrawal is higher than the rate at contribution. False: traditional wins when the later rate is lower

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Withdrawals are at the end of each month.

1. Find the balance needed to withdraw \$1,000 a month for 20 years at 4 percent compounded monthly.
   Show the full solution

   \( i = \dfrac{0.04}{12} \), \( n = 240 \). The factor is \( \dfrac{1 - (1.003333)^{-240}}{0.003333} = 165.0219 \), so \( PV = 165{,}021.86 \). \$165,021.86

2. Find the balance needed to withdraw \$500 a month for 10 years at 6 percent compounded monthly.
   Show the full solution

   \( i = 0.005 \), \( n = 120 \). The factor is \( \dfrac{1 - (1.005)^{-120}}{0.005} = 90.0735 \), so \( PV = 500 \times 90.0735 = 45036.73 \). \$45,036.73

3. By the 4 percent rule, find the annual and monthly withdrawal from \$600,000.
   Show the full solution

   \( 0.04 \times 600000 = 24000 \) a year, and \( \dfrac{24000}{12} = 2000 \) a month. \$24,000 a year; \$2,000 a month

4. By the 4 percent rule, what balance supports \$3,000 a month?
   Show the full solution

   The annual need is \( 3000 \times 12 = 36000 \), and the balance is \( \dfrac{36000}{0.04} = 900000 \). \$900,000

5. A balance of \$400,000 earns 5 percent compounded monthly. Find the interest in the first month.
   Show the full solution

   \( 400000 \times \dfrac{0.05}{12} = 1666.67 \). \$1,666.67

6. With no interest, how long does \$120,000 last at \$2,000 a month?
   Show the full solution

   \( \dfrac{120000}{2000} = 60 \) months, or 5 years. 60 months

7. Find the balance needed to withdraw \$1,500 a month for 30 years at 5 percent compounded monthly.
   Show the full solution

   \( i = \dfrac{0.05}{12} \), \( n = 360 \). The factor is 186.2816, so \( PV = 1500 \times 186.2816 = 279422.43 \). Total withdrawn: \$540,000. \$279,422.43

8. Find the balance needed to withdraw \$2,500 a month for 25 years at 6 percent compounded monthly.
   Show the full solution

   \( i = 0.005 \), \( n = 300 \). The factor is \( \dfrac{1 - (1.005)^{-300}}{0.005} = 155.2069 \), so \( PV = 2500 \times 155.2069 = 388017.16 \). \$388,017.16

9. How long does \$200,000 last at \$1,500 a month at 5 percent compounded monthly?
   Show the full solution

   \( 1 - \dfrac{200000 \times 0.004167}{1500} = 1 - 0.5556 = 0.4444 \). Then \( n = \dfrac{-\ln 0.4444}{\ln 1.004167} = \dfrac{0.810930}{0.004158} = 195.03 \) months, about 16.25 years. About 16.25 years

10. A student says that \$2,000 a month for 25 years requires \( 2000 \times 12 \times 25 = 600000 \) in the account. Find the error and the correct amount at 5 percent compounded monthly.
    Show the full solution

    The multiplication counts the total withdrawn and ignores the interest the balance earns while it waits. The account pays much of the income itself. The correct balance is \( 2000 \times 171.06 = 342120.09 \), which is \$257,879.91 less than the student's figure. \$342,120.09

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the interest for one month on a balance of \$1,800 at an APR of 19.5 percent.
   Show the full solution

   \( 1800 \times \dfrac{0.195}{12} = 1800 \times 0.01625 = 29.25 \). \$29.25

2. Find the monthly periodic rate for an APR of 18 percent.
   Show the full solution

   \( \dfrac{0.18}{12} = 0.015 \). 1.5 percent

3. Find the interest for one month on \$500 at an APR of 24 percent.
   Show the full solution

   \( 500 \times \dfrac{0.24}{12} = 500 \times 0.02 = 10.00 \). \$10.00

4. Find the daily periodic rate for an APR of 24 percent.
   Show the full solution

   \( \dfrac{0.24}{365} = 0.000658 \). 0.0658 percent a day

5. Find the interest for one month on \$3,200 at an APR of 29.99 percent.
   Show the full solution

   \( 3200 \times \dfrac{0.2999}{12} = 3200 \times 0.024992 = 79.97 \). \$79.97

6. A cardholder pays the full statement balance by the due date. How much interest is charged on that month's purchases, and why?
   Show the full solution

   None. The grace period means purchases carry no interest when the full statement balance is paid on time. The APR only matters when a balance is carried. \$0.00

7. A balance of \$1,000 at 18 percent compounded monthly is left with no payments for 12 months. Find the amount owed.
   Show the full solution

   \( 1000 \times (1.015)^{12} = 1000 \times 1.195618 = 1195.62 \). The debt grew by \$195.62, about 19.6 percent. \$1,195.62

8. A balance of \$1,250 at an APR of 19.2 percent receives a \$200 payment. Find the interest and the new balance.
   Show the full solution

   Interest: \( 1250 \times \dfrac{0.192}{12} = 1250 \times 0.016 = 20.00 \). New balance: \( 1250 + 20 - 200 = 1070 \). \$20.00; \$1,070.00

9. A cash advance of \$1,000 at 29.99 percent accrues interest from the day it is taken, with no grace period. Find the interest for 20 days by the daily method.
   Show the full solution

   \( 1000 \times \dfrac{0.2999}{365} \times 20 = 16.43 \). \$16.43

10. A student says, "An APR of 24 percent means I pay 24 percent a year." Find the effective rate when it compounds monthly, and explain.
    Show the full solution

    \( (1.02)^{12} - 1 = 0.268242 \), so the effective rate is 26.82 percent. Interest each month is added to the balance and then earns interest itself, so a year costs more than the APR. 26.82 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the first month's interest on \$1,500 at an APR of 18 percent.
   Show the full solution

   \( 1500 \times 0.015 = 22.50 \). \$22.50

2. Find the minimum payment on a \$3,000 balance under the larger-of-\$25-or-2-percent policy.
   Show the full solution

   Two percent of \$3,000 is \$60, which exceeds \$25. \$60.00

3. A \$1,500 balance at 18 percent is paid off with a fixed \$75 a month. Find the number of months and the total interest.
   Show the full solution

   \( n = \dfrac{-\ln(1 - 22.50/75)}{\ln 1.015} = \dfrac{-\ln 0.7}{0.014889} = 23.96 \), so 24 payments, the last smaller. The interest totals \$296.71. 24 months; \$296.71

4. A \$1,000 balance at 24 percent is paid with \$50 a month or \$100 a month. Find the months and interest for each.
   Show the full solution

   At \$50: 26 months and \$289.88 of interest. At \$100: 12 months and \$127.04. Doubling the payment cut the time by more than half and the interest by more than half. 26 months, \$289.88; 12 months, \$127.04

5. A \$500 balance at 24 percent is paid with \$10 a month. What happens to the balance?
   Show the full solution

   The monthly interest is \( 500 \times 0.02 = 10.00 \), exactly the payment. The balance never falls and the debt is never paid. The balance stays at \$500

6. A \$5,000 balance at 19.99 percent is paid with \$200 a month, or \$300 a month. Find the months and interest for each.
   Show the full solution

   At \$200: 33 months and \$1,521.01. At \$300: 20 months and \$906.24. The extra \$100 a month saves 13 months and \$614.77. 33 months, \$1,521.01; 20 months, \$906.24

7. In the worked example, find the total paid with the \$150 plan.
   Show the full solution

   \( 2400 + 493.71 = 2893.71 \). \$2,893.71

8. On the \$2,400 balance, compare the \$100 and \$200 plans.
   Show the full solution

   \$100: 33 months, \$845.31 interest. \$200: 14 months, \$353.50 interest. The difference is \( 845.31 - 353.50 = 491.81 \) dollars. \$200 saves \$491.81 and 19 months

9. In the first month of the minimum plan, what percent of the \$48.00 payment reduced the balance?
   Show the full solution

   Principal is \( 48.00 - 45.98 = 2.02 \), and \( \dfrac{2.02}{48.00} \times 100 = 4.2 \). The other 95.8 percent was interest. 4.2 percent

10. A student says that paying \$48 a month on a \$2,400 balance clears it in \( 2400 \div 48 = 50 \) months. Find the error.
    Show the full solution

    The division treats every dollar of the payment as repaying the balance. Each month starts with about \$46 of interest, so only about \$2 reduces the debt. The schedule takes 942 months under the minimum policy, about 19 times the student's estimate. 942 months, not 50

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Cycles are 30 days unless stated.

1. A balance is \$500 for 15 days and \$800 for 15 days. Find the average daily balance.
   Show the full solution

   \( \dfrac{500 \times 15 + 800 \times 15}{30} = \dfrac{19500}{30} = 650 \). \$650.00

2. A balance is \$1,000 for 10 days and \$1,400 for 20 days. Find the average daily balance.
   Show the full solution

   \( \dfrac{10000 + 28000}{30} = 1266.67 \). \$1,266.67

3. Find the interest on an average daily balance of \$1,200 at an APR of 18 percent, using the monthly rate.
   Show the full solution

   \( 1200 \times \dfrac{0.18}{12} = 1200 \times 0.015 = 18.00 \). \$18.00

4. A balance of \$1,000 is constant for the whole cycle. Find the average daily balance.
   Show the full solution

   \( \dfrac{1000 \times 30}{30} = 1000 \). \$1,000.00

5. A 31-day cycle has a balance of \$900 for 16 days and \$1,300 for 15 days. Find the average daily balance and the interest at an APR of 20 percent by the daily method.
   Show the full solution

   \( \dfrac{900 \times 16 + 1300 \times 15}{31} = \dfrac{33900}{31} = 1093.55 \). The interest is \( 1093.55 \times \dfrac{0.20}{365} \times 31 = 18.58 \). \$1,093.55; \$18.58

6. A card starts a 30-day cycle at \$400. Purchases of \$200 are made on day 11 and \$100 on day 21, with no payments. Find the average daily balance.
   Show the full solution

   Days 1 to 10 at \$400, days 11 to 20 at \$600, days 21 to 30 at \$700. The average is \( \dfrac{4000 + 6000 + 7000}{30} = 566.67 \). \$566.67

7. A balance of \$2,000 falls to \$1,500 after a \$500 payment. Find the average daily balance if the payment is made on day 6 (balance \$2,000 for 5 days) and if it is made on day 26 (balance \$2,000 for 25 days).
   Show the full solution

   Early: \( \dfrac{2000 \times 5 + 1500 \times 25}{30} = 1583.33 \). Late: \( \dfrac{2000 \times 25 + 1500 \times 5}{30} = 1916.67 \). \$1,583.33 early; \$1,916.67 late

8. At an APR of 24 percent, find the interest for each timing in the last problem and the saving from paying early.
   Show the full solution

   Early: \( 1583.33 \times 0.02 = 31.67 \). Late: \( 1916.67 \times 0.02 = 38.33 \). The saving is \( 38.33 - 31.67 = 6.66 \) dollars for making the same payment 20 days sooner. \$31.67 and \$38.33; \$6.66 saved

9. A \$600 payment is made on a balance of \$1,800 on day 1, or on day 30. Find the average daily balance in each case.
   Show the full solution

   Day 1: the balance is \$1,200 for all 30 days, an average of \$1,200.00. Day 30: the balance is \$1,800 for 29 days and \$1,200 for 1, so \( \dfrac{1800 \times 29 + 1200}{30} = 1780 \). \$1,200.00 and \$1,780.00

10. A student uses the ending balance of \$1,100 instead of the average daily balance of \$1,263.33 in the worked example. Find the interest each way at 1.8 percent a month and the error.
    Show the full solution

    Ending balance: \( 1100 \times 0.018 = 19.80 \). Average daily balance: \( 1263.33 \times 0.018 = 22.74 \). The student understated the interest by \$2.94. The lender charges on what was owed on each day, not on the last day's figure. \$22.74, not \$19.80

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A foreign transaction fee of 3 percent applies to a \$420 purchase. Find the fee.
   Show the full solution

   \( 0.03 \times 420 = 12.60 \). \$12.60

2. A cardholder pays late three times in a year, at \$35 each. Find the fees.
   Show the full solution

   \( 3 \times 35 = 105 \). \$105.00

3. An annual fee is \$95. Find the monthly cost.
   Show the full solution

   \( \dfrac{95}{12} = 7.92 \). \$7.92

4. Find the cash advance fee on \$500 and on \$100 at 5 percent with a \$10 minimum.
   Show the full solution

   On \$500: \( 0.05 \times 500 = 25 \). On \$100: \( 0.05 \times 100 = 5 \), which is below the minimum, so the fee is \$10. \$25.00 and \$10.00

