Texas A&M University-Commerce



Math 179 – Finance Practice Test KEY

1. The equation [pic] models an investment of $ _1000_, invested at _6.5_% simple interest, for __6___ years. The balance in the account at that time will be $__1390__. (Simple interest: No TVM Solver)

2. The equation 10,000[pic] models a loan of $_10,000____, borrowed at _5.2_% interest, compounded _quarterly, to be paid _quarterly for _3_ years. (NAP ÷ P/Y)

The size of payment needed is $_905.42_.

N = 12 I% = 5.2 PV = 10000 Pmt = ? FV = 0 (loan) P/Y = C/Y = 4 END

3. The equation 1,000,000 = Pmt [pic] models a retirement annuity account paying _10__% interest, compounded _month__ly. Payments of $__4881.74__ will be made __month___ly for

__10___ years. (NAP 120 ÷ P/Y 12) The balance at retirement will be $_1,000,000_____.

N = 120 I% = 10 PV = 0 Pmt = ? FV = 1000000 P/Y = C/Y = 12 END

4. Interpret the following TVM Solver screen.

|N = 16 |This screen models an investment or loan (circle one) of $ _8000_____, at __6___% interest, compounded _quarter__ly for __4__ years. |

|I% = 6 |(N = 16 quarters) Interest is paid / charged on a _month_ly basis. The balance at the end is $ __10,151.88____. |

|PV = - 8000 | |

|Pmt = 0 | |

|FV = ? | |

|P/Y = 4 | |

|C/Y = 4 | |

5. The equation FV = 120[pic] models an annuity with _week__________ly payments of $_120_________ made into an account earning _4.5__% interest, compounded _week______ly, for ___2___ years. (104 ÷ 52) The account will have a balance of $__13,052.93____at that time.

N = 104 I% = 4.5 PV = 0 Pmt = -120 FV = ? P/Y = C/Y = 52 END

|6. |4.1% cpd monthly |On the calculator: Finance ↕Eff(%, C/Y). Homescreen – Use Annual Yield formula |

| | |Eff(4,52) = 4.079% annually; Eff(4.1, 12)= 4.178% annually (Higher) |

|7. |Pmt = $ 83.92 |Add-on loan →simple interest→NO TVM Solver. FV = 950(1 + 0.06 X 1) = 1007. |

| | |Pmt = 1007 ÷ 12 |

|8. |Bal = $ 2672.54 |N = 6x1=6 I% = 4.1 PV = 2100 Pmt = 0 FV = ? P/Y = C/Y = 1 (annually) END |

|9. |Bal = $ 8756.65 |N = 8x12 = 96 I% =4.8 PV = 0 Pmt = -75 FV = ? P/Y = C/Y = 12 END |

|10. |Deposit $ 2,583.36 |N = 25x365 = 9125 I% = 13.2 PV = ? Pmt = 0 (deposit now = no pmt) |

| |Interest = $ 67,416.64 |FV = 70000 P/Y = C/Y = 365 END Interest = 70000 – 2583.36 |

|11. |Bal = $ 340,768.85 |N = 30 I% = 9 PV = 0 Pmt = -2500 FV = ? P/Y = C/Y = 1 (annually) END |

| |Interest = $ 265,768.85 |Interest = FV – Paid = 340768.85 – (2500 x 30) |

|12. |Pmt = $ 590.55 |N = 30x12 = 360 I% = 7.2 PV = 87000 Pmt = ? FV = 0 (loan) P/Y = C/Y = 12 END |

| |Interest = $ 125,598 |Interest = Paid – PV (loan) = (360 x 590.55) - 87000 |

|13. |Deposit $ 59.14 per month (or $ 60)|N = 18x12 = 216 I% = 8.5 PV = 0 Pmt = ? FV = 30000 P/Y = C/Y = 12 END |

|14. |Bal = $ 2235.67 |Simple interest → NO TVM Solver. FV = 1900 ( 1 + .053 x 40 ÷ 12) |

| | |Time must be in years → 40 months ÷ 12 months per year |

|15. |OMIT | |

| |Fall 2011 | |

|16. |Car priced at |N = 5x12= 60 I% = 4.2 PV = ? Pmt = -350 FV = 0 P/Y = C/Y = 12 END |

| |$ 20,911.85 |Ken can afford a loan (PV) of $ 18,911.85. Loan + Down payment = Price of car |

17.

|Pmt number |Payment Size |Interest Pmt |Principal pmt |Balance |

|0 |X |X |X |70,000 |

|1 |547.98 |415.63 |132.36 |69,867.65 |

|2 |547.98 |414.84 |133.14 |69,734.51 |

|3 |547.98 |414.05 |133.93 |69,600.58 |

For pmt # 1:

Calculate the interest owed using I = Prt → I = 70000 x 0.07125 x [pic] = 415.63 ([pic] because time is in years and one month is [pic] of a year.)

Calculate the amount that goes toward the loan (principal pmt) by subtracting the interest pmt from the pmt size (the check Kolby mailed in). 547.98 – 415.63 = 132.36

Subtract the principal pmt from the old balance to find the new balance: 70000 – 132.36 = 69867.65

For pmt # 2:

Same calculations, but use the new balance in step one. I = 69867.65 x 0.07125 x [pic] = 414.84

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