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# 3.Quiz3Sol - compute an amortization schedule for the...

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ADM3340A Fall 2008 Quiz #3 Student Name____________________________ Student ID: ______________________ Question 1 Quality 9000 International Inc., which began operations in 1996, sells 20,000 units of its product each year under the following warranty: defective units will be fixed free of charge during the calendar year of purchase and the next two calendar years. (This means it is best to buy from this company early in the year.) Only 1% of units sold have required warranty service in the past. The average cost has been \$200 per unit for servicing. Units require service only once and the likelihood of a unit requiring service is the same during each year in the warranty period. What is the balance in the warranty liability account at December 31, 1999? As of Dec. 31/99, the warranty for 1996, 1997 units is expired; Dec. 31/99 liability = For 1998 sales: 1/3(20,000)(\$200)(.01) = \$13,333 For 1999 sales: 2/3(20,000)(\$200)(.01) = 26,667 Total liability at Dec. 31/99 \$40,000 Question 2 It is often necessary to compute the book value of a bond issue several years into its term. Rather than
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Unformatted text preview: compute an amortization schedule for the entire term, it is possible to directly compute the net bond liability at any interest date under either the interest method or straight-line method. Assume that \$100,000 of 8% bonds were issued to yield 10% on January 1, 2000, the bond date. The bonds pay interest each December 31 and are scheduled to mature in ten years. Answer the following questions without producing an amortization schedule. Show all supporting calculations. (a) What is the book value of the bonds on January 1, 2006 if the firm uses the straight-line method? (b) What is the book value of the bonds on January 1, 2006 if the firm uses the interest (effective interest) method. (a) Original issue price = \$100,000(PV1,.10,10)(.38554) + \$8,000(PVA,.10,10)(6.14457) = \$87,711. Original discount = \$12,289. At Jan. 1/06, 4 years of term are left so BV = \$100,000 − .4(\$12,289) = \$95,084. (b) BV = 100,000(PV1,.10,4)(.68301) + \$8,000(PVA,.10,4)(3.16987) = \$93,660....
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