5. Find the interest on a \$500 cash advance for 30 days at 29.99 percent, and the total cost with the \$25 fee.
   Show the full solution

   \( 500 \times \dfrac{0.2999}{365} \times 30 = 12.32 \), and \( 25 + 12.32 = 37.32 \). \$12.32 interest; \$37.32 total

6. A balance transfer of \$5,000 has a fee of 3 percent with a \$5 minimum. Find the fee.
   Show the full solution

   \( 0.03 \times 5000 = 150 \). \$150.00

7. A card gives 1.5 percent cash back on \$6,000 of spending. The holder also carries an average balance of \$800 at 24 percent. Find the rewards, the interest and the net.
   Show the full solution

   Rewards: \( 0.015 \times 6000 = 90 \). Interest: \( 0.24 \times 800 = 192 \) per year. Net: \( 90 - 192 = -102 \). The interest cost more than the rewards earned. A net loss of \$102.00

8. A card charges a \$95 annual fee and gives 1.5 percent cash back on \$4,000 of spending. Find the net value of the card.
   Show the full solution

   Rewards \( 0.015 \times 4000 = 60 \). Net \( 60 - 95 = -35 \). The card loses \$35 a year at this spending level. A loss of \$35.00

9. A \$4,000 balance on a 22.99 percent card is moved to a zero-rate card for 12 months with a 3 percent fee. Find the net saving if the balance is otherwise left alone.
   Show the full solution

   Interest avoided: \( 4000 \times [(1.019158)^{12} - 1] = 1022.96 \). Fee: \( 0.03 \times 4000 = 120 \). Net saving \( 1022.96 - 120 = 902.96 \). \$902.96

10. A student says, "A \$35 late fee is small compared with my \$2,000 limit." Find the fee as a percent of the \$25 minimum payment that was missed, and correct the comparison.
    Show the full solution

    The fee is \( \dfrac{35}{25} \times 100 = 140 \) percent of the payment that was late. The right comparison is with what the fee was charged on, not with the limit. Paying the \$25 a few days early would have avoided a charge larger than the payment. 140 percent of the missed payment

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A card has a \$600 balance and a \$2,000 limit. Find the utilization.
   Show the full solution

   \( \dfrac{600}{2000} \times 100 = 30 \). 30 percent

2. A card has a \$900 balance and a \$3,000 limit. Find the utilization.
   Show the full solution

   \( \dfrac{900}{3000} \times 100 = 30 \). 30 percent

3. Two cards have limits of \$1,000 and \$4,000 and balances of \$250 and \$1,250. Find the total utilization.
   Show the full solution

   Balances total \$1,500 and limits total \$5,000: \( \dfrac{1500}{5000} = 30 \) percent. 30 percent

4. A card with an \$8,000 limit has a \$3,200 balance. How much must be paid to reach 30 percent utilization?
   Show the full solution

   Thirty percent of the limit is \( 0.30 \times 8000 = 2400 \). The payment is \( 3200 - 2400 = 800 \). \$800

5. The five score factors have weights of 35, 30, 15, 10 and 10 percent. Find the sum and name the largest.
   Show the full solution

   \( 35 + 30 + 15 + 10 + 10 = 100 \). The largest is payment history, 35 percent. 100 percent; payment history

6. A person has a \$1,500 balance on total limits of \$5,000 and is approved for a new card with a \$2,000 limit, with no change in balance. Find the new utilization.
   Show the full solution

   The new total limit is \$7,000: \( \dfrac{1500}{7000} \times 100 = 21.43 \), down from 30 percent. 21.43 percent

7. A card has a \$5,000 limit. What balance is 40 percent utilization?
   Show the full solution

   \( 0.40 \times 5000 = 2000 \). \$2,000

8. A \$900 balance sits on a \$1,000 card, and the person's total limits are \$6,000. Find the utilization on the card and overall.
   Show the full solution

   On the card: \( \dfrac{900}{1000} = 90 \) percent. Overall: \( \dfrac{900}{6000} = 15 \) percent. The overall figure looks healthy but the single card is maxed out, which scoring models also penalize. 90 percent on the card; 15 percent overall

9. A person has two cards with limits of \$1,000 and \$4,000 and a total balance of \$500 on the smaller card. Find the utilization before and after closing the \$4,000 card.
   Show the full solution

   Before: \( \dfrac{500}{5000} = 10 \) percent. After closing the larger card: \( \dfrac{500}{1000} = 50 \) percent. Closing the card removed \$4,000 of limit and raised utilization five times over. 10 percent, then 50 percent

10. A student says, "My income is high, so my utilization is low." Find the error.
    Show the full solution

    Utilization divides the balance by the credit limit, and income does not appear in it. A person earning any amount who owes \$900 on a \$1,000 card has a utilization of 90 percent. Utilization depends on the balance and the limit only

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the monthly payment and total interest on \$10,000 for 48 months at 5.0 percent and at 9.5 percent.
   Show the full solution

   At 5.0 percent: \( M = 230.29 \), total paid \$11,053.92, interest \$1,053.92. At 9.5 percent: \( M = 251.23 \), total paid \$12,059.04, interest \$2,059.04. \$230.29 and \$251.23; interest \$1,053.92 and \$2,059.04

2. In the last problem, find the extra interest the lower-score borrower pays.
   Show the full solution

   \( 2059.04 - 1053.92 = 1005.12 \). \$1,005.12

3. Find the monthly payment and total interest on \$18,000 for 60 months at 6.5 percent.
   Show the full solution

   \( M = 352.19 \); total paid \( 352.19 \times 60 = 21131.40 \); interest \( 21131.40 - 18000 = 3131.40 \). \$352.19; \$3,131.40

4. Find the monthly payment on \$8,000 for 60 months at 13.5 percent, and the total interest.
   Show the full solution

   \( M = 184.08 \); total paid \$11,044.80; interest \$3,044.80, which is 38.06 percent of the amount borrowed. \$184.08; \$3,044.80

5. Find the extra interest paid on \$12,000 for 36 months at 13.5 percent rather than 9.5 percent.
   Show the full solution

   At 9.5 percent: \( M = 384.40 \), interest \$1,838.40. At 13.5 percent: \( M = 407.22 \), interest \$2,659.92. The extra is \( 2659.92 - 1838.40 = 821.52 \). \$821.52

6. On a \$250,000 30-year mortgage, find the payment at 6.5 percent and at 7.0 percent.
   Show the full solution

   At 6.5 percent: \$1,580.17. At 7.0 percent: \$1,663.26. The monthly difference is \$83.09. \$1,580.17 and \$1,663.26

7. Over the 360 payments, find the extra cost of the half-point higher rate.
   Show the full solution

   \( 83.09 \times 360 = 29912.40 \). A difference of half a percentage point on a mortgage costs nearly \$30,000. \$29,912.40

8. A borrower in the 640-to-699 range raises her score into the 700-to-759 range before taking the \$20,000, 60-month loan. Find the interest she saves and the saving per month.
   Show the full solution

   Moving from 9.5 percent to 6.5 percent saves \( 5202.40 - 3479.20 = 1723.20 \) dollars, which is \( \dfrac{1723.20}{60} = 28.72 \) dollars a month. \$1,723.20 saved; \$28.72 a month

9. What percent more interest does the lowest-score borrower pay than the highest in the example?
   Show the full solution

   \( \dfrac{7612.00 - 2645.20}{2645.20} \times 100 = 187.8 \). 187.8 percent more

10. A student says, "My score only matters for getting approved, not for what I pay." Use the example to correct this.
    Show the full solution

    Approval is only the first effect. Among approved borrowers, the score sets the rate, and the rate sets the cost. On the same \$20,000 loan the borrower with the lowest approved score pays \$4,966.80 more interest than the borrower with the highest, about \$82.78 a month. The score sets the price, not just the decision

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Refer to the debts in the worked example.

1. Which debt does the avalanche method target first?
   Show the full solution

   The one with the highest rate, the 26 percent debt of \$1,200. The 26 percent debt

2. Which debt does the snowball method target first?
   Show the full solution

   The one with the smallest balance, the \$800 debt at 12 percent. The \$800 debt

3. Find the monthly budget if the minimums are \$35, \$90 and \$25 and the extra is \$150.
   Show the full solution

   \( 35 + 90 + 25 + 150 = 300 \). \$300.00

4. Find the total interest charged in the first month on the three debts.
   Show the full solution

   \( 26.00 + 71.25 + 8.00 = 105.25 \). \$105.25

5. Under the avalanche, find the balance of the 26 percent debt after the first month.
   Show the full solution

   \( 1200 + 26 - 35 - 150 = 1041 \). \$1,041.00

6. Under the snowball, find the balance of the \$800 debt after the first month.
   Show the full solution

   \( 800 + 8 - 25 - 150 = 633 \). \$633.00

7. Find the interest the avalanche saves over the snowball.
   Show the full solution

   \( 1584.05 - 1454.62 = 129.43 \). \$129.43

8. In which month does each method clear its first debt?
   Show the full solution

   The avalanche clears the 26 percent debt in month 8. The snowball clears the \$800 debt in month 5. The snowball gets its first win three months sooner. Avalanche month 8; snowball month 5

9. Compare the \$150-extra plan with paying only minimums in months and interest.
   Show the full solution

   Minimums only: 76 months and \$4,897.84 in interest. The avalanche plan: 27 months and \$1,454.62. The plan saves 49 months and \$3,443.22. 49 months and \$3,443.22 saved

10. A borrower owes \$2,000 at 24 percent (minimum \$50) and \$6,000 at 12 percent (minimum \$120) and can add \$100. Putting the extra on the 24 percent debt clears everything in 37 months for \$1,771.61 of interest; putting it on the 12 percent debt takes 39 months and costs \$2,313.17. Find the saving and explain it.
    Show the full solution

    \( 2313.17 - 1771.61 = 541.56 \) dollars and 2 months. Extra dollars reduce a balance that charges 24 percent instead of one that charges 12 percent, so each dollar avoids twice the interest. The avalanche always applies the money where it saves the most. \$541.56 and 2 months saved by targeting the higher rate

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Payments are monthly.

1. Find the payment and total interest on \$5,000 at 8 percent for 24 months.
   Show the full solution

   \( i = 0.006667 \). \( M = 5000 \times \dfrac{0.006667}{1 - (1.006667)^{-24}} = 226.14 \). Total paid \( 226.14 \times 24 = 5427.36 \), interest \$427.36. \$226.14; \$427.36

2. Find the payment and total interest on \$9,000 at 6 percent for 36 months.
   Show the full solution

   \( i = 0.005 \). \( M = 9000 \times \dfrac{0.005}{1 - (1.005)^{-36}} = 273.80 \). Total paid \$9,856.80, interest \$856.80. \$273.80; \$856.80

3. Find the payment and total interest on \$15,000 at 7.5 percent for 60 months.
   Show the full solution

   \( i = 0.00625 \). \( M = 300.57 \). Total paid \( 300.57 \times 60 = 18034.20 \), interest \$3,034.20. \$300.57; \$3,034.20

4. Find the payment and total interest on \$2,500 at 12 percent for 12 months.
   Show the full solution

   \( i = 0.01 \). \( M = 2500 \times \dfrac{0.01}{1 - (1.01)^{-12}} = 222.12 \). Total paid \$2,665.44, interest \$165.44. \$222.12; \$165.44

5. Find the payment and total interest on \$30,000 at 5.9 percent for 72 months.
   Show the full solution

   \( i = 0.004917 \). \( M = 495.77 \). Total paid \( 495.77 \times 72 = 35695.44 \), interest \$5,695.44. \$495.77; \$5,695.44

6. Find the payment and total interest on \$8,000 at 6.9 percent for 36 months.
   Show the full solution

   \( i = 0.00575 \). \( M = 246.65 \). Total paid \$8,879.40, interest \$879.40. \$246.65; \$879.40

7. A buyer can pay \$350 a month for 60 months at 7.2 percent. Find the most she can borrow.
   Show the full solution

   The loan is the present value of the payments: \( P = M \times \dfrac{1 - (1 + i)^{-n}}{i} = 350 \times \dfrac{1 - (1.006)^{-60}}{0.006} = 350 \times 50.2621 = 17591.75 \). \$17,591.75

8. How many payments does it take to repay \$12,000 at 6.9 percent if the borrower pays \$300 a month?
   Show the full solution

   \( n = \dfrac{-\ln(1 - Pi/M)}{\ln(1 + i)} = \dfrac{-\ln(1 - 69/300)}{\ln 1.00575} = \dfrac{0.261365}{0.005734} = 45.59 \). That is 46 payments, the last one smaller. 46 payments

9. A student estimates the payment on a \$12,000, 48-month loan at 6.9 percent with simple interest: \( \dfrac{12000 + 12000 \times 0.069 \times 4}{48} = 319.00 \). Find the error.
   Show the full solution

