Mathematics 2580 6.0A Test Mathematics 2580 6.0A Test #4 Answers

There were two versions of the test. The questions were identical in the two versions. The tables were different.

  1. NPV = -1200000-800000(1.10)-1+550000a5|0.10(1.10)-1 = -$31,879.34. Since NPV < 0, the investment is NOT worthwhile.

  2. K1 = 60+60/((1.07)4-1) = $295.23. K2 = X+X/((1.07)8-1) = 2.3923966X. K1 = K2 gives X = 295.23/2.3923966 = $123.40. Thus you should be willing to pay $123.40 for an eight-year battery.

    Remark: Many solutions to this found the extra amount $53.40 that should be paid. If you claimed this was the battery price (or did not clarify the point) you lost 2 marks.

  3. (a) 28000(1-d)10 = 8000 so 1-d = (2/7)0.1. B5 = 28000(2/7)(0.5) = $14,966.63.

    (b) R = (28000-8000)/10 = $2000.00 per year.. B5 = 28000-5(2000) = $18,000.00.

  4. P(at least one dies) = 1 - P(all 20 survive) = 1 - (P(one survives))20 = 1-(l48/l18)20. Using the tables, the answer is 0.385 (in one version) or 0.669 (in the other version).

  5. 0.51/20 = 0.965936329. Let 1+j = (1.03)/0.965936329 so j = 0.066322871. Then P = 45a20|j+1000(1+j)-20 = $767.50 is the price the investor should be willing to pay.

  6. (a) 10000D50/D40 = $6032.11 (one version) or $5916.28 (other version).

    (b) 10000(l50/l40)(1.09)-10 = $4150.47 (one version) or $4070.77 (other version).

  7. PV of premiums = PV of benefits so R(N28-N53)/D28 = 100000M28/D28. Thus R = 100000M28/(N28-N53). This gives $625.90 per year (one version) or $929.92 per year (other version).

  8. xlxdxpxqx
    10110002000.80.2
    102800 2400.7 0.3
    103560 168 0.7 0.3
    104392 196 0.5 0.5
    105196196 01


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On 10 May 2001, 16:02.