   Simple interest charges 6.9 percent on the full \$12,000 for all four years. A loan balance falls with every payment, so interest is charged only on what is still owed. The correct payment is \$286.80, and the estimate is \$32.20 a month too high. \$286.80

10. A student rounds the monthly rate in the worked example from 0.00575 to 0.0058 and gets a payment of \$287.13. How much does the rounding cost over the loan?
    Show the full solution

    The exact payment is \$286.80, so the rounded one is \$0.33 high. Over 48 payments that is \( 0.33 \times 48 = 15.84 \) dollars. Keep the full rate until the final answer. About \$15.84

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A \$10,000 loan at 6 percent has a monthly payment of \$304.22 for 36 months. Find the interest in month 1.
   Show the full solution

   \( 10000 \times \dfrac{0.06}{12} = 10000 \times 0.005 = 50.00 \). \$50.00

2. For that loan, find the principal paid and the balance after month 1.
   Show the full solution

   Principal \( 304.22 - 50.00 = 254.22 \). Balance \( 10000 - 254.22 = 9745.78 \). \$254.22; \$9,745.78

3. Find the interest in month 2 of that loan.
   Show the full solution

   \( 9745.78 \times 0.005 = 48.73 \). \$48.73

4. A \$3,000 loan at 12 percent over 6 months has a payment of \$517.65. Find month 1: the interest, principal and balance.
   Show the full solution

   Interest \( 3000 \times 0.01 = 30.00 \). Principal \( 517.65 - 30.00 = 487.65 \). Balance \( 3000 - 487.65 = 2512.35 \). \$30.00; \$487.65; \$2,512.35

5. For the same loan, find the interest in month 2.
   Show the full solution

   \( 2512.35 \times 0.01 = 25.12 \). \$25.12

6. A \$20,000 loan at 6 percent for 60 months has a payment of \$386.66. What percent of the first payment is interest, and of the last?
   Show the full solution

   The first interest is \( 20000 \times 0.005 = 100.00 \), which is \( \dfrac{100}{386.66} = 25.86 \) percent. In the last month the interest is \$1.92, which is \( \dfrac{1.92}{386.66} = 0.50 \) percent. 25.86 percent; 0.50 percent

7. A \$200,000 mortgage at 6.5 percent for 30 years has a payment of \$1,264.14. Find the interest and principal in month 1, and the interest as a percent of the payment.
   Show the full solution

   Interest \( 200000 \times \dfrac{0.065}{12} = 1083.33 \). Principal \( 1264.14 - 1083.33 = 180.81 \). The interest share is \( \dfrac{1083.33}{1264.14} = 85.7 \) percent. \$1,083.33; \$180.81; 85.7 percent

8. On that mortgage, the balance after 12 payments is \$197,764.50. Find the principal paid in the first year.
   Show the full solution

   \( 200000 - 197764.50 = 2235.50 \). After a year of payments totaling \( 12 \times 1264.14 = 15169.68 \), only \$2,235.50 has reduced the debt. \$2,235.50

9. After 120 payments the mortgage balance is \$169,551.54. Find the principal paid in ten years and the percent of the loan.
   Show the full solution

   \( 200000 - 169551.54 = 30448.46 \), which is \( \dfrac{30448.46}{200000} \times 100 = 15.2 \) percent. One third of the term has retired only 15 percent of the loan. \$30,448.46; 15.2 percent

10. A student says that in an amortized loan the principal paid is the same each month. Find the error using the \$5,000 example.
    Show the full solution

    The payment is the same each month, but the interest falls as the balance falls, so the principal rises. In the example the principal was \$399.76 in month 1 and \$433.97 in month 12, each month's principal being the last one times 1.0075. Payment is level; principal rises

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A \$10,000 loan at 8 percent has payments of \$452.27 over 24 months and \$244.13 over 48 months. Find the total interest for each.
   Show the full solution

   24 months: \( 452.27 \times 24 - 10000 = 854.48 \). 48 months: \( 244.13 \times 48 - 10000 = 1718.24 \). \$854.48 and \$1,718.24

2. A \$200,000 mortgage at 6.5 percent has payments of \$1,742.21 over 15 years and \$1,264.14 over 30 years. Find the total interest for each.
   Show the full solution

   15 years: \( 1742.21 \times 180 - 200000 = 113597.80 \). 30 years: \( 1264.14 \times 360 - 200000 = 255090.40 \). \$113,597.80 and \$255,090.40

3. In the last problem, find the monthly payment difference.
   Show the full solution

   \( 1742.21 - 1264.14 = 478.07 \). \$478.07 a month

4. In the same problem, find how much more interest the 30-year loan costs.
   Show the full solution

   \( 255090.40 - 113597.80 = 141492.60 \). The 30-year loan costs 2.25 times the interest of the 15-year loan. \$141,492.60

5. Find the payment and total interest on \$18,000 at 7.5 percent over 48 months.
   Show the full solution

   \( i = 0.00625 \). \( M = 435.22 \). Total paid \( 435.22 \times 48 = 20890.56 \), interest \$2,890.56. \$435.22; \$2,890.56

6. A \$15,000 loan at 6 percent has payments of \$289.99 over 60 months and \$219.13 over 84 months. Find the total interest for each and the extra interest for the longer term.
   Show the full solution

   60 months: \( 289.99 \times 60 - 15000 = 2399.40 \). 84 months: \( 219.13 \times 84 - 15000 = 3406.92 \). The extra is \( 3406.92 - 2399.40 = 1007.52 \) dollars. \$2,399.40 and \$3,406.92; \$1,007.52 more

7. In the worked example, find the total interest on the 72-month loan as a percent of the amount borrowed.
   Show the full solution

   \( \dfrac{5688.56}{25000} \times 100 = 22.75 \). 22.75 percent

8. A buyer can afford at most \$450 a month on the \$25,000 loan. Which terms in the example qualify, and which is cheapest?
   Show the full solution

   Payments of \$771.93 and \$598.66 are too high. The 60-month payment of \$495.03 also exceeds \$450. Only the 72-month payment of \$426.23 fits, so it is the only choice and costs \$5,688.56 in interest. 72 months only

9. Find the price of each dollar of monthly relief in the \$15,000 loan when the term moves from 60 to 84 months.
   Show the full solution

   The monthly saving is \( 289.99 - 219.13 = 70.86 \). The added interest is \$1,007.52. The price per dollar of relief is \( \dfrac{1007.52}{70.86} = 14.22 \) dollars. \$14.22 of interest per dollar of monthly relief

10. A student says, "The 72-month loan is cheaper because the payment is \$345.70 lower." Correct the claim with the total interest.
    Show the full solution

    A lower payment over more months is not a lower cost. The 72-month loan charges \$5,688.56 in interest against \$2,789.48 for the 36-month loan, which is \$2,899.08 more. The payment is smaller only because it is spread over twice as many months. It costs \$2,899.08 more

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use the results of the worked example where needed.

1. With an extra \$50 a month, how many months sooner does the \$25,000 loan end?
   Show the full solution

   \( 60 - 54 = 6 \). 6 months

2. Find the interest saved by an extra \$100 a month.
   Show the full solution

   \( 4701.82 - 3762.80 = 939.02 \). \$939.02

3. A single extra \$2,000 in month 12 saves how much interest and how many months?
   Show the full solution

   Interest \( 4701.82 - 4090.73 = 611.09 \); months \( 60 - 55 = 5 \). \$611.09; 5 months

4. A \$12,000 loan at 6.9 percent for 48 months costs \$1,766.32 in interest. With an extra \$25 a month it ends in month 44 with \$1,602.38 in interest. Find the saving.
   Show the full solution

   \( 1766.32 - 1602.38 = 163.94 \) dollars and \( 48 - 44 = 4 \) months. \$163.94; 4 months

5. In month 1 of the \$25,000 loan, how much of a \$495.03 payment plus a \$50 extra is principal?
   Show the full solution

   Interest is \$145.83, so the principal is \( 495.03 + 50 - 145.83 = 399.20 \). \$399.20

6. On the \$200,000 mortgage, an extra \$100 a month saves \$55,944.38 and ends the loan in month 293. Find the years and months saved.
   Show the full solution

   \( 360 - 293 = 67 \) months, which is 5 years and 7 months. 5 years 7 months

7. An extra \$200 a month on that mortgage ends it in month 250 and saves \$90,074.66. Find the total paid in extras and the interest saved per dollar of extra.
   Show the full solution

   The extras are about \( 200 \times 249 = 49800 \) dollars (the last month is smaller). The saving per dollar is \( \dfrac{90074.66}{49800} = 1.81 \) dollars of interest for each extra dollar paid. About \$1.81 of interest saved per extra dollar

8. Is prepaying a 7 percent loan better than keeping the money in a 4.5 percent savings account? Explain.
   Show the full solution

   Each extra dollar paid on the loan avoids 7 percent interest, a sure return of 7 percent. The savings account returns 4.5 percent, taxed. The prepayment is better by 2.5 points, assuming the borrower has an emergency fund and no higher-rate debt. Prepaying earns 7 percent against 4.5 percent

9. Find the total paid in the \$50-extra plan, given the interest of \$4,179.19.
   Show the full solution

   \( 25000 + 4179.19 = 29179.19 \), against \$29,701.82 with no extras. \$29,179.19

10. A student says, "An extra payment is taxed with interest first, so it only partly reduces the loan." Correct this using month 1.
    Show the full solution

    The month's interest, \$145.83, is covered by the regular payment of \$495.03 before the extra is considered. The extra \$50 then goes entirely to principal, which falls from \$349.20 to \$399.20. An extra payment is never reduced by interest. The whole extra reduces the principal

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the daily interest on a \$10,000 loan at 5.5 percent.
   Show the full solution

   \( \dfrac{10000 \times 0.055}{365} = 1.51 \). \$1.51 a day

2. Find the interest accrued on an \$8,000 unsubsidized loan at 6 percent over 6 months.
   Show the full solution

   \( 8000 \times 0.06 \times 0.5 = 240 \). \$240.00

3. A student borrows \$15,000 at 5 percent unsubsidized and stays in school 4 years. Find the capitalized balance.
   Show the full solution

   \( 15000 \times (1 + 0.05 \times 4) = 15000 \times 1.20 = 18000 \). \$18,000

4. Find the payment and total interest on \$18,000 at 5 percent over 10 years.
   Show the full solution

   \( i = 0.004167 \), \( n = 120 \): \( M = 190.92 \). Total paid \( 190.92 \times 120 = 22910.40 \), interest \$4,910.40. \$190.92; \$4,910.40

5. Find the payment on \$7,000 at 4.5 percent over 10 years.
   Show the full solution

   \( i = 0.00375 \). \( M = 7000 \times \dfrac{0.00375}{1 - (1.00375)^{-120}} = 72.55 \). \$72.55

6. Compare 10-year and 20-year repayment of \$30,000 at 6 percent: payments and total interest.
   Show the full solution

   10 years: \( M = 333.06 \), total interest \( 333.06 \times 120 - 30000 = 9967.20 \). 20 years: \( M = 214.93 \), total interest \( 214.93 \times 240 - 30000 = 21583.20 \). The longer term lowers the payment by \$118.13 and more than doubles the interest. \$333.06, \$9,967.20; \$214.93, \$21,583.20

7. A student pays the \$550 of interest each year on a \$10,000 unsubsidized loan. Find the total paid in school and the balance at graduation.
   Show the full solution

   Four payments of \$550 total \( 550 \times 4 = 2200 \) dollars. The balance stays at \$10,000. \$2,200 paid; balance \$10,000

8. In the worked example, compare the total paid by the student who pays the interest in school with the student who does not.
   Show the full solution

   The student who pays in school pays \( 2200 + 13023.60 = 15223.60 \). The student who waits pays \$15,888.00. Paying the interest while in school saves \( 15888.00 - 15223.60 = 664.40 \) dollars, by avoiding interest on interest. \$664.40 less

9. What is the capitalized balance if \$10,000 accrues simple interest at 5.5 percent for 4 years, and what percent larger is it?
   Show the full solution

   \$12,200, which is \( \dfrac{2200}{10000} \times 100 = 22 \) percent larger than the amount borrowed. \$12,200; 22 percent larger

10. A student says, "A subsidized and an unsubsidized loan for the same amount cost the same, because the rate is the same." Correct the claim with numbers.
    Show the full solution

    The rate is the same, but the subsidized loan does not accrue interest in school, so repayment starts from \$10,000. The unsubsidized loan starts from \$12,200. Total paid is \$13,023.60 against \$15,888.00, a difference of \$2,864.40. The unsubsidized loan costs \$2,864.40 more

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A \$100 loan for 7 days has a \$15 fee. Find the APR.
   Show the full solution

   \( \dfrac{15}{100} \times \dfrac{365}{7} \times 100 = 15 \times 52.14 = 782.14 \). 782.14 percent

2. A \$500 loan for 14 days has a \$75 fee. Find the APR.
   Show the full solution

   \( \dfrac{75}{500} \times \dfrac{365}{14} \times 100 = 15 \times 26.07 = 391.07 \). 391.07 percent

3. A \$300 loan for 30 days has a \$45 fee. Find the APR.
   Show the full solution

   \( \dfrac{45}{300} \times \dfrac{365}{30} \times 100 = 15 \times 12.17 = 182.50 \). 182.50 percent

4. A \$400 loan with a \$60 fee is rolled over 6 times. Find the total fees paid.
   Show the full solution

   The first fee plus six rollovers is seven fees: \( 7 \times 60 = 420 \). (If the original fee is counted with the six extensions, the borrower has paid \$420.) \$420.00

5. Find the amount due on a \$400 loan with a \$60 fee.
   Show the full solution

   \( 400 + 60 = 460 \). \$460.00

6. Find the interest on \$400 for 14 days on a credit card at 28 percent, with no fee.
   Show the full solution

   \( 400 \times \dfrac{0.28}{365} \times 14 = 4.30 \). It is one fourteenth of the payday fee. \$4.30

7. Find the total cost of a \$400 card cash advance for 14 days at 29.99 percent with a 5 percent fee.
   Show the full solution

   Fee \( 0.05 \times 400 = 20 \). Interest \( 400 \times \dfrac{0.2999}{365} \times 14 = 4.60 \). Total \$24.60. \$24.60

8. A \$1,000 loan for 30 days has a \$150 fee. Find the APR.
   Show the full solution

   \( \dfrac{150}{1000} \times \dfrac{365}{30} \times 100 = 15 \times 12.17 = 182.50 \). 182.50 percent

9. A fee of \$15 per \$100 is charged for 14 days. Find the cost of borrowing \$400 and of borrowing \$400 for a full year by rolling it over 26 times.
   Show the full solution

   One loan: \( 4 \times 15 = 60 \). Twenty-six of them: \( 26 \times 60 = 1560 \), which is 390 percent of the principal. Borrowing \$400 for a year costs nearly four times the amount borrowed. \$60.00; \$1,560.00

10. A student says, "The fee is only 15 percent, which is less than a lot of credit cards charge." Find the error.
    Show the full solution

    Credit card rates are annual. The 15 percent is for 14 days, so the fair comparison is the annualized figure, \( 15 \times 26.07 = 391 \) percent, which is more than ten times a card's 29.99 percent. A rate must always be stated for the same period before it is compared. 15 percent for 14 days is about 391 percent a year

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the monthly payments on \$10,000 for 48 months at 6.0 percent and at 6.5 percent, and the difference.
   Show the full solution

   At 6.0 percent: \$234.85. At 6.5 percent: \$237.15. The difference is \( 237.15 - 234.85 = 2.30 \) a month, or \( 2.30 \times 48 = 110.40 \) over the loan. \$234.85 and \$237.15; \$110.40 over the loan

2. An origination fee is 1 percent of a \$10,000 loan. Find the fee.
   Show the full solution

   \( 0.01 \times 10000 = 100 \). \$100.00

3. Find the payments and total interest on \$8,000 for 36 months at 5.9 percent and at 6.4 percent.
   Show the full solution

   At 5.9 percent: \$243.01 and \( 243.01 \times 36 - 8000 = 748.36 \). At 6.4 percent: \$244.83 and \( 244.83 \times 36 - 8000 = 813.88 \). \$243.01, \$748.36; \$244.83, \$813.88

4. In the last problem, find the extra interest on the higher-rate loan.
   Show the full solution

   \( 813.88 - 748.36 = 65.52 \). \$65.52

5. Find the payment on \$16,000 for 48 months at 6 percent.
   Show the full solution

   \( i = 0.005 \). \( M = 16000 \times \dfrac{0.005}{1 - (1.005)^{-48}} = 375.76 \). \$375.76

6. A \$20,000, 60-month loan at 5.5 percent has no fee. Another at 4.9 percent has a \$500 fee. Find the total cost of each.
   Show the full solution

   At 5.5 percent: interest \$2,921.20. At 4.9 percent: interest \$2,590.60 plus the fee \$500, which is \$3,090.60. The lower-rate loan costs \$169.40 more. \$2,921.20 and \$3,090.60

7. Rank Offers A, B and C in the worked example by monthly payment, lowest first.
   Show the full solution

   B is \$291.39, A is \$358.50 and C is \$465.90. B, A, C

8. Rank the same offers by total cost, lowest first.
   Show the full solution

   C is \$1,772.40, B is \$2,483.40 and A is \$2,508.00. C, B, A

9. With a \$400 payment limit, which offer is cheapest, and why is the ranking by payment different from the ranking by cost?
   Show the full solution

   C is ruled out by the limit. Of A and B, B costs \$2,483.40 and A \$2,508.00, so B. The rankings differ because the terms differ: the longest loan has the lowest payment and a higher cost than the shortest. Offer B

10. A student chooses the loan with the lowest stated rate, 4.9 percent, ignoring its \$500 fee. Use the \$20,000 comparison to explain the mistake.
    Show the full solution

    The fee is part of the price of the loan. Including it, the 4.9 percent loan costs \$3,090.60 against \$2,921.20 for the 5.5 percent loan with no fee, so the student pays \$169.40 more. The APR, which includes fees, exposes this. The lowest stated rate costs \$169.40 more

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A car costs \$18,200 and the sales tax is 8.25 percent. Find the tax, and the out-the-door price with \$300 of fees.
   Show the full solution

   Tax \( 0.0825 \times 18200 = 1501.50 \). Out the door \( 18200 + 1501.50 + 300 = 20001.50 \). \$1,501.50; \$20,001.50

2. A driver travels 15,000 miles a year at 25 miles per gallon with fuel at \$3.60. Find the yearly fuel cost.
   Show the full solution

   \( \dfrac{15000}{25} = 600 \) gallons, and \( 600 \times 3.60 = 2160 \). \$2,160.00

3. Find the yearly running cost when fuel is \$1,520, insurance \$1,680, maintenance \$900 and registration \$312.
   Show the full solution

   \( 1520 + 1680 + 900 + 312 = 4412 \). \$4,412.00

4. A car costs \$21,000 out the door, runs at \$3,900 a year, and sells after 3 years for \$12,500. Find the cost of ownership and the cost per month.
   Show the full solution

   \( 21000 + 3 \times 3900 - 12500 = 20200 \). Over 36 months, \( \dfrac{20200}{36} = 561.11 \). \$20,200; \$561.11 a month

5. A driver uses 400 gallons a year at \$3.80. Find the annual fuel cost.
   Show the full solution

   \( 400 \times 3.80 = 1520 \). \$1,520.00

6. A driver covers 12,000 miles a year. Fuel is \$3.80 a gallon. Find the yearly saving from a car that gets 40 miles per gallon instead of 25.
   Show the full solution

   At 25: \( \dfrac{12000}{25} \times 3.80 = 480 \times 3.80 = 1824 \). At 40: \( \dfrac{12000}{40} \times 3.80 = 300 \times 3.80 = 1140 \). The saving is \$684. \$684.00 a year

7. Find the fuel saving over 5 years in the last problem.
   Show the full solution

   \( 5 \times 684 = 3420 \). \$3,420.00

8. In the worked example, find the fees and tax as a percent of the sticker price.
   Show the full solution

   \( 2021.25 + 450 = 2471.25 \), and \( \dfrac{2471.25}{24500} \times 100 = 10.09 \). 10.09 percent

9. Car A costs \$22,000 out the door and \$5,100 a year to run, and resells after 5 years for \$9,000. Car B costs \$25,000 and \$3,900 a year, and resells for \$11,000. Find each five-year cost and the difference.
   Show the full solution

   Car A: \( 22000 + 5 \times 5100 - 9000 = 38500 \). Car B: \( 25000 + 5 \times 3900 - 11000 = 33500 \). Car B costs \$5,000 less, even though its price is \$3,000 higher. A: \$38,500; B: \$33,500; B is \$5,000 cheaper

10. A student compares the two cars in the last problem by sticker price alone and chooses Car A. Explain the error.
    Show the full solution

    The sticker price is one part of the cost. Car A's higher running costs (\$1,200 a year more) and lower resale value (\$2,000 less) outweigh its \$3,000 lower price. Over five years the total cost is what a buyer pays, so the comparison must include all of it. Compare total cost of ownership, not price

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A \$30,000 car loses 20 percent in its first year. Find its value.
   Show the full solution

   \( 30000 \times 0.80 = 24000 \). \$24,000.00

2. A \$20,000 car loses 15 percent a year. Find its value after 3 years.
   Show the full solution

   \( 20000 \times (0.85)^3 = 20000 \times 0.614125 = 12282.50 \). \$12,282.50

3. A \$16,000 car loses 12 percent a year. Find its value after 4 years.
   Show the full solution

   \( 16000 \times (0.88)^4 = 16000 \times 0.599695 = 9595.13 \). \$9,595.13

4. What percent of its value does a car keep after 5 years at 15 percent a year?
   Show the full solution

   \( (0.85)^5 = 0.443705 \). 44.37 percent

5. A \$30,000 car loses 20 percent a year. Find the dollar loss in year 2.
   Show the full solution

   Year 1 value \$24,000; year 2 value \( 24000 \times 0.80 = 19200 \). The loss is \( 24000 - 19200 = 4800 \). \$4,800.00

6. How many years does a car take to lose half its value at 15 percent a year?
   Show the full solution

   Solve \( (0.85)^t = 0.5 \): \( t = \dfrac{\ln 0.5}{\ln 0.85} = \dfrac{-0.693147}{-0.162519} = 4.27 \). 4.27 years

7. A car is worth \$16,000 and the loan balance is \$17,000. Find the equity.
   Show the full solution

   \( 16000 - 17000 = -1000 \). The owner is \$1,000 underwater. Negative \$1,000.00

8. The \$1,000 shortfall is rolled into a new \$4,000 loan at 6.9 percent over 60 months. Find the payment.
   Show the full solution

   \( i = 0.00575 \): \( 4000 \times \dfrac{0.00575}{1 - (1.00575)^{-60}} = 79.02 \). Paying for the old car's shortfall adds that amount to the next payment. \$79.02 a month

9. A \$24,500 car sells after 5 years for \$11,200. Find the straight-line loss per year, and compare with the loss in year one at 15 percent.
   Show the full solution

   Straight line: \( \dfrac{24500 - 11200}{5} = 2660 \) a year. At 15 percent the first year's loss is \( 0.15 \times 24500 = 3675 \), which is \$1,015 more, and the fifth year's loss is \$1,918.37, which is \$741.63 less. \$2,660 on average; \$3,675 in year one

10. A student says that a car loses the same dollars every year, so \$4,900 in year one means \$4,900 in year five. Correct the claim.
    Show the full solution

    Depreciation is a percent of the current value, and the value shrinks. In year one the loss is 20 percent of \$24,500, or \$4,900. In year five it is 15 percent of \$12,036.85, or \$1,805.53. Dollar losses fall each year. Year five's loss is \$1,805.53

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the money factor for an APR of 3.6 percent.
   Show the full solution

   \( \dfrac{3.6}{2400} = 0.0015 \). 0.0015

2. A \$30,000 car has a residual of \$18,000, a 36-month term and a money factor of 0.0015. Find the monthly payment.
   Show the full solution

   Depreciation: \( \dfrac{30000 - 18000}{36} = 333.33 \). Rent charge: \( (30000 + 18000) \times 0.0015 = 72.00 \). Payment \( 333.33 + 72.00 = 405.33 \). \$405.33

3. Find the residual value of a \$28,000 car if it is 60 percent.
   Show the full solution

   \( 0.60 \times 28000 = 16800 \). \$16,800.00

4. A lease charges \$0.25 a mile over its limit. Find the charge for 1,800 miles over.
   Show the full solution

   \( 1800 \times 0.25 = 450 \). \$450.00

5. A money factor of 0.0021 is what APR?
   Show the full solution

   APR in percent is the money factor times 2400: \( 0.0021 \times 2400 = 5.04 \). 5.04 percent

6. A lease has payments of \$320 for 36 months and \$1,500 due at signing. Find the net cost.
   Show the full solution

   \( 36 \times 320 + 1500 = 13020 \). \$13,020.00

7. In the worked example, find the total of the 36 lease payments alone.
   Show the full solution

   \( 36 \times 350.62 = 12622.32 \). With the \$2,000 due at signing the cost is \$14,622.32. \$12,622.32

8. In the worked example, find the residual as a percent of the price.
   Show the full solution

   \( \dfrac{14700}{24500} \times 100 = 60 \). 60 percent

9. A lessee drives 40,500 miles on a lease with a 36,000-mile limit and a \$0.20 charge per extra mile. Find the charge.
   Show the full solution

   Excess miles \( 40500 - 36000 = 4500 \), and \( 4500 \times 0.20 = 900 \). \$900.00

10. A student computes the money factor for 4.8 percent as \( \dfrac{0.048}{2400} \) and finds a rent charge of 78 cents on the \$24,500 lease. Find the error.
    Show the full solution

    The money factor uses the APR in percent, 4.8, not as a decimal. The correct factor is \( \dfrac{4.8}{2400} = 0.002 \) and the rent charge is \( 39200 \times 0.002 = 78.40 \). The student's lease would look \$77.62 a month cheaper than it is. Use 4.8, not 0.048

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A renter's gross monthly income is \$4,200. Find the most rent by the 30 percent guideline.
   Show the full solution

   \( 0.30 \times 4200 = 1260 \). \$1,260.00

2. Rent of \$1,400 is paid by a worker earning \$54,000 a year. Find rent as a percent of gross income.
   Show the full solution

   Annual rent \( 12 \times 1400 = 16800 \), and \( \dfrac{16800}{54000} \times 100 = 31.11 \). 31.11 percent

3. Rent of \$1,400 rises 4.5 percent. Find the new rent.
   Show the full solution

   \( 1400 \times 1.045 = 1463 \). \$1,463.00

4. Rent of \$1,200 rises 2.5 percent a year. Find the rent after 5 increases.
   Show the full solution

   \( 1200 \times (1.025)^5 = 1200 \times 1.131408 = 1357.69 \). \$1,357.69

5. Three roommates share \$2,400 of rent and \$210 of utilities equally. Find each share.
   Show the full solution

   \( \dfrac{2400 + 210}{3} = 870 \). \$870.00

6. Renter's insurance is \$18 a month. Find the yearly cost.
   Show the full solution

   \( 18 \times 12 = 216 \). \$216.00

7. A lease asks for a security deposit of two months' rent on \$1,650 rent. Find the deposit and the cash needed on day one with the first month's rent.
   Show the full solution

   Deposit \( 2 \times 1650 = 3300 \). With the first month, \( 3300 + 1650 = 4950 \). \$3,300.00; \$4,950.00

8. Rent is \$1,500 for year one and rises 3 percent for year two. Find the two-year total.
   Show the full solution

   Year one: \( 12 \times 1500 = 18000 \). Year two's rent is \( 1500 \times 1.03 = 1545 \), so \( 12 \times 1545 = 18540 \). Total \$36,540.00. \$36,540.00

9. What annual gross income is needed for rent of \$1,650 to be 30 percent of it?
   Show the full solution

   Monthly income \( \dfrac{1650}{0.30} = 5500 \), and \( 12 \times 5500 = 66000 \). \$66,000.00

10. A renter says, "My rent is 30 percent of my income." Her gross is \$5,500 and her net is \$4,400, and she pays \$1,650. Find the rent as a share of each and explain.
    Show the full solution

    Of gross: \( \dfrac{1650}{5500} = 30 \) percent. Of net: \( \dfrac{1650}{4400} = 37.5 \) percent. The statement is true only of gross pay. She actually spends more than a third of the money that reaches her account on rent. 30 percent of gross; 37.5 percent of net

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find a 15 percent down payment on a \$400,000 home.
   Show the full solution

   \( 0.15 \times 400000 = 60000 \). \$60,000

2. A \$300,000 home is bought with 20 percent down. Find the loan and the 30-year payment at 6 percent.
   Show the full solution

   Loan \( 300000 \times 0.80 = 240000 \). \( i = 0.005 \), \( n = 360 \): \( M = 240000 \times \dfrac{0.005}{1 - (1.005)^{-360}} = 1438.92 \). \$240,000; \$1,438.92

3. A \$380,000 home is bought with 20 percent down at 6.5 percent. Find the payment on a 15-year and a 30-year loan.
   Show the full solution

   Loan \( 0.80 \times 380000 = 304000 \). 15 years: \$2,648.17. 30 years: \$1,921.49. The 15-year payment is \$726.68 higher. \$2,648.17 and \$1,921.49

4. On the \$280,000 loan at 6.5 percent, find the interest and principal in the first payment of \$1,769.79.
   Show the full solution

   Interest \( 280000 \times \dfrac{0.065}{12} = 1516.67 \). Principal \( 1769.79 - 1516.67 = 253.12 \). \$1,516.67 interest; \$253.12 principal

5. Compare the payment on \$280,000 at 6.5 percent (\$1,769.79) with the payment at 7.0 percent (\$1,862.85).
   Show the full solution

   \( 1862.85 - 1769.79 = 93.06 \) a month, which is \( 93.06 \times 360 = 33501.60 \) over the loan. \$93.06 a month; \$33,501.60 over 30 years

6. Find the annual PMI on a \$315,000 loan at 0.6 percent.
   Show the full solution

   \( 0.006 \times 315000 = 1890 \), or \$157.50 a month. \$1,890.00

7. At what loan balance does PMI end on a \$350,000 home?
   Show the full solution

   When the loan reaches 80 percent of the price: \( 0.80 \times 350000 = 280000 \). \$280,000

8. How much down payment is needed for 20 percent on a \$425,000 home?
   Show the full solution

   \( 0.20 \times 425000 = 85000 \). \$85,000

9. Find the interest on the 30-year loan in the worked example as a multiple of the amount borrowed.
   Show the full solution

   \( \dfrac{357124.40}{280000} = 1.275 \). The interest is more than the loan itself. 1.28 times the amount borrowed

10. A student budgets for a \$280,000 loan using a teaser rate of 4 percent, which gives a payment of \$1,336.76, but the rate resets to 6.5 percent. Find the error.
    Show the full solution

    The payment at 6.5 percent is \$1,769.79. A budget built on the teaser rate is \( 1769.79 - 1336.76 = 433.03 \) dollars a month too low once the rate resets. A loan must be budgeted at the rate that will apply. Budget at \$1,769.79, not \$1,336.76

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the closing costs at 2.5 percent on a \$300,000 home.
   Show the full solution

   \( 0.025 \times 300000 = 7500 \). \$7,500.00

2. Find the monthly property tax at 1.1 percent on a \$420,000 home.
   Show the full solution

   \( \dfrac{0.011 \times 420000}{12} = \dfrac{4620}{12} = 385 \). \$385.00

3. Find PITI when principal and interest are \$1,650, taxes \$380 and insurance \$120 a month.
   Show the full solution

   \( 1650 + 380 + 120 = 2150 \). \$2,150.00

4. Find the monthly maintenance budget at 1 percent a year on a \$280,000 home.
   Show the full solution

   \( \dfrac{0.01 \times 280000}{12} = 233.33 \). \$233.33

5. A buyer puts down \$60,000 on a \$300,000 home with 2.5 percent closing costs. Find the cash to close.
   Show the full solution

   Closing costs are \$7,500. \( 60000 + 7500 = 67500 \). \$67,500.00

6. In the worked example, find how much more owning costs each month than renting for \$1,650.
   Show the full solution

   \( 2512.29 - 1650 = 862.29 \). \$862.29 a month

7. Find the annual cost of owning in the worked example.
   Show the full solution

   \( 12 \times 2512.29 = 30147.48 \). \$30,147.48

8. Find the full monthly cost with a \$180 HOA charge.
   Show the full solution

   \( 2512.29 + 180 = 2692.29 \). \$2,692.29

9. A home appreciates 3 percent in a year. Find the gain on a \$350,000 home, and say whether it covers the extra cost of owning over renting.
   Show the full solution

   The gain is \( 0.03 \times 350000 = 10500 \). The extra cost of owning over renting is \( 12 \times 862.29 = 10347.48 \), so a 3 percent gain about pays for it, and the owner also builds equity through the loan. In a year when prices fall, the owner loses on both. \$10,500 against \$10,347.48 of extra cost

10. A student says the cost of owning the \$350,000 home is the \$1,769.79 mortgage payment. Find the error and the size of the understatement.
    Show the full solution

    The loan payment omits property tax (\$320.83), insurance (\$130.00) and maintenance (\$291.67). The full cost is \$2,512.29, so the student understates it by \( 2512.29 - 1769.79 = 742.50 \) dollars a month. \$742.50 a month

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find 28 percent of a gross monthly income of \$6,000.
   Show the full solution

   \( 0.28 \times 6000 = 1680 \). \$1,680.00

2. With other debts of \$500 a month and income of \$6,000, find the back-end limit on housing.
   Show the full solution

   \( 0.36 \times 6000 = 2160 \), less \$500 is \$1,660. \$1,660.00

3. A housing cost of \$1,980 on gross monthly income of \$5,500: find the front-end ratio.
   Show the full solution

   \( \dfrac{1980}{5500} = 0.36 \), or 36 percent, above the 28 percent guideline. 36 percent

4. With other debts of \$420, find the back-end ratio for that housing cost.
   Show the full solution

   \( \dfrac{1980 + 420}{5500} = \dfrac{2400}{5500} = 0.4364 \). 43.64 percent

5. What annual income makes a \$2,100 housing cost 28 percent of gross income?
   Show the full solution

   Monthly \( \dfrac{2100}{0.28} = 7500 \), and \( 12 \times 7500 = 90000 \). \$90,000.00

6. Find the loan supported by a principal and interest payment of \$1,500 at 6 percent for 30 years.
   Show the full solution

   The factor is \( \dfrac{1 - (1.005)^{-360}}{0.005} = 166.7916 \), and \( 1500 \times 166.7916 = 250187.42 \). \$250,187.42

7. Find the loan supported by \$1,700 at 7 percent for 30 years.
   Show the full solution

   The factor is \( \dfrac{1 - (1.005833)^{-360}}{0.005833} = 150.3076 \), and \( 1700 \times 150.3076 = 255522.87 \). \$255,522.87

8. How much more can a payment of \$1,700 borrow at 6.5 percent than at 7 percent?
   Show the full solution

   At 6.5 percent: \( 1700 \times 158.2108 = 268958.40 \). The difference from the 7 percent loan is \( 268958.40 - 255522.87 = 13435.53 \). \$13,435.53

9. The back-end limit for a buyer with income of \$5,500 is 36 percent. Find it in dollars.
   Show the full solution

   \( 0.36 \times 5500 = 1980 \). This is the most the buyer can spend on housing and other debts together. \$1,980.00

10. A student checks only the front-end ratio and says a buyer is fine, but the buyer has \$900 in other monthly debts, income of \$5,500 and a housing cost of \$1,540. Find the back-end ratio.
    Show the full solution

    The front-end ratio is \( \dfrac{1540}{5500} = 28 \) percent, which passes. The back-end ratio is \( \dfrac{1540 + 900}{5500} = 44.36 \) percent, which exceeds 36 percent. The buyer fails the back-end test, so both ratios must be checked. Back-end 44.36 percent: fails

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A home is worth \$410,000 and the loan balance is \$285,000. Find the equity.
   Show the full solution

   \( 410000 - 285000 = 125000 \). \$125,000

2. Find the equity in the last problem as a percent of the value.
   Show the full solution

   \( \dfrac{125000}{410000} \times 100 = 30.49 \). 30.49 percent

3. A \$200,000 balance has 25 years left. Find the payment at 7 percent and at 6 percent, and the monthly saving.
   Show the full solution

   At 7 percent: \$1,413.56. At 6 percent: \$1,288.60. The saving is \( 1413.56 - 1288.60 = 124.96 \). \$1,413.56 and \$1,288.60; \$124.96 saved

4. The refinance in the last problem costs \$3,500. Find the break-even time.
   Show the full solution

   \( \dfrac{3500}{124.96} = 28.01 \) months, so she must stay at least 29 months. 28.01 months

5. A cash-out refinance increases a \$250,000 loan to \$290,000 at 6.25 percent over 25 years. Find the payment.
   Show the full solution

   \( 290000 \times \dfrac{0.0052083}{1 - (1.0052083)^{-300}} = 1913.04 \), which is \$263.87 more than the \$1,649.17 payment on \$250,000. \$1,913.04

6. A HELOC of \$20,000 at 9 percent is interest-only. Find the monthly payment and whether the balance falls.
   Show the full solution

   \( 20000 \times \dfrac{0.09}{12} = 150 \) a month. An interest-only payment leaves the \$20,000 unchanged, so the balance does not fall until principal payments begin. \$150.00; the balance does not fall

7. The homeowner refinances \$250,000 at 6.25 percent over a new 30 years. Find the payment and the total interest, and compare with the 25-year refinance.
   Show the full solution

   The 30-year payment is \$1,539.29, which is \$109.88 less than the 25-year payment. Total interest is \( 1539.29 \times 360 - 250000 = 304144.40 \), which is \$59,393.40 more than the \$244,751.00 on the 25-year loan. \$1,539.29; \$304,144.40

8. In the worked example, how many years to break even?
   Show the full solution

   \( \dfrac{20.17}{12} = 1.68 \) years. 1.68 years

9. If the homeowner sells after 18 months, find the net result of the refinance.
   Show the full solution

   Savings \( 18 \times 198.31 = 3569.58 \), less the \$4,000 cost, is \( -430.42 \). A net loss of \$430.42

10. A student says, "A lower payment always means a refinance saves money." Use the 30-year refinance to show the problem.
    Show the full solution

    Stretching the term to 30 years lowers the payment to \$1,539.29, but the total interest rises to \$304,144.40, which is \$59,393.40 more than the 25-year refinance and close to the \$304,244.00 of the original loan. The payment fell because the loan was made longer, and the saving was given back in interest. A lower payment can cost more in total

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A plan charges \$275 a month. Find the annual premium.
   Show the full solution

   \( 275 \times 12 = 3300 \). \$3,300

2. A plan has a \$1,000 deductible. A patient's bill is \$800. Find her payment.
   Show the full solution

   The bill is below the deductible, so she pays all of it. \$800

3. The plan has a \$1,000 deductible, 20 percent coinsurance and a \$4,000 maximum. Find the payment on a \$6,000 bill.
   Show the full solution

   \( 1000 + 0.20 \times 5000 = 2000 \), below the maximum. \$2,000

4. With the same plan, find the payment on a \$30,000 bill.
   Show the full solution

   Uncapped: \( 1000 + 0.20 \times 29000 = 6800 \). The maximum limits it. \$4,000

5. A plan charges a \$30 copay per visit and \$15 per prescription. Find the cost of 3 visits and 12 prescriptions.
   Show the full solution

   \( 3 \times 30 + 12 \times 15 = 90 + 180 = 270 \). \$270

6. With the plan in problem 3 and a \$275 monthly premium, find the year's total cost for a \$6,000 bill and what the insurer pays.
   Show the full solution

   Total \( 3300 + 2000 = 5300 \). The insurer pays \( 6000 - 2000 = 4000 \). \$5,300 total; insurer pays \$4,000

7. At what bill is the \$4,000 maximum reached for that plan?
   Show the full solution

   \( 1000 + 0.20 \times (B - 1000) = 4000 \), so \( B - 1000 = 15000 \) and \( B = 16000 \). \$16,000

8. Find the worst-case cost of a year on that plan.
   Show the full solution

   The premium plus the maximum: \( 3300 + 4000 = 7300 \). \$7,300

9. In the worked example, find the insurer's payment on the \$12,000 bill.
   Show the full solution

   \( 12000 - 3600 = 8400 \). \$8,400

10. A student says a \$12,000 bill on the worked example's plan costs \( 0.20 \times 12000 = 2400 \). Find the error.
    Show the full solution

    The deductible comes first. The patient pays the first \$1,500 in full and 20 percent of the remaining \$10,500, which is \$2,100, for \$3,600. Applying 20 percent to the whole bill understates the cost by \$1,200. \$3,600, not \$2,400

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A coin flip wins \$10 on heads and loses \$6 on tails. Find the expected value.
   Show the full solution

   \( 0.5 \times 10 + 0.5 \times (-6) = 5 - 3 = 2 \). \$2.00

2. A 5 percent chance of a \$4,000 loss: find the expected loss.
   Show the full solution

   \( 0.05 \times 4000 = 200 \). \$200

3. A 0.4 percent chance of a \$250,000 loss: find the expected loss.
   Show the full solution

   \( 0.004 \times 250000 = 1000 \). \$1,000

4. A game pays \$20 with probability 0.3 and nothing otherwise. Find the fair price.
   Show the full solution

   \( 0.3 \times 20 = 6 \). \$6.00

5. An event pays \$0 with probability 0.6, \$500 with probability 0.3 and \$2,000 with probability 0.1. Find the expected value.
   Show the full solution

   \( 0.6 \times 0 + 0.3 \times 500 + 0.1 \times 2000 = 0 + 150 + 200 = 350 \). \$350.00

6. A policy costs \$900 a year for a 3 percent chance of a \$25,000 loss. Find the expected payout and the insurer's margin.
   Show the full solution

   Expected payout \( 0.03 \times 25000 = 750 \). Margin \( 900 - 750 = 150 \), which is 20 percent of the expected payout. \$750; \$150

7. An insurer sells 10,000 of those policies. Find the premiums collected and the expected claims.
   Show the full solution

   Premiums \( 10000 \times 900 = 9{,}000{,}000 \). Expected claims \( 10000 \times 750 = 7{,}500{,}000 \). \$9,000,000 and \$7,500,000

8. Find the insurer's expected surplus and explain why it is reliable.
   Show the full solution

   \( 9000000 - 7500000 = 1500000 \). With 10,000 customers the number of claims is close to 300, plus or minus about 17, so the total cost varies only a few percent. A single customer's loss is unpredictable; thousands together are not. \$1,500,000

9. A warranty costs \$150 and covers a repair of \$600 that happens 15 percent of the time. Find the expected repair cost and whether the warranty is a good buy.
   Show the full solution

   The expected cost is \( 0.15 \times 600 = 90 \). The warranty costs \$60 more than that, so on average the buyer loses money. It is worth buying only if a \$600 bill would be a hardship. \$90 expected against \$150 price

10. A student says, "If insurance costs more than it is expected to pay, nobody should buy it." Give the reason the statement is wrong.
    Show the full solution

    Expected value measures the average, but a person bears one outcome. Suppose a 3 percent chance of a \$25,000 loss that the person could not pay. The \$900 premium is \$150 above the expected loss, but it removes the risk of a loss that would be ruinous. The value of insurance is in removing the worst case, not in beating the average. It protects against a loss the buyer could not absorb

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A collision claim is \$7,200 and the deductible is \$1,000. Find the payment.
   Show the full solution

   \( 7200 - 1000 = 6200 \). \$6,200

2. A home costs \$250,000 to rebuild and the policy has an 80 percent clause. Find the amount required.
   Show the full solution

   \( 0.80 \times 250000 = 200000 \). \$200,000

3. Find the ratio of carried to required for \$150,000 carried and \$200,000 required.
   Show the full solution

   \( \dfrac{150000}{200000} = 0.75 \). 0.75

4. With \$150,000 carried and \$200,000 required, find the payment on a \$40,000 loss with a \$500 deductible.
   Show the full solution

   \( 0.75 \times 40000 = 30000 \), and \( 30000 - 500 = 29500 \). \$29,500

5. Find the payment on the same loss if \$200,000 is carried.
   Show the full solution

   The ratio is 1: \( 40000 - 500 = 39500 \). \$39,500

6. A driver with a liability limit of \$50,000 per accident is found responsible for \$70,000 of damages. Find what she must pay herself.
   Show the full solution

   The insurer pays up to the limit, \$50,000. She owes \( 70000 - 50000 = 20000 \). \$20,000

7. A claim of \$2,400 is paid with a \$500 deductible or a \$1,000 deductible. Find the payment in each case.
   Show the full solution

   \( 2400 - 500 = 1900 \) and \( 2400 - 1000 = 1400 \). \$1,900 and \$1,400

8. The higher deductible saves \$180 a year in premium. Find the monthly saving.
   Show the full solution

   \( \dfrac{180}{12} = 15 \). \$15 a month

9. A car accident causes \$3,800 of damage and the owner has no collision coverage. What does the owner pay?
   Show the full solution

   Without the coverage, nothing is paid for damage to the owner's own car. \$3,800

10. A home needing \$240,000 is insured for \$120,000, with a \$60,000 loss and a \$1,000 deductible. Find the payment and the percent of the loss the owner bears.
    Show the full solution

    The ratio is \( \dfrac{120000}{240000} = 0.5 \). Payment \( 0.5 \times 60000 - 1000 = 29000 \). The owner bears \( 60000 - 29000 = 31000 \), which is 51.7 percent of the loss. \$29,000; the owner bears 51.7 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use Plans A and B from the worked example.

1. Find the annual premium of Plan A.
   Show the full solution

   \( 210 \times 12 = 2520 \). \$2,520

2. Find the customer's payment under Plan A on a \$2,000 bill.
   Show the full solution

   The bill is below the \$4,000 deductible, so the customer pays \$2,000. \$2,000

3. Find the customer's payment under Plan B on a \$10,000 bill.
   Show the full solution

   \( 800 + 0.10 \times 9200 = 1720 \). \$1,720

4. Find each plan's total cost of the year on a \$10,000 bill.
   Show the full solution

   Plan A: \( 2520 + 5200 = 7720 \). Plan B: \( 5160 + 1720 = 6880 \). \$7,720 and \$6,880

5. At what bill does each plan reach its out-of-pocket maximum?
   Show the full solution

   Plan A: \( 4000 + \dfrac{7000 - 4000}{0.20} = 4000 + 15000 = 19000 \). Plan B: \( 800 + \dfrac{3500 - 800}{0.10} = 800 + 27000 = 27800 \). \$19,000 and \$27,800

6. Find the worst-case cost of a year under each plan.
   Show the full solution

   Plan A: \( 2520 + 7000 = 9520 \). Plan B: \( 5160 + 3500 = 8660 \). \$9,520 and \$8,660

7. Find each plan's cost of a year with no bills, and the difference.
   Show the full solution

   Plan A \$2,520 and Plan B \$5,160, a difference of \$2,640. \$2,520 and \$5,160; \$2,640

8. Find the bill at which the two plans cost the same, and the common total.
   Show the full solution

   Solve \( 2520 + b = 5160 + 800 + 0.10(b - 800) \): \( 0.90b = 3360 \), \( b = 3733.33 \). The total is \( 2520 + 3733.33 = 6253.33 \). \$3,733.33; \$6,253.33

9. Find how much more Plan A costs than Plan B in the worst year.
   Show the full solution

   \( 9520 - 8660 = 860 \). Plan A saves \$2,640 in a healthy year but costs up to \$860 more in the worst. \$860

10. A student says, "Plan A is cheaper because its premium is \$2,640 lower." Use a \$10,000 bill to correct this.
    Show the full solution

    The premium is only part of the cost. On a \$10,000 bill the customer pays \$5,200 under Plan A and \$1,720 under Plan B, a \$3,480 difference that more than cancels the \$2,640 premium saving. Totals are \$7,720 and \$6,880, so Plan B is cheaper by \$840. Plan B is cheaper by \$840

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find 10 times an annual income of \$45,000.
   Show the full solution

   \( 45000 \times 10 = 450000 \). \$450,000

2. Debts are \$25,000, the mortgage is \$180,000, 8 years of a \$50,000 income are to be replaced, education is \$60,000, and assets are \$40,000. Find the coverage to buy.
   Show the full solution

   Need \( 25000 + 180000 + 400000 + 60000 = 665000 \). Less assets: \( 665000 - 40000 = 625000 \). \$625,000

3. Find the annual and monthly premium for \$500,000 of term coverage at \$0.42 per \$1,000 a year.
   Show the full solution

   \( 500 \times 0.42 = 210 \) a year, and \( \dfrac{210}{12} = 17.50 \) a month. \$210; \$17.50

4. A disability policy replaces 60 percent of a \$48,000 income. Find the monthly benefit.
   Show the full solution

   \( 0.60 \times 48000 = 28800 \) a year, and \( \dfrac{28800}{12} = 2400 \). \$2,400 a month

5. The policy has a 90-day elimination period and the worker's expenses are \$2,400 a month. Find the cash cushion for the wait.
   Show the full solution

   Ninety days is three months: \( 3 \times 2400 = 7200 \). \$7,200

6. Disability insurance costs 1.5 percent of a \$60,000 income. Find the yearly premium.
   Show the full solution

   \( 0.015 \times 60000 = 900 \). \$900

7. A 20-year term policy costs \$480 a year and a whole life policy for the same death benefit costs \$3,600 a year. Find the total premiums over 20 years and the difference.
   Show the full solution

   Term: \( 480 \times 20 = 9600 \). Whole life: \( 3600 \times 20 = 72000 \). The difference is \$62,400, which could be saved and invested separately. \$9,600 and \$72,000; \$62,400 more

8. A worker earns \$52,000 and already has \$150,000 of coverage. A rule of thumb says 10 times income. How much more should she buy?
   Show the full solution

   The rule gives \( 10 \times 52000 = 520000 \). She needs \( 520000 - 150000 = 370000 \) more. \$370,000

9. In the worked example, find the premium as a percent of income.
   Show the full solution

   \( \dfrac{440}{60000} \times 100 = 0.73 \). 0.73 percent

10. A single 25-year-old with no dependents buys \$500,000 of term coverage "because insurance is responsible." Explain the flaw.
    Show the full solution

    Life insurance replaces income that other people depend on. With no dependents and no debts that someone else would have to pay, no one is harmed financially by her death, so the coverage has no beneficiary who needs it. The premium would be better saved. The exceptions are debts a co-signer would inherit and funeral costs. No dependents: little need

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A premium saving of \$180 a year is how much a month?
   Show the full solution

   \( \dfrac{180}{12} = 15 \). \$15

2. A higher deductible saves \$240 a year and adds \$1,000 of exposure. Find the break-even probability.
   Show the full solution

   \( \dfrac{240}{1000} = 0.24 \). 24 percent

3. With an 8 percent chance of a claim, find the expected extra cost of a \$1,000 higher deductible and the net saving of \$240.
   Show the full solution

   Expected extra cost \( 0.08 \times 1000 = 80 \). Net saving \( 240 - 80 = 160 \). \$80; \$160

4. A driver has made one claim in six years. Estimate the yearly chance of a claim and the expected extra cost of an added \$500 deductible.
   Show the full solution

   The chance is about \( \dfrac{1}{6} = 0.1667 \), so the expected extra cost is \( 0.1667 \times 500 = 83.33 \). 16.7 percent; \$83.33

5. Raising a deductible from \$500 to \$2,000 cuts the premium from \$1,680 to \$1,310. Find the saving, the exposure and the break-even probability.
   Show the full solution

   Saving \( 1680 - 1310 = 370 \). Exposure \( 2000 - 500 = 1500 \). Break-even \( \dfrac{370}{1500} = 0.2467 \). \$370; \$1,500; 24.67 percent

6. In the last problem, find the years of savings needed to recoup the exposure.
   Show the full solution

   \( \dfrac{1500}{370} = 4.05 \) years. If there will probably be no claim for about four years, the higher deductible pays. 4.05 years

7. A claim causes \$6,000 of damage. Find the insurer's payment with a \$500 deductible and with a \$1,000 deductible.
   Show the full solution

   \( 6000 - 500 = 5500 \) and \( 6000 - 1000 = 5000 \). \$5,500 and \$5,000

8. Find the total cost of a year with one claim under each policy in the worked example (premium plus deductible).
   Show the full solution

   Low deductible: \( 1420 + 500 = 1920 \). High deductible: \( 1240 + 1000 = 2240 \). In a year with a claim the low deductible is \$320 cheaper. \$1,920 and \$2,240

9. After six claim-free years and then a claim, find the net result for the higher deductible.
   Show the full solution

   Savings \( 6 \times 180 = 1080 \), less the extra \$500 on the claim: \$580 ahead. \$580 ahead

10. A student says, "I'll take the \$500 deductible because I might need to make a claim." Use expected value to evaluate this.
    Show the full solution

    The possibility is not the issue; its probability is. At a 10 percent yearly chance of a claim, the extra \$500 costs an expected \$50 a year, while the lower premium saves \$180. The \$500 deductible is worth choosing only if a claim is more likely than 36 percent in a year, or if the \$1,000 could not be paid when needed. The high deductible has the better expected value below 36 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the cost of 50 shares at \$34.
   Show the full solution

   \( 50 \times 34 = 1700 \). \$1,700

2. The 50 shares are sold at \$39.60. Find the gain and the percent return, ignoring dividends.
   Show the full solution

   Proceeds \( 50 \times 39.60 = 1980 \). Gain \( 1980 - 1700 = 280 \). Return \( \dfrac{280}{1700} \times 100 = 16.47 \). \$280; 16.47 percent

3. An investor buys 100 shares at \$28 and sells at \$21.50. Find the loss and the percent.
   Show the full solution

   \( 100 \times (21.50 - 28) = -650 \), and \( \dfrac{-6.50}{28} \times 100 = -23.21 \). A loss of \$650; 23.21 percent

4. That stock paid \$0.80 per share in dividends. Find the total return.
   Show the full solution

   Dividends \( 100 \times 0.80 = 80 \). Return \( \dfrac{-650 + 80}{2800} \times 100 = -20.36 \). Negative 20.36 percent

5. An investment returns 40 percent in total over 5 years. Find the annualized return.
   Show the full solution

   \( (1.40)^{1/5} - 1 = 0.0696 \). 6.96 percent a year

6. In the worked example, the broker charges \$8 on each trade. Find the total return after commissions.
   Show the full solution

   The cost is \( 2100 + 8 = 2108 \) and the proceeds net \( 2352 - 8 = 2344 \). Gain plus dividends: \( 2344 + 48 - 2108 = 284 \), and \( \dfrac{284}{2108} = 13.47 \) percent. 13.47 percent

7. How many whole shares can \$3,000 buy at \$47.50?
   Show the full solution

   \( \dfrac{3000}{47.50} = 63.16 \), so 63 whole shares, costing \$2,992.50. 63 shares

8. What gain is needed to recover from a 25 percent fall, and from a 50 percent fall?
   Show the full solution

   After a 25 percent fall the value is 0.75, so the gain is \( \dfrac{1}{0.75} - 1 = 33.33 \) percent. After 50 percent it is \( \dfrac{1}{0.5} - 1 = 100 \) percent. 33.33 percent; 100 percent

9. A stock goes from \$12 to \$15 and pays \$0.60 in dividends. Find the total return.
   Show the full solution

   \( \dfrac{3 + 0.60}{12} = 0.30 \). 30 percent

10. A student says that a stock that falls 50 percent and then rises 50 percent is back where it began. Test this with \$100.
    Show the full solution

    \( 100 \times 0.50 = 50 \), then \( 50 \times 1.50 = 75 \). The investor is down 25 percent, because the rise applies to the smaller amount. \$75, not \$100

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the annual coupon on a \$1,000 bond with a 5 percent coupon rate.
   Show the full solution

   \( 0.05 \times 1000 = 50 \). \$50

2. That bond is bought at \$1,025. Find the current yield.
   Show the full solution

   \( \dfrac{50}{1025} \times 100 = 4.88 \). 4.88 percent

3. Find the total interest and face value received over 10 years from that bond.
   Show the full solution

   Coupons \( 50 \times 10 = 500 \), plus the face value \$1,000, is \$1,500. \$1,500

4. A bond with a 6 percent coupon is bought at \$1,050. Find the current yield.
   Show the full solution

   \( \dfrac{60}{1050} \times 100 = 5.71 \). 5.71 percent

5. Is a bond bought at \$1,050 with a face value of \$1,000 bought at a premium or a discount?
   Show the full solution

   The price is above face value, so it is a premium, and \$50 will be lost at maturity. A premium

6. Find the approximate yield to maturity for that 5-year bond with a 6 percent coupon.
   Show the full solution

   The yearly loss is \( \dfrac{1000 - 1050}{5} = -10 \). Then \( \dfrac{60 - 10}{(1000 + 1050)/2} = \dfrac{50}{1025} = 4.88 \) percent, below the current yield of 5.71 percent. 4.88 percent

7. Find the price of a 10-year, 5 percent coupon bond when market rates are 6 percent and when they are 4 percent.
   Show the full solution

   At 6 percent: \( 50 \times 7.3601 + \dfrac{1000}{(1.06)^{10}} = 368.00 + 558.39 = 926.40 \). At 4 percent: \( 50 \times 8.1109 + 675.56 = 1081.11 \). \$926.40 and \$1,081.11

8. An investor holds 20 bonds of \$1,000 face with a 3.5 percent coupon paid twice a year. Find each payment.
   Show the full solution

   Yearly interest per bond is \$35, so \$17.50 each six months. For 20 bonds, \( 20 \times 17.50 = 350 \). \$350 every six months

9. In the worked example, the market rate rises from 4 percent to 5 percent. Find the percent change in the bond's price.
   Show the full solution

   The price falls from \$1,000 to \$956.71: \( \dfrac{956.71 - 1000}{1000} = -4.33 \) percent. An investor who must sell then takes the loss. Down 4.33 percent

10. A student says bonds are safe, so their prices never change. Correct this.
    Show the full solution

    The payments are fixed, so held to maturity they are predictable. The market price changes whenever interest rates change: the example bond is worth \$956.71 at a 5 percent market rate and \$1,045.80 at 3 percent. A bond sold before maturity can earn a loss. Prices move with interest rates

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the yearly fee on \$12,000 at an expense ratio of 0.75 percent.
   Show the full solution

   \( 0.0075 \times 12000 = 90 \). \$90

2. A fund earns 7.2 percent before fees and charges 0.85 percent. Find the net return.
   Show the full solution

   \( 7.20 - 0.85 = 6.35 \). 6.35 percent

3. A 5 percent front load is charged on a \$5,000 purchase. Find the load and the amount actually invested.
   Show the full solution

   Load \( 0.05 \times 5000 = 250 \). Invested \( 5000 - 250 = 4750 \). \$250; \$4,750

4. Find the fee on \$50,000 at 1.00 percent and at 0.05 percent.
   Show the full solution

   \( 0.01 \times 50000 = 500 \) and \( 0.0005 \times 50000 = 25 \). \$500 and \$25

5. What is the ratio of the two fees in the last problem?
   Show the full solution

   \( \dfrac{500}{25} = 20 \). 20 times

6. Find the value of \$10,000 after 20 years at net returns of 6.35 percent and 6.85 percent.
   Show the full solution

   \( 10000 \times (1.0635)^{20} = 34257.04 \) and \( 10000 \times (1.0685)^{20} = 37626.21 \). Half a point is worth \$3,369.17. \$34,257.04 and \$37,626.21

7. Find the value of \$10,000 after 40 years at net returns of 6.00 and 6.95 percent.
   Show the full solution

   \( 10000 \times (1.06)^{40} = 102857.18 \) and \( 10000 \times (1.0695)^{40} = 146970.97 \). The low-fee fund ends 43 percent higher. \$102,857.18 and \$146,970.97

8. On \$25,000 for 25 years, an index fund nets 7.45 percent and an active fund nets 6.4 percent. Find the final balances.
   Show the full solution

   \( 25000 \times (1.0745)^{25} = 150695.58 \) and \( 25000 \times (1.064)^{25} = 117891.02 \). The difference is \$32,804.56. \$150,695.58 and \$117,891.02

9. In the worked example, the monthly saver deposits \$200 for 30 years. Find the total deposited and the share of Fund A's balance that is growth.
   Show the full solution

   Deposited \( 200 \times 360 = 72000 \). Growth \( 241600.63 - 72000 = 169600.63 \), which is \( \dfrac{169600.63}{241600.63} = 70.2 \) percent of the balance. \$72,000; 70.2 percent

10. A student says, "A 1 percent fee is only 1 percent of my money, so it barely matters." Use the worked example to correct this.
    Show the full solution

    The fee is taken every year from the whole balance, including the growth, so it lowers the growth rate from 6.95 to 6.00 percent for thirty years. The balance ends at \$57,434.91 instead of \$75,062.61, a loss of \$17,627.70, or 23.5 percent. A small percent of a large balance, repeated for decades, is a large amount. The fee costs 23.5 percent of the final balance

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the mean of yearly returns 6, −2, 9 and 3 percent.
   Show the full solution

   \( \dfrac{6 - 2 + 9 + 3}{4} = 4 \). 4 percent

2. Find the sample standard deviation of those returns.
   Show the full solution

   The deviations are 2, −6, 5 and −1. The squares sum to \( 4 + 36 + 25 + 1 = 66 \) and \( s = \sqrt{\dfrac{66}{3}} = \sqrt{22} = 4.69 \). 4.69 percent

3. Find the range of returns of Asset A and of Asset B in the worked example.
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   A: \( 20 - (-5) = 25 \). B: \( 15 - (-8) = 23 \). 25 and 23 points

4. Find the return of a 70/30 mix of A and B in year 1.
   Show the full solution

   \( 0.70 \times 12 + 0.30 \times (-3) = 8.4 - 0.9 = 7.5 \). 7.5 percent

5. The 70/30 mix has yearly returns 7.5, −0.5, 11.6, 10.1 and 0.4. Find its mean.
   Show the full solution

   \( \dfrac{7.5 - 0.5 + 11.6 + 10.1 + 0.4}{5} = \dfrac{29.1}{5} = 5.82 \). 5.82 percent

6. That mix has a standard deviation of 5.56. Compare it with A alone and the 50/50 mix.
   Show the full solution

   A alone is 10.24 and the 50/50 mix is 3.82, so the 70/30 mix is in between. More weight on the more variable asset brings more risk. Its mean, 5.82, is also between the 50/50 mix's 5.3 and A's 6.6. Between 3.82 and 10.24

7. A portfolio is 60 percent stock and 40 percent bonds, worth \$10,000. In a bad year stocks fall 20 percent and bonds rise 4 percent. Find the dollar change and the percent.
   Show the full solution

   Stocks: \( 6000 \times (-0.20) = -1200 \). Bonds: \( 4000 \times 0.04 = 160 \). Net \( -1040 \), which is \( \dfrac{-1040}{10000} = -10.4 \) percent. A loss of \$1,040; 10.4 percent

8. Find the loss of the same \$10,000 if it were all in stocks, and compare.
   Show the full solution

   \( 10000 \times (-0.20) = -2000 \), a 20 percent loss. The mix lost about half as much. A loss of \$2,000

9. After a 30 percent loss, what gain is needed to recover?
   Show the full solution

   \( \dfrac{1}{0.70} - 1 = 0.4286 \). 42.86 percent

10. A student says that diversifying guarantees no losses. Correct the claim with the worked example.
    Show the full solution

    Diversification lowers the swings; it does not remove them. The 50/50 mix in the example had no losing year, but that is a five-year sample of two assets. In a year when all assets fall together, as in a broad market decline, a diversified portfolio falls too, as the 60/40 mix fell 10.4 percent. It reduces risk, not eliminates it

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A fund gains 10 percent and then loses 10 percent. Find the growth factor.
   Show the full solution

   \( 1.10 \times 0.90 = 0.99 \), a loss of 1 percent, although the average return is zero. 0.99

2. Find the compound average of +20 and −20 percent.
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   \( (1.20 \times 0.80)^{1/2} - 1 = (0.96)^{1/2} - 1 = -0.0202 \). Negative 2.02 percent a year

3. A balance of \$100 grows 30 percent and then falls 20 percent. Find the final value.
   Show the full solution

   \( 100 \times 1.30 \times 0.80 = 104 \). \$104

4. Find the compound average yearly return for that \$100 over the two years.
   Show the full solution

   \( (1.04)^{1/2} - 1 = 0.0198 \). 1.98 percent

5. A fund falls 20 percent and then rises 25 percent. Find the growth factor.
   Show the full solution

   \( 0.80 \times 1.25 = 1.00 \). The 25 percent gain exactly offsets the 20 percent loss, because it is taken from a smaller base. 1.00

6. An investment of \$10,000 grows to \$15,000 in 6 years. Find the compound annual return.
   Show the full solution

   \( (1.5)^{1/6} - 1 = 0.0699 \). 6.99 percent

7. A fund returns +30, −20, +30 and −20 percent. Find the final value of \$10,000 and the compound average.
   Show the full solution

   \( 10000 \times 1.30 \times 0.80 \times 1.30 \times 0.80 = 10816 \). The compound average is \( (1.0816)^{1/4} - 1 = 0.0198 \), although the arithmetic average is 5 percent. \$10,816.00; 1.98 percent

8. Find the value of \$10,000 growing steadily at 8 percent for four years, and compare with the last problem.
   Show the full solution

   \( 10000 \times (1.08)^4 = 13604.89 \). The volatile fund, with an arithmetic average of only 5 percent, ended \$2,788.89 below the steady 8 percent. \$13,604.89

9. Which figure tells an investor what happened to the money, the arithmetic or the compound average, and why?
   Show the full solution

   The compound average, because each year's return applies to the balance the year before produced. Only the geometric mean reproduces the final value when compounded for the same number of years. The compound average

10. A student says a fund that averaged 8 percent over four years turns \$10,000 into \$13,604.89. Explain why that may be wrong.
    Show the full solution

    Growing at 8 percent each year gives that figure, but an average of 8 percent does not mean every year was 8. If the returns swing, as +30, −20, +30 and −20 do, the balance follows the product of the factors. A variable path with a lower compound average ends lower, here at \$10,816. Only a steady 8 percent gives \$13,604.89

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A stock at \$60 pays \$1.50 a year. Find the dividend yield.
   Show the full solution

   \( \dfrac{1.50}{60} \times 100 = 2.5 \). 2.5 percent

2. A stock sells for \$45 with EPS of \$3. Find the P/E.
   Show the full solution

   \( \dfrac{45}{3} = 15 \). 15

3. A stock has EPS of \$4.25 and a P/E of 18. Find the price.
   Show the full solution

   \( 18 \times 4.25 = 76.50 \). \$76.50

4. A company earns \$3.60 a share and pays \$0.90. Find the payout ratio.
   Show the full solution

   \( \dfrac{0.90}{3.60} \times 100 = 25 \). 25 percent

5. An investor owns 200 shares paying \$2.10 each. Find the dividend income.
   Show the full solution

   \( 200 \times 2.10 = 420 \). \$420

6. Find the earnings yield of a stock with a P/E of 25.
   Show the full solution

   \( \dfrac{1}{25} = 0.04 \). 4 percent

7. In the worked example, EPS rises to \$7 at a P/E of 14. Find the new price and the percent gain.
   Show the full solution

   \( 14 \times 7 = 98 \), and \( \dfrac{98 - 84}{84} \times 100 = 16.67 \). \$98; 16.67 percent

8. The price falls to \$70 with the dividend unchanged at \$2.10. Find the dividend yield.
   Show the full solution

   \( \dfrac{2.10}{70} \times 100 = 3.00 \). 3.00 percent

9. A dividend yield of 2.5 percent is combined with earnings growth of 6 percent a year. Find the approximate return if the P/E does not change.
   Show the full solution

   \( 2.5 + 6 = 8.5 \) percent. With the P/E constant, the price grows with earnings, and the dividend is paid on top. About 8.5 percent

10. A student says a stock with a 9 percent dividend yield is always a better buy than one with a 2.5 percent yield. Explain the flaw.
    Show the full solution

    The yield rises when the price falls. A stock that has lost half its price with an unchanged dividend shows a yield twice as high, and it may reflect a troubled company about to cut the dividend. A payout ratio above 100 percent means the dividend exceeds profit and cannot be kept up. The yield must be read with the payout ratio and the reason for the price. A high yield can signal trouble

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A scheme promises 1 percent a week, compounded. Find the yearly rate.
   Show the full solution

   \( (1.01)^{52} - 1 = 0.6777 \). 67.77 percent a year

2. Another promises 2 percent a week. Find what \$10,000 becomes in one year.
   Show the full solution

   \( 10000 \times (1.02)^{52} = 28003.28 \). \$28,003.28

3. Find the number of people at level 10 of a pyramid in which each member recruits 6 others.
   Show the full solution

   \( 6^{10} = 60466176 \). 60,466,176

4. A scheme promises 5 percent a month. Find the yearly growth factor and the annual rate.
   Show the full solution

   \( (1.05)^{12} = 1.7959 \), so the annual rate is 79.59 percent. 1.7959; 79.59 percent

5. Use the rule of 72 to estimate how long money takes to double at 10 percent a month.
   Show the full solution

   \( \dfrac{72}{10} = 7.2 \) months. About 7.2 months

6. Compare \$10,000 over 10 years at a guaranteed 12 percent with 7 percent.
   Show the full solution

   At 12 percent: \( 10000 \times (1.12)^{10} = 31058.48 \). At 7 percent: \( 10000 \times (1.07)^{10} = 19671.51 \). A guarantee of 12 percent would beat the long-run market return by a wide margin, which no honest seller can guarantee. \$31,058.48 against \$19,671.51

7. A scheme has 1,000 investors who each put in \$5,000 and are promised 10 percent a month. Find the total invested and the monthly payout owed.
   Show the full solution

   Invested \( 1000 \times 5000 = 5{,}000{,}000 \). Payout \( 0.10 \times 5{,}000{,}000 = 500{,}000 \) a month. \$5,000,000; \$500,000 a month

8. If the scheme earns nothing, how much new money must arrive each month just to pay the promised returns, and for how many months does the original \$5,000,000 last if no money comes in?
   Show the full solution

   New money of \$500,000 a month must arrive. With no inflow, the original money would last \( \dfrac{5000000}{500000} = 10 \) months if the promoters paid returns from it, with nothing left to return to investors. \$500,000 a month; 10 months

9. Name four warning signs of an investment fraud.
   Show the full solution

   Any four of: guaranteed or unusually high returns; pressure to decide immediately; secrecy or a vague explanation of how profits arise; an unregistered seller; trouble withdrawing money; payment for recruiting others. See the list

10. A student says, "My statement shows gains every month, so the scheme must be real." Explain what is wrong.
    Show the full solution

    A statement is whatever the operator prints. In a Ponzi scheme the gains are invented, and the money for withdrawals comes from new deposits. The arithmetic test is stronger than the statement: a promised 10 percent a month is 213.84 percent a year, which cannot be earned legitimately, so the statement proves nothing. Statements can be fabricated; impossible arithmetic cannot

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Assets total \$25,000 and liabilities total \$31,000. Find net worth.
   Show the full solution

   \( 25000 - 31000 = -6000 \). Net worth can be negative, as it often is for a new graduate with student loans. Negative \$6,000

2. Find total assets: \$1,500 checking, \$4,200 savings, \$9,800 investments and a \$15,000 car.
   Show the full solution

   \( 1500 + 4200 + 9800 + 15000 = 30500 \). \$30,500

3. Find total liabilities: a \$800 card, a \$6,000 car loan and \$14,500 of student loans.
   Show the full solution

   \( 800 + 6000 + 14500 = 21300 \). \$21,300

4. Find net worth for the last two problems.
   Show the full solution

   \( 30500 - 21300 = 9200 \). \$9,200

5. Net worth rises from \$16,800 by \$5,000. Find the new value and the percent change.
   Show the full solution

   \( 16800 + 5000 = 21800 \), and \( \dfrac{5000}{16800} \times 100 = 29.76 \). \$21,800; 29.76 percent

6. Liquid assets are \$9,100 and monthly essential expenses are \$2,400. Find the months covered.
   Show the full solution

   \( \dfrac{9100}{2400} = 3.79 \). 3.79 months

7. A car loses \$2,000 in value. What happens to net worth?
   Show the full solution

   An asset falls by \$2,000 and nothing else changes, so net worth falls by \$2,000. It falls by \$2,000

8. A person pays \$3,000 of a loan out of savings. What happens to net worth?
   Show the full solution

   Assets fall by \$3,000 (the savings) and liabilities fall by \$3,000 (the loan). Net worth is unchanged, although the person's interest costs fall. No change

9. A person borrows \$10,000 and spends it on a vacation. Find the change in net worth.
   Show the full solution

   The loan adds a \$10,000 liability, and the vacation leaves no asset behind. Net worth falls by \$10,000. Down \$10,000

10. A student says that borrowing \$10,000 to buy a car worth \$10,000 makes him \$10,000 richer because he now owns a car. Find the error.
    Show the full solution

    The car adds \$10,000 to assets and the loan adds \$10,000 to liabilities, so net worth is unchanged. It then falls as the car depreciates, for example by 20 percent, or \$2,000, in the first year. Net worth does not change at the moment of purchase

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Inflow is \$2,800 and outflow is \$2,650. Find the net cash flow.
   Show the full solution

   \( 2800 - 2650 = 150 \). \$150

2. Find the savings rate if that \$150 is saved.
   Show the full solution

   \( \dfrac{150}{2800} \times 100 = 5.36 \). 5.36 percent

3. Find 10 percent of a \$3,120 net income.
   Show the full solution

   \( 0.10 \times 3120 = 312 \). \$312

4. Find the yearly saving of 20 percent of a \$2,600 net monthly income.
   Show the full solution

   \( 0.20 \times 2600 \times 12 = 6240 \). \$6,240

5. A raise adds \$200 a month to net pay, and spending does not change. The surplus was \$150. Find the new surplus.
   Show the full solution

   \( 150 + 200 = 350 \). \$350

6. Saving \$4,680 a year from a gross income of \$52,000: find the savings rate on gross income.
   Show the full solution

   \( \dfrac{4680}{52000} \times 100 = 9.0 \). 9.0 percent

7. A worker saves all of a \$570 surplus from net income of \$3,120. Find the savings rate.
   Show the full solution

   \( \dfrac{570}{3120} \times 100 = 18.27 \). 18.27 percent

8. How many months does it take to save \$5,955 at \$570 a month?
   Show the full solution

   \( \dfrac{5955}{570} = 10.45 \), so 11 months. 11 months

9. Monthly debt payments are \$600 and gross monthly income is \$4,800. Find the debt-to-income ratio.
   Show the full solution

   \( \dfrac{600}{4800} \times 100 = 12.5 \). 12.5 percent

10. A student computes a savings rate of \$300 against \$2,550 of spending, getting 11.76 percent. Find the error and the correct rate on \$3,120 of income.
    Show the full solution

    The rate is saving divided by income, not by spending. Dividing by spending makes the ratio look larger. The correct rate is \( \dfrac{300}{3120} \times 100 = 9.62 \) percent. 9.62 percent